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cs/0006046
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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 connecting it to MATH, so MATH and MATH. Further, to avoid cycles, each connected component in MATH must form a tree, and if such a component has MATH vertices, it must have MATH edges leaving it, and MATH else we could apply REF . So, MATH. If MATH, MATH. And if MATH, then again MATH. However, each leaf in MATH has at most two edges outside MATH, or MATH would not be maximal, so MATH.
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cs/0006046
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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 are adjacent to the MATH leaves. We assign each vertex in MATH unit weight, which we divide equally among the trees it belongs to. Then, suppose for a contradiction that some tree MATH in MATH has grandchildren with total weight more than five. Then, its grandchildren must form three forks, and at least five of its six grandchildren must have unit weight; that is, they belong only to tree MATH. Note that each vertex in MATH must have degree three, or we could have added it to the bushy forest, and all its neighbors must be in MATH, or we could have added it to MATH. The unit weight grandchildren each have one neighbor in MATH and two other neighbors in MATH. These two other neighbors must be one each from the two other forks in MATH, for, if to the contrary some unit-weight grandchild MATH does not have neighbors in both forks, we could have increased the number of trees in MATH by removing MATH and adding new trees rooted at MATH and at the missed fork. Thus, these five grandchildren each connect to two other grandchildren, and (since no grandchild connects to three grandchildren) the six grandchildren together form a degree-two graph, that is, a union of cycles of degree-three vertices. But after applying REF to MATH, it contains no such cycles. This contradiction implies that the weight of MATH must be at most five. Similarly, if the weight of MATH is more than three, it must have at least one fork, at least one unit-weight grandchild outside that fork, and at least one edge connecting that grandchild to a grandchild within the fork. This edge together with a path in MATH forms a cycle, which must contain a high degree vertex. We are not quite done, because the assignment of grandchildren to trees in MATH is fractional and non-disjoint. To form the desired forest MATH, construct a network flow problem in which the flow source is connected to a node representing each tree MATH by an edge with capacity MATH if MATH contains a high degree vertex and capacity MATH otherwise. The node corresponding to tree MATH is connected by unit-capacity edges to nodes corresponding to the vertices in MATH that are adjacent to MATH, and each of these nodes is connected by a unit-capacity edge to a flow sink. Then the fractional weight system above defines a flow that saturates all edges into the flow sink and is therefore maximum (REF , middle top). But any maximum flow problem with integer edge capacities has an integer solution (REF , middle bottom). This solution must continue to saturate the sink edges, so each vertex in MATH will have one unit of flow to some tree MATH, and no flow to the other adjacent trees. Thus, the flow corresponds to an assignment of vertices in MATH to adjacent trees in MATH such that each tree is assigned at most MATH vertices. We then simply let each tree in MATH consist of a tree in MATH together with its assigned vertices in MATH (REF , bottom).
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cs/0006046
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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 different, forcing the root to have one particular color (REF , right). In these cases the only remaining vertex after translation to a MATH-CSP instance and application of REF will be the child of MATH, so in each such case MATH accumulates a further cost of MATH. In the three cases in which MATH and MATH are colored the same (REF , left), we must also take an additional factor of MATH for MATH itself. One of these MATH factors goes to a high degree vertex, while the remaining work is split among the remaining eight vertices. The cost per vertex in this case is then at most MATH. If MATH has fewer than five grandchildren, we choose a color for the root of the tree as described at the start of the section. The worst case occurs when the number of grandchildren is either three or four, and is MATH.
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cs/0006046
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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 to two colors, while the remaining vertices form a MATH-CSP instance and can be colored using our general MATH-CSP algorithm. Let MATH denote the number of vertices that are roots in the bushy forest; MATH denote the number of non-root internal vertices; MATH denote the number of bushy forest leaves; MATH denote the number of vertices adjacent to bushy forest leaves; and MATH denote the number of remaining vertices, which must all be degree-three vertices in the height-two forest REF . Then the total time for the algorithm is at most MATH. We now consider which values of these parameters give the worst case for this time bound, subject to the constraints MATH, MATH, MATH (from the definition of a bushy forest), MATH (from the maximality of the forest), and MATH REF . We ignore the slightly tighter constraint MATH since it only complicates the overall solution. Since the work per vertex in MATH and MATH is larger than that in the bushy forests, the time bound is maximized when MATH and MATH are as large as possible; that is, when MATH. Further since the work per vertex in MATH is larger than that in MATH, MATH should be as large as possible; that is, MATH and MATH. Increasing MATH or MATH and correspondingly decreasing MATH, MATH, and MATH only increases the time bound, since we pay a factor of REF or more per vertex in MATH and MATH and at most MATH for the remaining vertices, so in the worst case the constraint MATH becomes an equality. It remains only to set the balance between parameters MATH and MATH. There are two candidate solutions: one in which MATH, so MATH, and one in which MATH, so MATH. In the former case MATH and the time bound is MATH. In the latter case MATH and the time bound is MATH.
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cs/0006046
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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 instances: one in which we replace the five edges by the two edges MATH and MATH, and one in which we replace the five edges by the two edges MATH and MATH; in each case we add a constraint between the two new edges.
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cs/0006046
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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.
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cs/0006046
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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.
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cs/0006046
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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 thus at most MATH. By REF , we can rewrite this bound as MATH. Since MATH, this time bound is maximized when MATH is maximized, which occurs when MATH and MATH. For this value, all the work occurs within REF , and gives the stated time bound.
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gr-qc/0006049
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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 of the exterior derivative, MATH . To proceed further we introduce coordinates on MATH as follows. Let MATH and let MATH be coordinates for MATH on a neighbourhood MATH of MATH. Given a frame MATH at a point MATH, we may express each MATH as MATH . The determinant of the matrix MATH is nonzero, so we may use MATH as coordinates for MATH on MATH. The coordinate expressions for the horizontal vector fields MATH and the vertical vector fields MATH are then MATH where MATH are the NAME symbols of MATH in the coordinates MATH CITE. Now the NAME tensor frame components MATH are related to the coordinate components MATH by MATH where MATH is the inverse of MATH. Applying REF to REF and inserting into REF then gives the desired result.
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gr-qc/0006049
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The last term in REF may be written as MATH which is clearly dominated by the second term if some MATH.
