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math/9909191 | Since any image flat is parallel to an image flat that is the limit of a sequence of MATH - compact image flats, it suffices to prove the lemma for MATH - compact image flats. Let MATH be a rank MATH abelian subgroup which acts as a lattice on MATH. For any MATH, write MATH under the splitting MATH. Then MATH is a translation on MATH. This is true because, for any geodesic MATH in MATH, MATH is the projection of some geodesic MATH in MATH. Since MATH maps MATH to a parallel geodesic in MATH, MATH is parallel to MATH. Let MATH be a translation in MATH. As MATH commutes with MATH for all MATH, MATH is also invariant under MATH. Since we have uniform bound in NAME distance between MATH or MATH and MATH, the preimage flat, we know that the NAME distance between MATH and MATH must be bounded by a uniform constant. Therefore, MATH must be contained in MATH, since otherwise we can choose a translation in MATH that sends MATH to a parallel MATH with arbitrarily large MATH. |
math/9909191 | Since any maximally singular geodesic MATH is an intersection of MATH - flats, there exist MATH - flats MATH with MATH. Let MATH be an image flat of MATH, MATH. Then REF of Theorem C says that there exists a geodesic MATH which is parallel to all MATH, and any geodesic MATH parallel to all MATH must be parallel to MATH. In the product space MATH, MATH is parallel to MATH if and only if MATH is parallel to MATH and MATH is parallel to MATH. Here the subscripts denote the projections. So if MATH is not parallel to a leaf of MATH or MATH, then any line in the plane MATH would be parallel to all those MATH, which is impossible. So, MATH must be parallel to either MATH or MATH. That is, MATH is contained in the disjoint union of MATH and MATH. From this and the definition of MATH, it is clear that for each MATH, MATH is a product foliation. Next we claim that for MATH, MATH and MATH are perpendicular. Take any geodesic MATH with MATH. Then any leaf MATH is contained in the neighborhood MATH by REF of Theorem C. Here MATH stands for the union of all geodesics parallel to MATH. So if the projection MATH of MATH onto MATH is not a point, any point in MATH will be within distance MATH to another point in MATH where there is a geodesic parallel to MATH passing through. That is, MATH, where MATH is the union of all geodesics in MATH that is parallel to MATH. Since MATH is geodesically complete, and MATH is convex, we know that MATH. That is, MATH is foliated by geodesics parallel to MATH, so MATH, hence MATH will contain a Euclidean factor, a contradiction. So geodesics from MATH to any point in MATH must be perpendicular to MATH. So MATH must be contained in MATH, the leaf through MATH of the orthogonal complement of MATH in MATH, where MATH by the Sandwich lemma. Since MATH is convex, we have MATH, hence MATH is perpendicular to MATH. On the other hand, since any geodesic in MATH is parallel to an image flat, which is contained the span of MATH through MATH, so MATH is spanned by those MATH, MATH. |
math/9909191 | Write MATH and MATH. Then we have MATH. Let MATH be the mid point of the geodesic segment from MATH to MATH. Then since MATH is NAME, any MATH will satisfy MATH . Since this is true for any MATH, the far right hand side of the above inequality must be bigger than MATH, hence MATH . Add this with the similar inequality obtained by reversing the role of MATH and MATH, we get MATH . If one replace MATH by the points on the geodesic segment from MATH to MATH that are very closed to MATH or MATH, then the coefficient MATH in the right hand side can be removed. |
math/9909191 | The geodesic segment from MATH to MATH gives a proper homotopy between MATH and the identity map. Hence, MATH induces the identity map on the top dimensional cohomology group with compact support, MATH, MATH-dim-MATH. The result follows. |
math/9909191 | First, notice that MATH does not contain any NAME translations. This is because the set of NAME translations in MATH forms a normal abelian subgroup. By the Main Theorem of CITE, MATH does not contain any nontrivial normal abelian subgroup, since MATH has no Euclidean NAME factor. Now let us assume that MATH for MATH. We want to derive a contradiction. Note that MATH is not a point, for otherwise MATH would contain a translation by the NAME Theorem. Denote by MATH and MATH the projection of MATH on MATH and MATH, respectively. Since MATH is a NAME group, the proof of Lemma A in CITE says that MATH is discrete. Let MATH be the kernel of MATH. The proof of REF says that either MATH is discrete or MATH contains a NAME translation. Since MATH is assumed to be irreducible, MATH can not be discrete. Since MATH contain no NAME translation, so does the subgroup MATH. This completes the proof. (Note that in this argument, REF is used. But it is obviously valid in the singular case.) |
math/9909191 | (compare proof of REF) Let MATH denote the limit set of MATH. By REF, MATH is not empty. Let MATH, then by REF there exist a unique point MATH such that MATH can be joined to MATH. If MATH is a point that can be joined to MATH then by REF MATH and therefore MATH. So MATH admits NAME translations along the geodesics joining MATH to MATH by Sandwich Lemma. |
math/9909191 | If MATH is not simple, we may take a lattice MATH and consider MATH equipped with MATH symmetric and MATH equipped with MATH. By the previous arguments, we know that MATH will be a product space of MATH, where those MATH are the simple factors of MATH. Therefore we may assume that MATH is simple. First consider the case when the rank of MATH is MATH. In this case, the geodesic sphere in MATH is a single MATH orbit, so it coincides with a geodesic sphere with the same center under MATH, the symmetric metric. For any two points MATH, MATH, let MATH be the midpoint under the metric MATH. Then there are two geodesic spheres under MATH, centered at MATH and MATH, respectively, so that MATH is their unique intersection point. This implies that MATH must be on the geodesic segment from MATH to MATH with respect to the metric MATH. So the geodesics in MATH and MATH have the same images. Fix MATH and MATH with MATH. Let MATH be the symmetric metric on MATH such that MATH. Let MATH be a one parameter subgroup of MATH such that MATH is the unit speed geodesic under MATH with MATH. By considering MATH, we know that MATH for any rational number MATH, whose denominator is a power of two. Hence, it also holds for any MATH by continuity. For any MATH, and for any MATH such that MATH and such that MATH lies in the image of MATH, we have MATH. So MATH is symmetric in this case. Now assume MATH is simple with rank MATH. We need the following. Suppose MATH is a connected, centerless, semisimple NAME group of noncompact type, and MATH a maximal compact subgroup. Let MATH be the NAME decomposition of the NAME algebras. Suppose MATH is a maximal abelian subalgebra in MATH, and MATH. Denote by MATH the centralizer of MATH in MATH, and denote by MATH the fixed point set in MATH of MATH. Then MATH is a totally geodesic submanifold in MATH containing the maximal flat MATH, where MATH. Denote by MATH the rank of MATH. CASE: MATH when and only when MATH is the product of MATH copies of the hyperbolic plane MATH. CASE: If MATH is simple, then either MATH, or MATH, depending on whether MATH is Hermitian symmetric or not. Now let us use NAME to finish the proof of REF . Let MATH be a maximal flat under MATH, with MATH the isotropy subgroup at MATH. First we claim that MATH is MATH - convex. Let MATH be the NAME decomposition of the NAME algebras of MATH and MATH, and MATH a maximal abelian subalgebra contained in MATH. Write MATH so that MATH. As in NAME, denote by MATH the centralizer of MATH in MATH, and denote by MATH the fixed point set of MATH in MATH. Then MATH is totally geodesic in MATH under MATH. Note that MATH is a symmetric space of nonpositive curvature, with MATH and MATH. Since MATH is the fixed point set of MATH - isometries, so it is MATH - convex, and MATH acting transitively on MATH as both MATH and MATH - isometries. By NAME, either MATH, in which case MATH is a MATH - flat, or MATH. In this case, the arguments in the previous subsections implies that MATH must split as the product of MATH surfaces, each is symmetric since MATH has rank MATH. So MATH is a symmetric metric on MATH, and MATH is a MATH - flat. Now consider any maximal flat MATH passing through MATH. The restriction of MATH to each flat MATH is a Euclidean metric which is invariant under the NAME group, which acts irreducibly on MATH since MATH is assumed to be simple. So MATH. Since any two flats through MATH can be joined by finitely many flats through MATH so that each one intersects the next at more than one point, MATH must be constant, and MATH is symmetric. This completes the proof of REF assuming NAME. |
