paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9903187 | This follows directly from REF . |
math/9903187 | The first assertion is clear because the eigenspaces of MATH are invariant subspaces under the action of MATH. Next we prove the second assertion. Let MATH and MATH be in a same fiber of MATH above MATH, and set MATH and MATH. Then REF holds for MATH, and also for MATH replaced by MATH. There exists MATH in MATH such that MATH. Hence REF also holds for MATH and MATH replaced by MATH and MATH respectively. Thus MATH and MATH. But the equality MATH implies that MATH. |
math/9903187 | Let MATH be as in REF. Direct verification yields MATH. The lemma follows now from REF (with MATH replaced by MATH). |
math/9903187 | Let MATH be a large integer. For MATH in MATH, we consider the subset MATH of MATH consisting of all points MATH in MATH such that MATH and MATH. Note that MATH, because MATH. Thus MATH with MATH, since MATH is integrable on MATH. By the first assertion of REF and by REF below, for MATH in MATH large enough with respect to MATH, we have for all MATH that MATH is stable at level MATH and that MATH. Hence MATH with MATH (because of REF). The lemma follows now, since MATH is isomorphic to MATH, the MATH-action on MATH being the diagonal one, and the image of MATH in MATH is equal to MATH (it is here that we use the fact that we work in MATH instead of MATH). |
math/9903187 | Since REF is a direct consequence of assertion REF in the proof of REF , taking MATH, and REF follows from REF , it remains to prove REF . By the first assertion in the statement of REF , MATH is cylindrical at level MATH, taking MATH large enough. In order to prove REF , we may assume that MATH is a locally closed subvariey of MATH. The inverse image of MATH under the natural map MATH is locally closed in MATH and is equal to MATH by REF and the fact that MATH is cylindrical at level MATH, for MATH large enough. Hence MATH is a locally closed subvariety of MATH, and MATH is a locally closed subvariety of MATH. Next we prove the following assertion: CASE: The stabilizer of MATH acting on MATH is trivial at every point of MATH. Let MATH and set MATH. Since MATH on MATH, we have MATH. Hence MATH is contained in MATH. Thus, since MATH is cylindrical, REF implies that MATH is contained in MATH when MATH is large enough. This concludes the proof of REF . Our next step is to construct a section of the morphism MATH. Let MATH be a section of the projection MATH such that the restriction of MATH to MATH is a piecewise morphism. The existence of such a section MATH has been shown in the proof of REF . Note that MATH is contained in MATH, since MATH is cylindrical at level MATH. Denote by MATH the map MATH and set MATH where MATH is the projection. Clearly MATH is a section of MATH. One proves that MATH is a piecewise morphism by exactly the same argument as for REF in the proof of REF , replacing MATH, MATH, and MATH by their quotient under the action of MATH. By REF , the natural morphism MATH is étale. We consider the fiber product MATH . The strategy of proof is to construct a MATH-equivariant morphism MATH, such that the following diagram is commutative, MATH then to show it may be endowed with the structure of a piecewise vector bundle of rank MATH, and finally to conclude by étale descent. We first construct the mapping MATH. Let MATH be a point in MATH. It follows from REF that there exists a lifting MATH in MATH of MATH such that MATH. Furthermore, by REF , the lifting MATH is uniquely determined by MATH. We set MATH . Clearly, the graph of MATH is constructible, hence, by REF , MATH is a piecewise morphism. We shall show later that, as soon as MATH is a morphism and MATH is smooth, MATH is an actual morphism. Now take a point MATH in MATH. We have MATH and MATH is a lifting of MATH. Hence the conditions for a point MATH to be in the fiber MATH are that MATH and MATH. Rewriting the first condition as MATH, with a unique MATH in MATH, the fiber MATH can be determined by rewriting the condition MATH using the NAME expansion of MATH at MATH. In this way, using again that MATH is large enough and that MATH is cylindrical at level MATH, we find that MATH where MATH is a system of linear homogeneous equations whose coefficients are regular functions of MATH. We refer to CITE for more details. Moreover the solution space of this linear system has dimension MATH, since the jacobian matrix of MATH at any point in MATH is equivalent over MATH to a diagonal matrix with diagonal elements MATH, MATH, , satisfying MATH, compare CITE . In order to prove REF , we may assume that MATH is a locally closed smooth subvariety of MATH and that MATH is a morphism, provided that from now on we only assume MATH is cylindrical at level MATH and that we do not anymore increase MATH, which could destroy the property of MATH to be a morphism. When MATH, we see from our previous discussion about MATH, that MATH is locally bianalytically isomorphic to MATH. Hence MATH is smooth for any MATH. Now let us prove that MATH is a morphism. When MATH, it is easy to see that MATH is continuous, hence is a morphism, since its domain is smooth and it is a piecewise morphism. Thus by the NAME principle, it follows that MATH is a morphism, for any MATH. The fact that it may be endowed with the structure of a vector bundle of rank MATH follows from the above description of the fibers. Now by étale descent (NAME 's REF , see, for example, CITE p. REF), we deduce that MATH may be endowed with the structure of a vector bundle of rank MATH. |
math/9903187 | The first statement is a direct consequence of REF with MATH, MATH replaced by MATH, MATH. The second follows then, using the decomposition MATH and the fact that MATH. |
math/9903187 | We will use the functor MATH of CITE which to a variety MATH over MATH associates an object MATH of the homotopy category MATH of bounded complexes of objects in MATH, such that MATH is the NAME characteristic of MATH. Consider the functor MATH which to an object MATH associates the complex in MATH which is zero in non zero degree and is equal to MATH in degree REF. It follows from the identity MATH in MATH and the definitions that MATH is isomorphic to MATH in MATH. By REF, MATH is a direct factor of MATH in MATH. The functor MATH being fully faithful and MATH being indecomposable, it follows that MATH is zero or equal to MATH. Using a realization, for instance the NAME realization, one obtains that MATH, and the result follows. |
math/9903187 | This is proved in exactly the same way as REF using REF . |
math/9903187 | Straightforward exercise, using REF . |
math/9903187 | We may assume that MATH is irreducible. Set MATH . We may assume there exists a closed subscheme MATH of MATH with MATH such that MATH is contained in MATH, because otherwise MATH and MATH have measure zero. Since MATH is measurable and MATH is contained in cylindrical subsets MATH of MATH with MATH arbitrary small, we see that, for every MATH, there exists cylindrical subsets MATH, MATH, of MATH, such that MATH, MATH, and MATH for all MATH. Moreover, when MATH is strongly measurable we can take MATH. Hence MATH. This implies the theorem, since by REF below, MATH is cylindrical and, for MATH, MATH is contained in a cylindrical subset MATH of MATH with MATH. |
math/9903187 | CASE: First assume that MATH. Then, since MATH is cylindrical, we have MATH and MATH is contained in some MATH, with MATH a closed subscheme of MATH, with MATH. This yields REF when MATH. Now suppose that MATH. Take MATH in MATH large enough to insure that MATH. We may assume that MATH is contained in MATH. Now we choose MATH large enough with respect to MATH to insure that MATH is cylindrical at level MATH and that MATH and MATH are cylindrically stable at level MATH. Set MATH and note that MATH is cylindrical at level MATH, since MATH is constructible. Moreover, MATH is contained in MATH, since MATH is contained in MATH and MATH. Hence MATH is cylindrically stable at level MATH. Thus MATH . Since MATH, we have, for MATH large enough and for MATH large enough with respect to MATH, that MATH and hence MATH. Together with REF , this yields REF . CASE: Using resolution of singularities, we may assume that MATH is smooth. By REF , there exists MATH in MATH such that MATH is contained in MATH and MATH is bounded on MATH. REF follows now from the first part of REF . |
math/9903187 | Reasoning as in the proof of REF , we reduce to the case where MATH is cylindrical and satisfies MATH, with MATH . For this reduction we use the assumption that MATH is strongly measurable to insure that the cylinder MATH in REF is contained in MATH, so that the restriction of MATH to MATH is injective. Next we can reduce to the case where MATH is smooth, using resolution of singularities. Since MATH, it follows from REF that there exists MATH in MATH such that MATH is contained in MATH and that MATH is bounded on MATH. Thus we may as well assume that MATH has constant value MATH on MATH and the theorem follows now from REF . |
