paper stringlengths 9 16 | proof stringlengths 0 131k |
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cs/9907024 | By linearity of expectation, REF prove that the expected cost at one level is MATH. Since the expected height of the structure is MATH, we get the claimed result. (The analysis is similar to the ananlysis for skip lists CITE.) |
cs/9907024 | Easy corollary of REF . |
cs/9907030 | Imagine constructing the quadtree bottom-up, by starting with a regular grid of squares and then consolidating quadruples of squares of one size to make squares of the next larger size. We color the initial grid by a regular pattern of three colors, depicted in REF . Then, when we consolidate four squares of one size to make squares of the next larger size, each larger square has only two colors among its smaller neighbors REF , forcing it to take the third color. Connected sets of larger squares then end up colored by the same regular pattern used to color the smaller grid, so we can repeat this process of consolidation and coloring within each such set. |
cs/9907030 | We form the desired quadtree by splitting squares one at a time; at each step we split the largest square possible. Thus the four smaller squares formed by each split are, at the time of the split, among the smallest squares in the quadtree. As we perform this splitting process, we maintain a valid four-coloring of the quadtree. When we split a square, we color the four resulting smaller squares. We give the upper right and lower left squares the same color as their parent. Each of the other two squares has at most four neighbors, two of which are the same color. Therefore each has at most three neighboring colors, and at least one color remains available; we give each of these two squares one of the available colors. |
cs/9907030 | An unbalanced quadtree is depicted in REF , with some of its squares labeled. A simple case argument shows that it has no three-coloring: suppose for a contradiction that we are attempting to color it red, blue, and green. Since squares MATH, MATH, and MATH are mutually adjacent, we may assume without loss of generality that they are colored red, blue, and green respectively. Since MATH is adjacent to MATH and MATH, it must be blue, and since MATH is adjacent to MATH and MATH, it must be red. Since MATH is adjacent to MATH and MATH, it must be green. But then MATH is adjacent to a red square (MATH), a green square (MATH), and a blue square (MATH), so it can not be given any of the three colors. Thus, four colors are required to color this quadtree. |
cs/9907030 | A balanced quadtree is depicted in REF . A simple case argument shows that it has no four-coloring: choose four different colors for the four squares MATH, MATH, MATH, and MATH meeting in the center vertex. Then, choose a color for one of the diagonal neighbors, MATH and MATH, of the two small center squares. Now repeatedly apply the following two coloring rules: CASE: If some square MATH has three differently colored neighbors, assign the remaining fourth color to MATH. CASE: If some square MATH has a corner shared by three other squares, each of which is adjacent to squares of some color MATH, assign color MATH to MATH since no other choice leaves enough free colors to the other squares sharing the corner. REF show the results of a partial application of these rules, for two choices of color for MATH. The third possible choice is symmetric with REF . No matter what color is chosen for MATH, these rules lead to an inconsistency at MATH: REF applies in two different ways, forcing MATH to have two different colors. Therefore the overall quadtree can not be colored. |
cs/9907030 | We form the adjacency graph of the squares in the quadtree, and apply the greedy algorithm: remove a minimum degree vertex from the graph, color recursively, then add back the removed vertex and give it a color different from its neighbors. If the maximum degree of a vertex removed at any step is MATH, this uses at most MATH colors. We can find the minimum degree vertex by maintaining for each MATH a doubly linked lists of the vertices currently having degree MATH; as we show below, at least one list will be nonempty, and it is straightforward to update these lists in constant time per step. Therefore, the overall time will be linear. Our bound of six colors then follows from the following lemma. Let MATH be a subset of the squares in a (not-necessarily balanced) quadtree. Define a big box to be a square that is not the smallest in MATH, that has at most five neighbors which are also not the smallest in MATH REF . Define a hanging box to be a square MATH that is not the smallest in MATH, that has at most three neighbors incident to the upper left corner, and at most two below or to the right; the below-right neighbors must also not be the smallest in MATH REF . Define a good chain to be a set of one or more squares all the smallest in MATH, with the following properties REF : Each square in the chain must have at most one neighbor below it; except for the bottommost square in the chain, this neighbor must be another square in the chain, adjacent at the bottom left corner. The bottommost square in the chain can be adjacent to a square MATH below it and outside the chain, but only if MATH is larger than the squares in the chain. Similarly, each square in the chain must have at most one neighbor to the right of it; except for the topmost square in the chain, this neighbor must be another square in the chain, adjacent at the top right corner. The topmost square in the chain can be adjacent to a square MATH to the right of it and outside the chain, but again only if MATH is larger than the squares in the chain. If the chain has exactly one square in it, it may have neighbors both below and to the right, as long as both neighbors are larger. Finally, define a good configuration to be any one of these three patterns: a big box, a hanging box, or a good chain. Note that all three of these configurations give a degree-five square or squares. Let MATH be any subset of the squares of a quadtree. Then MATH has a good configuration. We use induction on the number of levels in MATH. Let MATH be formed by replacing each smallest square in Q by its parent. (We think of MATH as being formed by splitting some squares in MATH and removing some of the resulting children.) Let MATH be a good configuration in MATH. First, suppose MATH is a big box in MATH. Then it is also a big box in MATH since none of its neighbors can be subdivided. Next, suppose MATH is a hanging box in MATH. If none of its neighbors is subdivided to form MATH, it is a big box in MATH. If one of its neighbors is subdivided and has a child neighboring MATH and not incident to the upper left corner of MATH, that child is a (singleton) good chain (its only below-right adjacency is to MATH itself). If MATH's neighbors are subdivided but the only children neighboring MATH are on the corner, MATH remains a hanging box in MATH. Finally, suppose MATH is a good chain in MATH. If some square of MATH is subdivided, and its lower right child is in MATH, that child is a (singleton) good chain in MATH. If not, but some squares are subdivided and have upper right or lower left children, any maximal contiguous sequence of such children is a good chain in MATH. If neither of these two cases holds, but some squares are subdivided and have only their upper left children in MATH, then some sequence of such children and of lower right children of neighbors of MATH forms a good chain in MATH. If no squares in MATH are subdivided and none of their upper or left neighbors are subdivided, each square in the chain becomes a big box in MATH. If no squares in MATH are subdivided, some upper or left neighbor is subdivided, and its lower right child is in MATH, that child is a singleton good chain. In the remaining case, any subdivided neighbor has neighboring children only on the upper left corners of squares in MATH, and all squares in MATH become hanging boxes in MATH. By the lemma above, any graph formed by a subset of the quadtree squares has a vertex of degree at most five, so the greedy algorithm uses at most six colors. This concludes the proof of REF . |
cs/9907031 | This follows by induction, as shown in REF , since the five such diamonds containing the five copies of MATH fit within the larger diamond defined by the Lemma. |
cs/9907031 | We show that, if MATH and MATH are non-adjacent vertices in the path, then there is some MATH forming an angle of at least MATH. We can assume that MATH and MATH are in different copies of MATH, since otherwise the result would hold by induction. But no matter where one places two points in different copies of the small diamonds containing the copies of MATH (depicted in REF ), we can choose one of the three interior vertices of MATH as the third point MATH forming an angle MATH. The result follows from the assumed inequality relating MATH to MATH. |
cs/9907031 | We have seen that we can choose a MATH such that the graphs MATH are MATH-skeletons. Since the endpoints of the path are at distance one from each other, the dilation of MATH is MATH. Each such graph has MATH vertices and dilation MATH. Since MATH, MATH. |
cs/9907031 | This can be shown by induction using the fact that the path from MATH to MATH is formed by concatenating the paths from MATH to MATH and MATH to MATH. |