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gr-qc/0006049
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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 action of any NAME transformation MATH, changing the curve according to MATH. Now the curvature scalar MATH is given by REF, which may be rewritten as MATH where MATH is the timelike index, greek indices go from REF to MATH, MATH is the curvature scalar and MATH is the scalar invariant MATH of MATH. Since MATH and MATH are scalar invariants, we only need to consider the last two terms in REF. Given the frame MATH at MATH, we apply NAME transformations in the MATH timelike planes spanned by MATH and MATH, MATH. After some algebra we find that REF is invariant if and only if MATH . From REF, the NAME tensor is given by MATH which is the condition for an NAME space CITE. Applying the first NAME identity to REF and permuting the indices shows that MATH. Thus REF implies MATH which means that the NAME tensor vanishes.
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gr-qc/0006050
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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 fixed MATH gives a frame field MATH, and a coframe field MATH dual to MATH. We also denote a MATH-derivative by a dot and write MATH for the tangent of MATH. The b- length functional REF can then be written as MATH . Note that if MATH is the horizontal lift of MATH for each fixed MATH, then the lift of MATH coincides with the canonical REF-form MATH, so REF agrees with REF of b- length for horizontal curves in MATH. Differentiating REF and evaluating at MATH, we get MATH where we have used that MATH since MATH is parameterised by b- length at MATH, and the indices MATH denote components in the frame MATH. Using that MATH and MATH, MATH since MATH and MATH. Rewrite MATH as an integral from REF to MATH and perform a partial integration on the second term. Then MATH . To proceed further we define MATH as the solution of the initial value REF . We can then partially integrate the first term in REF, which results in MATH . By the basic principle of variational calculus, we arrive at REF .
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gr-qc/0006050
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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 in MATH for all MATH. Then MATH is contained in MATH for all MATH, so MATH converges to MATH which contradicts the assumption on MATH. Since MATH is a cluster point of MATH, there is a sequence MATH such that-MATH. Put MATH. By the argument in the previous paragraph, MATH has no cluster point in MATH. Let MATH be a convex normal neighbourhood of MATH in MATH, let MATH be a cross-section of MATH over MATH, and let MATH whenever MATH. The action of MATH on MATH is free and transitive, so there are unique matrices MATH such that in MATH, MATH . MATH may be decomposed as MATH, where MATH and MATH are spatial rotations and MATH is a NAME boost by a hyperbolic angle MATH. Let MATH. If MATH had an upper bound MATH, then MATH would be contained in a compact subset of MATH, which is impossible since MATH has no cluster point. We therefore assume that MATH, the case when MATH being similar. Now MATH is compact, so there is a subsequence MATH of MATH such that-MATH, MATH and MATH as MATH. Let MATH and MATH . Then MATH as MATH. Since MATH is a constant rotational matrix, leaving the Euclidian norm invariant, it does not affect the length of MATH. From MATH it follows that MATH and so MATH where MATH, MATH, MATH and MATH, MATH, MATH and MATH are the standard horizontal vector fields on MATH CITE. Similarly, let MATH. Then MATH, MATH, MATH and MATH, so MATH . Let MATH be the integral curve of MATH through MATH. Then MATH is a null geodesic in MATH. We may assume that MATH is extended as far as possible as the horizontal lift of an unbroken null geodesic in MATH. We show that MATH is a limit curve of MATH. Let MATH be a point on MATH, let MATH be a neighbourhood of MATH in MATH, and let MATH be the tubular subset of MATH generated by all integral curves of MATH intersecting MATH. Since MATH, REF gives that there is a MATH such that if MATH then MATH is contained in MATH, that is, MATH does not leave MATH except possibly at the ends MATH. Now MATH does not contain any imprisoned incomplete curves since it is a convex normal neighbourhood, so MATH, having no endpoint in MATH, must leave MATH. Thus MATH intersects MATH for each MATH, which means that MATH is a limit point of MATH. Obviously, MATH is contained in MATH since it is a limit curve of MATH. It remains to show that MATH can be extended to an inextendible null geodesic cluster curve of MATH in the whole of MATH. Extend MATH as far as possible as an unbroken null geodesic in MATH, and let MATH be a point on MATH. Then the segment of MATH from MATH to MATH is closed and finite, so it can be covered by a finite sequence of convex normal neighbourhoods MATH with MATH. By the above argument, MATH is a limit curve of some subsequence MATH of MATH. Assume that MATH is a limit curve of a subsequence MATH for some MATH. Then any point MATH on MATH is a cluster point. Repeating the argument with MATH in place of MATH and MATH in place of MATH shows that MATH is a limit curve of some subsequence MATH as well. By induction, the whole curve MATH is a cluster curve of MATH.
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gr-qc/0006050
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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 contains subsequences converging to a point on MATH (see REF). But this can be dealt with by enlarging MATH around any such points. Thus REF gives us an inextendible null geodesic cluster curve MATH of MATH in MATH. Let MATH be the endless extension of MATH as a null geodesic in MATH and let MATH be a point on MATH. Then the segment of MATH from MATH to MATH is finite and so it can be included in a larger compact set MATH. Applying REF to MATH we find that the part of MATH in MATH is a cluster curve of MATH as well, and since MATH was arbitrary, the whole of MATH is a cluster curve in MATH.
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gr-qc/0006050
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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 the frame MATH. Then the length of MATH in the frame MATH is MATH, which tends to REF as MATH.
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gr-qc/0006050
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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 inextendible null geodesic cluster curve MATH of some subsequence MATH through MATH in MATH. Let MATH be the inextendible segment of MATH through MATH in MATH. We show that MATH is contained in MATH. Suppose that MATH is a point on MATH and let MATH be the segment of MATH from MATH to MATH. We may assume that MATH is parameterised by an affine parameter MATH ranging from MATH to MATH. As in the proof of REF , there is a pseudo-orthonormal frame MATH at MATH in which the tangent of MATH is MATH, and MATH intersects MATH at the frame MATH. Let MATH be the horizontal lift of MATH to MATH. Then the component vector MATH of MATH with respect to the standard horizontal vector fields on MATH is MATH . Since MATH is a rotation leaving MATH fixed and MATH is a boost in the MATH direction with hyperbolic angle MATH, the Euclidian norm of MATH is MATH so the b- length of MATH is MATH. It follows that the concatenation of MATH and MATH is a curve from MATH to MATH with length MATH, which tends to REF as MATH.
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gr-qc/0006050
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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 a contradiction. Since there are no totally imprisoned null geodesics in MATH, MATH must have an endpoint MATH on MATH. Then MATH is a limit point of MATH. Let MATH be a convex normal neighbourhood of MATH, sufficiently small for MATH and MATH not to be in MATH. Each curve MATH must enter and leave MATH for large enough MATH. By REF there is an endless null geodesic cluster curve of MATH through MATH in MATH. But every MATH is contained in MATH, so the cluster curve cannot leave MATH. We have thus obtained an extension of MATH in MATH, which contradicts that MATH is inextendible.
