paper stringlengths 9 16 | proof stringlengths 0 131k |
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quant-ph/0008047 | We have MATH . Since MATH, the result follows. |
quant-ph/0008047 | Suppose we had equality. A protocol MATH attaining this bound would certainly have to be p.p.t.; thus, if we apply this protocol to MATH, the output fidelity will take the form MATH for some constants MATH and MATH, or equivalently MATH for constants MATH, MATH. Evaluating this at MATH, MATH, we find: MATH . On the other hand, at MATH, we have MATH . Since the coefficients are both positive, we conclude that MATH, MATH. In particular, MATH must take MATH to MATH. Now, consider the action of MATH on the state MATH. Since MATH takes the pure state MATH to the pure state MATH, we conclude that MATH must take MATH to a state of the form MATH; by symmetry, we conclude that MATH. But then tracing away the other copy of MATH, we find that MATH takes MATH to MATH. On the other hand, we have MATH . We thus obtain a contradiction, and the theorem follows. |
quant-ph/0008055 | We proceed by assuming that another product state MATH orthogonal to all these exists, and showing that a contradiction results. Expanded in the basis above, MATH must have at least one non-zero term, which we will call MATH. That is MATH with MATH. Now, since simultaneous cyclic permutations of the MATH and MATH . NAME space leaves GenTilesREF invariant, we can always relabel things: MATH where MATH is understood to be mod MATH. If MATH this is the form we will proceed with. Otherwise, we transform instead to MATH, then we interchange MATH and MATH with a shift of REF (another symmetry of this tile set) to obtain MATH. In either case the state is written as MATH with MATH. Now the real work will consist of narrowing down what the . can consist of, given that this state must be orthogonal to all the states in the set GenTilesREF, and must be a product state. In order for MATH to be orthogonal to all the MATH states, it must be of the form MATH . The state MATH must have terms other than the ones shown, since otherwise it could not be orthogonal to MATH. We consider two cases: The easy case is B only has support on MATH. In this case we can write MATH for some MATH, MATH, and some MATH. But now MATH. So this case is ruled out; the other case involves considerably more work: B has support beyond MATH. In this case we can write MATH as MATH for some MATH, MATH and MATH (if there were only terms with MATH, MATH would not be a product state). In fact, since for a product state the MATH state must be independent of the result of a projection by MATH onto his basis, the state must therefore have the form MATH REF gives a graphical depiction of this state for MATH and MATH. The rest of the argument is easiest to follow using this series of pictures. REF shows the additional constraints on the state arising from the orthogonality with the set of states MATH (if MATH had been MATH, we would have used orthogonality with MATH to proceed instead). In symbols, this state is MATH but at this point it is much easier to understand graphically. Next we again impose the constraint that the state have a product form; graphically this requirement can be explained by saying that all the columns and rows have to be proportional to one another. Thus we ``fill out the rectangle" with MATH-s as in REF . In symbols, the state now has the description MATH . We now invoke orthogonality with MATH to fill in the additional MATH-s in REF . (We will now dispense with the algebraic expressions altogether.) Then again requiring a product state gives REF . To arrive at REF we invoke the orthogonality for MATH, MATH, except MATH. Requiring the product form brings us to REF . Now by enforcing orthogonality for MATH we find that MATH, so the entire state is constrained as in REF . But this state is just MATH, so it fails to be orthogonal to all the states. Therefore, an additional product state does not exist, and GenTilesREF is a UPB. |
quant-ph/0008055 | We show that this set is a UPB, for all MATH and MATH, by the same methods as before. Since GenTilesREF has less symmetry than GenTilesREF we will have to examine more cases, but the methods will be the same. We will number the following paragraphs to indicate the structure of the cases being considered. In all cases we begin by assuming that there is an additional product state MATH with nonzero amplitude on some basis state MATH; we will examine all possible values of MATH and MATH. CASE: NAME case. Suppose that MATH, so that the state is MATH . Orthogonality with MATH gives MATH . Note that the outcome would have been the same if we had started with assuming that the amplitude of MATH was nonzero. By relabeling of the NAME space, this case will cover any initial MATH which lies in a small tile. CASE: Consider the case where the MATH part of the product state is MATH. Orthogonality with MATH requires that REF have other nonzero terms: MATH . But then MATH, so this case is excluded. CASE: Consider the case where the MATH part of the product state is not just MATH. Then we know that MATH . Here MATH is REF or REF, and MATH. We now go down to two other subcases, depending on whether the additional term is in a small or a large tile. CASE: MATH is in a small tile (REF illustrates the case MATH, MATH). Orthogonality with respect to MATH gives REF ; then the product constraint, which requires rows REF and MATH to be proportional in the example shown, gives REF . Orthogonality with respect to MATH and MATH gives REF . The product state condition, which requires rows MATH and MATH to be proportional, gives MATH and brings us to REF . Orthogonality with MATH takes us to REF , another application of the product state condition gives REF . Finally, repeated application of orthogonality with MATH, MATH, and the product-state condition, fills in the whole state with MATH-s as in REF . But this is just MATH, so MATH cannot be orthogonal to all of REF in this case. CASE: MATH is in a large tile. REF shows the case MATH, MATH, it will be easy to see that all cases are equivalent. Orthogonality with MATH gives REF . The product state condition brings us to REF . Orthogonality with MATH and MATH gives REF , and the product state condition gives REF . Now, orthogonality with MATH requires MATH. The remainder of the reasoning follows the same track from REF ; this case is excluded. CASE: NAME case. That is, we assume that MATH has at least one non-zero element, and it is in a large tile. All large tiles are equivalent by relabeling of the NAME space, so we can pick one, say MATH, MATH. Then by orthogonality to MATH we get REF . There have to be more non-zero entries since MATH must be orthogonal to MATH. We consider different subcases corresponding to which other entry is nonzero. CASE: Component in MATH or MATH. This immediately brings us back to REF . Component in another short tile MATH. This is illustrated for MATH in REF , using the orthogonality to MATH. Invoking the product state condition gives REF ; invoking orthogonality to MATH (not shown) brings us back to the situation of REF . Component in another long tile; we illustrate this for MATH, invoking orthogonality to MATH, in REF . The product state condition brings us to REF ; invoking orthogonality to MATH (not shown), we again return to REF . This covers all cases; all methods of constructing the product state MATH lead to a contradiction. Thus, GenTilesREF is a UPB. |
quant-ph/0008059 | (See CITE for the original proof.) First, attach to the superposition of MATH the (two qubit state) MATH . Then, in superposition, add modulo MATH the values MATH to this register (step MATH). Finally, apply a general phase change MATH (step MATH). It is straightforward to see that this yields the desired phase change according to the equation: MATH for every MATH in the superposition. |
quant-ph/0008059 | See CITE for the original proof by NAME, and CITE for the single query version of it. |
quant-ph/0008059 | See the original article by CITE, or better yet, the excellent analysis of it by NAMECITE . |
quant-ph/0008059 | See the original article CITE, or any other standard text on combinatorial objects CITE. |
quant-ph/0008059 | First, we use the amplitude amplification process of NAME 's search algorithm CITE described in REF to create exactly the state MATH with no more than MATH queries to MATH. (See the article by CITE for a derivation of this upper bound. Obviously, no queries are required if MATH.) After that, following REF , one additional MATH-call is sufficient to insert the proper amplitudes, yielding the desired state MATH. |
quant-ph/0008059 | First, prepare the state MATH with MATH queries to the function MATH REF . After that, measure the state in the basis spanned by the vectors MATH. Because MATH is a weighing matrix, this basis is orthogonal and hence the outcome of the measurement gives us the value MATH (via the outcome MATH) without error. |
quant-ph/0008059 | We will prove these bounds by considering the decision trees that describe the possible classical protocols. The procedure starts at the root of the tree and this node contains the first index MATH that the protocol queries to the function MATH. Depending on the outcome MATH, the protocol follows one of the three outgoing edges to a new node MATH, which contains the next query index MATH. This routine is repeated until the procedure reaches one of the leaves of the tree. At that point, the protocol guesses which function it has been querying. With this representation, the depth of such a tree reflects the number of queries that the protocol uses, while the number of leaves (nodes without outgoing edges) indicates how many different functions the procedure can distinguish. For a probabilistic algorithm with error probability MATH, we need to use decision trees with at least MATH leaves. Because the number of outgoing edges cannot be bigger than MATH, a tree with depth MATH has maximally MATH leaves. This proves the first lower bound via MATH. For the second and third bound we have to analyze the maximum size of the decision tree as it depends on the values MATH and MATH. We know that for every index MATH, there are only MATH different functions with MATH. This implies that at every node MATH the joint number of leaves of the two subtrees associated with the outcomes MATH and MATH cannot be bigger than MATH. Hence, by considering the path (starting from the root) along the edges that correspond to the answers MATH, we see that a decision tree with MATH queries, can distinguish no more than MATH functions. The second bound is thus obtained by the resulting inequality MATH. (The case MATH is the strongest example of this bound.) In a similar fashion, we can use the observation that there are exactly MATH functions with MATH for every node MATH. Now we should consider the binary subtree that is spanned by the edges that correspond to the answers MATH and MATH. With depth MATH, this subtree has at most MATH leaves and MATH internal nodes. For the complete tree, each such internal node MATH gives at most MATH additional leaves, which are the functions with MATH. In sum, this tells us that the total tree (with depth MATH) has a maximum number of leaves of MATH, leading to the third result: MATH. |
quant-ph/0008059 | See the proof of REF in the appendix. |
quant-ph/0008059 | Consider, like in the proof of REF , a decision tree with nodes MATH and corresponding query indices MATH. For every index MATH there is exactly one function with MATH. For the tree this implies that every node MATH can only have two proper subtrees (corresponding to the answers MATH and MATH) and one deciding leaf (the case MATH). Hence, a decision tree of depth MATH can distinguish no more than MATH different functions. In order to be able to differentiate between MATH functions, we thus need a depth MATH of at least MATH. |
quant-ph/0008059 | Let MATH be the set of possible values of MATH at a given moment during the execution of the protocol. (Thus, initially we have MATH, and we want to end with the unique answer determined by MATH.) Below we will show that if MATH has at least MATH elements, then there always exists an index MATH such that all three possible answers to the query ``MATH?" lead to a reduced the set of options MATH with MATH. By repeating this procedure no more than MATH times, we can reduce the initial set MATH to four possibilities, which can then be checked with three additional queries. What follows is the existence proof of such an index for every possible subset MATH. Given a set MATH and an index MATH, we have a partition of MATH in three subsets according to the answer MATH to the query ``MATH?" MATH . Note that, depending on whether MATH is an element of MATH or not, MATH is either MATH or the empty set. Clearly, one of these three sets will be the reduced set MATH mentioned in the first part of the proof. Define the NAME matrix MATH by MATH, and let MATH be the characteristic vector MATH of the subset MATH. The product of MATH and MATH yields a new vector MATH with the following property for its MATH-th entry: MATH . By the near orthogonality property of the NAME sequence REF we know that for the matrix MATH we have MATH, where MATH is the `all ones' matrix of dimension MATH. The inner product MATH is the NAME weight of MATH and hence equals the size MATH of the set MATH. This implies for the inner product of MATH with itself: MATH. By the previous equation for MATH, we thus see that MATH . This proves that there exist at least one index MATH for which MATH, and hence MATH. In combination with the general bound MATH, this gives that for this MATH both MATH and MATH are less than MATH. For MATH this proves that indeed all three MATH, MATH and MATH have size less than MATH. |
quant-ph/0008059 | We exhibit the quantum algorithm in detail. We start with the superposition MATH . (The reason for the ``dummy" part of state that we use will be clear later in the analysis.) The first oracle call is used to calculate the different MATH values for the non-dummy states, giving a superposition of states MATH. At this point, we measure the rightmost register to see if it contains the value ``zero". If this is indeed the case (probability MATH), the state has collapsed to MATH which directly gives us the desired answer MATH. Otherwise, we continue with the now reduced state MATH on which we apply a conditional phase change (depending on the MATH values in the rightmost register). We finish the computation by `erasing' this rightmost register with a second call to MATH. (For the dummy part, we just reset the value to ``zero".) This gives us the final state MATH, depending on MATH, of the form MATH . What is left to show is that MATH forms a set of orthogonal vectors. REF tells us that for the inner product between two states MATH and MATH it holds that MATH if MATH, and MATH if MATH. In other words, the states MATH for MATH are mutually orthogonal. Hence, by measuring the final state in the MATH-basis, we can determine without error the shift factor MATH after only two oracle calls to the function MATH. |