math/9903189 | MATH . Notice that by REF , MATH. MATH since MATH and MATH link, clearly MATH. So MATH is well defined. We will distinguish the two cases: CASE: Suppose that MATH. Let us suppose that MATH. Set MATH. By the classical deformation lemma (see CITE) we have that for any MATH, there exists MATH such that MATH . But by the definition of MATH, there exists MATH such that MATH. Setting MATH we have MATH. This implies that MATH, so MATH but this contradicts the definition of MATH. CASE: Suppose MATH. We claim that MATH. Indeed, if by contradiction this was not the case. Then, since MATH . By REF , there exists MATH and MATH such that MATH on MATH and MATH for all MATH in MATH. But the definition of MATH implies that there exists MATH such that MATH . Let MATH, then by REF and the choice of MATH, the functional MATH. Now, since MATH and MATH link, there exists MATH such that MATH and then MATH and this contradicts (MATH). |
math/9903198 | Every NAME algebra that consists of point vector fields preserves the vertical distribution MATH, which is a one-dimensional subdistribution of the contact distribution MATH. Consequently, any reducible NAME algebra of contact vector fields also preserves a one-dimensional subdistribution of MATH. Conversely, let MATH be a NAME algebra of vector fields that preserves some one-dimensional subdistribution MATH of the contact distribution. If MATH and MATH are two functionally independent first integrals of MATH, then, as one can easily verify, the local diffeomorphism MATH is contact and transforms the vertical distribution MATH to MATH. It follows that the NAME algebra MATH preserves the vertical distribution and hence consists of point vector fields. |
math/9903198 | Let MATH be an irreducible NAME algebra of contact vector fields. For an arbitrary point MATH, we let MATH and define MATH and MATH. Then MATH is obviously an open subset in MATH. The NAME algebra MATH is transitive at point in general position if and only if MATH. Assume the contrary. Then the subspaces MATH form a completely integrable distribution MATH in MATH which is invariant under MATH. Consider the following two possibilities: MATH: MATH . Then the intersection MATH is one-dimensional at the points in general position, and this family of subspaces forms a one-dimensional subdistribution of the contact distribution which is invariant under MATH. MATH: MATH-In this case MATH can be locally embedded into a two-dimensional completely integrable distribution which, as follows from its construction, is also invariant under MATH. Then, arguing as in the previous case, we conclude that the NAME algebra MATH preserves a one-dimensional subdistribution of the contact distribution. |
math/9903198 | Fix a basis MATH in the space MATH. Then the action of the elements of MATH on MATH is given by the following matrices: MATH . Therefore, the NAME algebra MATH may be identified with MATH, and the MATH - module MATH with the natural MATH - module. Any irreducible subalgebra of MATH is conjugate to one and only one of the following subalgebras: MATH . If a subalgebra of MATH is nonsolvable, then it is either three-dimensional and coincides with MATH, or four-dimensional and is equal to the whole of MATH. Any solvable irreducible subalgebra is commutative. If it is one-dimensional, then, as follows from the classification of real NAME normal forms of MATH matrices, it is conjugate to the subalgebra REF . If MATH is two-dimensional, it coincides with the centralizer of one of the NAME normal forms, which implies that it is conjugate to the subalgebra REF . If we identify MATH and MATH, the subalgebras listed in REF are identified with the following subspaces MATH: MATH . Consider separately each one of these cases: CASE: It is easily verified that in this case we have MATH. Therefore MATH, and we arrive at the algebra which is listed in the theorem under the number MATH. CASE: Here we have MATH and the action of the subalgebra MATH on this space is irreducible. Therefore, the space MATH is either zero or coincides with the whole of MATH. In the former case we immediately find that MATH (subalgebra MATH). In the second case the subalgebras MATH and MATH coincide and are equal to the subalgebra MATH of the theorem. CASE: Here MATH, and the MATH - module MATH is irreducible. Hence either we have MATH and then MATH (subalgebra MATH), or MATH. In the latter case, however, the space MATH generates a finite-dimensional subalgebra. CASE: Here MATH, and the MATH - module MATH is a sum of two irreducible submodules MATH and MATH of the form MATH . The submodule MATH generates a finite-dimensional subalgebra, so that either MATH or MATH. In the former case MATH (subalgebra MATH), while in the latter the subalgebras MATH and MATH coincide and are equal to the subalgebra MATH of the theorem. |
math/9903198 | Suppose MATH for some MATH. For every MATH, consider the subspace MATH . It is easy to show that, MATH for all MATH. Thus, the subspace MATH may be identified with MATH, and since MATH, this identification is in agreement with the structure of the NAME algebras MATH and MATH. Hence, we have found an isomorphism of the NAME algebras MATH and MATH which is compatible with their filtrations. |
quant-ph/9903014 | Let MATH. Since MATH is a normal matrix, MATH can be written as MATH where MATH is a unitary matrix and MATH is the diagonal matrix of eigenvalues with the MATH-th eigenvalue having the form MATH CITE. If all eigenvalues in MATH are rotations through rational fractions of MATH, that is, MATH is rational, then let MATH where MATH is the denominator of MATH. Thus MATH and we are done. Otherwise, at least one eigenvalue is a rotation of unity through an irrational fraction of MATH. Let MATH be the number of these eigenvalues. For the other MATH eigenvalues compute MATH, just as above, and let MATH. The value of the MATH-th element on the diagonal of MATH is either MATH or MATH where MATH is some irrational real number. Consider taking MATH to some power MATH. The values that are MATH do not change, but the other MATH values that are of the form MATH where MATH, form a vector that varies through a dense subset in a MATH-dimensional torus. Hence, there exists MATH such that the MATH-dimensional vector is arbitrarily close to MATH. Thus, for any MATH there exists a MATH such that MATH. Hence MATH . Select MATH such that MATH to complete the proof. |
quant-ph/9903014 | The `if' direction follows from the fact that the transition function for a GFA is also a valid transition function for an MO-QFA that can accept the same language with certainty. For the `only if' direction, by contradiction, assume that there exists a language MATH that can be accepted by an MO-QFA with bounded error but cannot be accepted by a GFA. Since the class MATH is a subset of the regular languages, MATH must be regular. Let MATH be an MO-QFA that accepts MATH with bounded error. If two strings MATH and MATH take MATH into the same reachable configuration, then for any string MATH the probability of MATH accepting MATH is equal to the probability of MATH accepting MATH, which means that MATH if and only if MATH. Therefore, the space of reachable configurations of MATH's computation can be partitioned into a finite number of equivalence classes defined by the corresponding minimal DFA for MATH. Let MATH and MATH denote reachable configurations of MATH and let MATH be the right invariant equivalence relation induced by MATH. Since MATH cannot be accepted by a GFA, there must exist two distinct equivalence classes MATH and MATH, an equivalence class MATH, and a symbol MATH, such that MATH. If MATH is the transition matrix for symbol MATH, MATH and MATH then MATH and MATH. Since MATH accepts MATH with bounded error, let MATH be the margin. By REF there exists an integer MATH such that MATH and MATH. Hence, MATH because if MATH where MATH is an arbitrary unitary matrix, then by REF the probability of MATH being measured in a particular state is within MATH of MATH being measured in the same state; this probability is less than the margin. Similarly MATH. Hence MATH and MATH. We assumed that MATH and showed that MATH and MATH; therefore, MATH. Let MATH be the string that distinguishes MATH and MATH. Then the string MATH partitions MATH into at least two distinct equivalence classes, but this is a contradiction. Therefore, there cannot exist a language MATH that can be accepted by an MO-QFA with bounded error but not by a GFA. |
quant-ph/9903014 | Let MATH where MATH, MATH, MATH, and MATH is defined by the transition matrices MATH where MATH is an irrational fraction of MATH. Since MATH is a rotation matrix and MATH is an irrational fraction of MATH, the orbit formed by applying MATH to MATH is dense in the circle, and there exists only one MATH, such that MATH, namely MATH. This also holds for MATH. Thus, MATH if and only if the number of MATH rotations applied to MATH is equal to the number of MATH rotations, which is true if and only if the MATH. Otherwise, MATH has a non-zero probability of halting in state MATH. |
quant-ph/9903014 | Construct a free group of rotation matrices drawn from the group MATH as discussed by CITE. Let MATH be a MATH-state MO-QFA where MATH such that MATH is equal to the sum of the number of rotation matrices and their inverses, MATH is defined by the rotation matrices and their inverses, and MATH. The MO-QFA will accept identity words with certainty and reject non-identity words with a strictly non-zero probability, hence solving the word problem for the free group. |