math/9909191 | Let us start with REF . Without loss of generality, we may assume that MATH is simple. Assume MATH, we want to conclude that MATH is the hyperbolic plane. Let MATH be the rank of MATH. Since the NAME algebra MATH of MATH is trivial, MATH is a NAME subalgebra for MATH, the complexification of MATH. Denote by MATH the real points of MATH under the conjugation of MATH in MATH. Then MATH is the identity component of MATH. Denote by MATH the maximal compact subgroup of MATH with identity component MATH. Let MATH be the group of MATH - torsion elements in MATH. Then MATH is contained in MATH, the centralizer of MATH in MATH. Note that the number of connected components in MATH is bounded by the order of the outer automorphism group MATH. So the assumption MATH implies that MATH. By REF on page REF, we see that this is impossible unless MATH is MATH or MATH. In the MATH case, in fact in all four classical NAME algebra cases, it can be easily checked that MATH except for MATH. This completes the proof of REF . For REF , let us apply REF to the symmetric space (which may contain a Euclidean factor now) MATH, since MATH is trivial, by REF , we know that MATH. We want to show that MATH is either MATH or MATH if MATH is simple. Let MATH be the normalizer of MATH in MATH. Then MATH ( or MATH) consists of elements in MATH that stabilize MATH ( fix every point in MATH). MATH is a normal subgroup of finite index in MATH, and MATH is NAME group which acts irreducibly on MATH. For each MATH, since MATH, we have MATH. Now since any isometry of MATH preserves the Euclidean factor, we know that the MATH part of MATH is invariant under MATH. So either MATH or MATH. This completes the proof of NAME. |
quant-ph/9909012 | Let MATH and MATH denote respectively the initial superpositions of MATH on input MATH and of MATH on input MATH. Let MATH and MATH be the sets of all accepting configurations and rejecting configurations, respectively, of MATH on any string input of length at most MATH. For convenience, let MATH. Assume that MATH and MATH. We want to evaluate the value MATH. This term equals MATH, which is at most MATH. Obviously, this equals MATH, which is bounded above by MATH since MATH. We obtain MATH by the NAME inequality. Similarly, MATH. Hence, MATH is bounded above by MATH, which equals MATH. |
quant-ph/9909012 | Let MATH and MATH be respectively the initial superpositions of MATH on input MATH and on input MATH. Note that MATH. Let MATH and MATH be the time-evolution operators of MATH and MATH, respectively. For each MATH, let MATH, MATH, and MATH. Note that MATH. Thus, MATH equals MATH, which is at most MATH. This term equals MATH since MATH is unitary. The NAME inequality implies that MATH. Since MATH depends only on the configurations, in MATH, in which MATH is in a pre-query state with query words in MATH, we have MATH, and thus MATH. Since REF relativizes, we obtain MATH. The desired result therefore follows. |
quant-ph/9909012 | For any set MATH, let MATH if MATH and let MATH be undefined otherwise. This MATH satisfies that MATH. It is easy to show that MATH is in MATH by using the QTM that witnesses MATH. Conversely, assume that MATH witnessed by a certain polynomial-time well-formed QTM MATH. By our definition of accepting and rejecting configurations, the same QTM MATH witnesses MATH. Thus, MATH. |
quant-ph/9909012 | Clearly, MATH even though MATH is viewed as a class of partial functions. Let MATH. We can design a polynomial-time deterministic TM MATH with an appropriate polynomial MATH that satisfies the following conditions: for every string MATH, REF if MATH then MATH on input MATH outputs either MATH or MATH for all strings MATH and there exists a string MATH such that MATH on input MATH outputs MATH and REF if MATH then MATH on input MATH outputs MATH for all strings MATH. REF guarantees the existence of a reversible TM MATH that simulates MATH. Note that the running time of MATH depends only on the length of input. Consider the QTM MATH that carries out the following algorithm. On input MATH, where MATH is given on the first tape and MATH is on the second input tape, Observe the second input tape. If MATH is observed, copy MATH into a storage tape to avoid any future interference. Simulate MATH on input MATH. Since any reversible TM can be simulated on a certain well-formed QTM, MATH is a polynomial-time well-formed QTM. Clearly, MATH has MATH-amplitudes. Therefore, MATH belongs to MATH. |
quant-ph/9909012 | In this proof, we use the tape alphabet MATH. Let MATH be any function in MATH. Take a polynomial MATH and a polynomial-time deterministic TM MATH such that MATH for all MATH. From REF , we can assume that MATH is reversible and its running time depends only on the length of input. Define the new QTM MATH as follows. On input MATH, write MATH on a separate blank tape and apply MATH. Observe MATH on this tape and copy it into a storage tape to avoid any future interference. Simulate MATH on input MATH. Note that MATH has MATH-amplitudes since the above procedure can be conducted by a series of unitary operators with MATH-amplitudes. Clearly, MATH equals MATH. It suffices to set MATH and MATH. |
quant-ph/9909012 | CASE: Let MATH be any polynomial-time well-formed MATH-amplitude QTM that computes MATH with certainty and let MATH be any polynomial-time well-formed MATH-amplitude QTM whose acceptance probability MATH equals MATH. From REF , MATH can be synchronous with a single final state as well as quasi-stationary and in quasi-normal form on the output tape. Define the new QTM MATH as follows. On input MATH, simulate MATH. Note that the head of the output tape returns to the start cell. Observe the output tape after MATH enters a unique final state. When MATH is observed, simulate MATH on input MATH using a new set of blank tapes. Notice that the final superposition of MATH must have the form MATH, where MATH is the content of the output tape. Since MATH does not affect MATH's move, the acceptance probability MATH of MATH is exactly MATH. Thus, we obtain MATH. CASE: Let MATH for all MATH. Since MATH is a MATH-qubit source with MATH-amplitudes, let MATH be any polynomial-time well-formed clean MATH-amplitude QTM that produces qustring MATH on input MATH. Let MATH be another polynomial-time well-formed MATH-amplitude QTM witnessing MATH. Consider the QTM MATH that executes the following algorithm. On input MATH, copy MATH into a storage tape and then simulate MATH to produce MATH on a new blank tape. Observe string MATH on this tape. Copy MATH into a storage tape and then simulate MATH on input MATH. Obviously, MATH has MATH-amplitudes. Note that the probability of observing MATH is exactly MATH. Note that copying MATH prevents any further interference between two computations of MATH on input MATH and on different input MATH. Thus, MATH, which implies MATH. The second part follows from the fact that MATH is MATH-qubit source with MATH-amplitudes. In this case, we define MATH as MATH and MATH for each MATH and apply the first part. CASE: Since MATH, there exists a constant MATH such that MATH for all MATH. For a given MATH, let MATH be any polynomial-time well-formed MATH-amplitude QTM that witnesses MATH. By REF , MATH can be simulated over its tapes by a certain polynomial-time synchronous well-formed quasi-stationary quasi-normal-form MATH-amplitude QTM MATH with a single final state. Consider the following QTM MATH. On input MATH, compute MATH deterministically. Write MATH on a counter tape. Repeat the following by incrementing lexicographically string MATH written on the counter tape. Copy MATH and MATH into a new blank area of a work tape and then simulate MATH on input MATH. If all runs of MATH end with accepting configurations, then accept; otherwise, reject. Obviously, each run of MATH is independent because of the use of a new blank area each time. Thus, the acceptance probability of MATH equals MATH. Moreover, the number of runs of MATH on MATH is exactly MATH, which is polynomially bounded. Hence, MATH runs in polynomial time. CASE: Let MATH be any well-formed MATH-amplitude QTM that witnesses MATH in time polynomial MATH. REF yields the existence of a polynomial-time synchronous well-formed quasi-stationary quasi-normal-form MATH-amplitude QTM MATH, with a single final state, that simulates MATH over all the tapes of MATH. Since MATH, choose a polynomial MATH that satisfies MATH for all MATH. Consider the following algorithm. On input MATH, run MATH times and idle MATH steps MATH times to avoid the timing problem. Accept MATH if all the first MATH runs of MATH reach accepting configurations; reject MATH otherwise. Similar to REF, the above algorithm accepts MATH with probability exactly MATH since MATH. |
quant-ph/9909012 | Let MATH be any MATH-function, which is witnessed due to REF by a certain polynomial-time synchronous dynamic stationary unidirectional well-formed MATH-amplitude QTM MATH in normal form with a single final state. We also assume without loss of generality that MATH always outputs either MATH or MATH in the start cell. By the Reversal Lemma, there exists the reversing QTM MATH of MATH. Let MATH be any polynomial that bounds the running times of both MATH and MATH. For simplicity, assume that MATH and MATH share the same configuration space. Let MATH be any positive polynomial. Define MATH for all MATH. For convenience, we use integers between MATH and MATH instead of string of length MATH. By attaching a new blank storage tape to MATH, we obtain the simple expansion of MATH, say MATH. Note that MATH does not alter the content of this storage tape. We define the quantum algorithm MATH that starts with any superposition of MATH on input of length MATH. In this algorithm, we check only the cells indexed between MATH and MATH. Let MATH denote the identity operator. Apply MATH to the bit written in the start cell of the output tape of MATH. Simulate MATH on all the tapes except for the storage tape. Check if REF all the tapes except for the input and storage tapes are blank and REF the contents of the input tape and the storage tape agree. If so, apply MATH; otherwise, apply MATH. Finally, simulate MATH. Let MATH be any string of length