paper stringlengths 9 16 | proof stringlengths 0 131k |
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cs/0006046 | Divide MATH into two subsets MATH and MATH, where MATH consists of the vertices of degree four or more and MATH consists of the degree-three vertices. Let MATH denote the number of edges connecting sets MATH and MATH. Then each vertex in MATH must have at least one edge connecting it to MATH, and at most three edges co... |
cs/0006046 | We first show how to form a set MATH of non-disjoint trees in MATH, and a set of weights on the grandchildren of these trees, such that each tree's grandchildren have weight at most five. To do this, let each tree in MATH be formed by one of the MATH trees in MATH, together with all possible grandchildren in MATH that ... |
cs/0006046 | First, suppose that MATH has exactly five grandchildren. At least one vertex of MATH has high degree. Two of the children MATH and MATH must be the roots of forks, while the third child MATH is the root of a stick. We test each of the nine possible colorings of MATH and MATH. In six of the cases, MATH and MATH are diff... |
cs/0006046 | As described in the preceding sections, we find a maximal bushy forest, then cover the remaining vertices by height-two trees. We choose colors for each internal vertex in the bushy forest, and for certain vertices in the height-two trees as described in REF . Vertices adjacent to these colored vertices are restricted ... |
cs/0006046 | Let the given edge be MATH, and let its four neighbors be MATH, MATH, MATH, and MATH. Then MATH can be colored only if its four neighbors together use two of the three colors, which forces these neighbors to be matched into equally colored pairs in one of two ways. Thus, we can replace the instance by two smaller insta... |
cs/0006046 | Use a maximum matching algorithm in the graph induced by the edges with four neighbors. If the graph is REF-colorable, the resulting matching must contain at least MATH edges. Applying REF to an edge in a matching neither constrains any other edge in the matching, nor causes the remaining edges to stop being a matching... |
cs/0006046 | Assign a charge of MATH to each vertex of the graph, and redistribute this charge equally to each incident edge. Further assign an additional MATH charge to each four-neighbor edge. Then each edge receives a unit charge, so MATH. Subtracting MATH from both sides yields the result. |
cs/0006046 | We apply REF , resulting in a set of MATH constrained REF-edge-coloring problems each having only MATH edges. We then treat these remaining problems as REF-vertex-coloring problems on the corresponding line graphs, augmented by additional edges representing the constraints added by REF . The time for this algorithm is ... |
gr-qc/0006049 | Let MATH be the standard horizontal vector fields dual to MATH, fixed by the choice of basis for MATH. Similarly, let MATH be the fundamental vertical vector fields dual to MATH, corresponding to the choice of basis for MATH. Clearly MATH is a basis for MATH. Moreover, it is the unique dual of MATH. From the definition... |
gr-qc/0006049 | The last term in REF may be written as MATH which is clearly dominated by the second term if some MATH. |
gr-qc/0006049 | Let MATH be a horizontal curve ending at MATH such that the restriction to MATH is contained in MATH. Since MATH is extendible through MATH, the curvature scalar MATH must have a well defined limit along MATH as MATH. The restriction of MATH to MATH is a single point, so the limit of MATH must be invariant under the ac... |
gr-qc/0006050 | With a slight abuse of notation, we consider MATH to be a REF-parameter family of curves with variational parameter MATH, such that MATH corresponds to the original curve. The variational vector field MATH is assumed to be smooth with boundary conditions MATH and MATH. Parallel propagation of MATH along MATH for each f... |
gr-qc/0006050 | We may assume that MATH. Suppose that MATH has a cluster point MATH. Then there is a sequence MATH of real numbers such that MATH. Let MATH be an arbitrarily small neighbourhood of MATH in MATH. Then there is a small ball MATH around MATH in MATH such that MATH. But MATH, so there is a MATH such that MATH is contained ... |