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gr-qc/0006050
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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 . Using REF and integrating gives MATH . Combining REF, we find that MATH so the right hand side of REF is less than MATH. Solving REF for MATH and dividing by MATH we get MATH . Since MATH and MATH, MATH . Going back to REF and estimating from below results in MATH . The first term is positive, and expanding the square and applying the conditions on MATH and MATH gives MATH . From REF it follows that REF cannot be violated even at MATH. So unless MATH, the only remaining possibility is that MATH. The next step is to estimate MATH in terms of MATH. Using REF of MATH and the lower bound of REF gives MATH hence MATH . We can now provide bounds for MATH and MATH. Suppose that MATH. From REF, MATH where the inequality follows from REF. But MATH so using REF and integrating gives the desired bound on MATH. The argument for the case when MATH is similar.
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gr-qc/0006050
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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 which on integration gives MATH. Thus MATH so using REF gives MATH .
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hep-lat/0006027
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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, there doesn't necessarily have to be any MATH. In that case we define MATH. We thus always have MATH, with MATH, and MATH. Let us look at an arbitrary but fixed MATH, defining an irreducible affine complex algebraic curve MATH, running through the origin. Consider an arbitrary (complex) neighborhood of the origin, in which MATH is analytic. Any such neighborhood contains infinitely many points of MATH. These points represent locations of potential singularities of MATH, and thus must be canceled by zeroes of MATH. However, by NAME 's theorem, any two algebraic curves without common component have only finite number of points in common so this is only possible if MATH is the component of MATH. Consequently, MATH must be an irreducible factor of MATH. By the above argument, MATH is the desired common factor of MATH and MATH. From the unique factorization theorem it follows that MATH is the only factor (up to a multiplicative constant) of minimal degree. Also, its definition only depends on MATH as claimed.
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hep-lat/0006027
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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 factor of MATH but if it is, than the claim of the proposition is obviously true. CASE: Since MATH and MATH are mutually coprime, it follows from NAME 's theorem that the total number of their intersections MATH counted with multiplicities is even. CASE: Let MATH be the intersection point of MATH and MATH such that MATH. Then the intersection multiplicity MATH is even. Indeed, let MATH be arbitrary branch of MATH centered at MATH. Then MATH, or equivalently MATH . Because of the squares, we can conclude that upon the substitution, MATH is even. At the same time, since MATH, we have MATH, and thus MATH. But MATH, and so MATH is a positive even integer. Since this is true for all branches MATH of MATH centered at MATH, we can conclude that MATH is even as claimed. CASE: If MATH, then MATH. Indeed, MATH has a single branch centered at MATH, namely the line that can be parametrized as MATH. Since MATH and MATH we have MATH . For REF to be satisfied simultaneously, MATH and MATH must necessarily intersect at one additional point from the set of points where MATH and MATH meet. However, this set is precisely represented by the set of homogeneous coordinates from MATH. Consequently, MATH has to pass through one additional point from MATH as claimed.
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hep-lat/0006027
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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 factor MATH, then redefining MATH, MATH can only make the set of zeroes for new MATH smaller. As a consequence of the unique factorization theorem, the factors MATH, MATH are unique up to a trivial rescaling by complex number if this additional condition is imposed. Assuming the above, consider the transformation MATH, which does not change MATH. Since all possible nontrivial irreducible factors MATH of MATH satisfy MATH, while all irreducible factors MATH of MATH have MATH, it follows from the unique factorization theorem that MATH and MATH under the above reflection. The same is true under MATH and, as a result, MATH, MATH. This implies that we can change the variables MATH, MATH in the whole problem. In new variables, we have MATH with properties at the origin preserved. However, according to REF , MATH must have an additional zero on the set MATH. Consequently, MATH must vanish for at least two additional points from the set MATH which completes the proof.
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hep-lat/0006027
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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 REF and MATH, no NAME component of MATH has a pole away from the origin of the NAME zone. Consequently, the function MATH has also no poles away from the origin. Since MATH is ultralocal, we can use REF , to conclude that its NAME components have the following structure MATH where MATH is a symmetric polynomial, MATH an antisymmetric polynomial, and MATH a polynomial without definite symmetry properties. The classical continuum limit REF requires that MATH, MATH, and also MATH by virtue of its antisymmetry. The GW property MATH demands that the NAME components of chirally nonsymmetric part of the propagator (MATH and MATH) are analytic in some complex region containing the (real) NAME zone. We will concentrate on the consequences of analyticity in the vicinity of the origin. To this end, it is convenient to introduce a change of variables such as MATH which is invertible in the vicinity of the origin, maps the real region MATH onto the square MATH, and preserves analyticity in new variables except possibly on the boundary. Using REF , we have MATH where MATH is a symmetric polynomial such that MATH, MATH is an antisymmetric polynomial, and MATH is a polynomial such that MATH. It is also convenient to introduce symmetric polynomials MATH corresponding to MATH, and MATH respectively. With this notation, the GW property MATH implies that the functions MATH are analytic in the vicinity of the origin. Let us first consider MATH which is a rational function whose denominator vanishes at the origin. Consequently, according to REF , the polynomials MATH and MATH have a common polynomial factor MATH with zero at the origin, so that the possible singularity is removed. Let us fix MATH to be of minimal degree, and thus unique up to a constant multiplicative factor. Next, consider MATH. Since MATH and MATH are analytic and nonzero near the origin, the rational function MATH is also analytic and, moreover, the same denominator MATH is involved as in MATH. Consequently, the polynomial MATH divides MATH. In summary, we thus have that MATH divides MATH,MATH and MATH. Hence, from the form REF of MATH it follows that MATH also divides MATH. According to REF , an intriguing property of any polynomial MATH of the form REF is that if it is divisible by polynomial MATH vanishing at the origin, then MATH has at least two additional zeroes on the set of points MATH. Since MATH divides MATH, it follows that MATH also has these zeroes on MATH. Returning back to momentum variables, we thus came to the conclusion that in addition to the origin of the NAME zone, the function MATH of REF has at least two additional zeroes on the set MATH. However, this contradicts the conclusion of REF that MATH has no poles away from the origin of the NAME zone. That completes the proof.