quant-ph/0008059 | (For the quadratic character, these are simple instances of so-called NAME sums CITE; see for example REF.) Rewrite MATH with MATH. If MATH this sum equals MATH. Otherwise, we can use the fact that MATH (for MATH) to reach MATH . Earlier we noticed that MATH, and therefore in the above summation (where the value MATH is omitted) we have MATH. This confirms that indeed MATH which finishes the proof. |
quant-ph/0008095 | Let MATH, then the NAME representation of MATH is MATH where MATH. By simple calculation, MATH . Since MATH approximates MATH, MATH. |
quant-ph/0008095 | By induction. It is true for MATH. Assume the statement is true for all natural numbers smaller than MATH, and let's examine the case MATH. Pick a coordinate MATH such that both the subcubes of MATH and MATH have nonempty subsets MATH and MATH of MATH. Then MATH. We can assume without loss of generality that MATH. Then by simple calculation, MATH . |
quant-ph/0008095 | Let MATH be the number of true assignments. By REF , in the Boolean cube, the number of edges that connect two true assignments is less than MATH, and the number of edges that connect two false assignments is less than MATH. Therefore, MATH . |
quant-ph/0008095 | For each MATH, let MATH be the characteristic function of MATH, that is, MATH . Let MATH be the probability that MATH is observed as the output when the input is MATH. Then by REF , MATH is a nonnegative polynomial of degree no more than MATH, and MATH approximates MATH. Furthermore, for any MATH, MATH. For simplicity of notation, we shall use MATH in place for MATH, MATH for MATH, and MATH for MATH. Note that MATH and, MATH . Let MATH. We want to get a lower bound for MATH. By REF MATH . Summing over all MATH, and by REF , we get MATH . |
cs/0009006 | We form a bipartite graph, in which the vertices correspond to the variables and components of the instance. We connect a variable to a component by an edge if there is a (variable,color) pair using that variable and belonging to that component. The instance is solvable iff this graph has a matching covering all variables. |
cs/0009006 | We employ a backtracking (depth first) search in a state space consisting of MATH-CSP instances. At each step we examine the current state, match it to one of the cases above, and recursively search each smaller instance. If we reach an instance in which REF applies, we perform a matching algorithm and either stop with a solution or backtrack to the most recent branching point of the search and continue with the next alternative. A bound of MATH on the number of recursive calls in this search algorithm, where MATH is the maximum work factor occurring in our reduction lemmas, can be proven by induction on the size of an instance. To determine the maximum work factor, we need to set the parameter MATH. We used NAME to optimize MATH numerically, and found that for MATH the work factor is MATH. For MATH near this value, the largest work factors involving MATH are MATH, and MATH; the remaining work factors are below REF The true optimum value of MATH is thus the one for which MATH. As we now show, for this optimum MATH, MATH, which also arises as a work factor in another case. Consider subdividing an instance of size MATH into one of size MATH and another of size MATH, and then further subdividing the first instance into subinstances of size MATH, MATH, and MATH. This four-way subdivision has work factor MATH, and combines subdivisions of type MATH and MATH, so these three work factors must be equal. |
cs/0009006 | Randomly choose a subset of four values for each variable and apply our algorithm to the resulting MATH-CSP problem. Repeat with a new random choice until finding a solvable MATH-CSP instance. The random restriction of a variable has probability MATH of preserving solvability so the expected number of trials is MATH. Each trial takes time MATH. The total expected time is therefore MATH. |
cs/0009006 | As described, we find a maximal bushy forest MATH, then cover the remaining vertices by height-two trees. We choose colors for each internal vertex in MATH, and for certain vertices in the height-two trees. 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 MATH; MATH denote the number of non-root internal vertices; MATH denote the number of leaves of MATH; MATH denote the number of vertices adjacent to leaves of MATH; and MATH denote the number of remaining vertices, which must all be degree-three vertices in the height-two forest. We show that, if we assign cost MATH to each vertex adjacent to a leaf of MATH, then the cost of coloring each height-two tree, averaged over its remaining vertices, is at most MATH per vertex. Therefore, the total time for the algorithm is at most MATH. This bound is subject to the constraints MATH, MATH, MATH, MATH, and MATH. The worst case occurs when MATH, MATH, MATH, MATH, MATH, and MATH, giving the stated bound. |
cs/0009006 | We apply REF MATH times, resulting in a set of MATH constrained MATH-edge-coloring problems each having only MATH edges. We then treat these remaining problems as MATH-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. This is maximized when MATH and MATH. |
cs/0009013 | There are MATH edges. For each edge MATH, the complexity of the associated region can be at most MATH. Since any pair of constant degree polynomials intersect in a constant number of points, the overall complexity of MATH is given by MATH. |
cs/0009023 | According to CITE there are exactly two different rectilinear drawings of MATH, of which the convex hull is either a triangle or a quadrilateral. The former has no crossings and corresponds to the concave kite. The latter has one crossing and corresponds to the convex kite. Since the drawing is comprised of nested triangles, a kite originates at each of the three outer vertices. Since the vertices are non-collinear, each of the kites is either convex or concave. The drawing can have, zero (CCC), one (CCV), two (CVV), or three REF convex kites, with the rest being concave. |
cs/0009023 | Let MATH be a concave kite in a nested triangle drawing with the standard vertex labels MATH, MATH, MATH, and MATH. Since MATH is concave, the middle vertex MATH is within the triangle MATH. The vertices MATH and MATH determine a line that defines a half-plane MATH that does not contain MATH. Since the vertices MATH, MATH, and MATH comprise the inner triangle of the drawing and must be contained within the outer triangle, there must be an outer triangle vertex located in the half-plane MATH. Denote this vertex by MATH and note that a kite originates from it; hence, there are kite edges MATH and MATH. Thus, MATH is contained within the quadrilateral MATH. |
cs/0009023 | If both edges MATH and MATH each cross at least one kite edge, then we are done. Without loss of generality, assume that MATH does not cross any kite edges. Let MATH and MATH be the other two inner vertices, and consider the path MATH. Since edge MATH does not intersect the path, MATH creates a barrier on the other side of path MATH. The same argument with edge MATH applies to path MATH, hence two barriers are present, forcing two crossings. |
cs/0009023 | Since the two concave kites share the same middle vertex, there are two possible cases. Either the labels of the internal vertices match, in which case we are done. Otherwise, the left and right labels are interchanged. By way of contradiction, assume that they are interchanged; this implies that the kites are disjoint, that is, do not overlap. Consequently, they cannot share the middle vertex that is inside both of the kites; this is contradiction. |
cs/0009023 | Either vertex MATH is contained in kite MATH or not. If MATH is inside MATH, then, because kite MATH is concave, a barrier path MATH is created between vertex MATH and vertex MATH. Hence, edge MATH must cross the path MATH, intersecting one of the path's two edges. If vertex MATH is not contained in kite MATH, then assume, that vertex MATH is on the left side of kite MATH (clockwise with respect to MATH). The edge MATH defines a half-plane that separates vertex MATH from vertex MATH. Furthermore, the segment defining the half-plane located within the sector MATH corresponds to part of the edge MATH. Since the edge MATH must be within the sector MATH, it must cross edge MATH. If vertex MATH is on the right, by a similar argument, the edge MATH will cross edge MATH. |
cs/0009023 | Using the kite edges we construct two polygons MATH and MATH. Since both polygons contain region MATH and since the only shared edge, is a middle edge, edge MATH must cross into both polygons, contributing at least one kite edge crossing from each. |
cs/0009023 | Select a pair of green vertices and remove all other green vertices from the drawing. This forms a MATH with exactly one MATH-rgedge crossing that is uniquely identified by the two green vertices. Since there are MATH pairs of green vertices, there must be MATH-rg edge crossings. |
cs/0009023 | That the convex hull of an optimal rectilinear drawing of MATH is a triangle has been shown in CITE and CITE. Using a counting technique similar to CITE, consider a drawing composed of a red triangle that contains a green convex quadrilateral that contains two blue vertices. By the MATH principle there are MATH-rg crossings. At least two MATH-ggcrossings are present because a convex quadrilateral cannot be concentric with a triangle. Selecting one green and one blue vertex at a time and applying the MATH principle yields, MATH-rg crossings. Six MATH-gg crossings are due to the red-blue edges entering the green quadrilateral. Applying the MATH principle to the blue vertices yields one MATH-rb crossing. There are MATH-gb and MATH-gg crossings; the green quadrilateral is initially partitioned into four parts by one MATH-gg crossing, adding the first blue vertex creates two MATH-gg and adding the second vertex creates two more MATH-gg crossings and two MATH-gb crossings. This totals REF crossings. An additional eight MATH-gb and MATH-gg crossings occur inside the green quadrilateral, four per blue vertex, totaling REF crossings, which is greater than the optimal REF. By a similar argument any drawing whose second hull is not a triangle will also be non-optimal; see REF. |
cs/0009023 | Select two of the three red, green, and blue triangles. These two triangles form a nested triangle drawing of MATH with three REF-colour crossings. Hence, there are three REF-colour edges of each type. |
cs/0009023 | Let the outer triangle be red and the inner green. By the MATH . Principle REF there are three MATH-rg edge crossings. If the two triangles are non-concentric then there is at least one MATH-gg crossing. |
cs/0009023 | The red triangle contains the green triangle and the green triangle contains the blue triangle. Therefore, every red-blue edge must cross into the green triangle. Since there are nine red-blue edges, there are nine MATH-gg crossings. |
cs/0009023 | The are three green and three blue vertices, thus there are nine unique green-blue pairs of vertices. By the MATH principle, each pair contributes exactly one MATH-rb crossing. Hence, a nested triangle drawing of MATH has exactly nine MATH-rb crossings. |
cs/0009023 | We make use of the fact that the green and blue triangles form a MATH and that any rectilinear drawing of MATH falls into one of the five configurations: CCC, VVV, CVV, binary CCV, and unary CCV. The proof is by case analysis on the green-blue MATH sub-drawing. The green-blue MATH is drawn in one of the five configurations: CCC configuration: Since each of the blue vertices is a middle vertex of a concave kite, and all middle labels are distinct, by REF each of the nine red-blue edge crosses one green-blue edge, hence there are nine MATH-gb crossings. VVV or CVV configuration: If the drawing is in a VVV configuration, by the Barrier Lemma REF there are two MATH-gb crossings per blue vertex. Adding the three MATH-bb crossings yields nine. In the CVV configuration one of the blue vertices is responsible for at least three MATH-gb crossings rather than two; adding the two MATH-bb crossings yields the required result. Binary CCV configuration: Note that two of the blue vertices are responsible for three MATH-gb crossings, and the third vertex is responsible for two. Adding the single MATH-bb crossing yields nine. Unary CCV configuration: In the case of the unary CCV configuration, the drawing is partitioned into a heavy and light part by extending the blue edges incident on the middle vertex of the convex kite; see REF . A red-blue kite whose origin vertex is in the heavy side of the drawing is responsible for four or six MATH-gb crossings while a red-blue kite originating in the light side of the partition is responsible for three crossings if it is empty, and one crossing if it is full; the six edge crossings occur if there is an empty red-blue kite between the two concave kites. In order for the green triangle to be nested within the red, by the containment argument, at least one of the red-blue kites must originate in the heavy partition. This implies that in order to get fewer than eight MATH-gb crossings, two of the red-blue kites must be full and contain the green-blue kite in the light partition. This implies that the third red-blue kite must be an empty kite between the two concave red-green kites. Since this kite is responsible for six crossings, it follows that there are at least eight MATH-gb crossings and therefore at least nine internal crossings. |
cs/0009023 | The first part of the statement is proven in CITE and the counting argument in REF . Putting REF , and REF together accounts for REF of REF crossings in an optimal drawing. REF states that there are at least nine internal crossings. Since MATH, the number of MATH-gg and MATH-bb crossings must be zero; this implies concentricity. |
cs/0009023 | By REF , an optimal drawing of a MATH has REF crossings. Referring to REF , an optimal drawing has at least REF non-internal edge crossings (REF , and REF). By REF , there are at least nine internal edge crossings and hence, an optimal drawing has exactly nine internal edge crossings. NAME MATH-bb crossing occur if the green-blue MATH part of the drawing has configuration VVV. However by a Barrier argument similar to REF the configuration VVV creates nine MATH-gb crossings plus three MATH-bb crossings, which totals REF internal crossings and cannot occur in an optimal drawing of MATH. Consequently at most two MATH-bb crossings may occur. |