cs/9907031 | Define a ``leaf vertex" to be the vertex opposite the hypotenuse of a leaf triangle in MATH. We prune the tree one step at a time until each vertex appears at most twice as a leaf vertex. At each step, the path corresponding to MATH (and with length at most MATH) will visit all the leaf vertices in tree order (as well as possibly visiting some other vertices coming from interior nodes of the tree). Suppose some vertex MATH appears three or more times. Then we prune MATH by removing all subtrees descending from the path between the first and last appearance of MATH (occurring between the two appearences in tree order), and we shorten the corresponding path by removing the portion of it between these two appearances of MATH. At each step, the change to MATH comes from subtracting some triangle short side lengths corresponding to the subtrees removed from MATH, as well as adding some hypotenuses of triangles from the same subtrees. Each subtracted side length that is not cancelled by an added hypotenuse corresponds to one of the edges removed from the path, so the total reduction in MATH is at most as great as the total reduction in the length of the path, and the invariant that MATH bounds the path length is maintained. After this pruning, there will be no leaves between the two appearances of MATH, and no new leaves are created elsewhere in the tree, so the invariant that the path visits the leaf vertices in order is also maintained. This pruning process removes at least one appearance of MATH, and so can be repeated at most finitely many times before terminating. |
cs/9907031 | Since MATH does not depend on the ordering of tree nodes, we can assume without loss of generality that each node's child is on the left. For any such tree, we can increase MATH by performing a sequence of the following steps: CASE: If any triangle has an angle greater than MATH, change it to one having an angle exactly equal to MATH, without changing any other triangle shapes. CASE: If any triangle has a ratio of left to right side lengths less than some value MATH, split it into two triangles by adding a vertex on the right side. CASE: Add a child to the leaf of MATH. These steps are depicted in REF . The result of this sequence of transformations is the concatenation of many triangles with angles equal to MATH, very short left sides, and right sides with length close to that of the hypotenuse. In the limit we get a curve from MATH to MATH formed by moving in a direction forming an angle MATH to MATH, namely the logarithmic spiral REF . Integrating the distance traveled on this spiral against the amount by which the distance to MATH is reduced shows that it has the length formula claimed in the lemma. Since we reach this limit by a monotonically increasing sequence of tree lengths, starting with any finite one-leaf tree, any finite tree must have length less than this limit. |
cs/9907031 | We prove the result by induction on MATH; REF forms the base case. If there is more than one leaf in MATH, form a smaller tree MATH by removing from MATH each path from a leaf to the nearest ancestor with more than one child. These paths are disjoint, and each such removal replaces a subtree with one leaf by the edge at the root of the subtree, so using REF again shows that MATH. Each leaf in MATH has two leaf descendants in MATH, so the number of leaves in MATH is at most MATH and the result follows. |
cs/9907031 | We have seen REF that we can connect any pair of vertices in the skeleton by a path with length bounded by MATH, where MATH is a tree of triangles in which all angles are at least MATH, and where MATH has at most MATH leaves. By REF , the length of such a tree is at most MATH which has the form specified in the statement of the theorem. |
cs/9907032 | The proof is carried out by induction on the structure of MATH. For the base cases we have the following. MATH . Now, we assume that the lemma holds for MATH, MATH, MATH and MATH, for example, MATH, and show it holds for all combinations of operators or negated operators, for example . MATH, MATH, MATH, MATH. We consider the cases for MATH, MATH, MATH and MATH and note that proofs for the other operators are similar (where MATH, MATH, MATH and MATH are new proposition symbols). MATH where MATH from the induction hypothesis. MATH where MATH from the induction hypothesis. MATH . |
cs/9907032 | For any PLTL formula MATH, the first step in the transformation is to anchor MATH to the first moment in time, that is, MATH. From REF we have shown that MATH. Thus, as MATH holds at the first moment in time and the transformation implies that MATH holds at every moment in time, then MATH also holds now. |
cs/9907032 | We examine the structure of MATH. There are three main types of transformation that can be applied: the removal of classical operators, the renaming of complex subformulae and the rewriting of temporal operators applied to literals. We begin by considering the removal of classical operators. First, assume MATH is a conjunction MATH, that is, MATH . Applying the MATH translation we have MATH and so we must show there is a model MATH such that MATH . Now, as MATH for all MATH, then if MATH both MATH and MATH. That is MATH . So, by setting MATH equal to MATH we have such a model. The proofs are similar for the other classical logic operators. Next, we consider renaming transformations and assume MATH is of the form MATH where MATH is not a literal. Now, assume that there exists a MATH such that MATH . By applying the MATH transformation we have MATH where MATH is a new proposition symbol. Thus, we must show that there exists a model MATH such that MATH . First assume that MATH is never satisfied in MATH. A model MATH identical to MATH except it contains the variable MATH such that MATH is false everywhere will suffice. Otherwise let MATH be the first place that MATH is satisfied in MATH. As MATH for all MATH then MATH. Let MATH be the same as MATH except it contains a new proposition symbol MATH that is satisfied in all MATH and unsatisfied elsewhere that is, MATH. Thus, as MATH is identical to MATH, except for MATH, we have MATH for all MATH and from the definition of MATH we have for all MATH, MATH and, for all MATH, MATH. Thus, from the semantics of PLTL, MATH. Now, as MATH for all MATH then MATH from the semantics of MATH. Also, as MATH and by REF is the first place MATH is satisfied in MATH and therefore MATH, MATH. Further MATH as MATH from our choice of MATH. Hence MATH as desired. The proof of other renaming operations are similar. Finally we consider the removal of unwanted temporal operators. Again, we let MATH be MATH but this time assume that MATH is a literal. Assume that there exists a MATH such that MATH . By applying the MATH transformation we obtain MATH where MATH is a new proposition symbol. Thus, we must show that there exists a model MATH such that MATH . First assume that MATH is never satisfied in MATH. Similarly to the above, a model MATH identical to MATH except containing the variable MATH such that MATH is false everywhere will suffice. Otherwise let MATH be the first place that MATH is satisfied in MATH. Let MATH be the model that is identical to MATH except it contains the variable MATH such that for all MATH, MATH and for all MATH, MATH. Thus, as MATH is the same as MATH except for the valuation of MATH, and MATH then, we have MATH . We have assumed that MATH so for all MATH, MATH hence for all MATH, MATH. Thus, as MATH, where MATH is the first place that MATH holds and for all MATH, MATH we have MATH. Now as MATH is the first place that MATH holds and MATH for all MATH we have MATH and MATH. Also, as MATH, MATH then, due to our choice of MATH, for all MATH, MATH and so MATH. Hence MATH as required. |
cs/9907032 | Firstly note that if MATH then there is a model MATH such that MATH . The model MATH is identical to MATH except it includes the new proposition symbol MATH which is set to true where MATH and false everywhere else. Applying MATH to MATH, we obtain MATH . Now, from REF , and given that MATH has a model MATH every application of the MATH transformation can be satisfied in some new model. Hence, if MATH has a model then there exists a model that satisfies MATH. |
cs/9907032 | REF above show that if MATH is satisfiable in a model, then MATH is satisfiable in the same model. REF show that, given a model for MATH, then we can construct a model for MATH. |
cs/9907032 | If MATH has a model then, ignoring the value of each MATH at each moment gives a model for MATH. Conversely, if MATH has a model MATH, then MATH can be extended to a model MATH for MATH by giving MATH the same truth value as MATH in MATH in each state, and for each literal MATH. The model MATH clearly satisfies the formulae REF from REF above. The lemma follows easily from these two observations. |
cs/9907032 | If MATH is satisfiable then MATH has a model, and by REF it has a normal model MATH. The side conditions for temporal resolution guarantee that the loop resolvents that is .formulae REF given in REF hold in MATH, and so MATH is a (normal) model for MATH, that is, MATH is satisfiable. If MATH is unsatisfiable then the addition of NAME to produce MATH is also unsatisfiable. Hence MATH is well-behaved. |
cs/9907032 | This is established by induction on the length of the shortest path from an initial node to an arbitrary node in the behaviour graph of MATH. Let MATH be the length of the shortest path from an initial node to a node MATH. To show the base case we let MATH and show that any initial node in the behaviour graph of MATH is an initial node in the behaviour graph of MATH. Let MATH be the initial NAME of MATH and MATH the initial NAME of MATH. As MATH has been constructed by adding initial and/or step NAME to MATH, MATH. Take any initial node MATH in the behaviour graph for MATH. From the definition of the behaviour graph MATH must satisfy the right hand side of the NAME in MATH. As MATH then MATH must also satisfy the right hand side of the NAME in MATH. As the set of sometime NAME in MATH and MATH are unchanged, that is, as MATH satisfies the left hand side of the same sometime NAME in MATH and MATH the set MATH will be the same in each graph for MATH and thus the node MATH is also in the behaviour graph for MATH. Next we assume that if any node MATH, where the length of the shortest path from an initial node to MATH is MATH, is in the behaviour graph for MATH, it is also in the behaviour graph for MATH. We show that any node MATH in the behaviour graph for MATH whose shortest path length from an initial node is MATH, is also in the behaviour graph for MATH. Let MATH be the step NAME in MATH and MATH the step NAME in MATH. By assumption we have MATH. Consider some node MATH in the behaviour graph of MATH where the shortest path from an initial node to MATH is MATH. Let MATH be any node in the behaviour graph for MATH such that there is an edge from MATH to MATH and the shortest path from an initial node to MATH is of length MATH. By the induction hypothesis, we assume that MATH is also in the behaviour graph for MATH. Let MATH be the set of step NAME in MATH such that the left hand sides are satisfied by MATH and the right hand side satisfy MATH. Let MATH be the corresponding set of step NAME in MATH that is, where the left hand sides are satisfied by MATH and the right hand side satisfy MATH . As MATH we have MATH. Furthermore as no change has been made to the set of sometime NAME any eventualities outstanding from MATH or triggered by MATH will be the same in each graph. Thus MATH is also present in the behaviour graph for MATH. |