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hep-lat/0006027
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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 variable MATH in MATH, so that MATH, we can infer that MATH too has a radius of convergence strictly greater than MATH. The above conclusions prove that MATH is analytic in an annulus containing the unit circle of the complex plane. Consequently, MATH is a composition of two analytic maps, implying analyticity in the complex strip as desired in REF . Next we show that REF implies REF . Consider arbitrary MATH. We will integrate in the complex plane along the rectangular contour MATH with the following line segments: MATH from point MATH to point MATH, MATH from MATH to MATH, MATH from MATH to MATH, and MATH from MATH to MATH. Let us assume that MATH. According to NAME theorem, we have MATH . Since MATH is periodic with MATH, the contributions from the integrals along MATH and MATH will cancel each other and we thus obtain MATH . Finally, MATH is bounded on the path of the last integral due to analyticity, and we thus have MATH . Similarly if MATH, then we will use the rectangular integration contour in the upper half of the complex plane, yielding an analogous bound. Together, this then implies REF as claimed.
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hep-lat/0006027
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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 ordering of gamma - matrices in the definition of MATH, we can then rewrite the NAME decomposition of MATH in the form MATH . We will concentrate on contributions to MATH originating from subsets MATH, where MATH. Consider a single term in decomposition REF from this group, specified by the set of indices MATH. Then there exists element MATH, such that MATH. Under reflection MATH through the corresponding axis, we have MATH . NAME symmetry of MATH then requires that MATH . However, since MATH, the restricted variable MATH under MATH satisfies MATH, and hence MATH . Consequently, the only NAME element contributing from this group is MATH (when MATH), and the form REF follows. Analyticity, periodicity and relations REF for functions MATH follow from corresponding properties of unrestricted operator.
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hep-lat/0006027
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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 properties in REF imply that MATH, while MATH. Using the formulas for trigonometric functions of multiple arguments, namely MATH and MATH the forms REF directly follow, together with the exchange symmetry properties.
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hep-lat/0006027
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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 and so REF imply that MATH. That proves the claim of the proposition. To show REF , it is sufficient to concentrate on irreducible MATH. This is because such MATH is unique and all the reducible ones will thus contain it. The uniqueness of irreducible MATH follows from the fact that if there were at least two, then MATH would also have to be at least two, which is not the case. This also means that the corresponding MATH is symmetric in MATH. Indeed, if it were not, then the polynomial MATH would define another component of MATH running through the point MATH thus contradicting the uniqueness. Assuming the above, consider arbitrary branch MATH of MATH, centered at MATH. Due to symmetry in MATH there is a corresponding branch MATH centered at MATH. Using the fact that MATH and MATH we have MATH while at the same time MATH . The last two equalities relate to substituting MATH in MATH, and using REF which is valid for arbitrary branch MATH of MATH. It follows that MATH is even for arbitrary pair MATH, implying REF .
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hep-th/0006143
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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, let us put MATH. The endomorphisms MATH of MATH yield the MATH-module endomorphisms MATH of the sheaves MATH. They possess the properties required for MATH to be a fine sheaf. Indeed, for each MATH, MATH provides a closed set such that MATH is zero outside this set, while the sum MATH is the identity morphism.
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hep-th/0006143
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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.
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hep-th/0006143
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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 MATH (that is, MATH), there exists an exterior form MATH on MATH such that MATH. Moreover, a glance at the homotopy operators for MATH and MATH CITE shows that the jet order MATH of MATH is bounded for all exterior forms MATH of fixed jet order. Let us call this fact the finite exactness of the operator MATH. Given an arbitrary bundle MATH, the finite exactness takes place on MATH over any open subset MATH of MATH which is homeomorphic to a convex open subset of MATH. Let us prove the following. CASE: Suppose that the finite exactness of the operator MATH takes place on MATH over open subsets MATH, MATH of MATH and their non-empty overlap MATH. Then, it is also true on MATH. CASE: Given a family MATH of disjoint open subsets of MATH, let us suppose that the finite exactness takes place on MATH over every subset MATH from this family. Then, it is true on MATH over the union MATH of these subsets. If these assertions hold, the finite exactness of MATH on MATH takes place because one can construct the corresponding covering of the manifold MATH REF . Proof of REF . Let MATH be a MATH-exact form on MATH. By assumption, it can be brought into the form MATH on MATH and MATH on MATH, where MATH and MATH are exterior forms of finite jet order. Let us consider their difference MATH on MATH. It is a MATH-exact form of finite jet order which, by assumption, can be written as MATH where MATH is also of finite jet order. REF below shows that MATH where MATH and MATH are exterior forms of finite jet order on MATH and MATH, respectively. Then, putting MATH we have MATH on MATH where MATH is of finite jet order. Proof of REF . Let MATH be a MATH-exact form on MATH. The finite exactness on MATH holds since MATH on every MATH and, as was mentioned above, the jet order MATH is bounded on the set of exterior forms MATH of fixed jet order MATH.
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hep-th/0006143
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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 is zero on a neighborhood of MATH and, therefore, can be extended by REF to MATH. Let us denote it MATH. Accordingly, the exterior form MATH has an extension MATH by REF to MATH. Then, MATH is a desired decomposition because MATH and MATH are of finite jet order which does not exceed that of MATH.
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hep-th/0006245
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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.
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hep-th/0006245
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Easy computation .
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math-ph/0006001
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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 fraction-linear mappings between projective lines. Thus they are determined by images of any REF distinct points on MATH. Thus the line MATH is uniquely determined by MATH. Since MATH if MATH, MATH is uniquely determined by MATH. To show the existence take any parameterized quadric MATH, let MATH, MATH. Then no REF points out of MATH are on the same line, thus there is a projective mapping MATH such that MATH, MATH. Then MATH is the parameterized quadric we need.
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math-ph/0006001
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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.
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math-ph/0006001
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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 standard coordinate system on MATH. Similarly, leaves of MATH are given by the equation MATH.
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math-ph/0006001
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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 MATH, since the foliation is unique if it exists, thus gluing pieces together is not a problem. We may assume that MATH is an open subset of MATH, and that MATH. Say that a tangent vector MATH at MATH is MATH-compatible, MATH, if MATH and MATH is of the form MATH with an appropriate number MATH. Obviously, in an appropriate neighborhood of any point MATH there is exactly one MATH-compatible vector MATH for MATH. Define functions MATH by MATH. Then the fundamental relationship between commutator and NAME differential CITE MATH implies MATH. Since MATH, one can write MATH; here MATH is a MATH-form defined near MATH. Hence MATH . Thus MATH. On the other hand, MATH. Together with MATH this implies MATH. By the principal theorem of the theory of ODE, one can find local coordinates MATH such that MATH, MATH. Since MATH is orthogonal to MATH, MATH, this implies that MATH, thus MATH gives a foliation with the required properties.