cs/0009023 | By way of contradiction, assume that there exists a rectilinear drawing of MATH with REF crossings. Since each edge crossing comprises of four vertices, the sum of responsibilities of each vertex totals MATH. Therefore, the average responsibility of each vertex is MATH. Furthermore, each vertex in the drawing is responsible for exactly REF edge crossings. For if a vertex is responsible for more than REF edge crossings, then removing the vertex from the drawing yields a drawing of MATH with fewer than REF edge crossings, which contradicts MATH. Similarly, if the drawing has a vertex that is responsible for fewer than REF crossings, then by the averaging argument, there must be a vertex that is responsible for more than REF crossings, leading to the same contradiction. Therefore, each vertex is responsible for REF crossings. Thus, any drawing of MATH with REF crossings contains an optimal drawing of MATH. Starting with an optimal drawing of MATH we try to place the tenth vertex. We have two choices; either place it such that one of the hulls of the MATH drawing is a convex quadrilateral or the drawing comprises of nested triangles with a vertex in the inner triangle. In the latter case, the edge connecting the tenth vertex to one of the outer triangle vertices must intersect an inner triangle edge. Removing the inner triangle vertex that is opposite the intersected edge creates a drawing of MATH that fails the concentricity condition. Hence, the latter drawing will not be optimal. If the former situation arises there are two subcases. If the quadrilateral is the outer or the second hull, then removing an inner vertex creates a non-optimal MATH drawing, which is a contradiction. If the innermost hull is a convex quadrilateral, then a priori it is not concentric with the outer triangle. Let MATH be the vertex such that there is an edge from it to a vertex in the outer triangle that intersects the quadrilateral. Remove a vertex from the quadrilateral that is antipodal to MATH. This creates a non-optimal MATH drawing. The result follows. By an identical argument any rectilinear drawing of MATH cannot have fewer than REF crossings. |
cs/0009023 | At least two of the red-white edges must cross into the green triangle on distinct green-green edges as a consequence of the nested triangle requirement and the containment argument. Select two of the three red-white edges such that they cross into the green triangle on distinct green-green edges and such that the total number of MATH-gbcrossings is minimized. Let MATH and MATH be the number of MATH-gb crossings for which each of the two red-white edges is responsible and assume, without loss of generality, that MATH. The lower bound on the total number of MATH-gb crossings is MATH. We say that the red-white edge of lesser responsibility REF has weight two, and we say that the other red-white edge, of responsibility MATH, has weight one. Upon examining MATH-gb crossings the proof falls into three main cases corresponding to the numbers of MATH-gb crossings; if there are six or more MATH-gb crossings then we are done. We consider the cases when the number of MATH-gb crossings is MATH, MATH, and MATH, the latter being the most challenging. REF, or REF MATH-gb crossings By the Barrier Lemma, every blue vertex forces at least one MATH-gb crossing. Hence, there must be at least three MATH-gb crossings. CASE: MATH-gb crossings Considering only the MATH-gb crossings, the configuration that minimizes the number of MATH-gb crossings occurs when the red-white edge of weight two crosses zero green-blue edges and the red-white edge of weight one crosses three. However, we must consider blue-white edges also; by the Barrier principle one of the blue-white edges must cross at least two green-blue edges, and the other must cross at least one. This brings the total up to at least six. REF or REF MATH-gb crossings Assume there are at least four MATH-gb crossings. If there are two or more MATH-bw crossings then we are done. It remains to consider two subcases: that of zero or one MATH-bw crossings. CASE: MATH-bw crossings Assume there are zero MATH-bw crossings. This case can only occur when no green-blue edge intersects the blue triangle, that is, the green-blue kites are in a CCC configuration because there are no MATH-bb crossings. The white vertex is in the green-blue free zone; a free zone consists of all regions of a nested triangle drawing of MATH where a seventh vertex can be placed such that no kite edge blocks visibility of any inner vertices. Note that removal of the inner edges of all convex kites in a configuration creates a free zone. A free zone occurs naturally in a CCC configuration. If there is a green-blue edge intersecting the blue triangle, then there exists a green-blue-green path between two of the blue vertices that forces at least one MATH-bw crossing. Since the white vertex must be in the naturally occurring green-blue free zone, that is, a green-blue CCC configuration, by the CCC Lemma REF , this forces every red-white edge to generate at least two MATH-gb crossings. This yields a total of at least six crossings. We reach a count of five crossings of the required type. The remainder of the proof is devoted to producing one more edge crossing of one of the required types. CASE: MATH-bw crossing We now consider the MATH-bw crossings. Consider the red-blue kite configuration. Either the configuration is a CCC or not. CASE: NAME red-blue configuration Assume that the red-blue kite configuration is not in a CCC configuration. By the converse of the argument used in REF there is at least one MATH-bw crossing. Adding to the existing five yields at least six distinct crossings of the required type. This leaves only one case: the CCC red-blue configuration. CASE: CCC red-blue configuration We now consider the five subcases corresponding to the distinct green-blue configurations within the red-blue CCC configuration. CASE: CCC green-blue configuration If the green-blue kites are in a CCC configuration, then this case is covered by REF. CASE: CVV and VVV green-blue configurations For every green-blue edge that intersects the blue triangle, there is at least one MATH-bw edge crossing; see REF . Hence, if the green-blue kites are in a CVV or a VVV configuration then we have at least two MATH-bw crossings. This sums to at least six crossings. CASE: Unary CCV green-blue configuration If the green-blue configuration is a unary CCV configuration then the red and green triangles are not concentric; therefore, there is at least one MATH-gg crossing. Adding at least four MATH-bw crossings, and at least one MATH-bw crossing, by the same argument as in REFEF, yields at least six crossings. CASE: Binary CCV green-blue configurations We are now left with the case of a CCC red-blue kite configuration and a binary CCV green-blue kite configuration with the white vertex either inside the red-blue free zone or not. If the white vertex is not inside the red-blue free zone then there is at least one MATH-bw crossing, by the same argument used in REF, plus at least one MATH-bw crossing, by the same argument as in REFEF, plus at least four MATH-gb crossings. The sum of these crossings is at least six. Thus, assume that the white vertex is in the red-blue free zone. We will argue that there must always be either at least five MATH-gb crossings plus at least one MATH-bw crossing, or at least four MATH-gb crossings plus at least two MATH-bw crossings. Consider the drawing minus the single green-blue edge in the only convex green-blue kite, that is, the inner edge of the convex kite. This creates a green-blue free zone, inside of which there are no MATH-bw edge crossings. In order to cross into the green-blue free zone, a red-white edge must cross a green-blue edge. Furthermore, if a green-blue kite and a red-blue kite are both concave, and have their internal (blue) vertices labeled identically, then we may invoke REF (Kite Lemma). That is, the red-white edge, incident on the origin vertex (red) of the red-blue kite, must cross into the concave green-blue kite before crossing into the free zone. This produces an additional MATH-gb crossing. The white vertex is either inside the green-blue free zone or not. If the white vertex is inside the green-blue free zone, then the red-blue CCC configuration together with the pigeon-hole principle implies that we can match up a concave red-blue kite with each of the two concave blue-green kites. By REF , each of these match-ups contribute at least two MATH-gb crossings, and the third red-white edge contributes at least one MATH-gb crossing. Thus, if the white vertex is in the green-blue free zone there are five MATH-gb crossings. By the argument used in REFEF, the single convex green-blue kite contributes to at least one MATH-bw crossing. Thus we get at least six crossings. If the white vertex is outside the green-blue free zone, then we get at least one MATH-bw crossing by the same argument used in REF and at least one MATH-bw crossing by the same argument used in REFEF. Since we have at least four MATH-gb crossings REF , we get a grand total of at least six crossings. In all possible cases that can occur we have shown that the number of crossings of the required type is at least six. |
cs/0009023 | By way of contradiction, assume that there exits an optimal rectilinear drawing of MATH whose convex hull is not a triangle and REF edge crossings. By the same averaging argument used in REF , at least four of the vertices are responsible for REF edge crossings; removing any of them yields an optimal drawing of MATH with REF crossings. If any of the vertices with responsibility REF are not on the convex hull, then removing such a vertex yields a drawing of MATH with a non-triangular convex hull, which is a contradiction. Therefore, all the vertices of responsibility REF must be on the convex hull of the original drawing. Since we can always remove one of the four vertices such that the outer hull of the new drawing is not a triangle, this contradicts the original assumption. Hence, the first convex hull must be a triangle. Assume that the second hull is not a triangle. Either the second hull is a convex quadrilateral or the second hull has more than four vertices; assume the latter. Since at least four of the vertices must have responsibility REF and the outer hull is a triangle, at least one vertex of responsibility REF must either belong to the second hull, or be contained within it. In either case, removing said vertex creates a drawing of MATH that has REF crossings and whose second hull is not a triangle. This is a contradiction. Finally, assume that the second hull is a convex quadrilateral. If within the second hull there is a vertex of responsibility REF or higher, removing said vertex creates a drawing of MATH with REF or fewer vertices. By REF such a drawing should have at least REF crossings, contradiction. Hence, assume that all three vertices inside the second hull have responsibility REF. Consequently, the remaining REF vertices, must have responsibility REF. Since, the second hull is non-concentric with the first, by the same argument used in REF , we can always remove one of the vertices from the second hull such that the outer two hulls are non-concentric. This implies that we can create an optimal drawing of MATH whose outer two hulls are non-concentric, a contradiction of REF . Hence, the outer two hulls must be triangular. |
cs/0009023 | By REF the outer two hulls of the optimal drawing MATH must be triangles. We must still account for the four internal vertices. If the four vertices form a convex quadrilateral then we are done; otherwise, assume the tenth vertex is inside the third nested triangle. Colour the tenth vertex white. Now count the number of red-white and green-white edge crossings, starting with the green-white edge crossings. Each green-white edge must cross into the blue triangle; multiplying by three yields a total of three MATH-bb crossings. By the MATH principle there are three MATH-gb crossings. Each blue vertex has three incident red-blue edges that partition the green triangle into three regions. The white vertex must be in one of the regions; by the Barrier argument there is at least one MATH-rb crossing per blue vertex. The total of the green-white edge crossings sums to nine. Each red-white edge must cross into both the green and blue triangles, totaling six edge crossings. By the MATH principle, there are three MATH-rg crossings and three rw-rb crossings. This gives an additional REF crossings. By REF there are at least six additional crossings of the MATH-gb, MATH-bw, MATH-bw and MATH-gg type, of which at least three are MATH-gb crossings. Altogether, the number of white and MATH-gg crossings is REF. Since MATH, the number of edge crossings in the drawing of MATH with the white vertex in the blue triangle is, MATH. |
cs/0009023 | By way of contradiction assume that MATH. By REF the inner hull must be a convex quadrilateral. Repeat the argument from REF disregarding the MATH-bb and MATH-bb edge crossings (because there is no blue triangle). This gives us an initial count of MATH edge crossings. Let the entire inner convex quadrilateral be coloured blue. Inside the quadrilateral there will be one MATH-bb crossing (the diagonals). Furthermore, since the quadrilateral is neither concentric with the red triangle nor the green triangle, there will be a minimum of two MATH-bb edge crossings and two MATH-bb edge crossings. Summing the edge crossings yields MATH. |
cs/0009023 | NAME 's rectilinear drawing of MATH with REF edge crossings CITE is exhibited in CITE, and hence MATH. By REF MATH. The result follows. |
cs/0009023 | This drawing is coloured by only two colours: red and green. By the MATH principle there are MATH-rg crossings. Since the six green vertices comprise the second hull, there are MATH-gg crossings. REF crossings counted so far include all except the MATH-gg crossings. We now consider the MATH-gg crossings. Select four of the green vertices; these form a convex quadrilateral and at least one green vertex, the guilty vertex, has a red-green edge that intersects the quadrilateral. This edge partitions the green hull into two parts with one green vertex on one side of the green hull and three on the other, or two on each side. In the former case the red-green edge crosses three green-green edges that are incident on the single vertex. In the latter case, the red-green edge intersects four green-green edges that are incident on the two green vertices in one of the partitions. In both cases, there is an additional MATH-gg crossing due to the red-green edge crossing an edge of the quadrilateral. Hence, at minimum four MATH-ggcrossings are due to the single red-green edge. Since there are at least three guilty vertices in a hull on six vertices. There must be at least REF MATH-gg crossings. Therefore, the total number of crossings is at least MATH. |
cs/0009023 | As before, the single vertex inside the second hull is coloured blue. By the MATH principle there are MATH-rg crossings and MATH-rb crossings. By the same argument used in the previous lemma there are MATH-gg crossings. There are at least five MATH-gg crossings. Thus, we reach a count of REF crossings without having considered the MATH-gg, MATH-gg, and MATH-gb crossings. We count the MATH-gg, and MATH-gb crossings by the guilty vertex argument used in the previous lemma. A hull on five vertices will have at least two guilty vertices. Each guilty vertex is responsible for at least three MATH-gg crossings and, by the Barrier argument, at least one MATH-gb crossing. This yields an additional eight crossings, bringing the total up to REF. Finally, consider the MATH-gg crossings. At least three occur from the red-blue edges having to cross into the green hull. By the containment argument, at least one of these three edges has to cross two of the green-green diagonals within the green hull. This brings up the total to at least five MATH-gg crossings. Adding this to the running total yields MATH. |