cs/9907032 | To construct a model from a suitable path, MATH where each MATH, through the behaviour graph (that is, one which is infinite and all eventualities are satisfied) take the valuation MATH from each node MATH in the path (and delete any negated proposition symbols). Any proposition symbols that do not occur in MATH but are required in the model may be set arbitrarily. Details of how to construct models from behaviour graphs are given in REF . Take any model MATH for MATH. We show that this model can be constructed from a path through the behaviour graph. First delete any proposition symbols not in MATH from MATH to give MATH. As these proposition symbols do not occur in MATH they have no constraints on them so by setting these proposition symbols to true and false in the correct way we can recover MATH. Note that MATH is a model for MATH. By definition the behaviour graph for MATH is the reachable subgraph from the set of initial nodes. The behaviour graph has been constructed where the MATH component of each node consists of every possible valuation. Let MATH be the set of non-negated proposition symbols in MATH. As MATH is a model for MATH, MATH must satisfy the initial rules MATH. To construct the behaviour graph for MATH the initial nodes are those with valuations that satisfy MATH, for a particular MATH component. As nodes are constructed with each valuation and subset of eventualities there must be a node MATH where MATH. Next for some MATH in MATH assume that there is a node MATH in the behaviour graph for MATH such that MATH. We show that MATH for some node MATH in the behaviour graph for MATH. Let MATH be the set of step NAME in MATH. Take the set of step NAME MATH such that the left hand side of the NAME in MATH is satisfied by MATH. As MATH, MATH must satisfy the left hand side of the NAME in MATH. As MATH is a model for MATH, MATH must satisfy the right hand side of each NAME in MATH having deleted the next operator. From the construction of the behaviour graph, edges are drawn from MATH to nodes whose valuation satisfies the right hand side of each NAME in MATH having deleted the next operator (for some MATH component). As nodes have been constructed for all valuation/eventuality component combinations there will be one MATH such that MATH. Hence we can construct MATH using the valuations from each node and following a path through the behaviour graph for MATH. This can be extended to MATH by setting the additional proposition symbols as required. |
cs/9907032 | First assume we have performed a simplification step. We show that any node and edge that is in the behaviour graph for MATH is also in the behaviour graph for MATH. The proof of the converse is similar. The proof is by induction on the length of the shortest path from an initial node. For the base case the length of the path from an initial node to MATH is REF, that is, MATH is an initial node. If the simplification step has not been performed on an initial NAME that is, the set of initial NAME in MATH and in MATH are the same then MATH must also be in the behaviour graph for MATH. Otherwise we have performed a simplification step on an initial NAME that is, MATH contains MATH and MATH contains MATH where MATH. Each initial node MATH in the behaviour graph for MATH satisfies MATH by definition of the behaviour graph. As MATH node MATH also satisfies MATH so MATH is in the behaviour graph for MATH. Next assume the node MATH in the behaviour graph for MATH, whose shortest path distance from an initial node is MATH, is also in the behaviour graph for MATH. We show that any node of shortest path length MATH from an initial node is also in the behaviour graph for MATH. Take a node MATH in the behaviour graph for MATH whose shortest path length from an initial node is MATH. Consider MATH such that MATH is an edge in the behaviour graph from MATH where the shortest path length from MATH to an initial node is MATH. From the induction hypothesis MATH is also in also in the behaviour graph for MATH. Assume that a simplification step has been applied to REF to obtain MATH and that MATH satisfies MATH. Thus from the definition of the behaviour graph MATH must satisfy MATH. As we have performed a simplification step MATH and MATH so MATH also satisfies MATH and MATH satisfies MATH as the sets MATH and MATH are unchanged apart from this. Hence MATH and the edge MATH must also be in MATH. If the node MATH didn't satisfy MATH, or the simplification rule had been on an initial NAME then MATH would again be in the behaviour graph for MATH as the remaining rules are unchanged. The proof of the converse is similar. To show the proof holds for a subsumption step assume MATH contain rules MATH and MATH where MATH and MATH. Thus by a subsumption step MATH. The proof is similar to the above. |
cs/9907032 | There are two conditions for the deletions of nodes to form a reduced behaviour graph. Firstly nodes with no successors are deleted. No nodes are reachable from a node with no successors hence the lemma follows. Secondly nodes MATH that are deleted where MATH is an outstanding eventuality, that is, MATH but no reachable node satisfies MATH, that is, MATH. From the construction of the behaviour graph and from the conditions allowing us to delete MATH, any node MATH reachable from MATH must contain MATH as an outstanding eventuality, that is, MATH and but doesn't satisfy MATH. Thus any node reachable from MATH must also be deleted. |
cs/9907032 | Let MATH be a set of SNF NAME. An infinite path through the (unreduced) behaviour graph for MATH, starting at an initial node gives a sequence of valuations for the propositional symbols - that is, a PLTL model. By construction of the graph, this model satisfies the initial and step NAME of MATH. Furthermore, by REF any such model must arise from a path through the behaviour graph. However, not all paths give models for the full set of NAME MATH, since either the paths may not be infinite or they may fail to satisfy some eventualities (which occur within sometime NAME). If a node, MATH, has no successors, then there are no infinite paths through that node, so any model for MATH must arise from a path through the graph with MATH deleted. Thus the first deletion criterion can be applied without removing any potential models. Also, if a node MATH contains an eventuality MATH then any path through that node which is to yield a model for MATH must satisfy MATH either at MATH or somewhere later in the path. Thus, if a node contains an eventuality that cannot be satisfied then this node cannot be part of a model for the set of NAME, hence, we can apply the second deletion criterion without discarding potential models for MATH. The ``if" part of the proposition follows. To prove the ``only if" part, suppose the reduced behaviour graph for MATH, call it MATH, is non-empty. We will now use MATH to construct a model for MATH. First note that the set of initial nodes in MATH is non-empty, since, in the behaviour graph, every node is reachable from the initial nodes and any node reachable from a deleted node is also deleted (by REF ). Now, choose an initial node MATH. If MATH is non-empty, choose an ordering MATH for the literals in MATH. Since MATH has not been deleted, there is a path in MATH to a node MATH in which the eventuality MATH is satisfied. If the eventuality MATH is not present in MATH it must have been satisfied somewhere along the path. Otherwise, we can extend the path to a node MATH which satisfies MATH. Continuing in this way we can find a path MATH (which may consist simply of the node MATH if all of MATH are satisfied there) such that each element of MATH is satisfied at some point along MATH. Let MATH be a successor of the end point of MATH (it must have a successor since we have deleted all terminal nodes). Repeating our construction, we can find a path MATH beginning at MATH along which all the eventualities in MATH are satisfied. Let MATH be a successor of the end point of MATH. Repeat this construction until MATH for some MATH, which must happen eventually since MATH is finite. Let MATH be the path MATH. Then the path MATH has the property that, for each node in the path, each eventuality in that node is satisfied at some node later in the path. To see this, recall that if a node contains an eventuality MATH but does not satisfy MATH, then MATH is in the eventuality set of all immediate successors of MATH. So, either MATH is satisfied before we reach the next MATH or MATH is an eventuality in MATH and so is satisfied along MATH. Furthermore MATH is obviously an infinite path. It follows by the construction of the behaviour graph that the sequence of valuations given by MATH is a model for MATH. |