MATH. For readability, we write MATH for MATH. Consider the following two superpositions. Let MATH and MATH be respectively the superpositions of all final configurations of MATH in which MATH starts with input MATH given to both the input and storage tapes and halts with bit MATH and with bit MATH written on the output tape. Let MATH be the real number in MATH satisfying that MATH. Clearly, MATH. The algorithm MATH has two eigenvalues MATH and MATH with their corresponding eigenvectors MATH and MATH; namely, MATH and MATH. To approximate MATH, we need to estimate MATH. This is done by the following phase estimation algorithm. On input MATH of length MATH, copy MATH into the storage tape to remember MATH and then simulate MATH on input MATH. When MATH halts, produce MATH on a new blank memory tape and apply MATH to generate MATH. Observe MATH and apply MATH times to all the tapes except for this memory tape. Since MATH, we then obtain qustring MATH. Next, we apply MATH to the memory tape and observe this tape. After MATH, the sum of the squared norms of both MATH and MATH becomes at least MATH. Similarly, the squared norms of both MATH and MATH sum up to at least MATH. After observing MATH on the memory tape, we define MATH as MATH if MATH and MATH otherwise. The probability that either MATH or MATH is at least MATH, which is larger than MATH. Moreover, MATH and thus, MATH. It follows from MATH that MATH. At the end, we output a MATH-approximation of the value MATH. Unfortunately, our algorithm has a minor problem due to the fact that no QTM can carry out MATH exactly CITE. However, it is possible to replace MATH by a certain polynomial-time well-formed QTM that MATH-approximates MATH. Therefore, MATH is in MATH. |
quant-ph/9909012 | We show the contrapositive. Assume that MATH. Let MATH be any set in MATH. There exists a MATH-function MATH and a polynomial MATH such that, for every MATH, MATH implies MATH and MATH implies MATH. By REF , there exist two functions MATH and MATH such that MATH for all MATH. Let MATH be any polynomial satisfying MATH for all MATH. Since MATH, we can choose a function MATH such that MATH for all MATH. Let MATH. On one hand, if MATH then MATH, which implies MATH. On the other hand, if MATH then MATH. Thus, MATH. Clearly, MATH and MATH can determine the membership of MATH. Since they are both in MATH, MATH belongs to MATH. Therefore, MATH. Since MATH, we obtain MATH. |
quant-ph/9909012 | Let MATH be any function in MATH. There exists a polynomial-time well-formed QTM MATH with MATH-amplitudes such that MATH. By REF , we assume that MATH is synchronous and quasi-stationary on its output tape. Note that MATH coincides with MATH for every MATH. Thus, MATH. Since MATH and MATH are both MATH-functions, MATH belongs to MATH. Thus, MATH. Conversely, assume that MATH. It follows from REF that there exist two polynomial-time synchronous well-formed MATH-amplitude QTMs MATH and MATH, which are both quasi-stationary on their output tape, such that MATH. Now, consider MATH. Generally speaking, the halting timing of MATH may differ from that of MATH. Let MATH and MATH be polynomials that measure the running times of these machines MATH and MATH, respectively. Without loss of generality, we can assume that MATH and MATH have the same alphabet, states, and tapes. To synchronize the halting timing of these machines, we attach a counter tape that behaves like a clock (counting the number of steps). When the machines halt, we force them to idle until the counter hits MATH. Thus, we can assume that MATH and MATH halt at the same time. Now, consider the following QTM MATH whose tape alphabet includes MATH. On input MATH, first write MATH on a separate blank tape and apply MATH. Observe this tape. When either MATH or MATH is observed, simulate MATH on input MATH. Otherwise, simulate MATH on input MATH. Note that MATH has MATH-amplitudes since MATH. The acceptance probability MATH is exactly MATH. Thus, the gap MATH is exactly MATH, which equals MATH. Therefore, MATH. This concludes that MATH. |
quant-ph/9909012 | Let MATH be the QTM given in the lemma. We define the desired QTM MATH as follows. Firstly, given input MATH, MATH simulates MATH on the same input. Let MATH be the initial configuration of MATH on input MATH. Assume that MATH halts in a final superposition MATH, where MATH ranges over all (valid) configurations (except for the content of the output tape) of MATH and the last qubit MATH represents the content of MATH's output tape. The acceptance probability MATH thus equals MATH. Secondly, MATH applies MATH to MATH and then we obtain the superposition MATH. By the Reversal Lemma, there exists a polynomial-time synchronous dynamic normal-form unidirectional well-formed QTM MATH that reverses the computation of MATH. Note that MATH also has MATH-amplitudes since MATH. Now, MATH simulates MATH starting with MATH as its initial superposition. Note that if we run MATH on superposition MATH then MATH becomes the initial superposition MATH. In our notation MATH, MATH can be viewed as a unitary operator. Abusing this notation, we write MATH to mean the transposed conjugate of MATH. Observe that the inner product of MATH and MATH is MATH, which equals MATH. Finally, MATH outputs REF (that is, acceptance) if it observes exactly MATH; otherwise, MATH outputs REF (that is, rejection). The amplitude of the configuration MATH described in the lemma is exactly MATH, which equals MATH. Since MATH preserves the inner product, we have MATH. This completes the proof. |
quant-ph/9909012 | Let MATH be any MATH-function. Using REF , we obtain a polynomial-time synchronous dynamic normal-form unidirectional well-formed MATH-amplitudes QTM MATH with a single final state such that MATH for all MATH. Let MATH be any polynomial satisfying that MATH on input MATH halts at time MATH. It follows from the Gap Squaring Lemma that there exists another polynomial-time well-formed MATH-amplitude QTM MATH that starts on input MATH and halts in a final superposition in which the amplitude of configuration MATH is MATH, where MATH is a unique configuration of MATH that consists of MATH on the input tape, MATH on the output tape, and empty elsewhere. Thus, the acceptance probability of MATH equals MATH, which is obviously MATH. Therefore, MATH belongs to MATH. |
quant-ph/9909012 | The desired QTM MATH works as follows. On input MATH, MATH simulates MATH on input MATH; when MATH halts in a final configuration MATH, MATH starts another round of simulation of MATH in a different set of tapes. After MATH reaches a final configuration MATH, MATH deterministically checks if both final configurations MATH and MATH are identical. If MATH, then MATH outputs the blank symbol and halts. Now, assume that MATH. If this unique configuration MATH is an accepting configuration, then MATH outputs MATH; otherwise, it outputs MATH. On this computation path, we obtain the amplitude MATH, which equals MATH. For each MATH, let MATH denote the set of all final configurations, of MATH on MATH, in which the output tape consists of symbol MATH in the start cell. Thus, the sum MATH equals MATH, which is exactly MATH. Similarly, MATH equals MATH. |
quant-ph/9909012 | CASE: It easily follows that MATH. To show that MATH, let MATH and let MATH be any polynomial such that MATH for all MATH. Define MATH and MATH. The last set MATH is necessary to determine the length of MATH. We can show that MATH and MATH are both in MATH by simulating the QTM that computes MATH. Thus, MATH. Now, it is easy to show that MATH by making nonadaptive queries MATH to both MATH and MATH. It still remains to prove that MATH. Let MATH for a certain oracle MATH in MATH. Let MATH be any polynomial-time well-formed oracle QTM that, on input MATH, outputs MATH with certainty. The Canonical Form Lemma allows MATH to be in a canonical form with oracle MATH. Since MATH, by REF , MATH is recognized with probability MATH by a certain polynomial-time synchronous dynamic stationary normal-form unidirectional well-formed MATH-amplitude QTM MATH with a single final state. We further assume from the Squaring Lemma that MATH's final superposition consists entirely of a configuration, with amplitude MATH, in which MATH is in a final state, MATH's output tape holds only one bit in the start cell, and all other tapes are empty. Such a configuration can be identified with a bit written on the output tape. Consider the quantum algorithm MATH that simulates MATH on input MATH and, whenever it invokes a query MATH, simulates MATH on input MATH. This algorithm MATH can be implemented on a certain well-formed oracle QTM since MATH makes the same number of queries to oracle MATH with query words of the same length along each computation path on any input of fixed length. This implies that MATH. CASE: Similar to REF except for the proof of MATH. We can show MATH in a way similar to MATH CITE by amplifying the success probability of a QTM, which computes a given oracle set, from MATH to close to MATH so that the cumulative error is still bounded above by MATH after the polynomially-many runs of this QTM. CASE: Let MATH, which is witnessed by a certain polynomial MATH and a polynomial-time well-formed QTM MATH as in REF . Let MATH be any polynomial satisfying MATH for all MATH. We modify the definitions of MATH and MATH in REF as follows. Let MATH be the collection of all strings MATH, where MATH and MATH, such that there exist a string MATH and a qustring MATH satisfying that MATH on input MATH outputs MATH with probability at least MATH with the additional condition that the MATH-th bit of MATH must be MATH. The set MATH is defined similar to MATH but it checks if MATH on input MATH outputs MATH with MATH with probability at least MATH. It is easy to see that MATH and MATH are in MATH because of the choice of MATH. Similar to REF, making appropriate nonadaptive queries to MATH computes MATH in polynomial time. REF These proofs are similar to REF. |