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math-ph/0006001
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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.
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math-ph/0006001
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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.
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math-ph/0006001
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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 polynomial of degree REF in MATH. Thus MATH, including MATH. Moreover, MATH for any MATH, which implies the existence of MATH for MATH as well. Thus the second statement implies the first one. Since MATH is quadratic in MATH, MATH is a polynomial of degree REF in MATH. Thus the second statement is equivalent to the third one. By construction MATH are proportional to MATH, MATH, MATH, and MATH correspondingly. This implies that MATH for MATH. Consequently, the fourth statement is equivalent to the third one. Assume that MATH. Let MATH, MATH. Then MATH and MATH can be written as MATH . (It is clear that MATH for MATH.) Since MATH, the equation MATH is proportional to REF, which implies the last statement of the corollary.
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math-ph/0006001
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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 statement for MATH with an arbitrary MATH. By an appropriate choice of MATH we can ensure that MATH (and additionally MATH if we wish).
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math-ph/0006001
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By REF , MATH is MATH-admissible, thus it corresponds to a web MATH. Write the leaves of MATH as MATH, and apply REF again.
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math-ph/0006001
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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 non-degenerate functions MATH and MATH defined in a neighborhood of MATH in MATH, one can decrease the neighborhood so that the functions become gauge transforms of each other. This finishes the proof of REF .
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math-ph/0006001
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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 conditions MATH, MATH, MATH. However, MATH contradicts MATH, etc.
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math-ph/0006001
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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.
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math-ph/0006001
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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.
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math-ph/0006001
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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, and MATH. Thus one can include MATH and MATH into a chart-like subset of MATH. The next step is to show that the leaves intersected with MATH or MATH are connected. In the case of the ball it is enough to show that MATH is concave on MATH for MATH. It is enough to show that the Hessian MATH of MATH on MATH cannot have a large negative eigenvalue under an appropriate choice of constants MATH and MATH. This Hessian is a sum of a non-negative part MATH and of MATH. In turn, it is enough to show that MATH can be made bounded by REF/REF. Since MATH can be bounded by MATH, it is enough if we can bound second derivatives of MATH as MATH. However, the estimates on MATH, MATH, given above allow one to estimate second derivatives of MATH in terms of derivatives of MATH. This finishes the proof of MATH-convexity in the case of the ball. Investigate strict MATH-convexity in the case of the ball. It is clear that one can invert MATH and write MATH. It is enough to prove that the MATH-image of a small ball is convex, which follows from the following simple There is a number MATH (which depends on MATH only) such that given a function MATH on MATH such that MATH satisfies REF with MATH in MATH, then the image MATH is convex for MATH. Investigate the case of the polydisk. The stronger assumptions we have in the polydisk case allow ensuring MATH for any given MATH. Now the statement follows from the following Given MATH, there are numbers MATH and MATH which satisfy the following condition. Given a smooth function MATH defined on MATH and any numbers MATH, and MATH, if MATH satisfies MATH on MATH, then MATH is convex, and MATH is connected if non-empty for any MATH. The statement is obvious in the real case, so assume complex-analytic situation. Start with the case MATH. Put MATH, MATH. We may assume MATH, MATH, then MATH on MATH. Thus the direction of the tangent line MATH to the curve MATH rotates counterclockwise when MATH grows, with the angular velocity being close to REF. This implies convexity of MATH. The connectivity of MATH is obvious. In the case MATH the convexity follows from similar arguments: the boundary of the image of MATH is the curve MATH; here MATH are appropriate functions, MATH, and the direction of the tangent line the curve MATH behaves as in the case MATH. For connectivity proceed by induction in MATH. We may assume that MATH, MATH. Increasing MATH and MATH, one can ensure that MATH is given by MATH if MATH, and MATH satisfies REF with MATH taken instead of MATH. Thus MATH is connected if non-empty. On the other hand, MATH is diffeomorphic to MATH. Since MATH is convex, it is connected, thus MATH is connected as well. This finishes the proof of REF .
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math-ph/0006001
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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 assume that MATH. Then given an estimate REF for MATH, we can estimate MATH from below, and MATH from above in MATH, loosing REF units in MATH. In particular, MATH satisfies REF with MATH or MATH.
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math-ph/0006001
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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.
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math-ph/0006001
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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 neighborhood of MATH is NAME if MATH is NAME. Indeed, any bundle over a NAME manifold with a fiber isomorphic to a disk MATH is CITE. In the case MATH one needs to consider MATH functions MATH instead of one, and replaces MATH by MATH (using results of CITE.)
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math-ph/0006001
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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, and MATH to consists of leaves of MATH, MATH, which intersect MATH finishes the proof.
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math-ph/0006001
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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 space MATH with a NAME curve. Take MATH to be the symmetric power MATH, MATH, and let the NAME curve consists of MATH-st powers of elements of MATH. Then MATH can be identified with homogeneous polynomials of degree MATH of two variables (two coordinates on MATH), thus MATH provides such a polynomial MATH up to a constant. It is easy to check that MATH iff MATH does not vanish at the points of MATH in the direction of MATH. There are at most MATH such directions, and given such directions with appropriate multiplicities, one can find a polynomial MATH which vanishes at these points.
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math-ph/0006001
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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.
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math-ph/0006001
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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 curve MATH in MATH-plane the subset MATH is MATH; here MATH for the curves MATH in MATH-plane, MATH for the curves MATH in MATH-plane. It is clear that for a curve with any other direction MATH and MATH. In particular, it is so for the curve MATH given by MATH. Thus MATH. Thus MATH, MATH satisfy conditions of REF , thus one can glue the twistor transform MATH from two open subsets, one being a bundle over MATH, another over MATH. Moreover, MATH, and if MATH, then MATH. In such a case MATH can be glued from two open subsets, one being a bundle over MATH, another over MATH. To describe MATH, it is enough to describe the gluing function MATH, MATH, for small MATH. Taking MATH as the coordinate on MATH and MATH as the coordinate on MATH, one can describe this gluing function in the following way: take a point MATH on MATH, find the leaf of MATH which passes through MATH, and intersect this leaf with MATH. Then MATH is the MATH-coordinate of the point of intersection. Consider the surface MATH given by the equation MATH. The foliation MATH can be described by the equations MATH; here the derivative of MATH is given by REF . The curves cut out by this foliation on MATH have both MATH and MATH as normal vectors. Thus these curves are tangent to directions MATH (notations as in REF). One can easily check that the ODE of the theorem describes MATH-projections of these curves for MATH. Thus MATH. The only thing to prove is MATH. In fact MATH. To check this, it is enough to find the intersection of the leaf of MATH through MATH with MATH. As above, the direction of this curve is given by MATH. Again, it is easy to check that this agrees with REF.