quant-ph/9903014 | The second result follows from REF because every GFA is also a PFA. Since we can bilinearize MATH, MATH is a generalized cut-point event (GCE)CITE. Since the class of GCEs is equal to the class of probabilistic cut-point events (PCEs)CITE, which are accepted by PFAs, there exists a PFA that can accept MATH with some cut-point MATH. |
quant-ph/9903014 | Closure under complement follows from the fact that we can exchange the accept and reject states of the MM-QFA. This exchanges the probabilities of acceptance and rejection but does not affect the margin. Given an MM-QFA MATH and a homomorphism MATH we construct an MM-QFA MATH that accepts MATH. Let MATH and MATH. Assume that MATH and MATH are defined in terms of matrices MATH and MATH. Unlike the proof for MO-QFAs in CITE, the direct construction of MATH will not work because a measurement occurs between transitions, and combining transitions without taking this into account could produce incorrect configurations. After every transition some amount of probability amplitude is placed in the halting states and should not be allowed to interact with the non-halting states in the following transitions. This is achieved by storing the amplitude in additional states; this technique is also used in CITE. Assume without loss of generality that MATH where MATH and MATH. Let MATH and let MATH where MATH . Intuitively, we replicate the halting states MATH times; each replication is termed a halting state set. We construct MATH from the matrices of MATH. Let MATH be a unitary block matrix MATH where MATH . The matrix MATH is a unitary matrix that shifts the amplitudes in the halting set MATH to the halting set MATH and the amplitude in halting set MATH to halting set MATH. In analogy to the MO-QFA case where MATH, for MM-QFAs let MATH where MATH and MATH. After every MATH sub-transition the halting amplitude is shifted and stored in the MATH halting sets of states. When the sub-transition is done, the amplitude in halt state set MATH is zero, which is what is required to prevent unwanted interactions. A minimum of MATH sub-transitions must occur before halting set MATH contains non-zero amplitude, but no more than MATH sub-transitions will ever occur; therefore halting set MATH will never receive non-zero amplitude from halting set MATH. Since MATH has the same distribution as MATH, the margin will not decrease. Closure under word quotient follows from closure under inverse homomorphism and the presence of both end-markers. |
quant-ph/9903014 | Let MATH and define a homomorphism MATH to be MATH, MATH, and MATH. Since MATH can be accepted by an RFA, MATH CITE, but MATH, the result follows. |
quant-ph/9903014 | By contradiction, assume that MATH. Let MATH. Since the minimal DFA for MATH does not satisfy the partial order condition there exist states MATH and strings MATH as defined above and a distinguishing string MATH such that MATH if and only if MATH. Without loss of generality assume that MATH and MATH. Let MATH be the shortest string such that MATH. Let MATH. By REF , MATH. Define the homomorphism MATH as MATH where the last definition is for completeness. Let MATH. By REF MATH. But MATH, a contradiction. |
quant-ph/9903014 | Let MATH be a DFA and MATH be the corresponding minimal DFA. Assume by contradiction that MATH does not satisfy the partial order condition. Hence, MATH has two states that correspond to the equivalence classes MATH and MATH such that MATH and MATH. By the NAME theorem CITE, the equivalence classes partition the set of reachable states in MATH. Hence, for each equivalence class MATH there is a corresponding subset of MATH. Let MATH and MATH denote the subsets of MATH corresponding to the equivalence classes MATH and MATH and assign an arbitrary order on each subset. Select the first state, say MATH, and define the set MATH. If there exists a state MATH and string MATH such that MATH, then MATH does not satisfy the partial order condition, and this is a contradiction. Otherwise, there does not exist a MATH such that MATH for all MATH. In this case there is a partial order on MATH and on MATH because MATH will never be visited again if MATH reads a sufficient number of MATH-s. Remove MATH from MATH and repeat the procedure on MATH. After a finite number of iterations we will either find a MATH that satisfies our requirements, which means that MATH does not satisfy the partial order condition and is a contradiction, or none of the states in MATH will have the required characteristics, in which case MATH satisfies the partial order condition. Therefore, if MATH satisfies the partial order condition, so will its minimal equivalent MATH. |
quant-ph/9903014 | Let MATH be the minimal DFA accepting the language MATH and let MATH be the minimal DFA accepting the language MATH. We first construct an automaton MATH that accepts MATH by combining MATH and MATH using a direct product. Define MATH where MATH, MATH, MATH and MATH. We argue that if MATH and MATH satisfy the partial order condition, then so will MATH. Assume, by contradiction, that MATH does not satisfy the partial order condition. Then there exist two states MATH and MATH and strings MATH such that MATH, MATH and MATH if and only if MATH. In the first case assume that either MATH or MATH, and without loss of generality, assume the former. Then there exists state MATH and state MATH such that MATH, MATH. But this means that MATH does not satisfy the partial order condition, a contradiction. In the second case assume that MATH and MATH. This implies that MATH and hence there cannot exist a string MATH that distinguishes the two states, also a contradiction. Therefore MATH must satisfy the condition. Since MATH satisfies the partial order condition and accepts MATH, by REF the minimal automaton that accepts MATH satisfies the partial order condition, and hence MATH itself, satisfies the partial order condition. |
quant-ph/9903014 | By REF the intersection of two languages that satisfy the partial order condition is a language that satisfies the partial order condition. |
quant-ph/9903014 | We construct an MM-QFA MATH with MATH states that accepts MATH where MATH and MATH. For each link in the trigger chain we require a junk state and a non-halting state. We order the states to correspond with the description of the MATH matrices. Specifically, the first MATH states are the non-halting states, interleaved with junk states. Each triple of states MATH corresponds to a link of the trigger chain, of which there are MATH. State MATH is the decisive accept state and state MATH is the decisive reject state. The junk states are rejecting states. Let MATH and MATH where MATH . Define MATH by the transition matrices MATH. Each transition matrix MATH consists of a product of matrices: MATH where the matrices MATH implement the triggers. Define MATH to be MATH where MATH . The matrix MATH shifts the amplitude of MATH to the junk state MATH. This is the first trigger that is activated when MATH is read. Finally, let the transition matrix for the end-marker MATH be MATH where MATH and the matrix MATH . The matrix MATH sends all amplitude into the junk states. The matrix MATH sends some minimum amount of amplitude into an accept state if the amplitudes of states MATH and MATH differ. The initial configuration of the machine is MATH where MATH that is, the amplitude is evenly distributed among all non-halting states. The only decisive accepting state in the machine is MATH, and amplitude only flows into it when the end-marker is read. In order for it to get a non-zero amplitude, the amplitudes of states MATH and MATH must differ. Since all non-halting states start with the same amplitude, and since the amplitude of state MATH will not change during the execution of the machine until the end-marker is read, the amplitude of state MATH must change in order for the amplitude of state MATH to change. Following the same argument, state MATH will not change in amplitude, until state MATH changes in amplitude. Furthermore, the change in amplitude of state MATH is governed by the matrix components MATH and MATH. Hence, the initial change of amplitude of state MATH depends exclusively on a change in amplitude of state MATH and is governed by component MATH that is located in the transition matrix MATH. If any other transition matrix is applied, then the amplitude of state MATH will not change. Hence, MATH can read MATH without changing the amplitude of state MATH, but, as soon as MATH is read, component MATH will be applied and MATH will have a decreased amplitude, provided state MATH already had a decrease of its amplitude. Finally, the amplitude of any state MATH will never increase beyond its initial value, and once the amplitude of state MATH decreases, it will never increase beyond MATH. For the case of symbol MATH, the amplitude of state MATH is changed by matrix MATH to MATH and is the starting trigger. When the end-marker is read a minimum of MATH of amplitude is placed into the accepting state only if the amplitude of state MATH has decreased. The amplitude from MATH is channeled into a decisive reject state. The rest of the amplitude, from the remaining MATH non-halting states is channeled into junk states. If the amplitudes of MATH and MATH do not differ then all amplitude is channeled into junk and decisive reject states. The probability of MATH accepting a string not in the language is MATH, while the probability of MATH accepting a string in the language is at least MATH. We select the cut-point to be strictly between the two values. |