gr-qc/0006050 | If MATH is an incomplete endless curve partially imprisoned in a compact set MATH, then the intersection of MATH with the interior of MATH is a family of incomplete endless curves MATH with horizontal lifts whose lengths go to REF. The problem is the endpoints of MATH on MATH, and also the possibility that MATH contain... |
gr-qc/0006050 | Let MATH be a null geodesic from MATH to MATH with affine parameter MATH. Pick a pseudo-orthonormal frame MATH at MATH such that MATH at MATH, and parallel propagate MATH along MATH. The length of MATH in the frame MATH is MATH . Now let MATH be a boost in the MATH direction by hyperbolic angle MATH, that is, MATH in t... |
gr-qc/0006050 | Let MATH be a point in MATH and let MATH be a defining family of curves for MATH. Since we may remove any loops at MATH, the projections MATH to MATH are incomplete and endless curves in MATH. Also, MATH so no subsequence of MATH converges to a point. Thus REF with MATH in place of MATH implies that there is an inexten... |
gr-qc/0006050 | Let MATH and let MATH be a defining family of curves for MATH. Since MATH, no subsequence of MATH converges to a point. By REF , there is a null geodesic limit curve MATH of a subsequence MATH of MATH through MATH in MATH which is inextendible in MATH. We assume that MATH never reaches MATH and show that this leads to ... |
gr-qc/0006050 | We start by estimating MATH, given by REF. Let MATH be the largest number such that MATH satisfies MATH on MATH. Here MATH is the value of MATH at MATH. We show that either MATH or MATH at MATH. If we insert MATH in REF, we get MATH on MATH. Using REF and integrating then gives MATH . Next, from REF we have MATH . Usin... |
gr-qc/0006050 | We may assume that MATH is parameterised by b- length and that MATH for some curve MATH in MATH, with MATH. Then by REF, MATH so MATH. Since MATH is a curve in MATH, there is a curve MATH in the NAME algebra MATH such that MATH, where MATH is the exponential map MATH and MATH. Then by REF, MATH. It follows that MATH wh... |
hep-lat/0006027 | Consider the factorization of MATH into irreducible factors, and splitting these factors in two groups. In particular, we label the factors that vanish at the origin as MATH, and the factors that do not vanish at the origin as MATH. Using this, define MATH and MATH. Note that while there is always at least one MATH, th... |
hep-lat/0006027 | For convenience, let us denote MATH and MATH. We will concentrate on the intersections of MATH with MATH. It is sufficient to consider factors MATH that do not have common polynomial factor with MATH. Indeed, MATH can not be a factor of MATH, and hence it can not be a factor of MATH. On the other hand, MATH can be a fa... |
hep-lat/0006027 | Let us consider an arbitrary but fixed polynomial MATH of the prescribed form, and the set of all factorizations MATH with required properties. It will be sufficient to prove the statement for the case when MATH does not contain any nontrivial polynomial factor with nonzero constant term. Indeed, if MATH has such a fac... |
hep-lat/0006027 | We will proceed by contradiction, and thus first assume that there actually exists an element MATH such that the requirements MATH are satisfied. To this MATH we will assign its restriction MATH under MATH which, according to REF , has the form MATH with the corresponding propagator being MATH . Note that according to ... |
hep-lat/0006027 | We first show that REF implies REF . Consider the change of variables MATH. Then MATH where the function MATH is defined by MATH . We split this up as MATH where MATH REF simply says that MATH where MATH, implying that MATH has a radius of convergence strictly greater than MATH. Similarly, if we further change the vari... |
hep-lat/0006027 | Let us denote the set of indices MATH for arbitrary positive integer MATH. NAME basis can be subdivided into non - intersecting subsets MATH, where MATH, MATH, contains the elements that can be written as the product of MATH gamma-matrices. In particular, MATH, MATH, and so on. With the appropriate convention on orderi... |