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math-ph/0006001
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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.
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math-ph/0006001
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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 points of MATH are disks, with the exception of the points MATH, preimages of which are isomorphic to MATH. Cutting out far-away points of MATH together with an appropriate neighborhood on MATH, one may ensure that the resulting manifold is a disk bundle over MATH. Each blow-up decreases the degree of the normal bundle by REF, thus we reduced the statement to the case MATH, and MATH. Show that this leads to contradiction. Indeed, topological bundles with the fibers being oriented disks are isomorphic iff their boundaries are isomorphic as bundles with a fiber being oriented circles. In turn, any such bundle is isomorphic to a spherical bundle of a line bundle over MATH, which is determined by its degree up to an isomorphism. We conclude that the topological bundle MATH is isomorphic to a neighborhood of MATH-section in the total space of MATH, MATH. However, MATH has no continuous nowhere-REF sections: indeed, such a section would give a trivialization of the spherical bundle of MATH, thus, due to arguments given above, to an isomorphism of MATH with MATH.
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math-ph/0006001
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Take MATH, and apply REF to MATH and MATH.
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math-ph/0006001
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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 MATH. Put MATH, then MATH passes through MATH, MATH.
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math-ph/0006001
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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-modules with local sections being sections of MATH which vanish at MATH. By the NAME semicontinuity theorem CITE, the NAME characteristic MATH of MATH does not change when MATH changes, and the individual terms MATH are semicontinuous from above. Similar results hold for MATH considered as functions of MATH and MATH. Consider the exact sequence of sheaves MATH REF; here MATH is the skyscraper sheaf with the fiber over MATH being MATH. Since the mapping MATH of taking the value at MATH is surjective on global sections, the cohomological long exact sequence shows that MATH for MATH and MATH. This implies MATH for MATH if MATH, thus MATH does not depend on MATH and MATH. Now a consideration of the long exact sequence for MATH shows that MATH is surjective for MATH and MATH. This implies that MATH is a surjection. By REF , a neighborhood of MATH is the twistor transform of a web on an open subset MATH, MATH. Since MATH (and, by similar arguments, MATH for any MATH) is an injection, this web is separating. Prove airiness. One can find a neighborhood MATH of MATH in MATH such that MATH implies MATH. Indeed, let MATH. It is a neighborhood of MATH, thus one can apply REF to MATH instead of MATH. Obviously, the resulting neighborhood MATH of MATH satisfies the requirement above. Let MATH. Now any section MATH of MATH has a form MATH for MATH. Obviously, this implies also MATH. On the other hand, a section of a twistor transform of MATH induces a section of MATH, thus the restriction of the web on MATH is airy.
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math-ph/0006001
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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 twistor transform.) This provides a construction of MATH and MATH.
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math-ph/0006001
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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, zeros of this section correspond to points on MATH. Thus it is enough to construct a hyperplane in MATH which transversally intersects MATH in MATH points. Since all the NAME inclusions are projectively isomorphic, it is enough to consider one given by MATH. Let MATH. Then the functional MATH with coordinates MATH satisfies the condition above.
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math-ph/0006001
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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 zeros on MATH. Then MATH is a well-defined function on the unit circle with values in MATH, and MATH; here MATH are numbers of zeros of MATH (inside and outside of the unit circle correspondingly). In our case MATH, and MATH.
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math-ph/0006001
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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 bundle of MATH inside MATH. We know that MATH. On the other hand, MATH is isomorphic to MATH, thus is a trivial vector bundle over MATH. Thus both MATH and MATH are cohomologically trivial. The exact sequence MATH shows that MATH is also cohomologically trivial, and MATH. In other words, the NAME - NAME theory REF is applicable, and there is an mapping MATH and MATH, such that MATH, and the associated infinitesimal family MATH is a bijection. On the other hand, MATH is a deformation of a constant mapping to a point MATH, thus is a constant mapping itself. Denote the image-point of this constant mapping by MATH. It is clear that the derivative of MATH coincides with the composition MATH, thus MATH is a local diffeomorphism. Thus we can identify MATH with an open subset of MATH. We obtain a family of mappings MATH, MATH, such that MATH is the constant mapping to MATH. In other words, MATH is a section of MATH, thus induces a pair of functions MATH. This shows existence of solutions MATH, as well as the analytic dependence on parameters. Uniqueness follows from REF - NAME theory REF .
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math-ph/0006001
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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.