quant-ph/9909012 | Clearly, MATH. Since MATH by REF , it follows that MATH. We still need to prove that MATH. Let MATH be any set in MATH. It is easy to see that MATH. Since MATH, we obtain that MATH. By the Squared Function Theorem, MATH is in MATH. Since MATH, MATH also belongs to MATH. This yields the desired conclusion that MATH. Therefore, MATH. Similarly, we can show that MATH. |
quant-ph/9909012 | The implication from left to right is obvious. Let MATH be any function in MATH. Assuming that MATH, we want to show that MATH belongs to MATH. Let MATH be any polynomial that bounds the length of the value of MATH. Without loss of generality, we assume that MATH is length-regular since, otherwise, we can set MATH for all MATH. For simplicity, assume that MATH for all MATH. Define MATH, where MATH is the dot product of MATH and MATH. It follows from MATH that MATH is in MATH. By our assumption, MATH is also in MATH for a certain function MATH. Since REF relativizes, there exists a polynomial-time synchronous dynamic stationary normal-form unidirectional well-formed oracle QTM MATH, with a single final state, that recognizes MATH with oracle MATH with error probability MATH. We first amplify its success probability from MATH to MATH. For such a QTM, we apply the Squaring Lemma (for an oracle QTM) and obtain another QTM MATH. We modify this MATH so that, on input MATH, it produces a final superposition of configurations, one of which has only MATH written on the tapes with positive real amplitude MATH. Obviously, MATH makes only MATH queries. The new QTM MATH works as follows. On input MATH of length MATH, write MATH on a new blank tape and apply MATH. We then have MATH, where MATH. For each MATH, where MATH, run MATH to change MATH to MATH, where MATH is a certain qustring. At this moment, we obtain MATH. Apply MATH. The final superposition becomes MATH for a certain qustring MATH since MATH. Unfortunately, MATH is not known to be orthogonal to MATH. However, since MATH, we can observe MATH with probability at least MATH. Thus, MATH. |
quant-ph/9909012 | We say that a set MATH is good if, for every MATH, either MATH or MATH. For any set MATH and any string MATH, let MATH. To compute this function MATH, consider the following oracle QTM MATH with oracle MATH. On input MATH of length MATH, write MATH on a query tape and apply MATH. Copy the first MATH bits into a query list on a designated tape. Invoke an oracle query. Delete the query list. Again, apply MATH. Observe the first MATH bits on the query tape. Output MATH if MATH is observed, and output MATH otherwise. The deletion of each query list is possible since the query list contains the exact copy of the first MATH bits on the query tape. It follows by a simple calculation that MATH on input MATH outputs MATH with certainty if MATH is good. Thus, MATH belongs to MATH for any good oracle MATH. Subsequently, we construct a good oracle MATH such that MATH. For our construction, we need an effective enumeration of all MATH-time bounded nondeterministic TMs. Let MATH be such an enumeration and define MATH to be an enumeration of natural numbers (with possible repetition) such that each MATH halts within time MATH on all inputs of length MATH, independent of the choice of oracles. We construct the desired oracle MATH stage by stage. Initially, set MATH and MATH. At stage MATH of the construction of MATH, let MATH denote the minimal integer satisfying that MATH and MATH. Assuming MATH, we define MATH. Clearly, MATH. If MATH, then define MATH to be MATH. Assume otherwise. There exists a unique accepting computation path MATH of MATH on MATH. Let MATH denote the set of all words that MATH queries along this computation path MATH. Since MATH, there is a subset MATH of MATH such that MATH and MATH. For this MATH, MATH but MATH. Thus, we should set MATH. After all the stages, define MATH. This set MATH satisfies the proposition. |
quant-ph/9909012 | We begin with the definition of a test function MATH. For each string MATH, let MATH. Note that MATH. For completeness, whenever MATH, set MATH. The desired function MATH is defined as MATH for each MATH and MATH. Since MATH for all MATH and MATH, MATH is in MATH. To complete the proof, it suffices to construct a set MATH satisfying that MATH. Let MATH and MATH be respectively two effective enumerations of all polynomial-time well-formed oracle QTMs and of all polynomials such that each MATH halts within time MATH on all inputs of length MATH independent of the choice of oracles. We build by eries of disjoint sets MATH and then define MATH. This MATH satisfies the theorem. For convenience, set MATH and MATH. Consider stage MATH. Let MATH be the minimal integer such that MATH and MATH. In the case where MATH does not make valid nonadaptive queries to a certain oracle MATH with MATH, we set MATH as this MATH and go to the next stage. Hereafter, we assume that MATH makes nonadaptive queries to any oracle of the form MATH with MATH. Now, we want to show the existence of a set MATH such that MATH. Assume otherwise that MATH for any set MATH, and draw a contradiction. For readability, we omit subscript MATH in the following argument. Let MATH be the set of all strings MATH such that at least one of the query lists of MATH on input MATH include word MATH. Note that MATH does not depend on the choice of oracles since MATH makes nonadaptive queries to any oracles of the form MATH with MATH. We first claim that MATH since, otherwise, we can choose an appropriate oracle MATH such that MATH. For each MATH, let MATH be the sum of all squared magnitudes of MATH's configurations MATH in any superposition of MATH on input MATH where MATH has a query list containing word MATH. Note that each query list consists of at most MATH words. It thus follows that MATH, where MATH is the superposition of MATH's configurations at time MATH on input MATH. Recall that MATH is the query magnitude of string MATH of MATH on input MATH at time MATH. Let MATH be any string in MATH and fix MATH that satisfies MATH. Moreover, let MATH be MATH except that MATH. Note that MATH. It follows by our assumption that MATH. By REF , since MATH, we have MATH. Clearly, MATH for each MATH since MATH makes nonadaptive queries. Thus, MATH, which implies MATH. This immediately draws the conclusion that MATH since MATH. This contradicts the fact MATH since MATH. |
quant-ph/9909012 | REF implies REF Since MATH, there exist a polynomial-time deterministic TM MATH and a polynomial MATH such that, for every MATH, MATH iff MATH. Let MATH for every MATH. By modifying the proof of REF , we can show the existence of a unique function MATH satisfying that MATH for every MATH. Therefore, MATH iff MATH. Define MATH for all MATH. Clearly, MATH is in MATH. Since MATH, REF follows. REF implies REF Assume that there exist two MATH-functions MATH and MATH such that MATH. Using REF , take four functions MATH satisfying that MATH and MATH. Define MATH and MATH for all MATH. REF guarantees that MATH and MATH are in MATH. It is also obvious that MATH iff MATH. Thus, we have MATH. REF implies REF Assume that there exist two functions MATH and MATH in MATH such that MATH. Define MATH for all MATH. It follows from REF that MATH belongs to MATH. Moreover, by REF , there exists a function MATH in MATH such that MATH for all MATH. This implies that MATH iff MATH. From the MATH-characterization of MATH, it follows that MATH is in MATH. |
quant-ph/9909012 | We show the contrapositive. We omit script MATH for readability. Assume that MATH. Let MATH. There exists a function MATH satisfying that MATH. The Squared Function Theorem implies that MATH. Consider the function MATH defined by MATH if MATH and MATH otherwise. Let MATH. By the MATH-characterization of MATH, MATH belongs to MATH and thus, MATH is in MATH. It is easy to show that MATH is in MATH by making a single query MATH and then computing MATH (if necessary). By our assumption, MATH, which is MATH by REF . Note that, for every MATH, MATH implies MATH and MATH implies MATH. This concludes that MATH is in MATH. Therefore, MATH. Symmetrically, we can show that MATH. |
quant-ph/9909012 | CASE: Assume that MATH. Let MATH and MATH be any two functions in MATH and set MATH. Define MATH. By REF , the function MATH defined by MATH is in MATH. Since MATH, MATH belongs to MATH. By our assumption, MATH is also in MATH. It is obvious that MATH belongs to MATH, which is a subset of MATH. Since MATH is admissible, we can show that MATH similar to REF . Hence, MATH is in MATH. This implies that MATH is closed under MATH. Similarly, we can show the case for the minimality. CASE: Assume that MATH is closed under MATH and MATH. Let MATH be any set in MATH. By REF , MATH belongs to MATH and thus, there exists a quantum function MATH such that MATH. Let MATH. By our assumption, MATH is in MATH. Note that MATH is in MATH. Now, define MATH for all MATH. By the Squared Function Theorem, MATH is in MATH. Since MATH, take an appropriate polynomial MATH such that MATH for all MATH (this fact is implicitly used in the proof of REF ). Finally, we define MATH. Since MATH is in MATH, MATH also belongs to MATH by our closure assumption of MATH. This MATH satisfies that MATH for every MATH. Thus, MATH belongs to MATH. |