hep-lat/0006027 | Since MATH is ultralocal, the NAME components of MATH have finite number of NAME terms. Then there exists a non-negative integer MATH, such that when grouping together the NAME terms related by reflection properties in REF of REF , the NAME expansions can be written in the form MATH . Furthermore, the exchange properti... |
hep-lat/0006027 | We are dealing with the special case of REF , given by MATH, and all the arguments of the the proof in REF are valid here as well. Using the notation defined there and setting MATH we have in particular MATH . We will show below that for MATH it is now true in addition that MATH . However, denoting MATH we have MATH an... |
hep-th/0006143 | Though the MATH-modules MATH fail to be MATH-modules CITE, one can use the fact that the sheaves MATH are projections MATH of sheaves of MATH-modules. Let MATH be a locally finite open covering of MATH and MATH the associated partition of unity. For any open subset MATH and any section MATH of the sheaf MATH over MATH,... |
hep-th/0006143 | Since MATH is a strong deformation retract of MATH (see, for example, CITE), the first isomorphism in REF follows from the NAME - NAME theorem CITE, while the second one results from the well-known NAME theorem. |
hep-th/0006143 | Let the common symbol MATH stand for MATH and MATH. Bearing in mind decompositions REF - REF , it suffices to show that, if an element MATH is MATH-exact in the algebra MATH, then it is so in the algebra MATH. REF states that, if MATH is a contractible bundle and a MATH-exact form MATH on MATH is of finite jet order MA... |
hep-th/0006143 | By taking a smooth partition of unity on MATH subordinate to the cover MATH and passing to the function with support in MATH, one gets a smooth real function MATH on MATH which is REF on a neighborhood of MATH and REF on a neighborhood of MATH in MATH. Let MATH be the pull-back of MATH onto MATH. The exterior form MATH... |
hep-th/0006245 | To prove the previous two propositions, it is sufficient to prove it in the case MATH with MATH is as in REF . To illustrate this idea, we compute MATH with MATH, MATH . It is clear that is equivalent to prove the nilpotency of MATH in the abelian case. We use the same trick to prove the other assertions. |
hep-th/0006245 | Easy computation . |
math-ph/0006001 | Recall that given a point MATH, one can consider a projection MATH with the center at MATH, which sends MATH onto a projective space MATH of tangent directions at MATH. Here MATH is the direction of the line MATH. Consider compositions MATH, MATH. Since MATH is a stereographic projection, these compositions are fractio... |
math-ph/0006001 | Since MATH, MATH, are known for any MATH, by REF one can find MATH for any MATH and MATH. This uniquely determines MATH for any MATH. |
math-ph/0006001 | Instead of determining a web up to a local diffeomorphism near MATH it is enough to uniquely determine the diffeomorphic image MATH of this web, which is a web on a neighborhood of MATH. By REF it is enough to determine MATH, MATH. However, leaves of MATH, MATH, MATH are given by REF, MATH, MATH; here MATH is the stand... |
math-ph/0006001 | The ``only if" part is simple: in an appropriate neighborhood MATH of any given point MATH the foliation MATH can be written as MATH; here MATH is a function on MATH, and MATH for any MATH. Thus MATH for an appropriate function MATH on MATH, and MATH. For the ``if" part it is enough to show the existence locally on MAT... |
math-ph/0006001 | Any NAME curve in MATH is a projective transformation of the closure of the image of the mapping MATH. A consideration of the corresponding linear transformation of MATH provides polynomials MATH. It is enough to show uniqueness for the curve MATH. Obviously, MATH. Moreover, since MATH, MATH is a constant. |
math-ph/0006001 | Write MATH as MATH. Since MATH can be written as MATH, we know that MATH, MATH. This uniquely determines the quadratic polynomial MATH. Proceed similarly for MATH and MATH. |