quant-ph/9903014 | Let MATH and MATH be end-decisive MM-QFAs that accept MATH and MATH. Using these two MM-QFAs we construct an MM-QFA MATH that satisfies the above inequalities. Let MATH and MATH. The sets of halting states are defined as MATH and the transition function MATH is defined as MATH which is a tensor product of the transition functions MATH and MATH. Since MATH and MATH are end-decisive, that is, the accepting states will only have non-zero amplitude when the end-marker is read; thus the MM-QFA MATH will be end-decisive. By the tensor product construction, the probability of MATH accepting MATH is MATH . Since MATH multiplying out the probabilities yields REF , and REF. |
quant-ph/9903014 | Let MATH, MATH, MATH, and MATH be the respective cut-points and margins of MM-QFAs MATH and MATH. Since MATH, MATH, and MATH, the result follows from REF . |
quant-ph/9903014 | Using REF to compose MATH copies of MATH yields the result. |
quant-ph/9903014 | Assume that MATH accepts words in MATH with probability at least MATH and accepts words not in MATH with probability at most MATH. Similarly, assume that MATH accepts words in MATH with probability at least MATH and accepts words not in MATH with probability at most MATH. Using REF let MATH be the MATH-th tensor power of MATH and MATH be the MATH-th tensor power of MATH. Let MATH and MATH, where MATH and MATH. Let MATH and MATH be represented by the unitary matrices MATH and MATH respectively. Let MATH where MATH, MATH is represented by the matrices MATH, and MATH. The automata is initialized with the amplitude evenly divided between the states MATH and MATH, that is, each state contains MATH amplitude. Intuitively, MATH and MATH run in parallel, not interacting unless one of the two crashes. In that case the computation is over. If MATH then MATH if MATH then MATH if MATH then MATH and if MATH then MATH . The last case corresponds to MATH. By setting MATH and MATH appropriately, we can ensure that MATH . Hence, the MM-QFA MATH accepts MATH with bounded error. Furthermore, MATH is end-decisive because both MATH and MATH are end-decisive. |
quant-ph/9903014 | Since MATH, the same argument as in REF applies. |
quant-ph/9903014 | Let MATH be an end-decisive MM-QFA that accepts MATH with bounded positive one-sided error. Since MATH rejects all strings not in MATH with certainty, for every computation of MATH on MATH zero amplitude is placed into the accepting states of MATH. Let MATH, let MATH and assume that MATH. We use MATH to construct an end-decisive MM-QFA MATH to accept MATH with bounded error. Let MATH where MATH and the transition function MATH is extended in the following manner. For all symbols except the end-marker, the transition function for MATH is defined by the matrices MATH . The end-marker transition is defined by the matrix MATH where matrix MATH performs an averaging and cleanup operation. We define MATH in terms of MATH sub-matrices. Every accept state MATH in MATH becomes a reject state in MATH. Additionally, for each such state, REF additional states were added to MATH, MATH, MATH, and MATH, these are an accepting, a non-halting, and a rejecting state respectively. The matrix MATH operates on REF-tuples of states MATH. Each operation is localized to REF-tuple of states and hence can be described by a MATH matrix MATH. Assume that the order of rows and columns of the matrix correspond to the order in REF-tuple. Let MATH . Since MATH accepts with positive amplitude, the amplitude in state MATH will be non-negative. If the non-halting state MATH contains a fixed amount of amplitude MATH, and the old accept state MATH contains MATH amplitude. Then, the averaging operation places MATH amplitude in the accept state MATH. Then, the cleanup operation places any amplitude remaining in the non-halting state MATH into the reject state MATH. We initialize MATH in the same way as MATH except that a fraction of the amplitude is placed in the new non-halting states. These states behave as reservoirs until the end-marker is read. The amount of amplitude placed in the states is greater than the maximum amount of amplitude that any accepting state may ever contain. If MATH then at least one of the accept states of MATH will contain a minimum amount of positive amplitude. Hence, the amount of amplitude in at least one of the accept states of MATH will be strictly less than MATH, by some fixed amount. If MATH then all accept states of MATH will have exactly MATH amplitude in them. Hence, if MATH the probability of MATH accepting MATH will be strictly less than if MATH. Hence, MATH accepts MATH with bounded error. Since the accept states are only observed after the end-marker is read, MATH is end-decisive. |
quant-ph/9903014 | Let MM-QFA MATH accept MATH with cut-point MATH, margin MATH, and maximum margin MATH, and let MM-QFA MATH accept MATH with cut-point MATH, margin MATH, and maximum margin MATH. First, consider the inequalities in REF that occur when we compose the MM-QFAs MATH and MATH using the tensor technique. Since MM-QFA MATH accepts with bounded positive one-sided error, the inequalities are: MATH . These reduce to three cases: MATH . If we can guarantee that MATH then the tensor technique is sufficient to construct the intersection. Let MATH be the MATH-th tensor composition of MATH. By REF MATH accepts words in MATH with probability at least MATH and accepts words not in MATH with probability at most MATH. Construct MM-QFA MATH by composing MATH with MATH using the tensor technique; for sufficiently large constant MATH the inequality MATH will be satisfied. Thus, MM-QFA MATH accepts MATH end-decisively with bounded error. |
quant-ph/9903014 | Let MATH be a piecewise testable set. We first rewrite it in canonical form: MATH . By REF we can construct end-decisive MM-QFAs that accept partial piecewise testable sets, MATH with bounded positive one-sided error. Using these constructions and REF , we can construct end-decisive MM-QFAs that accept languages MATH and MATH with bounded positive one-sided error. The constructions in REF only channel non-negative amplitude into their accept states, furthermore, the constructions in REF do not negate amplitude. Consequently, the constructions for languages MATH and MATH only channel positive amplitude into their accept states. Hence, said constructions accept with positive amplitude. Since MATH is also accepted with bounded positive one-sided error, by REF , we can construct an end-decisive MM-QFA that can accept MATH with bounded error. Since MATH is accepted by an end-decisive MM-QFA with bounded positive one-sided error and MATH is accepted by an end-decisive MM-QFA with bounded error, by REF we can construct an end-decisive MM-QFA that accepts MATH with bounded error. Since the languages MATH can be accepted by end-decisive MM-QFAs with bounded error, by REF , we can construct an end-decisive MM-QFA that accepts MATH with bounded error. |
quant-ph/9903014 | Let MATH be an MO-QFA with left and right end-markers, effectively allowing MATH to start in any possible configuration. Define MATH from MATH. Let MATH be defined in terms of the transition matrices MATH. We define MATH from MATH in the following way: for every MATH let MATH and let MATH . Now consider what happens when MATH and MATH read a string MATH. Since MATH the probability of MATH accepting MATH is equal to the probability of MATH accepting MATH. Thus one end-marker on the right suffices, and by symmetry one left end-marker would also suffice. Therefore, an MO-QFA starting in configuration MATH can simulate an MO-QFA starting in any arbitrary configuration. |
quant-ph/9903014 | Let MATH be an MM-QFA that uses two end-markers and accepts MATH. Assume without loss of generality that MATH which facilitates a simpler description of MATH. We construct MATH that accepts MATH with only the right end-marker. Let MATH, MATH and MATH. Assume that MATH is defined in terms of transition matrices MATH. The construction of MATH is similar to that in the proof of REF . Let MATH represent an identity matrix of size MATH and MATH. We define MATH in terms of its unitary block matrices. For all MATH let MATH where MATH transfers (sweeps) all probability amplitude from states in the old halting states to the new halting states. The old halting states, those in MATH and MATH, are no longer halting states in MATH. The operation of MATH is similar to the operation of the QFA constructed in REF , The ``sweeping" operation saves the amplitude that was in the old states, while it performs the MATH operation in the new halting states (since otherwise the MATH would corrupt the amplitude stored in the original halting states). |
quant-ph/9903035 | Both items are proved in a similar way which has two parts. The first part shows that computing a function in MATH can be reduced to computing the parity of MATH other queries to MATH. The second part then proceeds by showing that using binary search one can compute the parity of MATH-queries with MATH adaptive queries to MATH. On the other hand, it is trivial to see that any computation with MATH adaptive queries can be simulated exhaustively with MATH non-adaptive oracle calls. |