cs/9907032 | The proof proceeds by induction on the number of nodes in the behaviour graph of MATH. First we consider the effect of simplification and subsumption rules on the behaviour graph for a set of NAME. Given a set of NAME MATH let the application of simplification and subsumption rules to MATH result in the set of NAME MATH. By REF the behaviour graph of MATH is identical to that of MATH. If the behaviour graph is empty, then the set of initial NAME in MATH is unsatisfiable. By the completeness of classical resolution, we can use step resolution on the set of initial NAME to derive the empty clause. Now suppose the behaviour graph MATH is non-empty. By REF , the reduced behaviour graph is empty and so there must be a node which can be deleted from MATH. If MATH has a terminal node MATH, let MATH be the set of step NAME whose left hand sides are satisfied by MATH. Then, having deleted the next operator, the right-hand side of the NAME in MATH form an unsatisfiable set MATH of propositional clauses. By completeness of classical resolution again, there is a refutation of MATH. Choosing an element of MATH corresponding to each element of MATH, we can ``mimic" this classical refutation by step resolution inferences to derive a step NAME MATH where each MATH is a literal which is satisfied by MATH. The temporal resolution procedure allows us to rewrite NAME REF as MATH . By REF , adding NAME REF (and any other resolvents derived along the way) to MATH produces a NAME set MATH whose behaviour graph MATH is a subgraph of MATH. (MATH is in fact a proper subgraph, since MATH has no node whose valuation is MATH. If MATH was an initial node it doesn't satisfy the initial NAME REF as MATH for MATH. If MATH was a non-initial node, as the left hand side MATH is satisfied by every node in MATH the successor of any node must also satisfy MATH. As we have MATH for MATH no edges can be drawn to MATH so MATH does not contain MATH.) Furthermore, MATH is well-behaved since it has exactly the same models as MATH. By induction, MATH, and hence MATH, has a refutation. If MATH does not have a terminal node, then it must contain a node MATH such that some eventuality MATH is not satisfied at any node reachable from MATH. Let MATH be the set of nodes reachable from MATH. For each MATH, let MATH be the set of step NAME in MATH whose left-hand sides are satisfied by MATH. Let MATH be a MATH NAME that is the result of applying the MATH merging operation to the NAME in MATH. Note MATH is the conjunction of the left hand side of the NAME in MATH and MATH is the conjunction of the right-hand sides of the NAME in MATH (contained in the next operator) and MATH satisfies MATH. Note MATH and MATH are simply classical propositional formulae. Then each MATH logically implies MATH since none of the MATH in MATH satisfy MATH. Each MATH leads to a node MATH satisfying MATH for some MATH. Thus MATH must satisfy MATH or MATH. By definition each successor of a node in MATH is also in MATH (as MATH is unsatisfied in all nodes reachable from MATH). As MATH is not satisfied by any node in MATH we have MATH is unsatisfiable and thus MATH is valid (in classical propositional logic). Also each MATH logically implies the disjunction of the MATH's corresponding to the successors of MATH. As each node MATH leads to a node MATH that satisfies MATH. By REF and MATH satisfies MATH. Thus MATH is unsatisfiable. Hence MATH. Hence, we can use MATH NAME of the form REF in an application of temporal resolution. Let MATH be the disjunction of the MATH. Then each MATH satisfies MATH. For each node MATH in MATH either there is a NAME MATH in MATH and the valuation at MATH satisfies MATH, or for each predecessor MATH of MATH the valuation at MATH satisfies MATH. Let MATH be the result of adding the loop resolvents REF from REF, and let MATH be the behaviour graph for MATH. Then MATH has no nodes from the set MATH. So MATH is a proper subgraph of MATH by REF and MATH is well-behaved by REF . Once again, it follows by induction that there is a refutation for MATH. |
cs/9907032 | Following the translation to normal form the set of NAME is augmented so no new proposition symbols are required during the proof. Hence we have a finite number of proposition symbols. Further, there are a finite number of right and left hand sides we may obtain as initial and step NAME modulo ordering of the conjunctions or disjunctions. Simplification rules mean that the left or right hand sides cannot grow indefinitely. Note that the number of sometime NAME does not change. Thus REF either generates MATH and terminates or we have tried to resolve each NAME with every other and obtained no new NAME that is, something that isn't in the set already (modulo ordering of conjunctions/disjunctions). The argument is similar for the termination of REF . Having augmented the set of NAME with the new proposition symbols needed to translate resolvents from temporal resolution into SNF, at some point no new resolvents will be generated as we have a finite set of possible NAME. |
cs/9907032 | The proof is by induction on the length of MATH. The base case is where MATH has length REF, that is, it has the form MATH, MATH, MATH, MATH, MATH. As illustrated in REF MATH produces one NAME, MATH produces two NAME and NAME produces two NAME (where const is MATH, MATH, MATH or MATH) and MATH produces one NAME. In each case if the number of NAME produced is MATH, MATH . For the inductive hypothesis we assume that the theorem holds for formula of length MATH and examine each case for length MATH. Again, by considering the proofs in REF, the maximum number of NAME from removing any operator (or negated operator) is REF (from MATH). MATH . The cases for the other operators are similar. |
cs/9907032 | Let MATH be a PLTL formula. To transform it into SNF we apply the MATH transformation that is . MATH . From REF we know the maximum number of NAME from MATH is MATH; hence, the maximum number for the translation of MATH is MATH. |
cs/9907032 | The proof is by induction on the length of MATH. The base case is where MATH has length REF, that is, it has the form MATH, MATH, MATH, MATH, MATH. Each of these produces no new proposition symbols so as MATH we are done. For the inductive hypothesis we assume that the theorem holds for formulae of length MATH and examine each case for length MATH. Again we examine some of the cases involved. MATH . The cases for the other operators are similar. |
cs/9907032 | Let MATH be a PLTL formula. To transform it into SNF we apply the MATH transformation that is . MATH . From REF we know the maximum number of new proposition symbols from MATH is MATH. Hence the maximum number for the translation of MATH is MATH. |
cs/9907032 | Consider a loop MATH in MATH formed from the set of NAME MATH. Let the disjunction of the left-hand side of the MATH NAME in MATH be MATH. As MATH is a loop the right-hand side of each MATH NAME in MATH implies both MATH and MATH. Assume there are MATH NAME in MATH. Each MATH NAME (or an equivalent MATH NAME) in MATH must be in the set MATH before deletions as MATH has been made by combining NAME in MATH. We next consider the deletion of any MATH NAME in MATH from MATH. REF will not remove any of the MATH NAME in MATH from MATH as it removes MATH NAME whose right-hand side do not imply MATH but, by assumption, each MATH NAME in MATH has a right-hand side that implies MATH. Assume we are about to remove a MATH NAME MATH, contained in MATH from the set MATH using REF. Let MATH be the disjunction of the left-hand sides of the MATH NAME remaining undeleted in MATH that are not in MATH. Thus MATH is being deleted as it is not the case that MATH. However we know that MATH, as MATH is a loop, so MATH must also hold giving a contradiction. Hence none of the MATH NAME in MATH can be deleted from MATH so the algorithm must return a set of MATH NAME containing MATH. Consider any set of MATH NAME MATH output by the algorithm. Each MATH NAME has been made by combining NAME in MATH. Each right-hand side implies MATH otherwise it would have been deleted by REF. Each right-hand side implies the disjunction of the left-hand side of the set of MATH NAME otherwise it would have been deleted by REF. The set of MATH NAME satisfies the side conditions for being a loop, hence this loop can be constructed by combining the relevant NAME in MATH. |
gr-qc/9907067 | The property MATH is immediate when one uses the fact that MATH is simple, which implies MATH. REF follows from direct calculation. The projector property MATH follows from MATH and the fact that MATH. Finally, MATH follows from MATH and the four-dimensionality of twistor space. |
gr-qc/9907067 | The proof is straightforward once we have made the following observation. From REF we obtain for any MATH and similarly for the other term, which equals MATH. Thus, MATH maps MATH which is obviously an isomorphism if MATH. |
gr-qc/9907067 | A further consequence of REF is MATH . Thus, MATH is a two-form with values in MATH and by REF it is of the form REF at points with MATH. The fact that MATH is hermitian and trace-free follows from hermiticity and vanishing trace of MATH. We extend this form for the curvature by continuity to the points with MATH. Then we obtain, that for regular MATH some parts of the local twistor curvature must vanish. This is in complete analogy to the conformal field equations where the regularity of MATH implies that the NAME curvature vanishes on MATH. The first form of the field equation for MATH follows from the NAME identity MATH, while the second form arises either by simple manipulation of the first, using the fact that the infinity twistor is covariantly constant, or by observing that MATH. |
hep-th/9907013 | As MATH is a MATH-cone we have MATH, and as MATH is diffeomorphic to MATH we have MATH. It is not difficult to show that the metric on MATH is isometric to the quotient of MATH with its flat Euclidean metric by the lattice MATH with basis MATH, MATH. The eigenvectors of MATH on MATH lift to functions of the form MATH on MATH which are invariant under lattice translations. Calculation shows that the only eigenvalue of MATH in MATH is REF, with multiplicity REF, and so MATH. Similarly, MATH has eigenvalue REF with multiplicity REF, and as MATH and the subgroup MATH of MATH preserving MATH is MATH with dimension REF, we see that MATH is rigid as MATH. |