cs/0009024 | We apply a randomized incremental arrangement construction algorithm. Each cell in the recursive decomposition is an arrangement cell at some stage of the construction. The bound on the representation of a halfspace comes from applying the methods of CITE to the zone of the boundary hyperplane. |
gr-qc/0009004 | Let MATH be REF-dimensional hyperbolic pair of pants described in the previous section. Turn the pair of pants up-side-down and think of MATH as being in the shape of a ``Y". Each endpoint of ``Y" represents a component of MATH isometric to the hyperbolic REF-manifold MATH described in the previous section. We call the three components of MATH, the left component MATH, the bottom component MATH, and the right component MATH. The covering projection MATH restricts to isometries MATH, MATH, MATH. We choose an orientation for MATH and orient MATH with the induced orientation from MATH. The covering projection MATH has a covering transformation MATH that cyclically permutes MATH, since MATH is regular. The isometry MATH is orientation preserving, since MATH has order three. Hence MATH preserves orientation when cyclically permuting MATH. This implies that we can orient MATH so that MATH are all orientation preserving. Let MATH be a positive integer greater than one. Glue MATH-copies, MATH, of MATH together, so that the right component MATH of MATH is glued to the bottom component MATH of MATH by the orientation reversing isometry corresponding to MATH for each MATH. More specifically, we glue MATH to MATH by identifying the point MATH of MATH with the point MATH of MATH, where MATH and MATH are the isometries that correspond to MATH and MATH, respectively. This gives a compact, connected, oriented, hyperbolic REF-manifold MATH, with a totally geodesic boundary, such that MATH is the disjoint union of MATH copies of MATH, labeled MATH where MATH corresponds to the bottom component of MATH, and MATH corresponds to the left component of MATH for MATH, and MATH corresponds to the right component of MATH. Moreover the natural injection of MATH into MATH is orientation preserving for each MATH, since the gluing map MATH is orientation reversing for each MATH. Assume MATH is odd and let MATH. Let MATH be a permutation of the set MATH. We construct a compact, connected, oriented, hyperbolic REF-manifold MATH from MATH by gluing the boundary component MATH to the boundary component MATH by the orientation reversing isometry corresponding to MATH, for each MATH. Then MATH has a totally geodesic boundary corresponding to the boundary component MATH of MATH. To emphasize that the boundary of MATH is isometric to MATH, we shall identify MATH with MATH. Let MATH and MATH be permutations of the set MATH and let MATH be an isometry that restricts to the identity map on MATH. We now show that MATH and MATH is the identity map. The manifolds MATH and MATH are obtained from the manifold MATH by gluing the boundary components MATH to the boundary components MATH in the order specified by the permutations MATH and MATH. Let MATH be the interior of MATH. Then MATH is an open dense submanifold of both MATH and MATH. If an isometry MATH of hyperbolic REF-space MATH restricts to the identity map on a hyperplane of MATH, then either MATH is the identity map or the reflection in the hyperplane. This implies that MATH restricts to the identity map on an open regular neighborhood of MATH in MATH. Therefore MATH restricts to the identity map on all of MATH, since MATH is analytic. This implies that MATH and MATH is the identity map, since MATH is continuous. Let MATH and MATH be permutations of the set MATH and let MATH, for MATH, be isometries such that MATH. Then MATH is an isometry which restricts to the identity map on MATH. By the previous argument, MATH and MATH is the identity map. Therefore MATH. Thus an isometry MATH depends only on the permutation MATH and the isometry MATH of MATH. The manifold MATH has a finite number MATH of isometries, since MATH is a closed hyperbolic REF-manifold. It now follows that for each permutation MATH of the set MATH, there are at most MATH permutations MATH of MATH such that MATH is isometric to MATH. Hence there are at least MATH nonisometric hyperbolic manifolds MATH. Now by NAME 's Formula MATH . Therefore MATH grows superexponentially with MATH, since MATH. Now MATH . Thus the number of compact, connected, orientable, hyperbolic REF-manifolds MATH of volume at most MATH, with totally geodesic boundary MATH, grows superexponentially with MATH. |
gr-qc/0009004 | Let MATH be as in REF and let MATH be as in the proof of REF . Let MATH be the compact hyperbolic REF-manifold obtained by doubling MATH along the union of the right and left components of MATH. Then MATH has a totally geodesic boundary consisting of two copies of MATH corresponding to two copies of the bottom component of MATH. We call MATH a REF-dimensional hyperbolic sleeve. By attaching enough sleeves end-to-end to the boundary components of MATH, we may assume without loss of generality that the diameter of MATH is less than the distance between the components of MATH. By the proof of REF , the number of compact, connected, orientable, hyperbolic REF-manifolds MATH of volume at most MATH, with totally geodesic boundary MATH, containing MATH as a submanifold such that the bottom component of MATH is equal to MATH, grows superexponentially with MATH. Now consider a hyperbolic gravitational instanton MATH obtained by doubling one of the above REF-manifolds MATH. Then MATH has MATH as an initial hypersurface. Suppose MATH has another initial hypersurface MATH isometric to MATH. Now MATH subdivides MATH into two isometric compact, connected, hyperbolic REF-manifolds MATH and MATH with boundary MATH. The submanifold MATH subdivides MATH in a similar fashion. Neither MATH nor MATH is contained in the complement of MATH in MATH, since the volume of MATH and MATH is half the volume of MATH and any compact connected subset of the complement of MATH in MATH has volume less than half the volume of MATH. Therefore MATH and MATH must intersect. Now since the diameter of MATH is less than the distance between the boundary components of MATH, we conclude that MATH must be contained in the interior of the compact submanifold MATH of MATH where MATH is the reflection of MATH across its bottom boundary component MATH. It follows from REF , and REFEF that MATH contains three noncoplanar closed geodesics. Let MATH be the length of the longest of these geodesics. Then MATH contains only finitely many closed geodesics of length at most MATH, since MATH is a compact hyperbolic manifold. Therefore there are only finitely many possibilities for MATH, since MATH is spanned by any three noncoplanar closed geodesics in MATH. Moreover, the number of possibilities for MATH is bounded by a number MATH that depends only on MATH. Therefore MATH can be marked by an initial hypersurface isometric to MATH in at most MATH ways depending only on MATH. Thus the proof of REF implies that the number of hyperbolic gravitational instantons MATH of volume at most MATH, with an initial hypersurface isometric to MATH, grows superexponentially with MATH. |
gr-qc/0009105 | In terms of the function REF , the lens map can be written in the form MATH . As MATH is an immersion transverse to MATH at MATH and MATH is a submersion, the differential of MATH at MATH has rank REF if and only if the differential of MATH at MATH has rank REF. This proves the first claim. For proving the second claim define, for each MATH, a map MATH by applying to each vector in MATH the differential MATH, parallel-transporting the result along the geodesic MATH to the point MATH and then projecting down to MATH. In the last step one uses the fact that, by transversality, any vector in MATH can be uniquely decomposed into a vector tangent to MATH and a vector tangent to the geodesic MATH. For MATH, this map MATH gives the differential of the lens map. We now choose a basis in MATH and a basis in MATH, thereby representing the map MATH as a MATH-matrix. We choose the first basis right-handed with respect to the natural orientation on MATH and the second basis right-handed with respect to the orientation on MATH that is adapted to MATH. Then MATH is positive as the parallel transport gives an orientation-preserving isomorphism. The function MATH has a single zero whenever MATH is a conjugate point of multiplicity one and it has a double zero whenever MATH is a conjugate point of multiplicity two. Hence, the sign of MATH can be determined by counting the conjugate points. |
gr-qc/0009105 | By contradiction, let us assume that there is a sequence MATH with pairwise different elements in MATH. By compactness of MATH, we can choose an infinite subsequence that converges towards some point MATH. By continuity of MATH, MATH, so the hypotheses of the proposition imply that MATH. As a consequence, MATH is a regular point of MATH, so it must have an open neighborhood in MATH that does not contain any other element of MATH. This contradicts the fact that a subsequence of MATH converges towards MATH. |
gr-qc/0009105 | The result MATH can be read directly from REF , choosing the regular value MATH which has exactly one pre-image point under MATH. This implies that MATH must be surjective since a non-surjective map has degree zero. So MATH being the continuous image of the compact set MATH under the continuous map MATH must be compact. It is well known (see, for example, CITE, p. REF ) that for MATH the existence of a continuous map MATH with MATH onto a compact oriented MATH-manifold MATH implies that MATH must be simply connected. As the lens map gives us such a map onto MATH (after changing the orientation of MATH, if necessary), we have thus found that MATH must be simply connected. Owing to the well-known classification theorem of compact orientable two-dimensional manifolds (see, for example, REF), this implies that MATH must be diffeomorphic to the sphere MATH. |
gr-qc/0009105 | We fix a trivialization for the bundle MATH and identify MATH with MATH. Then we consider the bundle MATH over MATH, where MATH is, by definition, the subspace of all lightlike directions that are tangent to past-oriented lightlike geodesics that leave MATH transversely at MATH. Now we choose for each MATH a vector MATH, smoothly depending on MATH, which is non-tangent to MATH and outward pointing. With the help of this vector field MATH we may identify MATH and MATH as bundles over MATH in the following way. Fix MATH, MATH and MATH and view the tangent space MATH as a natural subspace of MATH, where MATH. Then the desired identification is given by associating the pair MATH with the direction spanned by MATH, where the number MATH is uniquely determined by the requirement that MATH should be lightlike and past-pointing. - Now we consider the map MATH given by following each lightlike geodesic from a point MATH into the past until it reaches MATH, and assigning the tangent direction at the end-point to the tangent direction at the initial point. As a matter of fact, REF gives a principal fiber bundle with structure group MATH. To prove this, we first observe that the geodesic spray induces a vector field without zeros on MATH. By multiplying this vector field with an appropriate function we get a vector field whose flow is defined on all of MATH (see the second paragraph after REF for how to find such a function). The flow of this rescaled vector field defines a MATH-action on MATH such that REF can be identified with the projection onto the space of orbits. REF guarantee that no orbit is closed or almost closed. Owing to a general result of CITE, this is sufficient to prove that this action makes REF into a principal fiber bundle with structure group MATH. However, any such bundle is trivializable, see, for example, CITE, p. REF/REF. Choosing a trivialization for REF gives us the desired diffeomorphism MATH from MATH to MATH. The commutativity of REF follows directly from the definition of the lens map MATH. |
gr-qc/0009105 | In the proof of REF we shall adapt techniques used by CITE in their study of asymptotically simple and empty spacetimes. To that end it will be necessary to assume that the reader is familiar with homology theory. With the sphere bundle MATH, introduced in REF , we may associate the NAME homology sequence MATH where MATH denotes the MATH homology group of the space MATH with coefficients in a field MATH. For any choice of MATH, the NAME sequence is an exact sequence of abelian groups, see, for example, CITE, p. REF or, for the analogous sequence of cohomology groups, CITE, p. REF. By REF , MATH and MATH have the same homotopy type, so MATH and MATH are isomorphic. Upon inserting this into REF , we use the fact that MATH-trivial group consisting of the unit element only) for MATH and MATH for MATH because MATH and MATH. Also, we know that MATH and MATH since MATH and MATH are connected. Then the exactness of the NAME sequence implies that MATH and MATH . From REF we read that MATH is compact since otherwise MATH. Moreover, we observe that MATH has the same homology groups and thus, in particular, the same NAME characteristic as REF-sphere. It is well known that any two compact and orientable REF-manifolds are diffeomorphic if and only if they have the same NAME characteristic (or, equivalently, the same genus), see, for example, REF. We have thus proven REF. - To prove REF we consider the end of the exact homotopy sequence of the fiber bundle MATH over MATH, see, for example, CITE, p. REF, MATH . As MATH has the same homotopy type as MATH, we may replace MATH with MATH, so the exactness of REF implies that MATH, that is, that MATH is simply connected. If, for some MATH, the homotopy group MATH would be different from MATH, the NAME isomorphism theorem (see, for example, CITE, p. REF or CITE, p. REF .) would give a contradiction to REF . Thus, MATH for all MATH, that is, MATH is contractible. - We now prove REF . Since MATH is contractible, the tangent bundle MATH and thus the sphere bundle MATH over MATH admits a global trivialization, MATH. Fixing such a trivialization and choosing a contraction that collapses MATH onto some point MATH gives a contraction MATH. Together with the inclusion map MATH this gives us a homotopy equivalence between MATH and MATH. (Please recall that a homotopy equivalence between two topological spaces MATH and MATH is a pair of continuous maps MATH and MATH such that MATH can be continuously deformed into the identity on MATH and MATH can be continuously deformed into the identity on MATH . On the other hand, the projection MATH from REF , together with the zero section MATH gives a homotopy equivalence between MATH and MATH. As a consequence, REF tells us that the lens map MATH together with the map MATH gives a homotopy equivalence between MATH and MATH, so MATH is homotopic to the identity. Since the mapping degree is a homotopic invariant (please recall Property B of the mapping degree from REF), this implies that MATH. Now the product theorem for the mapping degree (see, for example, CITE, p. REF) yields MATH. As the mapping degree is an integer, this can be true only if MATH. In particular, MATH must be surjective since otherwise MATH. |