quant-ph/9903035 | For simplicity we only describe what is happening to the states that get effected by the oracle query. Construct the following initial state: MATH . Next, make the only query to MATH depending on the value of the first bit. Note that MATH will thus be queried in superposition for both MATH and MATH. Applying MATH establishes the following evolution on the two qubits: MATH . This results in the following outcome when applied to the initial state: MATH . Which means that if we apply a NAME transformation to the first register, we obtain MATH . Hence observing the first bit yields the correct answer MATH. |
quant-ph/9903035 | Without loss of generality we will assume that the queries are made to MATH, and that the predicate that is computable with MATH queries to MATH is MATH. Let MATH, MATH,, MATH be the queries that the computation of MATH makes. We will use the proof technique of REF (also called mind-change technique) which enables us to compute MATH by calculating the single bit MATH. Here the new formulae MATH can be computed in polynomial time from MATH, MATH, and MATH, but without having to consult MATH. Next, we use REF to compute the parity MATH for odd MATH REF with MATH non-adaptive queries to MATH. Finally we compute the parity of these answers, thus obtaining the necessary information for calculating MATH. |
quant-ph/9903035 | The proof is by induction on MATH. For MATH we have back the situation of REF . Let the predicate MATH be computable with MATH non-adaptive queries to MATH. As in the proof of REF we reduce the MATH queries MATH that MATH makes, to the calculation of the parity-bit MATH. Next, we construct MATH new formulae MATH according to: MATH . The construction of each such MATH can be done in polynomial time. To see this, consider the non-deterministic polynomial time NAME machine MATH that on input MATH, accepts if and only if it can find for MATH of the formulae a satisfying assignment. CITE - proving that MATH is MATH -complete for MATH - showed that any polynomial time non-deterministic NAME machine computation MATH in polynomial time can be transformed into a formula that is satisfiable if and only if MATH has an accepting computation. Let MATH be the result of this NAME reduction. Note the following two properties of those formulae MATH: CASE: The parity MATH is the same as the parity MATH. CASE: For every MATH we have MATH. Now we are ready to make the first query. We compute the parity of MATH and MATH. This can be done in one query using REF . By doing this we have at the cost of one query reduced the question of computing the parity of MATH formulae to computing the parity of MATH. These we can solve using MATH queries using the induction hypothesis. To see this observe the following. For convenience set MATH and MATH. Suppose the parity of MATH and MATH is odd. Hence MATH, MATH,, MATH are all satisfiable and MATH all un-satisfiable (using REF above). Also note that MATH is even, so the parity of MATH is the same as the parity of MATH (these are MATH many formulae). On the other hand assume that the parity of MATH and MATH is even. This means (again using REF above) that MATH are all either satisfiable or un-satisfiable and hence have even parity. So again the question reduces to the parity of the remaining formulae: MATH and MATH. Which happen to be MATH many formulae. |
quant-ph/9903035 | Fix MATH, the input MATH of length MATH and let MATH be the function in MATH. Suppose that MATH with MATH for some MATH depending on MATH. The goal is to obtain the following state: MATH . Since with this state one application of MATH will give us MATH (see REF ). Similar to REF we can obtain this state if we had access to a function MATH with the same property as the one in REF . The goal thus is to transform the function we have access to - MATH in our case - into one that resembles the one in REF . The way to do this is to make use of a quantum subroutine. Observe the following: the binary function MATH is in MATH because we can first compute MATH with MATH queries to MATH and then determine MATH. By REF this function is computable in MATH. Hence, when we use this adaptive MATH algorithm in superposition we have the desired function MATH. There is however one problem with this approach. The algorithm that comes out of REF leaves several of the registers in states depending on the input MATH and MATH. For example the algorithm that computes the parity of two function calls in one generates a phase of MATH depending on the value of the first function call (see REF ). These changes in registers and phase shifts obstruct our base quantum machine and as a consequence the sum computed in REF does not work out the way we want (that is, the interference pattern is different and terms do not cancel out as nice as before.) The solution to this kind of `garbage' problem is as follows: CASE: Compute MATH with MATH queries to MATH. CASE: Copy the outcome onto an extra auxiliary qubit (by setting the auxiliary bit MATH to the exclusive or of MATH and the outcome). CASE: Reverse the computation of MATH making another MATH queries to MATH. Observe that when we compute MATH in this way, all the phase changes and registers are reset and are in the same state as before computing MATH, except for the auxiliary qubit that contains the answer. Since the subroutine was exact (that is, in MATH) the answer bit is a classical bit and will not interfere with the rest of the computation. Note (see REF) that this corresponds exactly to one oracle call to MATH. Thus we simulated MATH call to MATH with MATH queries to MATH and hence have established a way of producing the desired state of REF . The second part of the theorem is proved in a similar way now using REF . |
quant-ph/9903035 | Let MATH be the following set: MATH. Using the the same approach as the proof of REF it is not hard to see that MATH can be computed relative MATH with only a single query. |
quant-ph/9903035 | Let MATH be the function we want to compute relative to MATH. Without loss of generality we assume that MATH for some MATH depending on MATH. As before we construct the following set: MATH . As in REF it follows that MATH is computable with one quantum query to MATH. Since MATH is closed under MATH-reductions and MATH, it follows that MATH. Furthermore, since MATH is MATH-complete for MATH it also follows that MATH. Thus the quantum query can be made to MATH itself instead of MATH. The proof of the second part of the theorem is analogous to the first. |
chao-dyn/9904005 | The existence of the manifold follows quite directly from the Center Manifold Theorem CITE. We would now like to compute MATH. First, we introduce the matrix MATH satisfying MATH, which exists because MATH and MATH have no eigenvalues in common CITE. In fact, it is given by MATH . The change of variables MATH yields MATH . This system admits a center manifold locally described by MATH, where MATH satisfies the partial differential equation MATH . For a vector field MATH, we denote by MATH the terms of order MATH of its NAME expansion around MATH. We know CITE that MATH, where MATH . The motion on the center manifold is thus given by REF, where MATH . To compute MATH, we need to solve REF for MATH. In fact, we will only need to know that MATH, where MATH is the coefficient of MATH in MATH. |
chao-dyn/9904005 | The normal form of REF-jet of REF can be written as MATH with MATH and MATH. The assertion has been proved in CITE, see also CITE. In fact, the origin is an unstable NAME - NAME point if MATH or if MATH. If MATH, it is asymptotically stable if MATH and MATH are both negative, and unstable if one of them is positive. |
chao-dyn/9904005 | Let us consider MATH . The first two conditions of REF are satisfied if we choose MATH and MATH. To satisfy the third one, we have to choose MATH in such a way that MATH, which is possible under our assumptions. The last condition looks more difficult to satisfy, but in fact, we have MATH, where the constant depends only on previously fixed quantities, so that MATH can always be made negative. |
chao-dyn/9904005 | REF is obtained by eliminating nonresonant terms and rescaling space and time. REF comes from the linear transformation MATH followed by a scaling. REF is obtained with the transformation MATH. |
chao-dyn/9904005 | REF can also be written as MATH . The transformation MATH yields a system whose linearization at the point MATH is the matrix MATH . The center manifold theorem implies the existence of a local invariant manifold MATH. Moreover, it is shown in REF , p. REF, that in some neighbourhood of the origin, any solution satisfies a bound of the form MATH for some positive MATH. Since MATH, we have MATH . When MATH, this equation coincides with the equation on the instantaneous center manifold. The assertion on MATH follows from REF . |
chao-dyn/9904005 | Use integration by parts once. |