quant-ph/9909012 | Take any set MATH in MATH. Note that MATH. REF guarantees the existence of two functions MATH and MATH satisfying MATH for all MATH. Define MATH as follows: for every MATH, MATH if MATH and MATH otherwise. Thus, MATH for all MATH. Clearly, MATH. Since MATH, MATH belongs to MATH. |
quant-ph/9909012 | Note that MATH for any oracle MATH because the proof of REF relativizes. It suffices to show that MATH for a certain oracle MATH. This immediately follows from the result of CITE, who proved that MATH for a certain oracle MATH. |
quant-ph/9909040 | We have, from REF , MATH . The conclusion follows. |
quant-ph/9909040 | Substituting REF into REF and simplifying, we obtain MATH . The lemma follows. |
quant-ph/9909040 | Straightforward verification. |
quant-ph/9909040 | Use the matrix representation REF and the definition of MATH. |
quant-ph/9909040 | This is the major theorem in REF ; see REF and particularly REF therein. Note also the work of NAME REF . |
quant-ph/9909080 | The right distributive law follows from MATH . It implies MATH. From this, we may deduce that addition is commutative, as follows. The quantity MATH satisfies MATH and we have MATH . |
quant-ph/9909080 | This follows directly from the definitions. |
quant-ph/9909080 | CASE: If MATH is invalid then MATH and by REF every MATH is in MATH. Hence MATH. REF are straightforward. CASE: Since MATH is commutative, every quantity is normal, hence MATH. Thus, since any two quantities commute, REF implies REF , that is, MATH is ideal. |
quant-ph/9909080 | CASE: We have to prove that for every valid selection MATH there is a maximal valid selection MATH containing MATH. But the form of REF implies that the set of selections has the chain property. Hence the assertion follows from NAME 's Lemma. CASE: Similarly, the set of valid ideal selections has the chain property, and NAME 's lemma applies. |
quant-ph/9909080 | Straightforward. |
quant-ph/9909080 | Let MATH be a selection containing MATH and MATH. If MATH then there is a quantity MATH such that MATH. Hence MATH. Since MATH, REF implies MATH. |
quant-ph/9909080 | Clearly, REF is satisfied. Two nonzero operators MATH and MATH commute iff MATH and MATH, and the assumptions on MATH then imply that MATH. Hence REF also holds. |
quant-ph/9909080 | This is a straightforward consequence of REF . |
quant-ph/9909080 | REF follows directly from REF , and REF from REF . |
quant-ph/9909080 | CASE: Clearly MATH is Hermitian, and MATH. Hence MATH is an event. The left equality in REF holds by definition. Since MATH, we have MATH. But if MATH then MATH, hence MATH, so that MATH. Hence REF holds. Finally, suppose that MATH. Then MATH by REF , and by REF , MATH . If MATH, we may multiply on the left by the complex number MATH and find the contradiction MATH. Hence MATH. CASE: Since MATH and MATH commute, MATH, hence MATH is Hermitian; and it is idempotent since MATH. Finally, MATH is an event. The assertions about expectations are immediate. CASE: Since, by REF , MATH for all MATH, we have MATH, and since MATH, we have MATH. CASE: Let MATH. If MATH then MATH for some MATH, hence MATH and MATH. Now MATH implies MATH, hence MATH, so that MATH. Therefore MATH. Equality holds since MATH satisfies MATH, hence MATH. Thus MATH. Replacing MATH by MATH and noting REF gives MATH. CASE: MATH follows from MATH . To get the expression for MATH when the events are independent, we first note that MATH . In the expectation of this sum we get MATH contributions of size MATH and MATH contributions of size MATH. Hence MATH . |
quant-ph/9909080 | By REF , MATH with MATH, and by REF , MATH. In particular, for a pure ensemble described by the unit vector MATH, we have MATH. |
quant-ph/9909080 | CASE: For arbitrary MATH we have MATH . We now choose MATH, and obtain for arbitrary real MATH the inequality MATH . Now MATH by REF . If MATH we can choose MATH and obtain MATH . After division by MATH, we find that REF holds. And if MATH then MATH since otherwise a tiny MATH produces a negative right hand side in REF . Thus REF also holds in this case. CASE: Since MATH, it is sufficient to prove the uncertainty relation for the case of quantities MATH whose expectation vanishes. In this case, MATH . The assertion follows since MATH and MATH . CASE: Again, it is sufficient to consider the case of quantities MATH whose expectation vanishes. Then MATH and REF follows. REF is an immediate consequence of REF , and REF follows easily from REF . |
quant-ph/9909080 | The quantities MATH and MATH are Hermitian and satisfy MATH where MATH. Now the assertion follows from MATH . |
solv-int/9909010 | REF is an immediate consequence of REF and the skew-symmetry of MATH. REF follows from REF and the NAME decomposition of MATH and MATH: MATH by the uniqueness of the NAME decomposition of MATH, we have MATH . Substituting MATH in the second equation and comparing it to the first one, yields MATH, which is REF . Substituting this relation into the first and second equations yields REF , which by REF and the definition of MATH and MATH, amounts to REF . Relation REF follows from REF , footnote REF and the multilinearity of determinant; or, in the semi-infinite case, from REF , using MATH: MATH . |
solv-int/9909010 | In REF , set MATH, MATH, MATH and MATH; then using MATH and REF , we have MATH . Setting first MATH and then MATH, one finds respectively, since odd MATH-functions vanish on MATH in view of REF : MATH and MATH . Taking the square root, with the consistent choice of sign, REF yields REF , and then REF upon setting MATH or MATH. |
solv-int/9909010 | These follow from REF by straightforward calculations. |
solv-int/9909010 | Relations REF are obtained by differentiating REF in MATH, setting MATH and identifying the coefficients of MATH. The first two relations in REF are obtained by differentiating REF in MATH and MATH (that is, applying MATH), setting MATH and identifying the coefficients of MATH. The last relation in REF is obtained by differentiating the first formula in REF in MATH and MATH, setting MATH, substituting MATH for MATH, and then identifying the coefficients of MATH. Finally, expanding both identities REF in MATH and MATH, for example, MATH and identifying the powers of MATH and MATH yields relations REF . |
solv-int/9909010 | REF follows from REF upon replacing MATH by MATH and MATH by MATH, using REF , with MATH, to eliminate MATH, MATH, MATH and MATH and, upon dividing both sides by MATH. Substituting MATH and MATH for MATH and MATH, respectively, into the left hand side of REF and NAME expanding it in MATH, we obtain MATH showing the equivalence of REF . |
solv-int/9909010 | The Fay identity REF follows from the bilinear identity REF by substitutions MATH . Indeed, since MATH, we have MATH and MATH so the first and second terms on the left hand side of REF , divided by MATH, become, respectively, MATH and MATH showing REF . Note that when MATH, MATH and MATH, denoting MATH for MATH, and multiplying both sides of REF by MATH, we obtain MATH . The differential Fay identity REF follows from REF by taking a limit (set MATH, divide by MATH and let MATH). Alternatively, we can prove REF directly from REF : Set MATH, MATH in REF and in the clockwise integral about MATH, set MATH, yielding MATH . The first integral has a simple pole at MATH and a double pole at MATH, while the second integral has a simple pole at MATH only, yielding, after substitution MATH, MATH or, after carrying out MATH on the left hand side, MATH . Shifting MATH and multiplying both sides by MATH yield REF . Since MATH by the definition of NAME operator, REF is the same as REF , which, as we have pointed out, are nothing but the coefficients of MATH in REF , or REF . It also follows from REF , since, for any power series MATH which satisfies MATH, MATH . Indeed, differentiating REF , which is equivalent to REF , in MATH, setting MATH and using MATH, MATH etc., we have MATH which, noting MATH, is a generating function for REF . |
solv-int/9909010 | This may be obtained, up to the sign, from the second identity in REF: MATH by setting MATH, MATH, MATH, taking the square roots of both sides and using REF . Rather than taking this route, here we prove REF by induction on MATH, using the bilinear Fay identity REF . First, REF is trivial when MATH. (Note also that it gives REF when MATH.) Suppose REF holds for MATH. Then we have, for every MATH, MATH . Multiplying both sides by MATH, summing it up for MATH, and using MATH and the identity MATH which follows from REF of the Pfaffian, we have MATH completing the proof of REF by induction. |
solv-int/9909010 | The theorem follows from REF by straightforward calculations: MATH . The corollary is shown by expanding MATH in MATH and MATH. Recall that MATH where MATH is the vertex operator in the KP theory CITE, and MATH with MATH the coefficients of similar expansion of MATH. Expanding MATH in REF as above leads to MATH . In particular, since MATH REF and MATH REF are linear combinations of each other CITE: MATH one sees for MATH, REF that MATH . |
solv-int/9909010 | Using REF and MATH the left hand side of REF becomes MATH . This equals the right hand side of REF , because MATH . To prove REF , we have MATH in terms of the skew-orthogonal polynomials REF . Multiplying this with an exponential and noting REF , one obtains REF . Summing up this telescoping sum yields REF : MATH . |