math-ph/0006001 | Obviously, MATH for any MATH and any MATH. By REF , MATH is equivalent to existence of a foliation to which MATH is normal. Thus by REF the first statement implies the second one. If MATH for any MATH, then by REF , the required in REF foliation exists for MATH. However, MATH is defined for MATH, and MATH is a polynomi... |
math-ph/0006001 | Indeed, a direct calculation shows that MATH, MATH, MATH, MATH, and MATH. Thus the statement holds for MATH. However, if MATH is a projective transformation, then MATH is MATH-admissible iff it is MATH-admissible. Since cross-ratio is invariant with respect to projective transformations, it is enough to prove the state... |
math-ph/0006001 | By REF , MATH is MATH-admissible, thus it corresponds to a web MATH. Write the leaves of MATH as MATH, and apply REF again. |
math-ph/0006001 | The first statement is obvious, and the second one is the corollary of the first since MATH if MATH is a gauge transform of MATH. The third and the fourth statements are reformulations of parts of REF . The last statement is a direct corollary of the first one and of the following obvious statement: Given MATH for two ... |
math-ph/0006001 | Given MATH put MATH, MATH, MATH. REF on MATH can be translated to an additional linear equation MATH on MATH, MATH, MATH. This equation is independent of MATH, thus there is a unique (up to proportionality) solution MATH of these two equations. What remains to check is that this solution does not contradict the conditi... |
math-ph/0006001 | To prove the first statement, apply REF . Since MATH is a solution, so is MATH for any MATH. Similarly, since MATH is a solution for any MATH and MATH, MATH is an eikonal solution as well. |
math-ph/0006001 | It is easy to check the first claim by a direct calculation. In the second claim we already know that MATH is a solution of the eikonal equation for MATH. Since characteristic cones coincide, MATH is also a solution of the eikonal equation for MATH. |
math-ph/0006001 | Transposing coordinates MATH, one can ensure that MATH, MATH. Changing MATH to MATH allows us to assume that MATH. Obviously, one can find MATH and MATH such that the condition above implies that in MATH one has MATH and MATH. Consequently, in MATH one can write any leaf of MATH which passes through MATH, MATH, as MATH... |
math-ph/0006001 | Proceed similarly to the proof of REF . One may assume that MATH. Let MATH. With the stronger conditions of the amplification one can ensure that MATH is sufficiently small in MATH. Then REF give absolute bounds on derivatives of MATH, both from above and from below. Multiplying MATH by an appropriate constant, we may ... |
math-ph/0006001 | Obviously, any ball or polydisk is MATH-convex for REF exceptional foliations MATH, MATH, MATH of the web. Other foliations of the web are given by MATH; here MATH is a non-degenerate solution of REF. Application of Amplification REF finishes the proof. |
math-ph/0006001 | Suppose that MATH. Consider any function MATH on a neighborhood of MATH such that the vertical derivative of MATH on MATH does not vanish. Put MATH. Then MATH gives the required identification of a neighborhood of MATH with a subset of MATH. The existence of such a function MATH follows from the fact that a neighborhoo... |
math-ph/0006001 | If MATH, there is a neighborhood MATH of MATH and a neighborhood MATH of MATH such that for MATH the leaves of MATH are not tangent to MATH at any point of MATH, and each leaf intersects MATH in at most one point. Since MATH is compact, one can decrease MATH so that this condition is satisfied for any MATH. Taking MATH... |
math-ph/0006001 | The statements of this lemma concern one tangent space MATH only. The tangent spaces MATH are orthogonal complements to directions MATH in MATH. Thus MATH is determined by MATH and the image of the curve MATH. This is a NAME curve, and any two such curves are isomorphic. Thus we may replace MATH by an arbitrary vector ... |