hep-th/9907013 | As MATH and MATH is a subgroup of MATH, we see that MATH is isomorphic to MATH, MATH or MATH. But using NAME duality ideas for manifolds with boundary, we can show that the map MATH must have image and kernel MATH, and this forces MATH. |
hep-th/9907013 | Let MATH or REF and MATH satisfy MATH and MATH, and let MATH be as in Conjecture REF. Suppose also that MATH is small enough that MATH, to ensure that MATH, and define MATH, MATH and MATH as in REF. As the MATH depend smoothly on MATH they are all diffeomorphic for fixed MATH, and so it is enough to prove the proposition for sufficiently small MATH. Using the exact sequence MATH and MATH, we find that MATH, and by excision and the diffeomorphism MATH we see that MATH. Thus MATH. Recall that MATH by REF . It is easy to show using this and facts about manifold topology that MATH is finite. As MATH is diffeomorphic to MATH we see that MATH is isomorphic to MATH, and is generated by MATH. But the argument in REF shows that MATH is homologous in MATH to MATH in MATH, and MATH is homologous to REF in MATH. Thus, the image of MATH in MATH is zero. Furthermore, no smaller positive multiple of MATH can have zero image in MATH, because MATH represents MATH, which by definition generates the kernel of MATH. Now consider the exact sequence MATH . We have shown above that MATH, which is finite, and that the image of MATH in MATH is isomorphic to the cyclic group MATH. Thus by exactness MATH, and so MATH is finite with MATH, as we have to prove. Since MATH is finite we have MATH, so MATH by NAME duality. But MATH is connected (this is part of our definition of manifold) and oriented, as it is special Lagrangian. So MATH is by definition a rational homology REF-sphere. |
hep-th/9907013 | As MATH is a cone on two disjoint copies of MATH we have MATH, and as MATH is diffeomorphic to MATH we have MATH. There are two connected components MATH of MATH, each of which is isometric to the unit sphere MATH in MATH with the round metric. For both MATH the only eigenvalue of MATH in MATH is REF, with multiplicity REF, and eigenvectors the restriction to MATH of linear functions on MATH. So MATH. Similarly, MATH has eigenvalue REF with multiplicity REF on each of MATH, and eigenfunctions the restrictions to MATH of harmonic homogeneous quadratic polynomials on MATH. As MATH and the subgroup MATH of MATH preserving MATH is isomorphic to SO REF with dimension REF, we see that MATH is rigid as MATH. |
hep-th/9907013 | As MATH are homology REF-spheres we have MATH, so the condition MATH in MATH in REF holds automatically. Thus, it follows from REF that for some MATH we can extend MATH to smooth families MATH of compact, nonsingular, embedded SL REF-folds in MATH for MATH. As these are smooth, connected families the MATH are clearly diffeomorphic and homologous to MATH. Now MATH as MATH is a homology REF-sphere, so MATH is rigid by REF , that is, it has no special Lagrangian deformations. Thus the families MATH are unique. To intersect transversely in one point is an open condition on pairs of submanifolds, so by making MATH smaller if necessary we can suppose that MATH and MATH intersect transversely in one point, which will be a positive intersection as MATH. |
hep-th/9907042 | This is clear from the construction of MATH. |
hep-th/9907042 | It is well known that MATH is a subgroup CITE. It can be written as MATH where MATH is the isotropy subgroup of MATH under the adjoint action MATH, MATH of MATH on itself. Because the intersection of an arbitrary set of closed sets is closed, it suffices to prove that all isotropy subgroups MATH are closed. The action of MATH on itself leads, restricted to a fixed element MATH, to the continuous map MATH. MATH is the preimage of a closed set (because MATH is NAME) under a continuous map, and hence closed. Now, MATH is a closed subset of a compact group and therefore compact. |
hep-th/9907042 | According to REF we have to show that the image of MATH - which is identical to the image of MATH - is closed. We will start by showing that the domain of definition of MATH, i. CASE: MATH, is compact. The elements of this space take values in a compact set, because we know in the first place from REF that MATH is compact. The generalized NAME fields take values in a compact manifold, too. Such a NAME field has several components given by the linear map MATH subject to the linear REF . Therefore, the values of MATH lie in a linear subspace of MATH, and exponentiating yields generalized NAME fields taking values in a compact submanifold of MATH. MATH can be constructed as a projective limit along the lines described in Subsection REF. The projective family consists of compact spaces, because the maps involved assign elements of compact spaces to a finite number of edges and vertices of graphs. Therefore, the projective limit is compact in its induced NAME topology which is equivalent to the induced topology as a subset of MATH. Now, as stated in Subsection REF, the NAME topology is equivalent to the NAME topology. But MATH is continuous in the NAME topology (being by construction a continuation of a continuous function on the dense subspace MATH), and so the image of MATH is compact. As a compact subset of a NAME space it is closed. |
hep-th/9907042 | According to REF we have to prove that the support of a distribution in the image of MATH contains only MATH-invariant connections. This amounts to showing that for any non-invariant generalized connection MATH there is a neighborhood MATH of MATH so that the restriction of any distribution MATH to MATH is the zero distribution. Because of REF MATH has a neighborhood which is entirely contained in MATH. If we restrict MATH to this neighborhood, it will be the zero distribution due to its very definition in REF : The pull back with MATH of any function supported on the neighborhood of MATH will be zero. |
hep-th/9907042 | Let MATH be a symmetric state. If MATH are two cylindrical functions that are identical when restricted to MATH-invariant connections, then MATH. Let us, therefore, introduce the following equivalence relation which respects the algebraic and topological structure of MATH. Two cylindrical functions MATH are equivalent if and only if MATH. Symmetric states can be seen as functions on the space of equivalence classes, and we need to show that, if MATH, there is a distribution MATH with MATH. Let us look now at the function MATH where MATH is not trivial. As noted earlier, it may not be cylindrical, not even integrable. But if it is cylindrical, it will correspond to a symmetric state obeying MATH. If it is not, we can approximate it by a sequence of cylindrical functions which is obtained by projecting it onto cylindrical subspaces, cylindrical with respect to graphs of an increasing net constructed as follows: If MATH will lie in MATH, but not in MATH, then it will be a countably infinite sum of terms, each being cylindrical with respect to a finite graph, the union of all these graphs being an infinite graph. The projections will be obtained by truncating to a finite number of these graphs, their number tending to infinity in the sequence mentioned. (This is reminiscent of the well known approximating sequences of the MATH-distribution or, more generally, of an approximate identity in an algebra without identity.) All the functions MATH in the sequence will fulfill MATH proving our assertion. Separation now follows from linearity, but can also be proved directly: We pick a representative for each equivalence class of cylindrical functions in MATH, and for each representative MATH the cylindrical function MATH, if it lies already in MATH, or else an appropriate element of the sequence approximating MATH. The term appropriate means, that any linear relation between the chosen functions reflects a linear relation between equivalence classes. An appropriate selection can always be done, because the sequences approximate the functions MATH which are different for different classes. Thereby, we obtain a class of distributions separating the equivalence classes. |
hep-th/9907042 | Let MATH be the space of functions on the space MATH of generalized gauge invariant and MATH-invariant connections, and MATH its topological dual. Due to the definition in REF of MATH, MATH can be identified with the space of MATH-symmetric states as defined in REF . According to REF , MATH is in separating duality with MATH. This implies that MATH is dense in MATH in the weak topology CITE, and, therefore, in the space of MATH-symmetric states. (Note, however, that the topology on MATH induced from the weak topology by the continuous map MATH is coarser than the topology in which MATH is completed to MATH.) In conclusion, the map MATH is injective (by construction of MATH - where the NAME constraint REF is assumed to be solved - and by the construction of MATH), and has a dense image in the space of MATH-symmetric states. |
hep-th/9907042 | We have already seen, that the MATH-relations MATH and MATH are represented properly. Because MATH and MATH are canonically conjugate, we have the NAME bracket MATH . The only non-vanishing matrix elements of the commutator MATH are given by MATH . We, therefore, have the relation MATH implementing the correct representation of the NAME structure. |