solv-int/9909010 | Upon performing the following operations MATH the bilinear identity REF yields MATH . Subtracting this expression (which is MATH), the left hand side of REF equals MATH ending the proof of the lemma. |
solv-int/9909015 | From REF it follows that MATH must be of the form MATH . We first show that without loss of generality one can put the proportionality function MATH equal to MATH. Applying the partial difference operator MATH to REF and using the first REF of the discrete NAME formulas we get MATH . The scalar products with MATH and with MATH give MATH which after simple manipulation gives MATH similarly we have MATH . Notice that due to REF the normal vector MATH defines the same lattice MATH, which shows that in REF we can put MATH. After such a change, REF can be rewritten in the form MATH which yields MATH . Similarly we obtain MATH . REF give, together with the NAME equations satisfied by MATH and MATH, the following equations MATH . This gives MATH which implies that MATH moreover MATH satisfies the NAME equation of MATH. Finally, REF with MATH and MATH given by REF can be put in the form of the NAME transformation REF. |
solv-int/9909015 | Lines of the W - congruence are represented by bi-vectors MATH . We will show that MATH satisfies the NAME equation. Because of REF we have MATH and therefore MATH where MATH . Using REF we get MATH which, due to REF, can be transformed to MATH . Identity REF gives, together with REF, that MATH similarly, MATH which yields MATH . Inserting REF to REF leads to conclusion that the bi-vector MATH satisfies the NAME equation. |
solv-int/9909015 | It is enough to verify directly that the lattice MATH given by REF is a solution of REF which define the NAME transformations of MATH by means of MATH, and that it is also a solution of REF which define the NAME transformations of MATH by means of MATH. |
solv-int/9909015 | Due to REF the lattices MATH and MATH can be given by MATH where MATH, MATH and MATH are like in REF; notice the change of sign in REF induced by the change of sign in REF. Transforming lattice MATH by REF by means of MATH and applying REF one obtains REF, likewise transforming lattice MATH by REF by means of MATH. |
solv-int/9909016 | First observe MATH and MATH . Second, MATH and MATH . |
solv-int/9909016 | For the even evolution we have MATH while for the odd time evolution MATH . |
solv-int/9909016 | We first compute MATH and then we calculate MATH . |
solv-int/9909016 | We begun with MATH and MATH . Therefore, the transformed NAME 's operator indeed is the solution of NAME 's equation. |
solv-int/9909025 | Let us calculate the MATH-th equation in MATH: MATH . The last equality in REF is due to the symmetry of MATH. Thus, clearly, MATH if and only if REF are satisfied and the equivalence of REF is established. Let us now calculate the total derivative of MATH with respect to MATH. MATH . The second term on the right hand side of the above equation has been rewritten by renaming indices. It contains precisely the cyclic REF . So, if one (and thus both) of REF are satisfied, then both terms in REF vanish. On the other hand, if MATH then terms at different powers of MATH in REF must be equal to zero, which implies both REF. |
solv-int/9909025 | The requirement MATH yields (compare REF ) MATH where the index MATH at the vector expression containing matrices MATH and MATH denotes its MATH-th component. The equality is satisfied identically with respect to MATH and so both sums must be separately equal to zero. It follows, that the MATH satisfy the cyclic conditions MATH and that MATH. The latter yields precisely REF since we assumed MATH. So REF are proved. The operator MATH acts as differentiation on the algebra of constants of motion, so that MATH (where MATH) which proves REF. |
solv-int/9909025 | If MATH is potential, then according to REF MATH and so MATH. The potential MATH exists provided that MATH. This yields exactly the NAME REF for MATH. |
solv-int/9909025 | Our qLN system REF is generated by either of the two functions MATH and MATH and so REF , that is, MATH, must be satisfied. This implies that MATH. This equation for the function MATH has solutions if and only if its compatibility condition MATH is satisfied. This yields a PDE for the function MATH which, after the substitution MATH and with use of the cyclic REF , yields that MATH satisfies REF . By inserting into this equation the explicit form of the polynomials MATH we obtain REF . On the other hand, REF implies also MATH, and its compatibility condition MATH gives a PDE which in terms of MATH must attain the form REF with interchanged entries of MATH and MATH (since the equation MATH becomes MATH when one exchanges MATH and MATH). Due to the skew-symmetry of coefficients of the equation for MATH with respect to the entries of matrices MATH (clearly seen from the form of REF ) the obtained equation for MATH differs from the equation for MATH by a minus sign on the right-hand side only. This proves that MATH and MATH both satisfy REF (notice, however, that this does not imply MATH). The existence of MATH (that is, the possibility of integrating REF in order to obtain MATH) follows from the fact, that the condition MATH together with MATH yields precisely the fundamental equation for MATH which is satisfied due to assumptions. One can similarly prove the existence of MATH. The second statement of the theorem can now be proved by checking that both pairs MATH and MATH given by REF satisfy REF and thus give rise to two systems of qLN equations. |
solv-int/9909025 | (of the recursion theorem) Consider a solution MATH of the fundamental equation and the functions MATH defined by REF . Then obviously MATH which immediately implies MATH. Let MATH. Then MATH where we used that MATH as follows from the NAME theorem. The last equality is due to REF above. Thus MATH up to a non-essential additive constant. This proves the first assertion of the theorem. If we now define the sequence MATH via the recursive procedure REF then a simple induction argument shows that each pair MATH satisfies REF and thus both MATH and MATH determine the same qLN system. Moreover, each MATH is a solution of the fundamental equation as REF states. Finally, to obtain REF it is enough to insert the formula MATH into the second equation in REF . |
solv-int/9909025 | An easy calculation shows that the fundamental equation associated with the matrices MATH and MATH differs from the fundamental equation associated with the matrices MATH and MATH by the multiplicative factor MATH on the right-hand side, that is, by the non-zero determinant of the transformation between MATH and MATH and so it is in fact the same equation. This shows REF. Assume now that REF is associated with a pair MATH. Consider the vector MATH of the coefficients of REF at the highest derivatives MATH respectively. Then MATH or MATH where MATH and MATH are three-dimensional vectors depending on MATH and MATH. Hence, for a fixed MATH both vectors MATH and MATH are orthogonal to MATH. The coefficients at MATH, MATH and MATH yield equations which are differential consequences of REF and so they do not impose any additional restrictions on MATH. Suppose now that there exist matrices MATH and MATH satisfying the cyclic REF and associated with the same fundamental equation. This means that REF has another solution that is, that MATH so that the vectors MATH and MATH are orthogonal to MATH and in consequence they are linear combinations of MATH and MATH: MATH with some coefficients that may depend on MATH and MATH. For the corresponding matrices it immediately follows that MATH . It remains to show that the coefficients MATH in fact do not depend on MATH nor MATH. This can be shown by inserting the explicit form REF of entries of matrices MATH, MATH, MATH and MATH into REF . This shows REF. |
solv-int/9909025 | Since MATH REF yields MATH, MATH, MATH, that is, it reproduces REF . The matrix MATH is antisymmetric and it is straightforward to verify that it satisfies the NAME identity in the phase space MATH. |
solv-int/9909025 | The conditions MATH and MATH (where ``cycl" means the cyclic permutation of expressions) hold identically due to the block structure of MATH. The condition MATH yields the symmetry of MATH: MATH. Further MATH . Let us denote the right hand side of the above equality by MATH. Notice that MATH can not depend on MATH and so we have MATH for some vector MATH which yields MATH for some vector MATH. By taking linear combinations of the conditions MATH and using the symmetry of MATH and the antisymmetry of MATH we get the following sets of REF MATH . REF show that MATH and MATH depend only on MATH and that MATH and MATH depend only on MATH. REF give MATH and so all terms in this expression must be equal to a constant MATH. Integration yields MATH where MATH and MATH are integration constants. Substituting REF into REF and integrating we get REF . If we now introduce the symmetric matrix MATH by the equality REF and use REF then REF will attain the form REF . It is straightforward to check that with the above forms of MATH and MATH the conditions MATH and MATH are satisfied identically. Further, the condition MATH after some calculations attains the form MATH which means that in the vector MATH the mixed derivatives of its components are equal and so this vector is equal to the gradient of some function MATH, that is MATH or MATH. Finally, by direct calculation we verify that MATH and so REF is proved. |