chao-dyn/9904005 | Notice that for small enough MATH, we have MATH. CASE: Simplification of the linear part. Consider the initial value problem MATH . Since MATH, its solution satisfies MATH. (In fact, one can prove that MATH). The transformation MATH yields MATH where MATH satisfies MATH, and MATH and MATH satisfy similar bounds as MATH and MATH. CASE: Simplification of the cubic part. Let us consider the effect of a change of variables MATH . It transforms REF into MATH where MATH denotes terms of order MATH. In particular, MATH with MATH. We see that the term in brackets of REF can be eliminated if MATH . This is a linear equation which can easily be solved. If MATH, one can choose the initial condition in such a way that MATH. When MATH, however, we cannot obtain such a good bound because the imaginary part of the term in brackets vanishes. Thus we do not attempt to eliminate this term. We obtain the equation MATH where MATH . Let us write MATH. CASE: Proof of the bound REF. The solution of REF has to satisfy MATH where MATH. We define the time MATH . By continuity, MATH. Using the bounds on MATH we can show that for MATH, MATH where MATH is estimated using REF. Using REF and the bounds on MATH, we obtain that MATH . Taking MATH sufficiently small, we have MATH. If we assume that MATH, we contradict the definition of MATH, which shows that MATH. Going back to the initial variables and using the bounds on MATH and MATH, we obtain the conclusion of the lemma. |
chao-dyn/9904005 | After subtracting MATH from REF and eliminating the term linear in MATH in the same way as in REF , we obtain MATH where MATH . Let MATH be a primitive of MATH. The solution of REF with initial condition MATH should satisfy MATH . Using the a priori estimate of REF and the bounds REF, we can apply REF to estimate the integral and obtain REF. To obtain REF, it is sufficient to subtract the particular solution MATH from the general solution, and to use the modulus as a NAME function. |
chao-dyn/9904005 | One can use a standard averaging result, see for instance REF . page REF. |
chao-dyn/9904005 | To simplify the notation, we consider the case MATH. CASE: Transformation of the equation. Let MATH be the positive solution of MATH. Then we have MATH . We decrease the order of the drift term MATH by the transformation MATH which implies that MATH, MATH and yields the system MATH where MATH . There is a time MATH such that MATH for MATH and MATH for MATH. Using REF and some algebra, we obtain the existence of a constant MATH such that MATH CASE: MATH. Consider the NAME function MATH . We have MATH and MATH . From REF we get the bound MATH . Let MATH and define the time MATH . For MATH we get from REF and a standard result on differential inequalities (see for instance CITE) that MATH for sufficiently small MATH, which proves that MATH and thus MATH. CASE: MATH. Using the fact that MATH we obtain MATH . We choose MATH such that MATH and define MATH . For MATH, we have MATH . Using NAME 's Lemma, we obtain that if MATH, MATH . Taking MATH, we obtain MATH and thus MATH. Transforming back to the original variables, we obtain MATH and MATH. |
chao-dyn/9904005 | The proof is similar to the proof of REF , so we only outline the differences. M. enumREF CASE: We have to eliminate quadratic terms from the equation as well. In order to get a normal form similar to REF, we start by eliminating quadratic terms, then we remove the term linear in MATH, and then only the nonresonant cubic terms. CASE: For MATH, we eliminate the real part of the linear term by the transformation MATH. We then change the direction of time, fix a MATH of order MATH and use REF . CASE: For MATH, we proceed exactly as in REF . |
chao-dyn/9904005 | CASE: Hamiltonian system. Consider, as a special case of REF, the Hamiltonian system MATH . REF shows the existence of a particular solution MATH, MATH. If MATH, the dynamics of MATH is governed by a Hamiltonian of the form MATH where MATH and MATH are bounded functions. Using REF, one can show that for sufficiently small MATH, there exists a constant MATH such that MATH . CASE: Normal forms. We write REF in the form MATH where MATH is the contribution of the Hamiltonian approximation REF, and MATH . This relation holds because the full system REF is a perturbation of size MATH of the Hamiltonian system REF. We now perform a number of changes of variables: a translation MATH, where MATH is the particular solution of REF ; a linear change of variables MATH, where MATH satisfies REF, which cancels the term linear in MATH; and a transformation to normal form MATH which yields the equation MATH where the functions MATH and MATH denote contributions of MATH and MATH, respectively. CASE: Bounds on the coefficients. We claim that MATH . The first claim can be checked by a direct calculation. We observe that the linearization of REF around the particular solution MATH has the form MATH, where MATH and MATH. Then we show that the function MATH occuring in the linear transformation is such that MATH. The second claim can be proved without lengthy calculations. By construction, MATH is polynomial in MATH. If we assume by contradiction that the leading term of MATH is of order MATH, we reach the conclusion that MATH would grow faster than allowed by the estimate REF. CASE: Final estimate. The NAME function MATH satisfies the equation MATH where MATH and MATH. We obtain the conclusion in a similar way as in REF . |
cs/9904019 | Set MATH. Consider the following algorithm: CASE: Apply exact search for MATH, each of which takes MATH queries. CASE: If no solution has been found, then conduct MATH searches, each with MATH queries. CASE: Output a solution if one has been found, otherwise output `no'. The query complexity of this algorithm is bounded by MATH . If the real number of solutions was in MATH, then a solution will be found with certainty in REF . If the real number of solutions was MATH, then each of the searches in REF can be made to have error probability MATH, so we have total error probability at most MATH. |
cs/9904019 | From REF we obtain the upper bound MATH . To prove a lower bound on MATH we distinguish two cases. CASE: MATH. By REF , we can achieve error MATH using MATH queries. Now (leaving out some constant factors): MATH . CASE: MATH. We can achieve error MATH using MATH queries, and then classically amplify this to error MATH using MATH repetitions. This takes MATH queries in total. Now: MATH . |
cs/9904019 | Let MATH be the sensitivity of MATH: the maximum, over all MATH, of the number of variables that we can individually flip in MATH to change MATH. Let MATH be an input on which the sensitivity of MATH equals MATH. Assume without loss of generality that MATH. All sensitive variables must be REF in MATH, and setting one or more of them to REF changes the value of MATH from REF to REF. Hence by fixing all variables in MATH except for the MATH sensitive variables, we obtain the OR function on MATH variables. Since OR on MATH variables has MATH CITE, it follows that MATH. It is known (see for instance CITE) that MATH for monotone MATH, hence MATH. |
cs/9904019 | Run the algorithms MATH and MATH of REF side-by-side until one of them terminates with a certificate. This gives a certificate-finding quantum algorithm for MATH with expected number of queries MATH. Run this algorithm for twice its expected number of queries and answer `don't know' if it hasn't terminated after that time. By NAME 's inequality, the probability of non-termination is MATH, so we obtain an algorithm for our zero-error setting with MATH queries. The classical lower bound follows from combining two known results. First, an AND-OR tree of depth MATH on MATH variables has MATH CITE (see also CITE). Second, for such trees we have MATH CITE. Hence MATH. |
cs/9904019 | A similar analysis as before shows MATH and MATH. For the quantum lower bound: note that if we set all variables to REF except for the MATH variables in the first subtree, then MATH becomes the OR of MATH variables. This is known to have zero-error complexity exactly MATH CITE, hence MATH. |
cs/9904019 | The quantum upper bound follows from REF and the CITE reduction. For the classical lower bound, suppose we have a classical zero-error protocol MATH for MATH with MATH bits of communication. We will show how we can use this to solve the Disjointness problem on MATH variables. (Given NAME 's input MATH and NAME 's MATH, the Disjointness problem is to determine if MATH and MATH have a REF at the same position somewhere.) Let MATH be the following classical protocol. NAME and NAME view their MATH-bit input as made up of MATH subtrees of MATH variables each. They add a dummy variable with value REF to each subtree and apply a random permutation to each subtree (NAME and NAME have to apply the same permutation to a subtree, so we assume a public coin). Call the MATH-bit strings they now have MATH and MATH. Then they apply MATH to MATH and MATH. Since MATH, after an expected number of MATH bits of communication MATH will deliver a certificate which is a common REF in each subtree. If one of these common REFs is non-dummy then NAME and NAME output REF, otherwise they output REF. It is easy to see that this protocol solves Disjointness with success probability REF if MATH and with success probability MATH if MATH. It assumes a public coin and uses MATH bits of communication. Now the well-known MATH bound for classical bounded-error Disjointness on MATH variables CITE implies MATH. |