math-ph/0006001 | Indeed, one can find MATH, MATH such that MATH is contained in a small neighborhood of REF, and MATH is contained in a small neighborhood of MATH. To finish the proof, note that MATH for any subset MATH which is compatible with a curve MATH passing through MATH. |
math-ph/0006001 | Consider the MATH-dimensional NAME web MATH associated to MATH such that the foliations MATH, MATH, MATH are associated to MATH, MATH, MATH. Take MATH, MATH to be the MATH-axis. Then the subset MATH (in notations of REF ) is MATH, since MATH is an intersection of a leaf of MATH and of a leaf of MATH. Similarly, for a c... |
math-ph/0006001 | Indeed, mappings with constant rank of the differential are submersions onto their images, thus preimages of points are foliations on MATH. The other statements are obvious. |
math-ph/0006001 | Suppose MATH. Let MATH be the points of intersection of MATH and MATH. Let MATH be blow-up of MATH at these points (make repeated blow-ups if needed to remove all the points of intersection). Removing proper preimages of MATH, MATH, from MATH, we obtain a manifold MATH with a mapping MATH to MATH such that preimages of... |
math-ph/0006001 | Take MATH, and apply REF to MATH and MATH. |
math-ph/0006001 | Let MATH be the values of MATH which correspond to exceptional foliations MATH, MATH, MATH of the web. Then MATH, MATH, are naturally identified with MATH. A choice of MATH, MATH, corresponds to a choice of REF leaves of these REF foliations, or, in other words, to a choice of coordinates MATH, MATH, MATH such that MAT... |
math-ph/0006001 | Deduce the first statement from REF . It is enough to calculate MATH. The condition on global sections is equivalent to MATH, thus all we need to show is that this rank does not change if we move to a nearby section MATH of MATH. Consider MATH as a sheaf of MATH-modules. For MATH denote by MATH the sheaf of MATH-module... |
math-ph/0006001 | Due to canonicity of MATH it is enough to prove this statement locally on MATH. Thus we may assume that MATH is weakly separating and separating. Consider the twistor transform of MATH. Since MATH may be identified with MATH, REF is applicable. (The condition on global sections is automatically satisfied if MATH is a t... |
math-ph/0006001 | It is enough to show that MATH; here the line bundle MATH is induced by the NAME inclusion MATH from the tautological line bundle on MATH (which is isomorphic to MATH). The fiber of MATH over MATH is the MATH-dimensional subspace of MATH corresponding to MATH. A linear function MATH on MATH induces a section of MATH, z... |
math-ph/0006001 | The first two statements are obvious. On the other hand, the normal bundle of MATH is canonically trivialized on MATH and on MATH, with the gluing function being MATH. To calculate the degree of a line bundle MATH it is enough to construct a section MATH in MATH and a section MATH in MATH. Suppose that MATH have no zer... |
math-ph/0006001 | Glue domains MATH and MATH together by gluing MATH to MATH for MATH, MATH, and MATH. Denote the resulting manifold by MATH, denote by MATH the mapping MATH, by MATH the natural projection MATH, and by MATH the section of MATH given by MATH. Consider the normal bundle MATH of MATH inside MATH. Let MATH be the normal bun... |
math-ph/0006001 | Indeed, one can write MATH . Then MATH can be rewritten as MATH, and MATH. The only thing one needs to prove is that MATH, which follows from MATH, MATH. |
math-ph/0006001 | Correctness follows from REF . Show that MATH is non-degenerate. Suppose that MATH for some value of MATH. Recall that MATH, MATH, is the value of MATH; here MATH is a NAME - NAME family of sections of MATH, and MATH, MATH, MATH are MATH. If MATH, this would mean that there is a one-parametric family of sections such t... |
math-ph/0006001 | The first statement is obvious, since MATH uniquely determines MATH. The second statement follows from MATH and MATH with MATH and MATH being REF or REF, and from the fact that MATH is an orthogonal basis in the NAME spaces MATH. |
math-ph/0006001 | Fix MATH. We may assume that MATH, for example, MATH. Let MATH, MATH. Then MATH . Since MATH, one can write MATH as MATH, here MATH and MATH have invertible holomorphic continuations into MATH and MATH correspondingly. Similarly, write MATH here MATH and MATH have holomorphic continuations into MATH and MATH correspond... |