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math-ph/0006001
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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 that the infinitesimal family is non-vanishing, but infinitesimal family vanishes for MATH. However, by REF , the infinitesimal family is a section of MATH, thus cannot vanish at REF distinct points. Show that MATH is MATH-admissible. Glue domains MATH and MATH together by gluing MATH to MATH for MATH, MATH. Call the resulting MATH-dimensional manifold MATH. It is equipped with a projection MATH to MATH and a section MATH of this projection. As in the proof of REF , one can show that degree of MATH is REF. Since MATH, MATH is cohomologically trivial, and fibers of MATH are generated by global sections. By REF , a neighborhood MATH of MATH in MATH is a twistor transform of a web of codimension REF on a manifold MATH. Since MATH, MATH. Again, MATH implies that MATH of this web spans a quadratic cone in MATH, thus this web is a NAME web. Taking a point MATH, MATH, gives a leaf of the foliation MATH on MATH. Since fibers of MATH over MATH, MATH, MATH and MATH are identified with subsets of MATH by the construction of MATH, this gives REF functions MATH, MATH, MATH, MATH on MATH, each constant on leaves of MATH. We may assume that MATH for an appropriate function MATH defined in a neighborhood of MATH. By definition, the latter function is MATH-admissible. On the other hand, a point MATH induces a section of MATH. A section of MATH which is close to MATH is determined by two functions MATH and MATH which satisfy REF. By REF this section is determined by MATH, MATH, and MATH, moreover, MATH. We conclude that MATH. This implies the first statement of the theorem. The second statement is a direct corollary of the first one. Given MATH, find MATH as in REF . By a projective transform of MATH one can make MATH, and MATH. By transformations MATH one can make MATH arbitrarily small. After this a transformation MATH with MATH would produce a triple MATH with desired properties. Given a non-degenerate solution MATH of the MATH-equation and MATH as found above, consider the MATH-dimensional NAME web defined by REF . Let MATH is the twistor transform of this web. Then MATH is (locally near MATH) isomorphic to MATH; here MATH is glued using the function MATH defined in REF . It is clear that MATH. Functions MATH, MATH, MATH, MATH on the manifold of this NAME web define local coordinates on the fibers MATH, MATH, thus on MATH. Denote these coordinates by the same symbols MATH, MATH, MATH, MATH. Investigate how these coordinates are related to coordinates MATH, MATH on two pieces of MATH it is glued of. Recall that REF defined the coordinate MATH on MATH by taking MATH-coordinates of the intersection point of leaves of the foliations with MATH, which is the MATH-axis. The leaves of the foliation MATH are MATH, thus the coordinates MATH and MATH on MATH coincide, and there is no translation of the argument MATH of the function MATH. The coordinate MATH over MATH is induced by taking MATH-coordinate of the intersection point of leaves with MATH. Thus the coordinates MATH and MATH on MATH coincide, and there is no translation of the argument MATH. Similarly, the coordinates MATH and MATH on MATH differ by the transformation MATH; here MATH, MATH are the equations of the curve MATH. Finally, the coordinate MATH on MATH is given by taking the value of MATH on the leaf of MATH. The leaf which corresponds to a given value of MATH passes through the point MATH, thus the corresponding value of MATH is MATH.
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math-ph/0006001
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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.
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math-ph/0006001
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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 correspondingly. Then MATH, MATH. Since operators MATH are linear and continuous in an appropriate topology, it is easy to see that MATH.
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math-ph/0006007
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It is clear that MATH if MATH, and one has: MATH, which proves REF . The proof of REF is similar.
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math-ph/0006007
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is straightforward using REF and the relation MATH .
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math-ph/0006007
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It suffices to check that in the case MATH one has: MATH, which is immediate.
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math-ph/0006007
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is straightforward using the facts that MATH, and MATH iff-MATH.
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math-ph/0006007
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It follows from REF that MATH . Since MATH, the proposition follows from REF b.
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math-ph/0006007
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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.
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math-ph/0006007
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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.
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math-ph/0006007
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REF is proved in REF are easily checked.
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math-ph/0006007
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The lemma follows from REF applied to the simple roots MATH, and REF applied to MATH, since, due to REF we have: MATH .
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math-ph/0006007
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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 assumption of the lemma. Hence REF holds for some non-negative integer MATH. Similarly, REF holds for some non-negative integer MATH. Similarly, applying REF to the union of sequences MATH and MATH , we conclude that MATH . Hence, adding up REF we get MATH . Now REF follows from REF .
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math-ph/0006007
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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 MATH. Due to REF , we may assume that MATH . Consider the sequence of odd roots MATH introduced in the proof of REF and let MATH, MATH, and notice that MATH . Let MATH be the highest weight vector of MATH with respect to MATH. Due to REF b, it remains to show that conditions listed by REF b imply that MATH . Recall that by REF we have: MATH where MATH. Let MATH for short. Then REF gives for some MATH, MATH, that MATH. In view of REF , we have: MATH . Hence, due to REF we get: MATH proving REF , since MATH.
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math-ph/0006007
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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 .
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math-ph/0006007
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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 .
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math-ph/0006007
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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 (respectively, MATH), and the subalgebra MATH of MATH acts on MATH via MATH, where the representation MATH of MATH on MATH is known to be integrable of level MATH (see CITE). Thus, the representation of MATH in each MATH is integrable. The irreducibility of MATH, provided that MATH, is proved using REF in exactly the same fashion as the proof of REF from CITE.
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math-ph/0006007
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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 Right-hand side is equal to MATH . Next, we have: MATH . These equalities prove REF .
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math-ph/0006007
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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, by ``completing the squares". The lemma is proved.
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math-ph/0006007
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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 MATH. Due to REF a, the local finiteness (respectively, integrability) with respect to MATH (respectively, MATH) implies that MATH (respectively, MATH). Hence, REF are necessary. Furthermore, it follows from REF , that in the cases MATH (respectively, MATH) the element MATH (respectively, MATH) is locally nilpotent. It remains to show that in the case of REF the element MATH is locally nilpotent iff REF holds, and in the case MATH, MATH is locally nilpotent iff REF holds. We shall concentrate on the first claim, the second being easier (compare also CITE). Introduce the following isotropic roots: MATH . We have: MATH for all MATH, and MATH . Let MATH, then MATH and we let MATH. Similarly MATH and we let MATH. We have: MATH . Let MATH denote the highest weight of MATH with respect to MATH. It can be computed by making use of REF . Introduce the following numbers: MATH . Using REF , MATH and REF , we get the following recurrent formula for the MATH's: MATH . In view of REF , the local nilpotency of MATH follows from MATH. This, clearly, holds if MATH (by REF ), which again shows that in this case there are no supplementary conditions. From now on we may assume that MATH. We may also assume that REF hold. We shall derive a recurrent formula for the MATH. Using that, by REF , MATH (respectively, MATH) if MATH (respectively, MATH) and that MATH, we obtain: MATH . For MATH, the recurrent formula involves numbers MATH which, using the above arguments, can be expressed in terms of the labels of MATH as follows: MATH . Then we have MATH: MATH where we let MATH . Note that MATH since MATH . Since MATH for MATH, formulae REF imply MATH . Furthermore, we have MATH . In order to show this, it suffices to prove that MATH, MATH. But, due to REF , MATH can take place for MATH only when MATH, which is impossible by REF . Also, MATH-respectively, MATH if MATH (respectively, MATH) cannot be positive since in this case MATH, which implies that MATH (see supplementary REF ). Suppose now that MATH is locally nilpotent. Then MATH (by REF ), hence we have from REF : MATH . Due to REF , we have the following three possibilities for some MATH and MATH: MATH . The possibilities REF of the supplementary REF correspond respectively to the following cases: CASE: MATH when MATH, where we put MATH, CASE: MATH when MATH, where we put MATH, CASE: MATH, where we put MATH, CASE: MATH, where we put MATH. We consider in detail only REF , the treatment of all other cases being similar. We have: MATH for some integer MATH such that MATH. Hence, by REF we have: MATH and, by REF , we have: MATH . The recurrent REF can be now rewritten respectively as follows: MATH . Summing up these equalities, we get: MATH, which, in view of REF , implies MATH. Suppose now that REF - REF hold. We have to show that MATH. As before, we may assume that MATH, hence, due to REF holds. Since REF hold, we again have only the possibilities MATH, MATH and MATH. In cases MATH and MATH, MATH, hence only case MATH remains. This case corresponds to REF iv when we have: MATH,MATH. Hence MATH and MATH (we have used here REF iv).