solv-int/9909025 | According to the proof of REF the matrix MATH satisfies all the NAME identities except possibly for MATH, since MATH differs from MATH by the form of MATH only. Like in the proof of REF we find that MATH yields that the mixed derivatives of the components of the vector MATH are equal and so MATH for some functions MATH and MATH. By comparing coefficients at different powers of MATH we get MATH and MATH and thus MATH. Further, MATH in the notation of REF so it is NAME. Easy calculation shows that MATH is NAME too. |
solv-int/9909025 | This proposition is a consequence of REF . If we modify the matrix MATH by substituting the matrix MATH by MATH and substituting MATH with MATH we obtain the matrix MATH where MATH . Due to REF the function REF is the NAME of REF . But REF is in fact equal to MATH since it can be verified that MATH and that MATH. |
solv-int/9909025 | (modification of the proof of NAME theorem CITE) Consider REF-dimensional manifold MATH in MATH. NAME brackets of all pairs of MATH induced by both structures MATH and MATH are equal to zero, since the functions MATH all belong to the same bi-Hamiltonian chain. For instance MATH with the second equality being a consequence of the bi-Hamiltonian structure of MATH. It follows that the NAME bracket MATH of both vector fields MATH and MATH is equal to zero, MATH, since the mappings MATH REF are NAME algebra homomorphisms between the NAME algebra of vector fields on MATH and the NAME algebra of all smooth functions on MATH with the NAME bracket defined by MATH. Moreover MATH and similarly MATH which proves that MATH is tangent to MATH. In the same way one can show that MATH is also tangent to MATH. Direct verification shows that MATH and MATH are linearly independent. We thus have a REF-dimensional sub-manifold MATH in MATH equipped with a pair of linearly independent, commuting vector fields MATH and MATH. We can now apply the construction of CITE and conclude that both MATH and MATH are completely integrable. |
solv-int/9909025 | Consider the vector field MATH from REF . Obviously MATH and so in the hyperplane MATH we have MATH which means that the hyperplane MATH is invariant with respect to the action of the vector field MATH. REF also shows that in the hyperplane MATH the vector field of the system REF is parallel to the vector field of the system REF and so their trajectories must coincide. |
solv-int/9909025 | Along each characteristic curve MATH given by REF we can consider REF as an ODE MATH with general solution MATH where MATH is a constant of integration. This can be verified by direct differentiation; the cyclic conditions imply that MATH and thus MATH . The constant of integration MATH can be different for different characteristic curves, so when we express the result in terms of MATH and MATH, MATH will depend on MATH (but not on MATH). |
solv-int/9909025 | REF implies MATH . Conversely, this expression can be integrated to give REF , where MATH is an arbitrary function of integration. By substituting MATH and simplifying the resulting expression using the cyclic conditions one obtains REF . Comparison with the general expression for the fundamental equation in REF proves the second statement of the lemma. |
solv-int/9909025 | We use the notation of REF . Let the system be generated by MATH with MATH. Since MATH, we can express MATH as MATH and calculate MATH by quadrature, as above. Now note that since the curves of constant MATH by definition have tangent MATH, we must have MATH for some function MATH, whose exact form depends on the choice of MATH. This gives MATH, and thus MATH . In order to complete the proof, we will show that MATH and MATH for some functions MATH and MATH, since this implies that the variables MATH and MATH separate. Explicitly, we can then find MATH from the quadrature MATH after which the inverse coordinate transformation gives us MATH. Notice, that for a given matrix MATH the characteristic coordinates MATH can be calculated explicitly so that the function MATH and thus MATH and MATH can be easily calculated and used in the quadrature REF above. The theorem covers however all the cases at once without any need of calculating MATH explicitly. To see that MATH, note that MATH implies that MATH satisfies the PDE MATH which has the same characteristic curves REF as REF . Along such a curve we determine MATH by integrating MATH which, taking into account MATH, gives MATH . The integration constant MATH can be different on different characteristic curves, so changing to MATH coordinates we obtain MATH as desired. Finally, we calculate the total derivative of the function MATH along a characteristic curve: MATH . Using the cyclic conditions, we find that this expression is identically zero. This implies that MATH is constant along the coordinate curves of constant MATH, that is, MATH. This completes the proof. |
solv-int/9909025 | We will compute explicitly the curves given by REF , which constitute the curves of constant MATH. (The curves of constant MATH are just horizontal lines, since MATH.) Inserting the explicit REF for the matrix MATH into REF , we obtain MATH . When solving these equations, we distinguish four different cases, depending on the values of the parameters in MATH. The case MATH. By setting MATH and MATH, which is just rescaling of the axes and translation of the origin, we transform REF into MATH . Subcase MATH (type REF). Either MATH, or MATH and MATH, where MATH and MATH are constants of integration. Eliminating MATH and writing MATH instead of MATH, we obtain MATH which represents a family of hyperbolas, each with asymptotes MATH and MATH. The solution MATH found above corresponds to the limiting cases MATH (see REF ). Subcase MATH (types REF). The substitution MATH, MATH yields MATH and thus MATH resulting in MATH . If MATH (type REF), this represents in the MATH-plane a family of hyperbolas centered around the MATH-axis, with asymptotes MATH and vertices MATH REF . If MATH (type REF), we obtain in the region MATH a family of hyperbolas with asymptotes MATH and vertices MATH, and in the region MATH a family of ellipses with vertices MATH and MATH REF . The corresponding curves in the MATH-plane are obtained by a shear in the MATH-direction with factor MATH REF . They are still hyperbolas and ellipses, but not aligned parallel with the MATH-axes. The case MATH. Subcase MATH (type REF). Translating the origin by MATH and MATH, we obtain MATH which yields MATH . With MATH,MATH, we obtain in the MATH-plane a family of parabolas MATH REF . The corresponding curves in the MATH-plane are parabolas obtained by a shear in the MATH-direction with factor MATH REF . Subcase MATH (type REF). Here we can assume that MATH, or else we get the degenerate case MATH. A simple calculation shows that MATH which is a family of translated parabolas seen on REF (or straight lines, if MATH). |
solv-int/9909025 | If such MATH and MATH exist, then MATH is an integral of motion for all MATH and MATH, and thus MATH is a two-parameter solution of REF. Conversely, if there is a two-parameter solution MATH of REF, then there are linearly independent integrals MATH and MATH with MATH and MATH. |
cond-mat/9910325 | We can suppose that the sections are normalized MATH. On a set MATH, and for a given MATH, define: MATH and for fixed MATH: MATH . In the theorem, we suppose that MATH . Thus MATH and then MATH. We now check that the sections MATH define a complex line bundle MATH. On an other set MATH, if MATH-and MATH then MATH is in the same line than MATH. We have thus obtain a homotopic deformation MATH, MATH between the line bundle MATH and MATH. They have thus the same NAME index MATH . |
cond-mat/9910325 | Consider the line bundle MATH defined by the zeroes REF , with local section noted MATH on each open set MATH. This bundle has MATH . NAME index. Consider the function MATH . From REF , the NAME index MATH of the bundle MATH is MATH . The zeroes of MATH are given by: MATH this gives MATH . We deduce from REF that MATH . |
cond-mat/9910325 | For every MATH, the line bundle is defined by the vector MATH. We have therefore to check that MATH , and that this line is periodic with respect to MATH. Because the space MATH is MATH dimensional, the condition MATH is MATH-dimensional. As soon as MATH, and for MATH-the condition MATH cannot be satisfied generically. Now MATH . So MATH and MATH-are proportional and define the same line MATH. |
cond-mat/9910325 | We note MATH, and MATH the line bundle defined in REF . The proof consists in constructing a homotopic deformation MATH from MATH to MATH for MATH. Because the NAME index MATH of MATH is constant, we therefore conclude that the NAME index of MATH is given by REF . For MATH, and MATH, define MATH where MATH is a standard coherent state on the plane, MATH, with MATH given by REF . We suppose moreover that the phases are chosen such that REF holds for MATH and MATH. Then MATH if and only if MATH and MATH. This last situation is non generic, and if it occurs, we can just choose MATH with arbitrary MATH to avoid this. So MATH fulfills REF , and defines a line bundle for every MATH. Because there are three parameters MATH we have to suppose now that MATH so that MATH. |