math-ph/0006007 | It is clear that MATH if MATH, and one has: MATH, which proves REF . The proof of REF is similar. |
math-ph/0006007 | is straightforward using REF and the relation MATH . |
math-ph/0006007 | It suffices to check that in the case MATH one has: MATH, which is immediate. |
math-ph/0006007 | is straightforward using the facts that MATH, and MATH iff-MATH. |
math-ph/0006007 | It follows from REF that MATH . Since MATH, the proposition follows from REF b. |
math-ph/0006007 | Due to REF holds if MATH. Let MATH. It is well-known (compare CITE) that MATH is integrable on MATH iff MATH . But we have just shown that REF holds for a NAME open set of MATH on the hyperplane MATH. Since REF is a polynomial condition, we conclude that it holds for all MATH on this hyperplane. |
math-ph/0006007 | We have: MATH, hence, by REF we have: MATH, etc. Thus, MATH for MATH. Therefore, by REF b , we have: MATH . Now the proposition follows from REF a. |
math-ph/0006007 | REF is proved in REF are easily checked. |
math-ph/0006007 | The lemma follows from REF applied to the simple roots MATH, and REF applied to MATH, since, due to REF we have: MATH . |
math-ph/0006007 | Consider the following two sequences of roots of MATH: MATH . It is clear by REF that MATH and MATH. Note that MATH (respectively, MATH) is equal to the left-hand side of REF (respectively, REF ). Note that MATH . If MATH for all MATH, using REF we would conclude, by REF , that MATH, in contradiction with the assumptio... |
math-ph/0006007 | In the case MATH, the only condition of integrability is local finiteness of MATH on MATH which is equivalent to MATH due to REF b. It follows from REF that in the case MATH, the conditions listed by REF b are necessary. In view of REF b, it remains to show that these conditions are sufficient for local nilpotency of M... |
math-ph/0006007 | Both statements follow from the corresponding OPE, which are easily derived from NAME 's formula. Below we give the less trivial OPE needed for the proof of REF . MATH . |
math-ph/0006007 | It is clear that, if MATH (respectively, MATH) MATH, then MATH consists of homogeneous polynomials of degree MATH in anticommuting operators MATH (respectively, MATH) and commuting operators MATH (respectively, MATH), applied to MATH. This proves REF (respectively, REF ), due to REF . |
math-ph/0006007 | Since the eigenspaces of MATH in MATH are finite-dimensional and MATH commutes with MATH, it follows that MATH is a direct sum of finite-dimensional MATH-modules, hence MATH acts locally finitely on MATH. Furthermore, we have: MATH where MATH (respectively, MATH) is the vertex algebra generated by the MATH (respectivel... |
math-ph/0006007 | If MATH (respectively, MATH), we have: MATH . Hence the LHS of REF is equal to MATH . Noticing that the second summand is zero and applying to the first summand the NAME triple product identity REF , we obtain REF . In the proof of REF we assume that MATH, the case MATH being similar: MATH . The second product on the R... |
math-ph/0006007 | Let MATH. We have: MATH, where MATH . Furthermore, we have: MATH . The first sum on the right is just MATH, which has asymptotics REF . But the second sum on the right has asymptotics of this form too since it can be written as a product of a power of MATH and a function MATH for some other affine linear function MATH,... |
math-ph/0006007 | In the case MATH, the theorem follows from REF . In general, the proof is based on similar arguments. Below we shall give details in the case MATH; in the rest of the cases arguments are the same. The even part of MATH is MATH and its simple roots are MATH for MATH and MATH,MATH for MATH. The simple roots of MATH are M... |
math-ph/0006007 | The only simple root of MATH which is not simple for MATH is MATH. Hence the proof of REF proves REF as well. |