cs/9904019 | For the lower bound, let MATH denote the degree of the unique multilinear multivariate polynomial MATH that represents a function MATH (that is, MATH for all MATH). CITE proves that MATH for every MATH. CITE prove that MATH for all monotone graph properties MATH. Combining these two facts gives the lower bound. Let MATH be the property ``the graph contains more than MATH edges". This is just a special case of the Majority function. Let MATH be Majority on MATH variables. It is known that MATH, where MATH is the number of REF in the binary expansion of MATH. This was first noted by CITE. It also follows immediately from classical results CITE that show that an item with the Majority value can be identified classically deterministically with MATH comparisons between bits (a comparison between two black-box-bits is the XOR of two bits, which can be computed with REF quantum query CITE). One further query to this item suffices to determine the Majority value. For MATH and MATH we have MATH and hence MATH. |
cs/9904019 | The quantum lower bound follows from MATH REF and MATH. Consider the property ``the graph contains a star", where a star is a node that has edges to all other nodes. This property corresponds to a REF-level tree, where the first level is an OR of MATH subtrees, and each subtree is an AND of MATH variables. The MATH variables in the MATH-th subtree correspond to the MATH edges MATH for MATH. The MATH-th subtree is REF iff the MATH-th node is the center of a star, so the root of the tree is REF iff the graph contains a star. Now we can show MATH and MATH analogously to REF . |
cs/9904019 | For MATH: MATH. |
cs/9904019 | A polynomial MATH with MATH and MATH for all integers MATH must have degree MATH. Since MATH for our MATH, we have MATH. Now MATH has degree MATH and MATH for all integers MATH . Applying REF to MATH (which is bounded by REF at integer points) with MATH we obtain: MATH . Now we rescale MATH to MATH (that is, the domain MATH is transformed to MATH), which has the following properties: MATH for MATH. Thus MATH is ``small" on all MATH and ``big" somewhere outside this interval (MATH). Linking this with REF we obtain MATH . Rearranging gives the bound. |
cs/9904019 | By induction on MATH. Base step. For MATH the bounds are trivial. Induction step (assume the lemma for MATH). Let MATH be the uniform MATH-level AND-OR tree on MATH variables. The root is an OR of MATH subtrees, each of which has MATH variables. We construct MATH as follows. First use multi-level NAME as in CITE to find a subtree of the root whose value is REF, if there is one. This takes MATH queries and works with bounded-error. By the induction hypothesis there exists an algorithm MATH with expected number of MATH queries that finds a REF-certificate for this subtree (note that the subtree has an AND as root, so the roles of REF are reversed). If MATH has not terminated after, say, REF times its expected number of queries, then terminate it and start all over with the multi-level NAME search. The expected number of queries for one such run is MATH. If MATH, then the expected number of runs before success is MATH and MATH will find a REF-certificate after a total expected number of MATH queries. If MATH, then the subtree found by the multi-level NAME will have value REF, so then MATH will never terminate by itself and MATH will start over again and again but never terminates. We construct MATH as follows. By the induction hypothesis there exists an algorithm MATH with expected number of MATH queries that finds a REF-certificate for a subtree whose value is REF, and that runs forever if the subtree has value REF. MATH first runs MATH on the first subtree until it terminates, then on the second subtree, etc. If MATH, then each run of MATH will eventually terminate with a REF-certificate for a subtree, and REF-certificates of the MATH subtrees together form a REF-certificate for MATH. The total expected number of queries is the sum of the expectations over all MATH subtrees, which is MATH. If MATH, then one of the subtrees has value REF and the run of MATH on that subtree will not terminate, so then MATH will not terminate. |
hep-th/9904002 | Consider any weight vector MATH whose entries are only MATH's and MATH's. The scalar product of such a vector with a simple root MATH is positive only if the vector has a MATH at the MATH-th position and a MATH at the MATH-st position. Then and only then will MATH also be a weight, according to REF . The effect of subtracting MATH from MATH will be to shift the MATH at the MATH-th position one step to the right. Doing this repeatedly one can transform the highest weight vector MATH, which has all the MATH's to the left and the MATH's to the right, into any other vector with MATH's. Furthermore one will never obtain a weight vector with more or less than MATH's. The number of different vectors with MATH's and the rest MATH's is equal to the number of ways one can choose MATH sites out of MATH and this gives the dimension of the MATH-th fundamental representation. |
hep-th/9904062 | This is an immediate consequence of the fact that REF is an affine bundle modelled over the pull-back vector bundle MATH. |
hep-th/9904062 | Given an arbitrary connection MATH on the fiber bundle MATH, the corresponding Hamiltonian map REF defines the form MATH which is exactly the Hamiltonian form REF . Since MATH is a MATH-valued basic REF-form on MATH, MATH is a horizontal density on MATH. Then the result follows from REF . Note that MATH iff REF . |
hep-th/9904062 | If MATH is closed, there is a contractible neighbourhood MATH of a point MATH which belongs to a holonomic coordinate chart MATH and where the local form MATH is exact. We have MATH on MATH. It is readily observed that the second term in the right-hand side of this equality is also an exact form on MATH. By virtue of the relative NAME lemma, it can be brought into the form MATH where MATH is a local function on MATH. Then the form MATH in REF reads MATH . Using REF , one can easily show that this is a Hamiltonian form on MATH. |
hep-th/9904062 | Given a Hamiltonian form MATH, let us consider the first order differential operator MATH on MATH where MATH is the canonical monomorphism REF . It is called the NAME operator associated with MATH. A glance at REF shows that this operator is an affine morphism over MATH of constant rank. It follows that its kernel is an affine closed imbedded subbundle of the jet bundle MATH. This subbundle has a global section MATH which is a connection on MATH. This connection obeys REF . |
hep-th/9904062 | The proof is based on REF . |
hep-th/9904062 | The relation REF takes the coordinate form MATH . Substituting REF , we obtain the relation REF . |
hep-th/9904062 | Let MATH be a vertical vector field on the affine jet bundle MATH which takes its values into the kernel of the tangent map MATH to MATH. Then MATH. |
hep-th/9904062 | Given a vector MATH, the value MATH is the same for all Hamiltonian maps MATH satisfying the relation REF . Then the results follow from the relation REF . |
hep-th/9904062 | Acting by the exterior differential on the relation REF , we obtain the relation MATH which is equivalent to the system of equalities MATH . Using these equalities and REF , one can easily see that MATH where MATH is the NAME - NAME - NAME operator REF . Let MATH be a section of MATH which lives in the Lagrangian constraint space MATH, and MATH. Then we have MATH . If MATH is a solution of the NAME equations, the exterior form MATH vanishes on MATH. Hence, the pull-back form MATH vanishes on MATH. It follows that MATH obeys the NAME REF - REF . We obtain from the equality REF that MATH, MATH. Hence, MATH is a solution of the NAME - NAME equations. |
hep-th/9904062 | The NAME REF hold by virtue of REF . Substituting MATH in the NAME REF and using the relations REF , we come to the NAME REF for MATH as follows: MATH . |
hep-th/9904062 | Such a Hamiltonian form MATH defines the global section MATH of the fibred manifold REF . Due to the relation REF , MATH and the constrained NAME equations can be written as MATH . Note that they differ from the NAME REF restricted to MATH which read MATH where MATH is a section of MATH and MATH is an arbitrary vertical vector field on MATH. A solution MATH of REF satisfies obviously the weaker REF . |
hep-th/9904062 | In accordance with the relation REF , the projection REF yields a surjection of MATH onto MATH. Given a section MATH of the fibred manifold REF , we have the morphism MATH . By virtue of REF , this is a surjection such that MATH . Hence, MATH is a bundle isomorphism over MATH which is independent of the choice of a global section MATH. Combining REF gives MATH that leads to the desired equivalence. |
hep-th/9904062 | The first of the relations REF is an immediate consequence of the relation REF . The latter follows from REF and the relation REF if we put MATH for some Hamiltonian form MATH associated with the almost regular Lagrangian MATH. |
hep-th/9904062 | Solutions MATH of the pointwise linear algebraic REF form an affine space modelled over the linear space of solutions of the linear algebraic equations MATH. At each point of MATH, these spaces are obviously connected. |
hep-th/9904062 | The map REF is a solution of the algebraic equations MATH . After pointwise diagonalization, the matrix MATH has some non-vanishing components MATH, MATH. Then a solution of REF takes the form MATH while the remaining components MATH, MATH, are arbitrary. In particular, there is a solution with MATH . It satisfies the particular relation MATH . Further on, we will take MATH to be the solution REF . If the Lagrangian REF is regular, the linear map REF is uniquely determined by REF . |