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math-ph/0006007
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The only simple root of MATH which is not simple for MATH is MATH. Hence the proof of REF proves REF as well.
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math-ph/0006007
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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), the lowest eigenvalue eigenspace being MATH (respectively, MATH), which is the trivial MATH-dimensional (respectively, the standard) representation of MATH. Provided that MATH are irreducible MATH-modules, REF follows. In order to prove irreducibility of MATH, pick elements MATH as in REF and define the field MATH as in that lemma. Let MATH be an element orthogonal to both MATH and MATH, and consider the field MATH, so that MATH. Since MATH commutes with MATH, we have by REF : MATH . Let MATH be an invariant with respect to MATH subspace. It follows from REF that MATH implies that MATH, MATH. Hence MATH is invariant with respect to all operators MATH, where MATH are such that MATH and MATH. Hence, provided that MATH, MATH contains a non-zero purely bosonic element, that is, an element obtained by applying a polynomial in the MATH to MATH. Thus we reduced the problem to the purely bosonic case, that is, the case when MATH. In this case the irreducibility was proved in CITE using the character formula for modular invariant representations of MATH from CITE and REF from CITE (the reference to REF is a misprint). The remaining cases, when MATH or MATH and MATH is even MATH can be reduced again to the purely bosonic case by a direct calculation. We give below details in the MATH case, the MATH case being similar. The simple root vectors of MATH are as follows: MATH . Then the simple root vectors of MATH are MATH and MATH. Any vector MATH of MATH can be uniquely written in the form: MATH where MATH are purely bosonic elements (that is, obtained by applying polynomials in the MATH's to MATH). Now, if MATH is a singular vector, that is, MATH for all MATH, then, in particular, MATH, and since all MATH commute with the MATH's, we get: MATH . It follows that all MATH are purely bosonic singular with respect to MATH vectors, hence, due to irreducibility of MATH for MATH mentioned above, we obtain that all MATH are linear combinations of elements MATH and MATH. Hence MATH . Using that MATH, we obtain: MATH which implies that MATH respectively, MATH if MATH. Thus, the only singular vectors in MATH (respectively, MATH) are scalar multiples of MATH (respectively, MATH). To conclude that the MATH-modules MATH are irreducible, note that MATH carries a unique non-degenerate Hermitian form MATH such that the square length of MATH is MATH and the adjoint operators of MATH and MATH are MATH and MATH, respectively. The absence of non-trivial singular vectors in MATH (respectively, MATH) implies that the MATH-submodules MATH (respectively, MATH) generated by MATH (respectively, MATH) is irreducible, hence the restriction of MATH to it is non-degenerate. Hence the orthogonal complement to MATH (respectively, MATH) is a complementary submodule which has no non-zero singular vectors, hence it is zero, and MATH are irreducible.
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math-ph/0006007
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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 integral level. But one knows CITE that all irreducible modules over such a vertex algebra are integrable MATH-modules. Using the complete reducibility of MATH-modules CITE, we deduce that any MATH-module, viewed as a MATH-module, is a direct sum of irreducible integrable MATH-modules, which proves the proposition.
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math-ph/0006007
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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 for some MATH. Using the NAME operator CITE, we obtain: MATH which is equivalent to MATH. But then REF implies that MATH, proving REF .
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math-ph/0006007
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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. In the MATH and MATH cases we note that MATH is isomorphic to the vertex algebra MATH (see REF ), MATH is its irreducible module, and these two modules produce all integrable highest weights of level MATH. The cases MATH and MATH are obvious since MATH is the only irreducible integrable module of level MATH (see REF ). It remains to show that MATH is a MATH-module in the MATH case. But MATH, hence the difference of this weight and MATH does not lie in the root lattice; we also have: MATH and level MATH. Hence we may apply REF .
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math-ph/0006021
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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 projection of MATH and MATH is the quotient map, then we get a bundle isomorphism MATH by MATH . Therefore, in this representation the lift MATH of MATH acts on MATH as MATH. Sections into an associated bundle MATH are just those sections of the bundle MATH which have the appropriate transformation property. By construction, MATH is a complex line bundle over MATH, but from MATH it inherits the vector bundle structure, so its sections fulfill: MATH . An analogous equation holds for the line bundle MATH over MATH. Finally, REF shows MATH . Here, all equalities are immediate from the definitions, besides the last but one, which may be seen as follows: MATH . So, both bundles are quotients of isomorphic bundles with respect to the same MATH-action.
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math-ph/0006021
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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.
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math-ph/0006021
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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 compact group MATH, and by the NAME theorem (or simpler: by the NAME theorem) they are dense in MATH. Since the operator norm of MATH and the supremum norm of MATH coincide the commutation relation follows for all MATH by continuity.
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math-ph/0006021
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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 manifold MATH it is essentially self-adjoint. MATH is a fiber of MATH CITE and therefore self-adjoint, thus both define the same unique self-adjoint extension MATH of MATH.
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math-ph/0006021
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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 MATH, so that there is an accumulation point MATH (MATH is compact), and MATH due to continuity. Since MATH is discrete MATH is an eigenvalue of MATH. The lift of an eigensection (which is smooth due to ellipticity) lies in MATH and therefore is bounded. Furthermore the lift satisfies the same eigenvalue equation because of REF.
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math-ph/0006021
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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 .
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math-ph/0006021
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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 MATH has no open subsets of NAME measure zero the trace is faithful. We can compute the scalar product that is defined by MATH for MATH as follows: MATH . Since MATH is dense in MATH with respect to the MATH-norm and therefore with respect to the norm generated by MATH, the GNS representation space for MATH is MATH. Hence, the module structures coincide.
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math-ph/0006021
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We get the fiber at MATH as GNS representation space of the state MATH. For the continuity structure see REF.
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math-ph/0006021
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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.
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math-ph/0006021
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Let MATH be an orthonormal basis of MATH, consisting of unitary elements of MATH. Then MATH .
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math-ph/0006021
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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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