gr-qc/0009105 | As usual, let MATH denote the chronological past of MATH in MATH, that is, the set of all points that can be reached from MATH along a past-pointing timelike curve in MATH. To prove REF , fix a point MATH. Choose a sequence MATH of points in MATH that converge towards MATH in such a way that MATH for all MATH. This implies that we can find for each MATH a past-pointing timelike curve MATH from MATH to MATH. Then the MATH are past-inextendible in MATH. Owing to a standard lemma (see, for example, REF ) this implies that the MATH have a causal limit curve MATH through MATH that is past-inextendible in MATH. We want to show that MATH is the desired lightlike geodesic. Assume that MATH is not a lightlike geodesic. Then MATH enters into the open set MATH (see REF ), so MATH enters into MATH for MATH sufficiently large. This, however, is impossible since all MATH have past end-point on MATH, so MATH must be a lightlike geodesic. It remains to show that MATH has past end-point at MATH. Assume that this is not true. Since MATH is past-inextendible in MATH this assumption implies that MATH is past-inextendible in MATH, so by REF MATH has past end-point on MATH and meets MATH transversely. As a consequence, for MATH sufficiently large MATH has to meet MATH which gives a contradiction to the fact that all MATH are within MATH. - To prove REF , we have to show that any sequence MATH in MATH has an accumulation point in MATH. So let us choose such a sequence. From REF we know that there is a past-pointing lightlike geodesic MATH from MATH to MATH in MATH for all MATH. By compactness of MATH, the tangent directions to these geodesics at MATH have an accumulation point in MATH. Let MATH be the past-pointing lightlike geodesic from MATH which is determined by this direction. By REF , this geodesic MATH and each of the geodesics MATH must have a past end-point on MATH if maximally extended inside MATH. We may choose an affine parametrization for each of those geodesics with the parameter ranging from the value REF at MATH to the value REF at MATH. Then our sequence MATH in MATH determines a sequence MATH in the interval MATH by setting MATH. By compactness of MATH, this sequence must have an accumulation point MATH. This demonstrates that the MATH must have an accumulation point in MATH, namely the point MATH. |
gr-qc/0009105 | In the first step we construct a MATH vector field MATH on MATH that is timelike on MATH, has MATH as an integral curve, and coincides with MATH on MATH. To that end we first choose any future-pointing timelike MATH vector field MATH on MATH. (Existence is guaranteed by our assumption of time-orientability.) Then we extend the vector field MATH to a MATH vector field MATH onto some neighborhood MATH of MATH. Since MATH is causal and future-pointing, MATH may be chosen timelike and future-pointing on MATH. (Here we make use of the fact that MATH is a closed subset of MATH.) Finally we choose a timelike and future-pointing vector field MATH on some neighborhood MATH of MATH that is tangent to MATH at all points of MATH. (Here we make use of the fact that the image of MATH is a closed subset of MATH.) We choose the neighborhoods MATH and MATH disjoint which is possible since MATH is completely contained in MATH and closed in MATH. With the help of a partition of unity we may now combine the three vector fields MATH into a vector field MATH with the desired properties. In the second step we consider the quotient space MATH. This space contains the open subset MATH whose boundary MATH is, by REF , a manifold diffeomorphic to MATH. We want to show that MATH is a manifold (which, according to our terminology, in particular requires that MATH is a NAME space). To that end we consider the map MATH which assigns to each point MATH the integral curve of MATH passing through that point. (NAME always means closure in MATH.) Clearly, MATH is continuous with respect to the topology MATH inherits as a subspace of MATH and the quotient topology on MATH. Moreover, MATH intersects each integral curve of MATH at most once, and if it intersects one integral curve then it also intersects all neighbboring integral curves in MATH; this follows from REF . Hence, MATH is injective and its image is open in MATH. On the other hand, REF implies that the image of MATH is closed. Since the image of MATH is non-empty and connected, it must be all of MATH. (The domain of MATH and, thus, the image of MATH is non-empty because MATH does not contain a closed timelike curve. The domain and, thus, the image of MATH is connected since MATH is connected.) We have, thus, proven that MATH is a homeomorphism. This implies that the NAME condition is satisfied on MATH and, in particular, on MATH. Since MATH is timelike and MATH contains no closed timelike curves, this makes sure that MATH is a manifold according to our terminology, see REF . In the third step we use these results to prove REF. Our result that MATH is a homeomorphism implies, in particular, that MATH has an intersection with MATH at some point MATH. Now REF shows that there is a past-pointing lightlike geodesic from MATH to MATH in MATH. This geodesic cannot contain conjugate points in its interior since otherwise a small variation would give a timelike curve from MATH to MATH, see REF , thereby contradicting MATH. The rest of REF is clear since all past-pointing lightlike geodesics in MATH that start at MATH are confined to MATH. In the last step we prove REF . To that end we choose on the tangent space MATH a NAME basis MATH with MATH future-pointing, and we identify each MATH with the past-pointing lightlike vector MATH. With this identification, the lens map takes the form MATH. We now define a continuous map MATH on the closed ball MATH by setting MATH for MATH and MATH. The restriction of MATH to the interior of MATH is a MATH map onto the manifold MATH, with the exception of the origin where MATH is not differentiable. The latter problem can be circumvented by approximating MATH in the MATH-sense, on an arbitrarily small neighborhood of the origin, by a MATH map. Then the mapping degree MATH can be calculated (see, for example, CITE, pp. REF) with the help of the integral formula MATH where MATH is any REF-form on MATH and the star denotes the pull-back of forms. For any REF-form MATH on MATH, we may apply this formula to the form MATH. With the help of the NAME theorem we then find MATH . However, the restriction of MATH to MATH gives the lens map, so on the left-hand side of REF we may replace MATH by MATH. Then comparison with the integral formula for the degree of MATH shows that MATH which, according to REF , is equal to MATH. For every MATH that is a regular value of MATH, the result MATH implies that the number of elements in MATH is finite and odd. By assumption, the worldline MATH meets neither the point MATH nor the caustic of the past light cone of MATH. The first condition makes sure that our perturbation of MATH near the origin can be done without influencing the set MATH; the second condition implies that MATH is a regular value of MATH, please recall our discussion at the end of REF. This completes the proof. |
gr-qc/0009105 | REF is obvious from REF was just established. The proof of the remaining two conditions is based on the fact that on MATH the MATH-lightlike geodesics coincide with the MATH-lightlike geodesics (up to affine parametrization). REF is satisfied since every lightlike geodesic in MATH has past end-point on MATH and future end-point on MATH. Moreover, the arrival on MATH must be transverse since MATH is MATH-lightlike. This shows that REF is satisfied as well. |
hep-th/0009124 | We show how to derive the first relation. Inserting the definition we find: MATH where we used NAME rule and rearranging of the summation of the second term in the second line and the the relation MATH on the last line. The second equation follows from a similar calculation. |
hep-th/0009145 | We can get rid of the term MATH in REF by an appropriate reparametrization MATH of the dilaton. With REF we see that MATH has to obey MATH . After one integration this may be verified to become MATH . We choose MATH to be orientation preserving, that is, MATH. Our assumption MATH and the regularity of MATH within MATH guarantee that the derivative of MATH with respect to MATH does not vanish or diverge. MATH is a (globally well-defined) diffeomorphism from some domain to the common domain of REF of the potentials. Now our claim is easily established by noting that MATH and that (using REF for the redefined field MATH and, in a second step, the null energy condition) MATH . |
hep-th/0009145 | The equations of motion of the class of models under consideration are given by MATH . By contracting this equation with the generator MATH of the horizon, we obtain MATH where this equation is assumed to be evaluated on the horizon. By integrating REF is now established as in the previous section provided the late-time assumption MATH holds. Let us now consider the more general case of the late-time assumption MATH. By introducing the function MATH we easily obtain the relations MATH as well as MATH . From REF and our assumptions concerning MATH we see that nonnegativity of MATH is equivalent to nonnegativity of MATH. REF , on the other hand, can be integrated for MATH using REF and our late-time assumption to yield MATH . Using the positivity of MATH, the nonpositivity of MATH and the null energy condition, we conclude from this equation that MATH is nonnegative. This is equivalent to the proposition of the theorem. |
math-ph/0009002 | The fact that MATH and MATH are annihilated by MATH follows trivially from the fact that MATH and MATH are annihilated by each pairwise interaction MATH. So, in fact these states are frustration-free ground states. Next, MATH by REF . We observe that each MATH is an orthogonal projection. Moreover MATH commutes with MATH for every MATH and MATH. So MATH . But MATH, which proves REF . |
math-ph/0009002 | First, MATH because MATH and MATH. It is also clear that MATH, and MATH, in agreement with REF . Because MATH and MATH are eigenvectors of the self-adjoint operator MATH, all that remains is to check that REF holds on MATH. But this is true by REF , since MATH and MATH on MATH. |
math-ph/0009002 | Define MATH. Define MATH an infinite matrix such that MATH. Let MATH be an orthonormal family in any NAME space, and let MATH. Then MATH and MATH. For simplicity let MATH, and let MATH. We consider MATH. Then we calculate MATH . Since MATH, this shows that MATH is bounded and MATH is invertible. Under the invertibility condition, it is true that MATH is also invertible on MATH, and considering this as its domain, MATH. If we let MATH and MATH operate on proper superspaces of MATH and MATH, then they will be identically zero on the orthogonal complements. But it is still true that MATH . In particular, if we let MATH be the orthogonal projection onto MATH, then MATH . This proves REF . To prove the second part, let MATH be a state in MATH. Let MATH. Then MATH . We calculate MATH . Breaking the sum into two pieces yields, for any MATH, MATH . So, using REF , we have MATH for any nonzero MATH. This proves REF . |
math-ph/0009002 | Partition MATH into MATH intervals MATH each of length MATH. If MATH then MATH . By REF , MATH . So MATH . In other words, MATH for some MATH. Since MATH, MATH. Let MATH, then REF holds. Note that for any orthogonal projection MATH and any operator MATH we have the decomposition MATH . If MATH is nonnegative, then MATH is as well. Hence MATH . On the other hand, it is obvious that MATH which implies MATH for any nonzero MATH. Moreover, MATH . In our particular case, where MATH and MATH, REF imply MATH . All that remains is to calculate MATH. Notice that MATH and that MATH commutes with MATH for all MATH except MATH and MATH. (We define MATH.) Straightforward computations yield MATH and MATH where MATH and MATH. It is easy to deduce that MATH is zero unless MATH. (MATH has a tensor factor MATH and MATH has a tensor factor MATH, which implies MATH is zero unless MATH. The term MATH is treated similarly.) Another straightforward computation yields MATH and MATH . So MATH where MATH. In particular MATH, so that MATH and MATH . Thus MATH, which along with REF proves REF . |
math-ph/0009002 | Since MATH, it is clear that MATH . So REF implies REF . To prove REF notice that for any operator MATH, any orthogonal projection MATH, and any nonnegative operator MATH, MATH . So, for any nonzero MATH, MATH . Setting MATH, MATH and MATH we have MATH . Since MATH the corollary is proved. |
math-ph/0009002 | We first prove that MATH . It is easy to see that, just as for the droplets on an interval, MATH where MATH. In fact, using the same tools as in REF, we can calculate exactly, for MATH, MATH . It is verifiable that this satisfies the bounds above. The other expectations MATH and MATH are similar. Applying REF , proves REF . Now we prove that, considering MATH acting on the invariant subspace MATH, MATH where MATH. To do this, we use REF . There exists a MATH and MATH such that, if MATH then for any MATH with MATH, REF guarantess the existence of a ``subinterval" MATH satisfying MATH, MATH, and MATH . We can take MATH and MATH. By ``subinterval", we mean that there exists an interval MATH, such that MATH. Without loss of generality, we assume MATH. Next, MATH and MATH. We estimate MATH, first. Of course, MATH, and since MATH, we see that MATH. Then using REF , MATH where MATH. But MATH where by MATH, we mean MATH, and by MATH, we mean MATH. We omit the calculations here. So MATH . Symmetrically, MATH where MATH denotes the spin-flip. But MATH. Also, MATH. So, for any MATH, MATH . REF together imply the corollary. |