hep-th/9904062 | The proof follows from a direct computation by means of the relations REF . |
hep-th/9904062 | By the very definitions of MATH and MATH, the Hamiltonian map REF satisfies REF . A direct computation shows that MATH. Then the relation REF also holds and, if MATH is a connection REF , the Hamiltonian form REF is associated with the Lagrangian REF . Let us write the corresponding NAME REF for a section MATH of the NAME bundle MATH. They are MATH . Due to the surjections MATH and REF , the NAME REF break in two parts MATH . Let MATH be an arbitrary section of MATH, for example, a solution of the NAME - NAME equations. There exists a connection REF such that the relation REF holds, namely, MATH where MATH is a connection on MATH which has MATH as an integral section. It is easily seen that, in this case, the Hamiltonian map REF satisfies the relation REF for MATH. Hence, the Hamiltonian forms REF constitute a complete set. |
hep-th/9904062 | Due to the splitting REF , we have the corresponding splitting of the vertical tangent bundle MATH of the NAME bundle MATH. In particular, any vertical vector field MATH on MATH admits the decomposition MATH such that MATH is a vertical vector field on the Lagrangian constraint space MATH. Let us consider the equations MATH where MATH is a section of MATH and MATH is an arbitrary vertical vector field on MATH. They are equivalent to the pair of REF . REF are obviously the NAME REF for MATH. Bearing in mind the relations REF , one can easily show that REF coincide with the NAME REF . The proof is completed by observing that, restricted to the Lagrangian constraint space MATH, REF are exactly the constrained NAME REF . |
hep-th/9904062 | CASE: Let us consider the affine difference MATH over MATH. We have MATH iff MATH. CASE: In the proof of REF , we have shown that, given MATH, there exists a connection REF which fulfills the relation REF . Let us consider the affine difference MATH over MATH. This is a local section of the vector bundle MATH over MATH. Let MATH be its prolongation onto MATH. It is easy to see that MATH is the desired connection. |
hep-th/9904062 | Similar to the well-known isomorphism between the fiber bundles MATH and MATH,MATH the isomorphism MATH can be established by inspection of the transformation laws of the holonomic coordinates MATH on MATH and MATH on MATH. |
hep-th/9904062 | The proof follows from a direct computation. We have MATH . Components of this connection obey the NAME REF and the equations MATH . |
hep-th/9904136 | We show how to derive the first relation. Inserting the definition we find: MATH where we used the relation MATH on the last line. The second equation follows from a similar calculation. |
math-ph/9904013 | The function MATH is positive, for MATH odd. |
math-ph/9904013 | The idea is to write MATH as the sum of a function MATH that is positive and a function MATH that absorbs the asymptotic behavior at infinity. Since MATH for MATH small, with MATH there exists MATH such that MATH for all MATH . Let MATH to be chosen below, and let MATH be the NAME step function, that is, MATH for MATH and MATH for MATH . Then, we define the function MATH by the equation MATH and we set MATH . In order to prove that MATH is positive, for MATH large enough, we write MATH where MATH and MATH . MATH is positive for MATH since in this case MATH and MATH is positive for MATH by definition of MATH . Next we consider MATH . For MATH we have that MATH. But MATH for all MATH and all MATH if MATH is sufficiently large, since MATH since MATH is bounded, and since MATH for MATH large enough. Finally, using the asymptotic properties of MATH and MATH we see that MATH for MATH if MATH is chosen large enough. We now estimate the function MATH . From the definition of MATH we get that MATH and therefore, since MATH is positive, we have the lower bound MATH for some constant MATH, from which REF follows. |
math-ph/9904013 | In terms of the functions REF - REF we get that, for MATH and therefore MATH, where MATH and where MATH and MATH are as defined above. Since MATH it remains to be shown that MATH . Using the differential equations for MATH and MATH we find that MATH where MATH as defined above, where MATH and where MATH . The functions MATH and MATH can be further decomposed as follows MATH where MATH and MATH are as defined above, and where MATH . It remains to be shown that MATH but this follows using the definitions. |
math-ph/9904013 | From REF it follows that either MATH or MATH . In the first case we get using REF that MATH and therefore MATH and in the second case we get using REF that MATH and therefore MATH . |
math-ph/9904013 | REF follow by using the triangle inequality and the asymptotic properties of the functions MATH, MATH, MATH and MATH. |
math-ph/9904013 | The case MATH follows from REF . Let now MATH . Since MATH for all MATH we find that MATH . Furthermore, since MATH near MATH . Since the function MATH is bounded, it follows that MATH . MATH with MATH . For MATH we improve this bound using additional properties of the function MATH . Namely, since MATH we have that MATH . |
math-ph/9904013 | Since MATH the solution of the integral equation will be dominated by MATH and, as we will see, MATH has been chosen such that MATH . The idea is therefore to show that, if MATH is large enough to make the nonlinear part of the map MATH small, and if MATH, then the map MATH contracts the ball MATH into itself, which by the contraction mapping principle implies the theorem. We first show that MATH maps the ball MATH into itself. For the contribution coming from the initial condition we have MATH and therefore MATH . We next estimate the norm MATH . Let MATH . Then, MATH and therefore MATH . It remains to be shown that the nonlinearity is also bounded by MATH for MATH large enough. For MATH we have, MATH . For MATH we get, since MATH and for MATH we get MATH and for the other cases we have MATH and therefore MATH if MATH is large enough. Using the triangle inequality we get that MATH which proves that MATH as claimed. We now show that MATH is NAME. Let MATH and MATH be in MATH . We have MATH and therefore we get, using the same estimates as for REF , that MATH provided MATH is large enough. This completes the proof of REF . |
math-ph/9904013 | We first prove that MATH is unique. Given a function MATH from MATH to MATH we define the function MATH. Assume that there are two values MATH such that the functions MATH and MATH are positive and satisfy MATH . We first show that the function MATH is positive for all MATH . Namely, if we assume the contrary, then because MATH, there must be a first MATH such that MATH . Furthermore, if MATH then MATH and therefore MATH a contradiction. Therefore MATH and as a consequence MATH are positive for all MATH from which it follows that MATH in contradiction with MATH . To prove the existence of a MATH for which MATH is positive and for which MATH, we use the so called shooting method. Note that, for any MATH, the initial value REF has a unique solution MATH and since MATH the function MATH is strictly decreasing on MATH for MATH small enough. We will show that for small enough MATH, the graph of MATH intersects the real axis and MATH becomes negative, whereas for MATH large enough, MATH has a minimum and then diverges to plus infinity. The (unique) point between those two sets is MATH. Define the two subsets MATH and MATH of MATH . We note that if MATH and MATH for some MATH, then MATH on any interval MATH on which MATH is defined, and a function MATH with MATH can therefore not converge to zero at infinity. Furthermore, since the function MATH is a solution of the differential REF , it follows, since solutions are unique, that MATH if MATH and therefore the intersection of MATH with MATH is empty. The sets MATH and MATH are open, by continuity of the solution MATH as a function of the initial data MATH. We now show that MATH is non empty and bounded, which shows that MATH . This MATH is neither in MATH nor in MATH, and therefore the function MATH is at the same time strictly positive and strictly decreasing, and therefore MATH. To prove that MATH is non empty, we fix any MATH positive and choose MATH small enough such that on MATH the solution MATH exists and is strictly decreasing. Then, MATH . Choose now MATH and let MATH be the corresponding solution. As before, we have that the function MATH and its second derivative MATH are positive on the interval MATH and therefore, since MATH we find that MATH . Using the definition of MATH we therefore find that MATH . Therefore MATH. We now prove that MATH is bounded. For MATH, let MATH be the largest value (possibly infinite) such that on MATH the solution MATH exists and is strictly positive. Then, MATH is positive on MATH and, therefore MATH for MATH. As a consequence, if the function MATH exists on MATH then MATH . Using again that MATH we then find that MATH for MATH, and therefore MATH on MATH, which implies that MATH which is positive if MATH . Therefore MATH must be equal to zero for some MATH . Any such MATH therefore belongs to MATH . If the function MATH ceases to exist before MATH it must have been diverging to plus infinity for some MATH which again implies that MATH must have been equal to zero for some MATH and the corresponding MATH is in MATH . |
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