math-ph/0006007 | The proof that MATH is a representation is, as usual, a straightforward use of NAME 's formula. The proof of integrability of MATH is the same as in the proof of REF . This establishes REF . Note that, as before, MATH commutes with MATH, and the spectrum of MATH on MATH (respectively, MATH) is MATH (respectively, MATH)... |
math-ph/0006007 | Denote by MATH the subalgebra MATH (respectively, MATH) of MATH (see REF). This is an affine NAME algebra. Denote by MATH the vertex subalgebra MATH of MATH. Since, by definition, MATH is an integrable MATH-module, it follows that it is MATH-irreducible CITE, hence MATH is a simple affine vertex algebra of non-negative... |
math-ph/0006007 | Let MATH denote the left ideal of MATH generated by elements REF . Then the MATH-module MATH is principal integrable, hence each of its singular weights MATH is integrable. Hence, if the condition of REF holds, the MATH-module MATH is irreducible, and therefore MATH. Furthermore, obviously, MATH, hence REF implies that... |
math-ph/0006007 | Note that in the MATH case MATH is a subalgebra of the vertex subalgebra MATH of MATH (constructed in REF), while the highest component of the MATH-module MATH restricted to MATH is MATH. Since the MATH exhaust all integrable highest weights of level MATH, by REF , they give a complete list of irreducible MATH-modules.... |
math-ph/0006021 | MATH is a MATH-principal fiber bundle, so we can use the lifted MATH-action to define MATH. Since this action is a lift of the MATH-action on MATH, MATH has a natural structure of a vector bundle over MATH. If MATH is the bundle projection of MATH, then the pull back by MATH is defined as MATH . If MATH is the bundle p... |
math-ph/0006021 | For MATH one has: MATH . Since MATH is a multiplication operator in each fiber it has fiber-wise norm MATH, and so have MATH and MATH. |
math-ph/0006021 | CASE: A decomposable operator commutes with the MATH-action, especially with MATH which is defined by MATH . By REF commuting with MATH is equivalent to commuting with MATH. CASE: To commute with the MATH-action means to commute with all MATH for MATH. Because of MATH the MATH are just the characters MATH of the compac... |
math-ph/0006021 | Given the remark above we have shown the decomposability already. MATH is a core for MATH, its image under MATH is contained in MATH and is a core for MATH, since MATH is an isometry. On this domain REF gives the action of MATH as asserted in the theorem. Since MATH is a symmetric elliptic operator on the compact manif... |
math-ph/0006021 | If MATH then, by the general theory for direct integrals, MATH has positive measure for every MATH. The fibers MATH are elliptic operators on a compact manifold and thus have discrete spectrum; the eigenvalues depend continuously on MATH (even piece-wise real- analytically; see below). We choose a sequence MATH with MA... |
math-ph/0006021 | MATH is obviously a MATH-submodule of MATH. Furthermore, by definition the scalar product is MATH and therefore continuous in MATH, since the last sum in REF is finite. The *-property is immediately clear, the MATH-linearity of the scalar product follows from MATH . |
math-ph/0006021 | Since MATH the closure of MATH in the MATH-norm is a subspace of MATH, and by definition a MATH-module. The integral with respect to a measure defines a trace. Since MATH is compact (MATH is discrete) it has finite volume with respect to NAME measure, so that the trace is finite, and all MATH are trace class. Since MAT... |
math-ph/0006021 | We get the fiber at MATH as GNS representation space of the state MATH. For the continuity structure see REF. |
math-ph/0006021 | For the generators of MATH one can easily show the relations MATH for MATH. Thus, from the trace property of MATH we have MATH . For all MATH we have MATH so that MATH is faithful: Set MATH, and note that MATH is a faithful trace on MATH. |
math-ph/0006021 | Let MATH be an orthonormal basis of MATH, consisting of unitary elements of MATH. Then MATH . |
math-ph/0006021 | If MATH is a projective NAME MATH-module then MATH is a direct summand of a free module MATH for a suitable NAME space MATH, and we can apply REF . |
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