math-ph/9907002 | Since there is no possible confusion, we drop the superscript MATH. Fix MATH and let MATH . By REF .ii and since MATH is a core for MATH, the domain of MATH contains the dense subset MATH. Now on MATH, one can write MATH . Thus for MATH small enough, one has MATH which implies that MATH is uniformly bounded in MATH. From REF one obtains MATH which is an operator uniformly bounded in MATH for all MATH small enough. Hence, the operator MATH is bounded. Now one has MATH . From REF, since MATH maps MATH into MATH, one obtains for the term in parenthesis in the right hand side of REF, in the weak sense MATH . Furthermore, by writing MATH and MATH one obtains, together with REF MATH . Then for all MATH, MATH . The constant MATH is independent of MATH. Taking the limit MATH, one gets for all MATH . This proves REF. Now, since we know that MATH maps MATH into MATH, one has for all MATH and MATH in MATH: MATH . Since the integrand in REF is uniformly bounded in MATH and since MATH is dense in MATH, one can extend REF to all MATH, which gives REF. |
math-ph/9907002 | We will drop the label MATH in the proof. Since MATH is self-adjoint and MATH is closed, we have for MATH CITE MATH where MATH and MATH is the NAME function. In order to bound the right-hand side of REF, we split the range of integration over MATH into two parts and fix MATH . Since MATH, one can bound the term REF from above by MATH . We now treat the term REF. Let MATH; then obviously, since MATH is in MATH, MATH and MATH are finite and from REF one has, by using the equality MATH . Therefore MATH is in the set MATH defined in REF . By writing MATH, one can prove with the same arguments as above that MATH. Then we get from REF MATH and, using REF .iii, REF is bounded above by MATH . To bound the term REF one writes MATH . Finally, for REF one has MATH . REF imply that there exists MATH independent of MATH such that MATH . As to the remaining contribution to REF, we write MATH . The first term in REF is bounded from above by MATH . According to the fact that MATH and MATH are both in the set MATH defined in REF, one can prove as above with the help of REF that each of the two terms in REF are bounded above by MATH . This inequality together with REF implies REF. |
math-ph/9907002 | Without loss of generality we assume that MATH and MATH. The proof is done in three steps. First we consider MATH with MATH, where MATH . Let us introduce the following notation MATH . If MATH is MATH-regular, we apply REF to get MATH . In case MATH is MATH-regular, we apply the adjoint of REF to obtain the bound REF with MATH replaced by MATH and MATH replaced by MATH on the right-hand side of REF . In any case one has for all MATH, with MATH, the estimate MATH . Second we derive the upper bound MATH for all MATH with MATH. To this end we note that MATH where MATH is defined as in REF. Then, MATH where we have used REF in the integral over MATH. For these MATH we proceed by bounding the resolvents according to MATH. This yields the term MATH in REF. For MATH we use MATH which is valid for all MATH with MATH, and the NAME REF , viz. MATH, giving the second term in REF. Finally we derive REF as follows. Given MATH, there exists MATH such that MATH . Thus we get, for some constants MATH and MATH . |
math-ph/9907002 | We assume here, without loss of generality, that MATH; then we obtain, for MATH, MATH and MATH . For fixed MATH and MATH we define MATH . Thus, according to REF , we get MATH . By using the inequality MATH in REF, we derive the upper bound MATH . Here MATH is a constant depending only on MATH and MATH. This gives the expected result for an appropriate dependence of MATH on MATH: for MATH such that MATH and if MATH, we take MATH. We thus obtain MATH. |
math-ph/9907002 | Let MATH, REF and MATH. We first apply MATH times the geometric resolvent equation to MATH with the increasing sequence of boxes MATH, MATH, MATH where MATH is the characteristic function of the support of MATH. Now, again with the help of the geometric resolvent equation, observe that for MATH . In each term of this last sum, the first factor is estimated by using MATH, and the second factor by using MATH. Thus, REF together with REF gives MATH for MATH large enough. Remark that here we have used the following inequality (see for example, CITE or CITE), for all MATH, MATH and MATH: MATH . |
math-ph/9907002 | Let MATH and MATH be such that MATH. We denote by MATH (respectively, MATH) the complementary events of MATH (respectively, MATH). We start by estimating the probability of the complement of the event that appears in REF MATH where MATH. The probability in the first term in REF is bounded from above by MATH where MATH and MATH, MATH are, respectively, the spectral family of MATH and MATH. Now REF is bounded from above by MATH . In REF we have used REF . The second term in REF is estimated as follows MATH for some uniform constant MATH depending only on MATH, MATH and MATH. In the last inequality we have used REF . Now, REF give MATH . |
math-ph/9907010 | Apply the universal property to the MATH-algebra MATH with the zero endomorphism. |
math-ph/9907010 | By the universal property of MATH and MATH, there is a unique MATH-algebra map MATH such that MATH. It suffices to show that MATH is a coalgebra map. First, we show that MATH is a map of augmented algebras, that is, MATH. By initiality of MATH, it suffices to show that the composite MATH is a map of MATH-algebras; in other words, that MATH. To prove this, apply MATH to the cocycle condition MATH. Using that MATH, and MATH (as observed above), this yields MATH. Thus MATH, as desired. Next, we show that the map MATH preserves coproducts. Observe that, by initiality of MATH, the square necesarily commutes as soon as all four maps are MATH-algebra homomorphisms. The map MATH is the only one for which this still has to be shown. But, we have just proved that MATH is a map of augmented MATH-algebras, so MATH by naturality. Since alse MATH, the map MATH is indeed a map of MATH-algebras. This completes the proof of the theorem. |
math-ph/9907010 | The second assertion for MATH follows from the first for MATH by adjointness. To see that the map MATH preserves the coproduct, simply apply initiality of MATH to the square exactly as in the proof of REF . |
math-ph/9907010 | Let MATH be a well-pointed object. Let MATH be the map MATH obtained from the unit MATH of this free algebra together with the canonical map MATH. We claim that MATH is the universal base-point preserving map from MATH into a MATH-algebra. Indeed, suppose MATH is any map into the underlying object MATH of a MATH-algebra MATH, with MATH. Since MATH is the free algebra, the restriction MATH extends uniquely to a MATH-algebra map MATH. It is easy to check that MATH for this map MATH. |
math-ph/9907010 | The map MATH is the one of REF . The map MATH is the unique map MATH of MATH-algebras, for the map MATH defined as follows. Since MATH has an augmentation MATH with MATH REF , we can write MATH where MATH maps MATH into MATH. Also, the free MATH-algebra MATH, briefly MATH, is augmented, hence splits as MATH. Now define MATH on these two summands separately: on MATH it is the composition MATH and on the other summand it is the map MATH where MATH is the restriction of MATH. Note that the map MATH thus defined satisfies the identities MATH where MATH is the universal map as in the proof of the previous lemma. We claim that MATH. By adjointness, it suffices to show MATH as maps of pointed objects. Now MATH as we have seen. Also, MATH is obtained from MATH by adjointness, hence MATH. Thus MATH. This shows that MATH and MATH are both maps of MATH-algebras on MATH, hence equal by initiality. It remains to observe that MATH respects the augmentation. Since MATH and MATH are both maps of MATH-algebras, so is the composite MATH. So MATH by initiality of MATH. This shows that MATH preserves the augmentation, and completes the proof. |
math-ph/9907022 | A direct calculation. Alternately, we can normalize MATH so MATH. Then MATH. |
math-ph/9907022 | Note that if MATH, and MATH, then MATH . Thus, by REF, MATH . By NAME 's inequality, MATH . Since MATH is maximized at MATH when it is MATH, we see that MATH is finite if MATH, so we can pick MATH with MATH and find (using the explicit value of MATH) in that case MATH which proves REF . |
math-ph/9907022 | Suppose that MATH. Let MATH . Then REF for each fixed MATH, MATH strongly. It follows that MATH . Thus, MATH, which implies that MATH. Now suppose MATH. Then MATH . Thus MATH uniformly in MATH which, given that MATH, implies that MATH. |
math/9907002 | For MATH, MATH is the line MATH and is therefore irreducible. For MATH, MATH also has dimension REF and it is not hard to check that MATH is irreducible over MATH. It therefore follows from REF below that MATH is irreducible. |
math/9907002 | Let MATH and MATH be two real varieties such that MATH. We shall see that MATH or MATH. At least one of the MATH must be of dimension REF. Assume for instance that this is true of MATH, and let MATH be a polynomial such that MATH. Since MATH has dimension REF, the complex variety MATH also has dimension REF. By irreducibility of MATH, MATH is irreducible. It follows that MATH and therefore MATH, that is, MATH. |
math/9907002 | If MATH is an odd number, the real varieties defined by MATH and MATH are distinct (they are in fact disjoint). Hence MATH is not irreducible in this case. If MATH is even, MATH has no real zeros. MATH is therefore irreducible by REF . |
math/9907002 | We shall prove that irreducibility is not definable in dimension REF. The result for higher dimensions follows immediately from this special case. Assume by contradiction that there exists a formula MATH of MATH such that for any real variety MATH of MATH, if we interpret MATH by membership to MATH then MATH if and only if MATH is irreducible. Let MATH be the formula obtained from MATH by replacing each instance MATH of MATH in MATH by MATH. By REF , MATH is true whenever MATH is an even integer, and is false whenever MATH is an odd integer. This is in contradiction with the o-minimality of the real exponential field (see CITE and the references there). |