math-ph/0009002 | The proof that MATH is essentially the same as in REF. One fact we should check is that for each MATH, MATH. We observe that MATH . But as before, MATH . An obvious fact is MATH . Taking MATH, yields the desired result. We have the usual orthogonality estimates MATH . In fact, the estimate of MATH follows by REF , taking the limit that MATH, and the other estimates are consequences. Applying REF proves REF . For the second part, suppose MATH. Then MATH . Furthermore MATH, and MATH by virtue of the fact that MATH, the total number of down spins in the state MATH, is finite. Essentially the same fact is restated as MATH, where MATH . Let us define MATH where MATH is the droplet state subspace for the finite chain. By REF , MATH . Since MATH in the norm-topology, as MATH, all we need to check is that MATH converges weakly to MATH. It helps to break up MATH into two pieces, MATH and MATH. Define MATH . Note that for any sequence MATH such that MATH, we have MATH . The reason is that MATH because the the left and right interfaces of the droplet in MATH are a distance at least MATH from the left and right endpoints of the interval MATH, and the probability of finding an overturned spin decays MATH-exponentially with the distance from the inteface. For the same reason, for any fixed MATH, MATH. These two facts imply that MATH converges weakly to MATH. Now MATH converges weakly to zero, because every state in MATH has over half its downspins concentrated in the annulus MATH, and the inner radius tend to infinity. This means that MATH, as claimed. Thus, taking the appropriate limits, MATH which finishes the proof of the theorem. |
math-ph/0009009 | NAME: Getting an upper bound for MATH in terms of MATH is relatively easy, as before. The problem is the lower bound. One might suppose that one decomposes MATH into small boxes as before, with NAME b.c. on each box, and in each box one approximates MATH by a constant. This will NOT work, even if MATH, because all the particles will then want to be in the box with the smallest value of MATH. The gradient term will vanish and we will not get MATH. The trick is to write the quantum MATH as MATH . This leads to a variational problem for MATH instead of MATH. Partial integration and the variational equation for MATH lead to the replacements: MATH and MATH . We have to show that, with these replacements, the energy is bounded below by MATH (up to small errors). Now we can effectively use the NAME box method on this MATH problem. There are still plenty of technical difficulties, but the back of the problem has been broken by recasting it in terms of MATH instead of MATH. |
math-ph/0009010 | The transform for MATH comes directly by taking the logarithmic derivative of MATH with respect to MATH. Then the result for MATH follows as it is adjoint to MATH. For MATH, convert MATH to MATH to find the stated result. |
math-ph/0009010 | We take several steps to pull the MATH-factors across all terms. CASE: Apply the NAME formula for sl REF : MATH with MATH. CASE: Recall the HW formula, REF MATH . Now the adjoint action gives MATH and hence from the above HW formula, MATH CASE: Next, since MATH acts a dilation on MATH, MATH CASE: Now for the MATH-factor, the adjoint action gives MATH and exponentiating, MATH CASE: And the HW NAME formula is the last step: MATH . Combining the factors involving MATH, MATH, and MATH yields the result stated. |
math-ph/0009010 | Apply the NAME formula in MATH and use the fact that appropriate elements of MATH and MATH are mutually adjoint, specifically, MATH and MATH. |
math-ph/0009010 | These follow readily from the commutation relations for the MATH and MATH operators. |
math-ph/0009010 | Use the hat-representation from REF in the dual form acting on the NAME function. Setting MATH, MATH, and MATH as in the statement of the Theorem, MATH and compute accordingly. |
math-ph/0009010 | Use the hat-representation found in REF . With MATH and MATH, it is readily checked that each of the sl REF operators commutes with MATH and MATH. |
math-ph/0009020 | Due to NAME 's formula (there is a typo at the NAME 's approximation (see REF ) in the NAME 's book) MATH holds MATH . Taking into account that MATH, and the first two assumptions of the NAME 's limiting process give MATH where the Right-hand side is MATH, which leads thanks to the third assumption of the limiting process to the claim of the Theorem. |
math-ph/0009020 | CASE: MATH . For the purpose of maximization MATH can be MATH-transformed, into MATH where MATH, MATH is gamma-function, and MATH. Necessary condition for maximum of MATH than is MATH since, according to the assumed adding-up constraint, MATH. First, it will be proved that MATH for any MATH, and any MATH. The first derivative of MATH can be written in a form of infinite series (see CITE) MATH . For MATH, denoted MATH, the series MATH reduces into a harmonic series MATH . Let MATH, where MATH. Then MATH so MATH . Since, difference of the major series converges to zero, MATH also MATH . Due to a known property of harmonic series MATH holds also MATH thus, adding REF up gives MATH respectively, for the derivative (recall that MATH) MATH what is just REF. Without loss of generality, for MATH we can restrict for sub-space of MATH, for MATH, if on the sub-space exists the maximum, so the conditions of maximum for MATH take, thanks to REF, form MATH . Due to REF , with MATH, system REF can be transformed into an equivalent one MATH REF MATH . Thus necessary conditions for maximum of relative entropy MATH, constrained by the respective adding-up constraint are MATH . Comparing REF completes the proof. |
math-ph/0009020 | MATH. |
math-ph/0009026 | Before we proceed to prove this theorem, let us consider in more details a matrix (spinor) representations of the antiautomorphisms MATH and MATH. According to NAME - NAME theorem the antiautomorphism MATH corresponds to an antiautomorphism of the full matrix algebra MATH: MATH, in virtue of the well - known relation MATH, where MATH is a symbol of transposition. On the other hand, in the matrix representation of the elements MATH, for the antiautomorphism MATH we have MATH. A composition of the two antiautomorphisms, MATH, gives an automorphism MATH, which is an internal automorphism of the algebra MATH: MATH where MATH is a matrix, by means of which the antiautomorphism MATH is expressed in the matrix representation of the algebra MATH. Under action of the antiautomorphism MATH the units of MATH remain unaltered, MATH; therefore in the matrix representation, we must demand MATH, where MATH also. Therefore, for the definition of the matrix MATH in accordance with REF we have MATH . Or, let MATH be a set consisting of symmetric matrices REF and let MATH be a set consisting of skewsymmetric matrices (MATH). Then the transformation REF may be rewritten in the following form: MATH . Whence MATH . Thus, the matrix MATH of the antiautomorphism MATH commutes with a symmetric part of the spinbasis of the algebra MATH and anticommutes with a skewsymmetric part. An explicit form of the matrix MATH in dependence on the type of the algebras MATH will be found later, but first let us define a general form of MATH, that is, let us show that for the form of MATH there are only two possibilities: CASE: MATH is a product of symmetric matrices or REF MATH is a product of skewsymmetric matrices. Let us prove this assertion another way: Let MATH be a product of MATH symmetric and MATH skewsymmetric matrices. At this point MATH. The permutation condition of the matrix MATH with the symmetric basis matrices MATH have a form MATH . From here we obtain a comparison MATH, that is, at MATH and MATH anticommute and at MATH commute. Analogously, for the skewsymmetric part we have MATH . From the comparison MATH, it follows that at MATH, MATH and MATH anticommute and at MATH commute. Let MATH, then from REF we see that at MATH and MATH anticommute, which is inconsistent with REF . The case MATH is excluded, since a dimensionality of MATH is even (in the case of odd dimensionality the algebra MATH is isomorphic to MATH, where MATH). Let suppose now that MATH, that is, let us eliminate from the product MATH one symmetric matrix, then MATH and in virtue of REF the matrices MATH that belong to MATH commute with MATH, but the matrix that does not belong to MATH anticommutes with MATH. Thus, we came to a contradiction with REF . It is obvious that elimination of two, three, or more symmetric matrices from MATH gives an analogous situation. Now, Let us eliminate from MATH one skewsymmetric matrix, then MATH and in virtue of REF MATH and all MATH commute with each other. Further, in virtue of REF the matrices MATH that belong to the product MATH commute with MATH, whereas the the skewsymmetric matrix that does not belong to MATH anticommute with MATH. Therefore we came again to a contradiction with REF . We come to an analogous situation if we eliminate two, three, or more skewsymmetric matrices. Thus, the product MATH does not contain simultaneously symmetric and skewsymmetric matrices. Hence it follows that the matrix of the antiautomorphism MATH is a product of only symmetric or only skewsymmetric matrices. Further, the matrix representations of the antiautomorphism MATH: MATH is defined in a similar manner. First of all, since under action of the antiautomorphism MATH we have MATH, in the matrix representation we must demand MATH also, or MATH . Taking into account the symmetric MATH and the skewsymmetric MATH parts of the spinbasis we can write the transformation REF in the form MATH . Hence it follows MATH . Thus, in contrast with REF the matrix MATH of the antiautomorphism MATH anticommutes with the symmetric part of the spinbasis of the algebra MATH and commutes with the skewsymmetric part of the same spinbasis. Further, in virtue of REF a matrix representation of the automorphism MATH is defined as follows MATH where MATH is a matrix representation of the volume element MATH. The antiautomorphism MATH, in turn, is the composition of the antiautomorphism MATH with the automorphism MATH; therefore, from REF it follows (recall that the order of the composition of the transformations REF is not important, since MATH): MATH, or MATH since MATH. Therefore, MATH or MATH. By this reason a general form of the matrix MATH is similar to the form of the matrix MATH, that is, MATH is a product of symmetric or skewsymmetric matrices only. Let us consider in sequence definitions and permutation conditions of matrices of the fundamental automorphisms (which are the elements of the groups MATH) for all eight types of the algebras MATH, depending upon the division ring structure. CASE: The type MATH, MATH. In this case according to NAME - NAME theorem there is an isomorphism MATH, where MATH. First, let consider a case MATH. In the full matrix algebra MATH, in accordance with the signature of the algebra MATH a choice of the symmetric and skewsymmetric matrices MATH is hardly fixed. MATH . That is, at this point the matrices of the first and second half of the basis have a square MATH and MATH, respectively. Such a form of the basis REF is explained by the following reason: Over the field MATH there exist only symmetric matrices with a square MATH, and there exist no symmetric matrices with a square MATH. Inversely, skewsymmetric matrices over the field MATH only have the square MATH. Therefore, in this case the matrix of the antiautomorphism MATH is a product of MATH symmetric matrices, MATH, or is a product of MATH skewsymmetric matrices, MATH. In accordance with REF let us find permutation conditions of the matrix MATH with the basis matrices MATH. If MATH, and MATH belong to the first half of the basis REF , MATH, then MATH . Therefore, we have a comparison MATH, whence MATH. Thus, the matrix MATH anticommutes at MATH and commutes at MATH with the basis matrices MATH. Further, let MATH, and MATH belong to the second half of the basis MATH, then MATH . Therefore at MATH, MATH commutes and at MATH anticommutes with the matrices of the second half of the basis. Let now MATH be a product of MATH skewsymmetric matrices, then MATH and MATH that is, at MATH commutes with the matrices of the first half of the basis REF and anticommutes with the matrices of the second half of REF . At MATH anticommutes and commutes with the first and the second half of the basis REF , respectively. Let us find permutation conditions of the matrix MATH with a matrix MATH of the volume element (a matrix of the automorphism MATH). Let MATH, then MATH . Whence MATH and, therefore at MATH, MATH and MATH commute, and at MATH anticommute. It is easy to verify that analogous conditions take place if MATH is the product of skewsymmetric matrices. Since MATH, then a matrix of the antiautomorphism MATH has a form MATH if MATH and correspondingly, MATH if MATH. Therefore, permutation conditions of the matrices MATH and MATH would be the same as that of MATH and MATH, that is, MATH and MATH commute if MATH and anticommute if MATH. It is easy to see that permutation conditions of the matrix MATH with the basis matrices MATH are coincide with REF - REF . Out of dependence on the choice of the matrices MATH and MATH, the permutation conditions between them in any of the two cases considered previously are defined by the following relation MATH that is, the matrices MATH and MATH commute if MATH and anticommute if MATH. Now, let us consider squares of the elements of the automorphism groups MATH, MATH, and MATH. For the matrices of the automorphisms MATH and MATH we have the following two possibilities: CASE: MATH, MATH. MATH . CASE: MATH, MATH. MATH . In virtue of REF , for the matrix of the automorphism MATH we have always MATH. Now, we are in a position to define automorphism groups for the type MATH. First of all, let us consider Abelian groups. In accordance with REF , the automorphism group is Abelian if MATH (MATH and MATH commute with each other). In virtue of REF the matrix MATH should be commuted with the first (symmetric) half and anticommuted with the second (skewsymmetric) half of the basis REF . From REF - REF it is easy to see that this condition is satisfied only if MATH and MATH. Correspondingly, in accordance with REF the matrix MATH should be anticommuted with the symmetric half of the basis REF and commuted with the skewsymmetric half of the same basis. It is obvious that this condition is satisfied only if MATH. Therefore, when MATH in accordance with REF , there exists an Abelian group MATH with the signature MATH if MATH, and MATH with the signature MATH if MATH. Further, in accordance with REF , the automorphism group is non - Abelian if MATH. In this case, from REF - REF it follows that the matrix MATH commutes with the symmetric half and anticommutes with the skewsymmetric half of the basis REF if and only if MATH is a product of MATH symmetric matrices, MATH. In its turn, the matrix MATH anticommutes with the symmetric half and commutes with the skewsymmetric half of the basis REF if and only if MATH. Therefore, in accordance with REF , there exist non - Abelian groups MATH with the signature MATH if MATH, and MATH with the signature MATH if MATH. In addition to the previously considered case MATH, the type MATH also admits two particular cases in relation with the algebras MATH and MATH. In these cases, a spinbasis is defined as follows MATH that is, a spinbasis of the algebra MATH consists of only symmetric matrices, and that of MATH consists of only skewsymmetric matrices. According to REF , for the algebra MATH the matrix MATH should commute with all MATH. It is obvious that we cannot take the matrix MATH of the form MATH, where MATH, since at MATH and MATH anticommute, which contradicts with REF , and at MATH and MATH that belong to MATH commute with each other, whereas MATH that do not belong to MATH anticommute with MATH, which again conradicts with REF . The case MATH is also excluded, since MATH is even. Therefore, only one possibility remains, that is, the matrix MATH is proportional to the unit matrix, MATH. At this point, from REF it follows that MATH and we see that REF are satisfied. Thus, the matrices MATH, MATH and MATH of the fundamental automorphisms MATH and MATH of the algebra MATH REF from an Abelian group MATH. Further, for the algebras MATH, in accordance with REF the matrix MATH should anticommute with all MATH. It is easy to see that we cannot take the matrix MATH of the form MATH, where MATH, since at MATH the matrix MATH and the matrices MATH that belong to MATH anticommute with each other, whereas MATH that do not belong to MATH commute with MATH, which contradicts with REF . Inversely, if MATH, MATH and MATH that belong to MATH commute, but MATH and MATH that do not belong to MATH anticommute, which also contradicts with REF . It is obvious that in this case MATH is excluded; therefore, MATH. In this case, according to REF the matrix MATH is proportional to the unit matrix. Thus, the matrices MATH, MATH and MATH of the automorphisms MATH and MATH of the algebra MATH REF from the group MATH. CASE: The type MATH, MATH. In virtue of the isomorphism MATH for the type MATH in accordance with the signature of the algebra MATH, we have the following basis MATH . Therefore, in this case the matrix of the antiautomorphism MATH is a product of MATH symmetric matrices, MATH or is a product of MATH skewsymmetric matrices, MATH. Let us find permutation conditions of the matrix MATH with the basis matrices MATH. Let MATH, then MATH and MATH that is, at MATH the matrix MATH anticommutes with the symmetric and commutes with the skewsymmetric part of the basis REF . Correspondingly, at MATH commutes with the symmetric and anticommutes with the skewsymmetric part of the basis REF . Analogously, let MATH, then MATH and MATH that is, at MATH the matrix MATH commutes with the symmetric and anticommutes with the skewsymmetric part of the basis REF . Correspondingly, at MATH anticommutes with the symmetric and commutes with the skewsymmetric part of REF . Further, permutation conditions of the matrices MATH and MATH are defined by the following relations: MATH . From a comparison MATH it follows that the matrices MATH and MATH commute with each other if MATH and anticommute if MATH. If we take MATH, then the relations MATH give analogous permutation conditions for MATH and MATH (MATH). It is obvious that permutation conditions of MATH (the matrix of the antiautomorphism MATH) with the basis matrices MATH and with MATH are analogous to REF - REF - REF , respectively. Out of dependence on the choice of the matrices MATH and MATH, permutation conditions between them are defined by a relation MATH that is, MATH and MATH commute if MATH and anticommute if MATH. For the squares of the automorphisms MATH and MATH we have following two possibilities: CASE: MATH, MATH. MATH . CASE: MATH, MATH. MATH . For the type MATH in virtue of REF a square of the matrix MATH is always equal to MATH. Now, let us consider automorphism groups for the type MATH. In accordance with REF - REF , the automorphism group MATH is Abelian if MATH. Further, in virtue of REF the matrix of the antiautomorphism MATH should commute with the symmetric part of the basis REF and anticommute with the skewsymmetric part of the same basis. From REF - REF , it is easy to see that this condition is satisfied at MATH if and only if MATH is a product of MATH skewsymmetric matrices (recall that for the type MATH, the numbers MATH and MATH are both even or both odd). Correspondingly, in accordance with REF , the matrix MATH should anticommute with the symmetric part of the basis REF and commute with skewsymmetric part of the same basis. It is obvious that this requirement is satisfied if and only if MATH is a product of MATH symmetric matrices. Thus, in accordance with REF , there exist Abelian groups MATH with the signature MATH if MATH and MATH, and with the signature MATH if MATH and MATH. Further, according to REF - REF , the automorphism group is non - Abelian if MATH. In this case, from REF - REF it follows that the matrix of the antiautomorphism MATH commutes with the symmetric part of the basis REF and anticommutes with the skewsymmetric part if and only if MATH is a product of MATH symmetric matrices. In its turn, the matrix MATH anticommutes with the symmetric part of the basis REF and commutes with the skewsymmetric part of the same basis if and only if MATH. Therefore in accordance with REF , there exists a non - Abelian group MATH with the signature MATH if MATH and MATH and MATH with the signature MATH if MATH and MATH. CASE: The type MATH, MATH. First of all, over the ring MATH there exists no fixed basis of the form REF or REF for the matrices MATH. In general, a number of the skewsymmetric matrices does not coincide with a number of matrices with the negative square REF as it takes place for the types MATH. Thus, the matrix MATH is a product of skewsymmetric matrices MATH, among which there are matrices with positive and negative squares, or MATH is a product of symmetric matrices MATH, among which also there are matrices with MATH and MATH squares. Let MATH be a number of the skewsymmetric matrices MATH of a spinbasis of the algebra MATH, MATH. Among the matrices MATH, MATH have MATH-square and MATH matrices have MATH-square. Let MATH and let MATH be a matrix of the antiautomorphism MATH, then permutation conditions of the matrix MATH with the matrices MATH of the symmetric REF and with the matrices MATH of the skewsymmetric REF have the respective form MATH that is, at MATH the matrix MATH commutes with the symmetric part and anticommutes with the skewsymmetric part of the spinbasis. Correspondingly, at MATH, MATH anticommutes with the symmetric and commutes with the skewsymmetric part. Further, let MATH be a product of the symmetric matrices, then MATH that is, at MATH the matrix MATH anticommutes with the symmetric part and commutes with the skewsymmetric part of the spinbasis. Correspondingly, at MATH commutes with the symmetric part and anticommutes with the skewsymmetric part. It is easy to see that permutations conditions of the matrix MATH with the basis matrices MATH are coincide with REF - REF . For the permutation conditions of the matrices MATH, MATH, and MATH we have MATH . MATH . Hence it follows that the matrices MATH, and MATH commute at MATH and anticommute at MATH. It is easy to verify that permutation conditions for the matrices MATH, MATH would be the same. In accordance with REF - REF , and also with REF - REF , the Abelian automorphism groups for the type MATH exist only if MATH and MATH, MATH. Let MATH and MATH be the quantities of the matrices in the product MATH, which have MATH and MATH-squares, respectively, and also let MATH and MATH be the quantities of the matrices with the same meaning in the product MATH. Then, the group MATH with the signature MATH exists if MATH and MATH (recall that for the type MATH we have MATH), and also, the group MATH with the signature MATH exists if MATH and MATH. Further, from REF , and REF - REF , it follows that the non - Abelian automorphism groups exist only if MATH and MATH, MATH. At this point the group MATH with the signature MATH exists if MATH and MATH. Correspondingly, the group MATH with the signature MATH exists if MATH and MATH. In absence of the skewsymmetric matrices, MATH, the spinbasis of MATH contains only symmetric matrices. In this case, from REF , it follows that the matrix of the antiautomorphism MATH should commute with all the basis matrices. It is obvious that this condition is satisfied if and only if MATH is proportional to the unit matrix. At this point, from REF , it follows that MATH and we see that REF is satisfied. Thus, we have the Abelian group MATH with the signature MATH. In other degenerate case MATH, the spinbasis of MATH contains only skewsymmetric matrices; therefore, the matrix MATH should anticommute with all the basis matrices. This condition is satisfied if and only if MATH. In its turn, the matrix MATH commutes with all the basis matrices if and only if MATH. It is easy to see that in this case we have the group MATH with the signature MATH. CASE: The type MATH, MATH. It is obvious that a proof for this type is analogous to the case MATH, where also MATH. For the type MATH we have MATH. As well as for the type MATH, the Abelian groups exist only if MATH and MATH, MATH. At this point the group MATH with MATH exists if MATH, and also the group MATH with MATH exists if MATH. Correspondingly, the non - Abelian group exist only if MATH is a product of MATH skewsymmetric matrices and MATH is a product of MATH symmetric matrices, MATH. The group MATH with MATH exists if MATH, MATH, and the group MATH with MATH exists if MATH, MATH. For the type MATH both the degenerate cases MATH and MATH give rise to the group MATH. CASE: The type MATH, MATH. In this REF dimensionality MATH is odd and the algebra MATH is semi - simple. Over the ring MATH the algebras of this type decompose into a direct sum of two subalgebras with even dimensionality. At this point there exist two types of decomposition CITE: MATH where each algebra MATH REF is obtained by means of either of the two central idempotents MATH and isomorphisms MATH . In general, the structure of the ring MATH in virtue of the decompositions REF - REF and isomorphisms REF - REF admits all eight kinds of the automorphism groups, since the subalgebras in the direct sums REF - REF have the type MATH or the type MATH. More precisely, for the algebras MATH of the type MATH, the subalgebras in the direct sum REF have the type MATH and only this type; therefore, in accordance with previously obtained conditions for the type MATH, we have four and only four kinds of the automorphism groups with the signatures MATH and MATH. Further, for the algebra MATH REF the subalgebras in the direct sum REF have the type MATH; therefore, in this case there exist four and only four kinds of the automorphism groups with the signatures MATH and MATH. In general case, MATH, the type MATH admits all eight kinds of the automorphism groups. CASE: The type MATH, MATH. In this case the algebra MATH is also semi - simple and, therefore, we have decompositions of the form REF - REF . By analogy with the type MATH, a structure of the double quaternionic ring MATH in virtue of the decompositions REF - REF and isomorphisms REF - REF is also admits, in a general case, all eight kinds of the automorphism groups, since the subalgebras in the direct sums REF - REF have the type MATH or the type MATH. More precisely, for the algebras MATH of the type MATH, the subalgebras in the direct sum REF have the type MATH and only this type; therefore, in accordance with previously obtained results for the quaternionic rings we have four and only four kinds of the automorphism groups with the signatures MATH and MATH. Analogously, for the algebras MATH (MATH), the subalgebras in the direct sum REF have the type MATH; therefore, in this case there exist four and only four kinds of the automorphism groups with the signatures MATH and MATH. In general case, MATH, the type MATH admits all eight kinds of the automorphism groups. CASE: The type MATH, MATH. For this type a center MATH of the algebra MATH consists of the unit and the volume element MATH, since MATH is odd and the element MATH commutes with all the basis elements of the algebra MATH. Moreover, MATH, hence it follows that MATH. Thus, for the algebras MATH of the type MATH, there exists an isomorphism MATH where MATH. It is easy to see that the algebra MATH in REF is a complex algebra with even dimensionality, where MATH is either even or odd. More precisely, the number MATH is even if MATH and MATH, and odd if MATH and MATH. In accordance with REF at MATH the algebra MATH admits the Abelian group MATH with MATH, and at MATH the non - Abelian group MATH with MATH. Hence it follows the statement of the theorem for this type. CASE: The type MATH, MATH. It is obvious that for this type the isomorphism REF also takes place. Therefore, the type MATH admits the group MATH if MATH and MATH, and also the group MATH if MATH and MATH. |
math-ph/0009026 | Let us consider first the types with the ring MATH. As follows from REF , the type MATH admits the Abelian automorphism groups MATH if MATH is the product of MATH skewsymmetric matrices REF and MATH is the product of MATH symmetric matrices (MATH). Therefore, MATH . Further, the type MATH admits the non - Abelian automorphism groups REF if MATH is the product of MATH symmetric matrices REF and MATH is the product of MATH skewsymmetric matrices (MATH). In this case, we have MATH . In the degenerate case MATH, MATH, we have MATH and MATH. Therefore, MATH is always symmetric and MATH. In other degenerate case MATH, MATH, we have MATH and MATH; therefore, MATH and MATH is always symmetric. Since for the type MATH we have MATH, or MATH and MATH for the degenerate cases (both degenerate cases correspond to the Abelian group MATH), it is easy to see that REF coincide with the first formula of REF . For the matrix MATH, we can unite REF into the formula which coincides with the second formula of REF . Indeed, the factor MATH does not change sign in MATH when MATH is even and changes sign when MATH is odd, which is equivalent to both REF . Further, the following real type MATH admits the Abelian automorphism groups if MATH and MATH, where MATH and MATH. Therefore, MATH . Correspondingly, the type MATH admits the non - Abelian automorphism groups if MATH and MATH, where MATH and MATH. In this case, we have MATH . It is easy to see that REF - REF are similar to REF - REF and, therefore, REF hold for the type MATH. Analogously, the quaternionic types MATH admit the Abelian automorphism groups if MATH and MATH, where MATH and MATH are even REF . Transposition of these matrices gives MATH . The non - Abelian automorphism groups take place for the types MATH if MATH and MATH, where MATH and MATH are odd. In this case we have MATH . As it takes place for these two types considered here, we again come to the same situation. Therefore, REF hold for the quaternionic types MATH. In virtue of the isomorphism REF , the matrices MATH and MATH for the types MATH with the ring MATH have the following form: MATH, MATH if MATH REF and MATH, MATH if MATH (MATH). It is obvious that for these types REF hold. Finally, for the semi - simple types MATH in virtue of the decompositions REF - REF we have REF - REF or REF - REF in case of the ring MATH REF - REF in case of the ring MATH (MATH). |
math-ph/0009031 | The statement is clearly true for MATH. Assume it to hold up to MATH. One calculates, using the lemma with MATH, MATH . Hence the proof follows by induction. |
math-ph/0009031 | Each normalized element MATH of MATH defines a covariant state MATH by REF . In particular, one has MATH with MATH . Using this result covariance follows from MATH . |