math/9907003 | The constant sequence MATH is in MATH since it is realized by taking MATH to be a singleton. The condition in REF is closed under addition. On the other hand, if MATH and MATH are exactly realized by systems MATH and MATH, then MATH exactly realizes MATH. For each MATH define a sequence MATH by MATH for MATH and MATH for MATH or MATH. Then MATH, where MATH. Since for each MATH the sequence MATH has only finitely many terms not equal to MATH, the product MATH is an element of MATH with infinitely many non-trivial factors. Finally, the sequence MATH is a non-trivial prime in MATH. |
math/9907003 | Assume that MATH with MATH, MATH, and that MATH. After multiplying the divisibility REF by the least common multiple of the denominators of the (rational) coefficients of MATH, we produce a polynomial with integer coefficients satisfying REF . It is therefore enough to assume that the coefficients MATH are all integers. Let MATH and MATH be the periodic points and least periodic points in the corresponding system MATH, and let MATH be any prime. By REF , MATH so MATH and therefore MATH divides MATH for all primes MATH, showing that MATH. Now let MATH be another prime, and recall that MATH . Since MATH, MATH and by REF MATH so MATH by REF . It follows that MATH for all primes MATH and MATH (since MATH is certainly divisible by MATH). So MATH taking the limit as MATH of MATH shows that MATH. This contradiction proves the first statement. There are many multiplicative sequences in MATH: if MATH is any multiplicative sequence, then so is the corresponding sequence MATH (see CITE). A multiplicative sequence MATH is completely multiplicative if MATH for all MATH. Assume that MATH is completely multiplicative, with MATH the realising sequence. For MATH a prime and any MATH, MATH by REF . It follows that MATH . With MATH this implies that MATH for all MATH, MATH. Now MATH for all MATH and MATH. It follows that MATH mod MATH for all MATH, so MATH for all MATH. It follows that MATH for all primes MATH, so MATH for all MATH. |
math/9907003 | First assume that MATH. Then, by the assumption, the sequence MATH is a multiple of the sequence MATH which is in MATH because the subshift of finite type corresponding to the matrix MATH realizes it (and therefore any multiple of it). Conversely, assume that MATH is a sequence in MATH satisfying REF . Write MATH for the sequence MATH and MATH for the sequence MATH both satisfying the recurrence REF . Notice that MATH for integers MATH and MATH. By REF , for any prime MATH . On the other hand, it is well-known that MATH mod MATH (where MATH is the NAME symbol), and MATH mod MATH (by the previous paragraph: MATH is in MATH). So REF implies that MATH for all primes MATH. We now claim that the NAME symbol MATH is MATH for infinitely many values of the prime MATH. This completes the proof of REF , since REF forces MATH and hence MATH is a multiple of MATH, namely MATH. To see the claim, choose MATH such that MATH and the NAME symbol MATH. Then by NAME, there are infinitely many primes MATH with MATH mod MATH and MATH mod MATH. For such primes, MATH, which completes the proof. |
math/9907003 | CASE: Assume that MATH and let MATH be the corresponding sequence of periodic points. Then MATH, so MATH is bounded. It follows that MATH is bounded, and hence MATH for all large MATH. This implies that MATH is bounded, and so MATH, which contradicts the assumption. CASE: For MATH this follows from REF . Suppose therefore that MATH. Then there is a sequence MATH with MATH, so for MATH a prime, MATH, and therefore MATH as MATH. Since MATH is an integer, it follows that MATH for all large MATH. Now let MATH be another large prime. Then MATH so MATH . Fix MATH large and let MATH tend to infinity to see that MATH which is impossible. The same argument shows that MATH cannot have any positive limit as MATH. For MATH, let MATH, where the product runs over prime divisors only. Then MATH so MATH . CASE: This is clear: for MATH the sequence is eventually MATH; for MATH the construction used in REF . works. |
math/9907003 | CASE: If MATH then MATH also, since MATH for all MATH. If MATH, then (for MATH large enough to have MATH) MATH . For each such MATH, choose MATH so that MATH and MATH. Then MATH . CASE: It is enough to show that if MATH then MATH also. For MATH, MATH . Let MATH be an upper bound for MATH and pick MATH. Choose MATH so that MATH . Then for MATH (so that MATH), MATH and the bracketed expression converges to MATH as MATH. Taking logs and dividing by MATH gives the result. CASE: Write MATH for the sequence of primes. Let MATH, and define a sequence MATH as follows. For MATH not of the form MATH, define MATH . For MATH of the form MATH define MATH according to the following scheme: MATH and so on. Then MATH off the MATH's clearly. Along the sequence MATH, MATH so MATH . On the other hand, along a subsequence of MATH's chosen to have MATH for a fixed MATH (which will exist by construction), we realize MATH as a limit point of the sequence MATH. |
math/9907003 | CASE: Let MATH be an upper bound for MATH. Then MATH . The converse is obvious. CASE: One direction is clear. Assume that MATH. Fix MATH; choose MATH so that MATH where MATH. Choose MATH so that MATH . Then for MATH, MATH . CASE: Assume that MATH. Then MATH along primes. For a fixed prime MATH, MATH as MATH. CASE: Assume that MATH. Then MATH along primes. For fixed prime MATH and MATH prime, MATH as MATH. |
math/9907004 | NAME 's result REF shows that MATH for any first order deformation MATH. The class MATH is of total degree two, and so MATH . |
math/9907005 | It is a consequence of the following Propositions. |
math/9907005 | Let us set MATH, and MATH. Then MATH, so that the hypothesis becomes MATH. Besides, as MATH, MATH is nonincreasing and convex, so that MATH, that is MATH, and the thesis is implied by MATH, but, as MATH, setting MATH, the thesis follows from MATH. So we have to prove that MATH implies MATH. Suppose on the contrary that there are MATH such that MATH, for MATH. Then, for MATH, we have MATH, which implies MATH, and MATH, so that MATH and MATH, which is absurd. |
math/9907005 | Recall that MATH, is positive, nondecreasing, concave, and MATH, MATH, as we can multiply MATH by a suitable positive constant without altering our statement. Then MATH is nondecreasing and concave, and MATH, so that MATH is positive, nonincreasing, MATH, MATH, and MATH. Finally MATH, so that MATH, and the hypothesis becomes MATH, that is for any MATH there is MATH such that MATH for all MATH. Observe now that MATH, so that MATH is equivalent to MATH, and, as MATH, our thesis will be proved as soon as we can show that MATH. So suppose, on the contrary, there are MATH such that MATH, for all MATH. From REF it follows that MATH for all MATH. As MATH, we can choose MATH, and MATH such that MATH, for all MATH. Then we have MATH, for all MATH, so that MATH from which it follows MATH . As REF it equivalent to MATH, for all MATH, it follows that MATH, when MATH. Therefore, with MATH, we have MATH, for all MATH. This implies MATH, which we integrate in MATH for MATH, to obtain MATH that is MATH, and choosing MATH small enough, we get MATH, for all MATH near MATH. But this contrasts with REF , so we are done. |
math/9907005 | It is a consequence of the following Propositions. |
math/9907005 | Recall that MATH, is positive, nondecreasing, concave, and MATH, MATH. Then MATH is nondecreasing and concave, and MATH, so that MATH is positive, nonincreasing, MATH, MATH, and MATH. Finally MATH, so that MATH, and the hypothesis becomes MATH . Observe now that MATH, so that MATH is equivalent to MATH, and, as MATH, our thesis will be proved as soon as we can show that MATH. So suppose, on the contrary, there are MATH such that MATH, for all MATH. From REF it follows that for all MATH there is MATH such that MATH, for all MATH, so that MATH for all MATH. As MATH, we can choose MATH, and MATH such that MATH, for all MATH. Then we have MATH, for all MATH, so that MATH, for all MATH, and this is absurd. |
math/9907005 | Setting MATH, as MATH is a decreasing and convex function, MATH, so that MATH implies the thesis. Let us assume on the contrary that there are MATH such that MATH, for all MATH. Then MATH, so that MATH, for some positive MATH. Then MATH, for all MATH, therefore MATH, for all MATH, so that MATH contrary to the hypothesis. |
math/9907005 | Let us set MATH, MATH, and assume MATH is never REF and MATH, otherwise the proof is obvious. Let MATH, MATH. Then MATH and MATH . Also MATH and MATH . Define, for any MATH, MATH, and, for any MATH, MATH. Then MATH and MATH and MATH. Indeed, defining MATH, and MATH, one proves that MATH and MATH, and the rest follows easily. Finally, since MATH one gets, taking the MATH, MATH . Analogously, since MATH one gets, taking the MATH, MATH . Putting all this together one gets MATH . The equality MATH is proved in the same way (using MATH and MATH). |
math/9907005 | Let MATH, MATH. Then MATH for MATH, that is, there exists a subsequence MATH such that MATH . Then, setting MATH, for any MATH we have MATH . By the arbitrariness of MATH, the limit is MATH, that is, MATH, which implies the thesis. |
math/9907005 | The result obviously follows by REF and the definition of the NAME measures. |
math/9907005 | By REF hence (compare REF ) MATH, therefore, by REF , MATH is eccentric and finally, by REF , we get the existence of a singular trace MATH. The trace property for the functional MATH is proved as in CITE. |
math/9907005 | MATH. As MATH, and MATH, for any MATH, we may restrict to the case MATH. By hypothesis we have that for any MATH for sufficiently large MATH. As a consequence, if MATH, MATH hence it is a summable sequence, which implies MATH, that is, MATH. Conversely, if MATH, MATH and this implies that MATH, that is, MATH. The thesis follows. MATH. For any MATH, let MATH be such that MATH, and write MATH for MATH and assume for simplicity that MATH. Then MATH . Taking the limit for MATH one gets that MATH exists, hence MATH by MATH, and also MATH namely MATH . Assume for the moment that MATH, and denote by MATH (the same as MATH of REF). Then, by a NAME theorem, MATH . Therefore MATH is eccentric if and only if such limit is one, that is, when MATH. If MATH, then denoting by MATH (the same as MATH of REF), the calculation above, suitably modified, shows that MATH is eccentric if and only MATH. |