math-ph/0009031 | Let MATH denote the linear space of continuous functions with compact support in MATH and with values in MATH. A sesquilinear form is defined by MATH for all MATH (MATH is the modular function of MATH). From the positivity of MATH follows that MATH is a positive form. Let us assume for simplicity of notations that MATH is not degenerated (it is easy to see that further definitions do not depend on the choice of the representative of the equivalent class given by the kernel of MATH). Then it is an inner product making MATH into a pre-Hilbert space. Let MATH denote its completion. Define MATH by MATH . By linearity MATH extends to a linear operator with domain MATH. If MATH then MATH . This implies that MATH is bounded. Since each element of MATH is a linear combination of positive elements one concludes that MATH is bounded for all MATH. It is obvious that MATH, that MATH is linear, and that MATH. Hence MATH is a *-representation of MATH in MATH. Next define a linear operator MATH by MATH . Note that one has MATH. A straightforward calculation gives MATH . This expression can be used to verify that MATH is unitary. Let MATH be defined by REF . Then a short calculation using REF gives MATH which is REF . Let MATH be an approximate unit of MATH. For each neighborhood MATH of MATH in MATH let MATH be a positive function vanishing outside MATH and satisfying MATH . Then the functions MATH form a NAME sequence in MATH, converging to some element MATH. From REF follows now that MATH for all MATH. The latter implies MATH . Comparison with REF shows that REF holds. From REF follows immediately that REF holds. Hence, REF asserts that MATH is a representation of the covariance system. Finally, we prove uniqueness up to equivalence of representations. Define MATH with domain MATH by MATH . It is straightforward to verify that MATH extends to an isometry of MATH into MATH. Because MATH is cyclic MATH is an isomorphism. That MATH intertwines MATH and MATH is obvious. For MATH and MATH one has MATH . Finally, it is obvious that MATH. |
math-ph/0009031 | Positivity is verified as follows. MATH . For each neighborhood MATH of the neutral element of MATH let MATH be a positive normalized function with support in MATH. Then MATH is an approximate unit of MATH. One has MATH . The latter tends to REF because of continuity of MATH and MATH in the vicinity of MATH. One concludes that MATH is a positive normalized linear functional on the involutive algebra MATH. By continuity it extends to a state of MATH. |
math-ph/0009033 | We assume MATH is real-valued; the proof for complex-valued MATH is similar. Let MATH with MATH. Further restrictions will be imposed later. Pick MATH, non-decreasing, with MATH, MATH if MATH, and MATH if MATH. The function MATH is a smooth version of MATH. By construction of MATH, MATH . Henceforth MATH and MATH are abbreviations for MATH and MATH respectively and MATH. We will also use the notation MATH. The operator MATH will play the role of the so called propagation observable. We will show that its NAME derivative MATH satisfies MATH for MATH, where MATH and MATH is sufficiently large. It will follow for MATH that MATH where, in the last step, we used that MATH, for all MATH. This estimate proves the theorem because MATH and because MATH by REF applied to MATH with MATH such that MATH. In order to prove REF we note that MATH, where by construction of MATH and by REF MATH and, with the notation MATH, MATH . In the last equation we used that MATH and then commuted one factor MATH to the left and the second one to the right. The commutators which arise cancel. By REF MATH where we used NAME 's inequality and MATH. Consider the factor MATH on the right-hand side of the last equation. From MATH on the support of MATH, the definition of MATH, and MATH it is easy to see that MATH . Next pick MATH with MATH and MATH where MATH, so that MATH and MATH. Then MATH implies that MATH . REF combined show that MATH for MATH large enough. Inserting this into REF we get MATH which, together with REF , shows that MATH . We choose now MATH so small that MATH. Then REF follows from REF with MATH. |
math-ph/0009033 | Since MATH is bounded uniformly in MATH it suffices to prove existence of REF for vectors MATH from the dense subspace MATH. We only consider creation operators, the proof for annihilation operators is similar. For given MATH let MATH where MATH. By NAME 's argument the existence of the limit REF follows if MATH . In the following we will use the notation MATH. A straightforward computation shows that MATH . Next we fix MATH and show that the corresponding term has norm which is integrable with respect to MATH. Choose MATH and so small that MATH. Pick then MATH, MATH, MATH, MATH, MATH if MATH and MATH for MATH. Let MATH and MATH henceforth. The term in the sum on the right-hand side of REF corresponding to the fixed MATH then has norm bounded by MATH . Consider first the MATH term. Since MATH, and since, by a stationary phase argument, MATH, for any MATH, the term with MATH on the right-hand side of REF is integrable with respect to MATH. Consider now the term with MATH. We first note, that MATH where MATH. In order to prove the last equation, write MATH and then commute the operator MATH to the left of MATH. The factor proportional to MATH on the right-hand side of REF arises from the commutator of these two operators. Since, by REF , MATH, it follows that the term with MATH on the right-hand side of REF is bounded by MATH. The term proportional to MATH is clearly integrable with respect to MATH and by the NAME inequality MATH . If we choose MATH this is finite by REF with MATH and MATH, because MATH by REF . |
math-ph/0009033 | CASE: The existence of the limit for MATH follows from REF . The operator bound REF follows from MATH and from the boundedness of MATH. Finally if MATH the existence of the limit follows by an approximation argument, because MATH is dense in MATH. CASE: Follows from REF and the CCR for MATH and MATH. CASE: For MATH and MATH because MATH as MATH. The proof of the second pull - through formula is similar. The proof of the essentially self - adjointness of MATH, in the case MATH, is very similar to the proof of the corresponding statement for the pseudo - relativistic model (see REF). CASE: From the spectral theorem we know that MATH if and only if the function MATH has an analytic extension MATH, for MATH, which satisfies MATH. Now pick MATH. Then we have, from REF, MATH . Since MATH and MATH are bounded from below, it follows that the function MATH has an analytic extension, given by, MATH, which satisfies MATH where we used the estimate REF . This proves that MATH. The statement for annihilation operators can be proved similarly. |
math-ph/0009033 | By REF and because of the assumption MATH, it follows that, for each MATH the vector MATH is in the range of a spectral projector MATH, for some MATH. Thus MATH is well defined and MATH . Now we want to prove the equality REF . We consider only the case with MATH creation operators, the other cases being similar. Assume first that MATH, for all MATH. We proceed then by induction over MATH. The statement for MATH follows from REF . Now we assume that the statement holds for any integer less than a given MATH and we prove that it holds also for the product of MATH creation operators. To this end we write MATH . The norm of the first term on the right-hand side of the last equation converges, by REF , to MATH as MATH. To handle the second term on the right-hand side of REF we insert the operator MATH just in front of the braces, and we commute the factor MATH through the operators within the braces. The second term on the right-hand side of REF becomes then MATH . Now the term in front of the braces is bounded uniformly in MATH. The first term inside the braces, and each factor in the first sum converge to MATH, as MATH, by induction hypothesis. Finally the terms in the last sum inside the braces have norm which is bounded by MATH . The first factor in REF converges to REF, as MATH. By REF it follows that the second factor in REF is bounded uniformly in MATH. We have thus shown that both terms on the right-hand side of REF converge to REF, as MATH. This proves REF if MATH, for all MATH. For MATH REF follows now by an approximation argument, because MATH is dense in MATH, by REF and because MATH provided MATH. The last estimate follows from the bound REF and from REF, which implies MATH . Finally REF follows from REF , and REF . This completes the proof of the theorem. |
math-ph/0009033 | We proceed by induction over MATH and assume the statement holds for MATH replaced by MATH. Let MATH where MATH MATH as MATH. Hence by REF the limit MATH exists and by the induction hypothesis MATH . Since MATH is closed this proves MATH and the first equation from MATH . The second equation follows from REF and the last one from the fact that, by REF , the convergence as MATH is uniform in MATH. |
math-ph/0009033 | We use MATH as an abbreviation for the operator MATH. For each MATH we have MATH which shows that MATH . The corollary now follows, because, by REF , MATH for MATH sufficiently large. |
math-ph/0009033 | This proof follows a similar pattern as the proof of REF . Some of the explanations given in that proof are not repeated here. We may assume MATH is real-valued. Pick MATH as in the proof of REF with the choice MATH. That is, MATH is non - decreasing, MATH, MATH if MATH and MATH for MATH. Here MATH and small. Again MATH is defined by MATH and obeys MATH . The propagation observable is given by MATH and the theorem follows if we show that MATH for MATH, where MATH and MATH is sufficiently large. By construction of MATH and by REF MATH . To compute MATH note that MATH for MATH where MATH is constant. Hence there exists a function MATH such that MATH for MATH. This makes REF applicable and leads to MATH . Using next that MATH and MATH one finds MATH by commuting one factor of MATH to the left of MATH, respectively to the right of MATH. Since MATH and by the NAME inequality we conclude MATH . To estimate the factor MATH we pick MATH with MATH and then commute MATH with MATH using MATH. This shows that MATH . Next write MATH and let MATH. From MATH for some constants MATH, it follows that MATH, and hence that MATH if MATH is small enough. These remarks in conjunction with REF show that MATH where we commuted MATH and MATH once again. We now insert this result in REF and we get MATH . From the last equation, and from REF , it follows that MATH . This implies REF , with MATH, and completes the proof of the theorem. |
math-ph/0009033 | Using the representation MATH we get MATH . The statement of the lemma follows from the last equation since the factor in the braces, on the right-hand side of the last equation, is equal to MATH. |
math-ph/0009033 | Since MATH is bounded uniformly in MATH it suffices to prove existence of REF for MATH. We only consider creation operators, the proof for annihilation operators is similar. For given MATH let MATH where MATH. By NAME 's argument the existence of the limit REF follows if MATH . In the following we will use the notation MATH. A straightforward computation shows that MATH where we expanded the commutator with MATH using REF , and where MATH. Since MATH we see that MATH . Next choose MATH, with MATH, MATH if MATH and MATH for MATH. Here MATH and so small that the propagation estimate REF holds for the cut - off function MATH and MATH. Let MATH and let MATH, MATH henceforth. Inserting MATH in the right-hand side of REF , we find MATH . Consider first the term with MATH. Since MATH and since, by a stationary phase argument, MATH for any MATH, the term with MATH on the right-hand side of REF is bounded by MATH and is therefore integrable with respect to MATH. Consider now the term with MATH. We first note that MATH where MATH. To see this write MATH and then commute the operator MATH to the left of MATH. The second term on the right-hand side of REF arises from the commutator of these two operators. Since, by REF , MATH, it follows that the term with MATH on the right-hand side of REF is bounded by MATH. The term proportional to MATH is clearly integrable with respect to MATH and by the NAME inequality MATH . If we choose MATH this is finite by REF with MATH, because MATH by REF . |
math-ph/0009033 | CASE: Follows from MATH dense, from MATH and because MATH is a form-core for MATH. CASE: Follows from REF and the CCR for MATH and MATH. CASE: For MATH, by REF, MATH . Now, by REF , MATH . Since MATH for MATH, by REF , it follows that MATH as MATH. This proves the first pull through formula, the proof of the second is similar. The essential self-adjointness of MATH follows from MATH and from NAME 's commutator theorem REF , since MATH is symmetric and bounded with respect to MATH. CASE: Follows in the same way as in the non - relativistic case (see REF). |
math-ph/0009033 | CASE: Suppose first MATH and pick MATH. Then existence of MATH is proved as in REF with only small modifications: rather than MATH as in REF one now has MATH for which we use the formula MATH . New is the factor MATH in front, which does not affect the subsequent estimates since MATH, and the additional term MATH which is dealt with similarly as the first term in braces. Next if MATH and MATH pick MATH with MATH. Existence of REF then follows by an approximation argument using a sequence MATH with MATH and REF . This establishes existence of MATH as a strong limit on the dense subspace MATH. REF now follows by an other approximation argument. REF follows easily from the NAME relations REF and from REF . CASE: From REF it follows that the mapping MATH is continuous. This implies that MATH is strongly continuous and hence so is MATH. NAME and the group structure follow from REF . Finally, because of the essential self-adjointness of MATH on MATH, it only remains to show that MATH for all MATH. By REF in the non-relativistic case, and by REF in the relativistic case, this is equivalent to MATH . But this follows from MATH which holds uniformly in MATH. Here we used REF again and REF . |
math-ph/0009033 | REF follows from MATH and the spectral theorem. REF follows from MATH . |
math-ph/0009033 | Let MATH, MATH and MATH . Since MATH we have MATH and hence it suffices to show that the lemma holds with MATH replaced by MATH. For the case MATH see for example, CITE. If MATH let MATH and note that MATH . Next, in each factor MATH commute as many factors of MATH to the right as possible. For MATH even one gets MATH and a similar formula holds for odd MATH. From REF , the commutator expansion REF for MATH, and MATH it follows that MATH which, together with REF proves the lemma. |
math-ph/0009033 | It is enough to show that MATH is form-bounded with respect to MATH. Since MATH by REF , and from MATH it follows that MATH for some MATH. Combined with the assumption on MATH this shows that MATH is form-bounded with respect to MATH and hence proves the lemma. |
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