math/9907005 | MATH . |
math/9907005 | Recall from REF , that MATH, as MATH, where MATH. Then from REF it follows MATH, as MATH, so that the thesis follows from REF . |
math/9907005 | From REF it follows that MATH . Therefore, if MATH, MATH and the thesis follows from REF . |
math/9907005 | MATH . Let MATH and observe that MATH, where MATH because MATH is a semigroup of class MATH on MATH. So that, with MATH, we get the thesis. |
math/9907005 | Set MATH for the supremum. Then for all MATH, there is MATH such that MATH, for all MATH, and, by previous proposition, MATH. Conversely MATH, for all MATH implies MATH. |
math/9907005 | We begin by proving that MATH. Recall that MATH where MATH. Then it is easy to see that MATH. For the reversed inequality, let MATH, MATH be such that MATH, and MATH be such that MATH. Let MATH be open neighbourhoods of MATH, respectively MATH, of finite measure such that MATH, for any MATH, and let MATH, MATH. Then MATH . So that MATH follows. Therefore to prove the thesis it suffices to show that MATH. As MATH is obvious, we show the opposite inequality. Let MATH, and MATH, MATH, MATH be as above. Then MATH where the last inequality follows from the continuity of MATH, if we choose MATH small enough. Analogously MATH. Then using the estimates proved above and NAME inequality, we obtain MATH and from the arbitrariness of MATH we get the thesis. |
math/9907005 | Assume on the contrary that there are MATH, MATH such that MATH. Then, by continuity, there is an open neighbourhood MATH of MATH such that MATH, for any MATH. Let MATH, so that MATH which is absurd. |
math/9907005 | In the following we use the notation MATH, where MATH to say that there are MATH, MATH such that MATH, for any MATH. As MATH and MATH are equivalent on MATH, and using REF , we get MATH . Therefore MATH . |
math/9907005 | In the following we use the notation MATH, as in the proof of REF . Let us denote by MATH the kernel of the integral operator MATH, and observe that MATH where the last relation follows from the fact that MATH is compact and MATH. Therefore MATH . Then, using REF , we get MATH . |
math/9907005 | First we choose a fundamental domain MATH in MATH and denote by MATH the projection given by the multiplication operator by the characteristic function of MATH, MATH. Then denote by MATH the ball of radius MATH in MATH, namely the set of elements which can be written as words of length MATH in terms of a prescribed set of generators for MATH. For any selfadjoint operator MATH acting on MATH and any MATH set MATH and note that MATH has finite propagation, hence it belongs to MATH. Observe then that if MATH is bounded, MATH is bounded too. Indeed MATH . Also, if MATH is periodic, MATH is periodic too. Indeed, for any MATH, MATH . Since MATH converges weakly to MATH, the thesis follows. |
math/9907005 | In this case MATH is given, for MATH, by MATH where MATH is the identity element, and we have chosen MATH. |
math/9907006 | Suppose that MATH. Then MATH is invertible in MATH, so MATH is invertible, and thus MATH, showing that MATH. As MATH pick a nonzero vector MATH. Then MATH, so MATH. And similar, as MATH, pick a nonzero MATH. Then MATH, so MATH, which proves the first assertion. For the second assertion suppose first that MATH and MATH. Then, by definition of spectrum, MATH is not invertible. But MATH is strictly positive, and therefore invertible, showing that MATH is invertible, a contradiction. Therefore there are no (strictly) negative numbers in MATH. Next suppose that the imaginary part of MATH is nonzero. Then we may write MATH for numbers MATH with MATH and MATH. Thus by spectral calculus CITE MATH belongs to MATH and clearly MATH which is impossible according to the previous part of the proof. Therefore the imaginary part of MATH cannot be nonzero, nor can MATH be negative, so MATH, as desired. |
math/9907006 | First note that MATH and MATH assure that MATH is a collection of nonempty subsets of MATH containing MATH. If MATH, then MATH, because MATH, so MATH. Let MATH. Note that MATH, so MATH holds for all MATH, and furthermore MATH as MATH. Therefore MATH being an idempotent intertwining the (strongly) irreducible MATH, has to be either MATH or MATH. Let MATH and suppose that MATH. Then MATH, because MATH and so MATH having MATH as the only solution within MATH. Thus MATH as well. If MATH, MATH, is the unit of MATH, then MATH, and thus MATH for any MATH. Hence MATH or MATH, because if MATH, then MATH, saying that MATH, and so MATH. We have thus shown that MATH is an ultrafilter over MATH. It remains to show that it is in fact MATH-complete. Suppose by ad absurdum that we have a countable (disjoint) partition MATH of MATH such that MATH, that is, MATH for all MATH. Define MATH by MATH for MATH, MATH, and MATH for MATH, MATH, whereas MATH for MATH and MATH. Then MATH and so MATH which shows that MATH for all MATH. But MATH and MATH with MATH. Hence we may identify MATH with MATH and MATH with MATH, and using spectral calculus and REF , we get for all MATH that MATH which is impossible, as the spectrum of any element in a NAME algebra is nonempty CITE, and as REF says, both MATH and MATH consist solely of positive numbers. |
math/9907006 | Let MATH be a partition of MATH, MATH. We must show that there exists MATH such that MATH. Define a collection of subsets of MATH by MATH . We claim that MATH is a MATH-complete ultrafilter over MATH: Clearly MATH implies that MATH, and MATH consists of nonempty subsets only. If MATH, then as MATH is a partition of MATH, we have MATH so MATH. If MATH and MATH, then MATH as MATH. Furthermore, if MATH then MATH, because MATH as MATH is a partition. Finally, if MATH is a countable partition of MATH, then MATH with MATH is a countable partition of MATH. Thus there exists a number MATH such that MATH, that is, MATH, so MATH is a MATH-complete ultrafilter over MATH. We conclude that MATH is a principal filter. Indeed, according to REF, the least cardinality for which there exists a nonprincipal MATH-complete ultrafilter is measurable, therefore MATH is principal, hence there exists MATH such that MATH as desired. |
math/9907006 | The first assignment is well defined, as MATH for two equivalent representations MATH and MATH of MATH. As for the second assignment notice that MATH is defined using MATH and a chosen collection of bases for each MATH. To prove that it is well defined, we therefore need to prove that MATH is independent of this choice of bases. Thus say that we have two collections MATH and MATH of bases for the spaces MATH and form the two associated irreducible representations MATH and MATH of MATH on MATH, respectively. We need to prove that they are equivalent. As MATH and MATH are collections of bases, we know that to each MATH there exists an invertible matrix MATH taking the basis in MATH from MATH to the basis in MATH from MATH. Define for any (finite dimensional complex) square matrix MATH the set MATH . The collection MATH of subsets of MATH, where the index MATH ranges over all quadratic matrices MATH, is clearly a partition of MATH into no more parts than there are real numbers. As MATH is a MATH-complete ultrafilter, it follows from REF that there exists a square matrix MATH such that MATH. Let MATH and consider the operators MATH, MATH or MATH, as matrices with respect to the standard bases REF . Furthermore, let MATH be as in the definition of the representations MATH. Then as MATH is a filter, we have MATH . As we noted just before REF , we thus get for any MATH, with MATH the matrices with respect to the bases for MATH from REF , that MATH . Hence we have proved that MATH and MATH are equivalent, and thus that the second assignment in the theorem is well defined. We now prove that these assignments are inverses to each other. If we start with MATH, form MATH and MATH, then we clearly end up with MATH. Thus we only need to see that going in the other direction gives us the identity map on MATH. Let MATH, and form MATH as prescribed; MATH where now MATH is an irreducible representation of MATH on MATH. To construct the irreducible representation MATH, we choose and fix a collection of bases for all the spaces MATH. Denote by MATH the natural number MATH associated to the MATH-complete ultrafilter MATH such that MATH. Define MATH by MATH . We complete the proof by showing that MATH, and we proceed in five steps: First REF we claim that MATH . To see this notice that for MATH, we have MATH thus MATH and MATH . Secondly REF, obviously MATH equals MATH when the operators are considered as matrices with respect to the bases for the spaces MATH and MATH. Thirdly REF we have MATH . For let MATH, define for MATH with respect to bases for MATH and MATH, the set MATH and consider the partition MATH of MATH. By adding MATH to this partition, we get a partition of MATH into no more parts than there are real numbers. It follows then from REF , that there exists a unique MATH such that MATH. Then MATH, because MATH . Now MATH, so MATH . From REF above we have MATH. By nontriviality of MATH, we get that MATH is isomorphic to MATH, as otherwise MATH would be a nontrivial ideal in the simple algebra MATH. Thus MATH is isomorphic to the algebra MATH, and as MATH is irreducible, MATH is isomorphic to MATH. We can now add the final step of the proof. By definition of the representation MATH we have for MATH, that MATH where MATH . But from above, we see that MATH by the identification of MATH with MATH, so MATH and MATH as wanted. |
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