cond-mat/9910325 | The proof is similar to that of REF . The bundle is well defined because, first MATH needs MATH conditions, generically not realized for arbitrary MATH, as soon as MATH. Secondly, MATH and MATH are proportional and define therefore the same line. In order to calculate the NAME index, consider the mapping MATH where MATH is a standard coherent state on the plane, MATH, and MATH we can therefore construct MATH and MATH. From REF , MATH define a line bundle with NAME index MATH. Consider now for MATH and MATH . Then for fixed MATH, MATH define a line bundle over MATH, because MATH nor MATH are generic (compare discussion in proof of REF ), and there is also periodicity with respect to MATH. The mapping MATH is a homotopic deformation from the line bundle MATH to the line bundle MATH. Their NAME index is constant and equal to MATH. |
cond-mat/9910505 | The defining relation of the resolvent is rewritten as MATH. By using the identity MATH this relation is deformed into MATH . Move the second term in the left side to the right side and recall the defining relations REF - REF . The representation REF is thus obtained. |
cond-mat/9910505 | The action of MATH on MATH is computed as MATH . By virtue of this relation we obtain MATH . This means MATH and therefore implies the lemma. The proof has been completed. |
cond-mat/9910505 | By definitions of MATH and MATH their dependences on MATH are MATH . Similarly the kernel MATH is connected with MATH as MATH . Using these recursion relations we obtain MATH . It thus follows that MATH . In the last equality the relation REF is used. Thus the integro-difference REF is obtained by removing the action of MATH. In the same way REF can be derived. |
cond-mat/9910505 | We show the differential REF . The other REF can be derived in the same way. According to the identity MATH the differentiation of REF is computed as MATH . By using the resolvent this is reduced to MATH . Because of REF we obtain REF . |
cond-mat/9910505 | Due to the relation REF it follows that MATH where MATH . Based on this recursion relation the NAME determinants of both sides are given by MATH . Here we note that MATH in the left side has no influence on the NAME determinant. Take the logarithm of MATH: MATH . The trace of MATH can be expressed in terms of MATH: MATH . REF with REF proves the lemma. |
cond-mat/9910505 | We start from the proof of REF . Using the definition of MATH and the canonical integral representation of MATH (see CITE) MATH we obtain MATH . Since MATH (which is clear from the integrable REF ) there exists the inverse of the conjugation matrix: MATH. The canonical integral representation of MATH is thus given by MATH . The identity MATH corresponds to REF . Let us compute the integro-difference relation of MATH with respect to MATH and prove REF . We introduce the following integral operator MATH . This obeys MATH with the conjugation matrix MATH . From the definitions of MATH, MATH and MATH, we see that MATH is independent of MATH. Hence applying NAME 's theorem we have MATH which implies MATH . By letting MATH act on this relation from the right the integro-difference REF is derived. Similarly REF is obtained. Thus REF has been proved. We show REF . In the same way as the FSFP case CITE and the impenetrable NAME gas case CITE, in the neighborhood of MATH, the integral operator MATH can be decomposed into MATH where MATH is a single-valued, invertible and analytic in the neighborhood of MATH. MATH is represented by MATH . Since MATH its logarithm-derivative is computed as MATH . Due to this relation and NAME 's theorem the logarithm-derivative of MATH is written as MATH with the coefficient MATH: MATH . By definitions of MATH and MATH the elements of MATH are expressed by MATH . Let MATH act on REF from the right and use this representation of MATH. The differential REF is thus obtained. Similarly REF can be derived. The proof of the theorem has been completed. |
cs/9910007 | Clearly MATH is in MATH. Regarding MATH-hardness for MATH, let MATH via transducer MATH, MATH set MATH, and MATH set MATH. Without loss of generality, on each input MATH, MATH asks exactly one question MATH to MATH, and one question MATH to MATH. Define sets MATH and MATH as follows: CASE: MATH accepts MATH if MATH is answered correctly, and MATH is answered ``no"MATH. CASE: MATH and the (one-variable) truth-table with respect to MATH of MATH on input MATH induced by the correct answer to MATH is neither ``always accept" nor ``always reject"MATH. Note that MATH, and that MATH, since MATH. But MATH via the reduction MATH. So clearly MATH, via the reduction MATH, where MATH and MATH are, respectively, reductions from MATH to MATH and from MATH to MATH. |
cs/9910007 | Suppose MATH. Let MATH, MATH, and MATH be MATH -complete for MATH, MATH, and MATH, respectively; such sets exist. From REF it follows that MATH is MATH -complete for MATH. Since (as MATH) MATH, and by REF, there exists a polynomial-time many-one reduction MATH from MATH to MATH. So, for all MATH: if MATH, then MATH if and only if MATH. Equivalently, for all MATH: REF : if MATH, then MATH . We can use MATH to recognize some of MATH by a MATH algorithm. The definitions of easy and hard used in this paper follow the easy and hard concepts used by CITE, NAME and NAME (CITE, see also CITE), and Beigel, CITE, modified as needed for our goals. In particular, we say that a string MATH is easy for length MATH if there exists a string MATH such that MATH and MATH where MATH. Let MATH be a fixed polynomial, which will be exactly specified later in the proof. We have the following MATH algorithm to test whether MATH in the case that (our input) MATH is an easy string for MATH. On input MATH, guess MATH with MATH, let MATH, and accept if and only if MATH and MATH. In light of REF above, it is clear that this is correct. We say that MATH is hard for length MATH if MATH and MATH is not easy for length MATH, that is, if MATH and for all MATH with MATH, MATH, where MATH. If MATH is a hard string for length MATH, then MATH induces a many-one reduction from MATH to MATH, namely, MATH, where MATH. Note that MATH is computable in time polynomial in MATH. We can use hard strings to obtain a MATH algorithm for MATH. Let MATH be a MATH machine such that MATH with oracle MATH recognizes MATH. Let the run-time of MATH be bounded by polynomial MATH, which without loss of generality satisfies MATH (as promised above, we have now specified MATH). Then MATH . If there exists a hard string for length MATH, then this hard string induces a reduction from MATH to MATH. Thus, with any hard string for length MATH in hand, call it MATH, MATH with oracle MATH recognizes MATH for strings of length MATH, where MATH is the machine that simulates MATH but replaces each query to MATH by the first component of MATH. It follows that if there exists a hard string for length MATH, then this string induces a MATH algorithm for MATH, and therefore certainly a MATH algorithm for MATH. However, now we have a MATH algorithm for MATH: On input MATH, the NP base machine of MATH executes the following algorithm: CASE: Using its MATH oracle, it deterministically determines whether the input MATH is an easy string for length MATH. This can be done, as checking whether the input is an easy string for length MATH can be done by one query to MATH, and MATH by our MATH hypothesis. CASE: If the previous step determined that the input is not an easy string, then the input must be a hard string for length MATH. So simulate the MATH algorithm induced by this hard string (that is, the input MATH itself) on input MATH (via our NP machine itself simulating the base level of the MATH algorithm and using the NP machine's oracle to simulate the oracle queries made by the base level NP machine of the MATH algorithm being simulated). CASE: If the first step determined that the input MATH is easy for length MATH, then our NP machine simulates (using itself and its oracle) the MATH algorithm for easy strings on input MATH. We need one brief technical comment. The MATH oracle in the above algorithm is being used for a number of different sets. However, as MATH is closed under disjoint union, this presents no problem as we can use the disjoint union of the sets, while modifying the queries so they address the appropriate part of the disjoint union. Since MATH is complete for MATH, it follows that MATH. |
cs/9910007 | Let MATH. Recalling that MATH, it is not hard to see that MATH. In particular, this holds due to REF , in light of the facts that REF MATH (due to CITE - see the discussion just before REF ), and REF MATH. So, since MATH is closed under complementation, we have MATH. However, this says, under the hypothesis of the theorem, that MATH, which itself, by REF , implies that MATH. |
cs/9910007 | Suppose MATH. Let MATH, MATH, and MATH be MATH -complete for MATH, MATH, and MATH, respectively; such languages exist, for example, via the standard canonical complete set constructions using enumerations of clocked machines. From REF it follows that MATH is MATH -complete for MATH. Since (as MATH) MATH, and by REF, there exists a polynomial-time many-one reduction MATH from MATH to MATH. So, for all MATH: if MATH, then MATH . We can use MATH to recognize some of MATH by a MATH algorithm. In particular, we say that a string MATH is easy for length MATH if there exists a string MATH such that MATH and MATH where MATH. Let MATH be a fixed polynomial, which will be exactly specified later in the proof. We have the following algorithm to test whether MATH in the case that (our input) MATH is an easy string for MATH. On input MATH, guess MATH with MATH, let MATH, and accept if and only if MATH and MATH. This algorithm is not necessarily a MATH algorithm, but it does inspire the following MATH algorithm to test whether MATH in the case that MATH is an easy string for MATH. Let MATH be languages in MATH such that MATH. Then MATH if and only if MATH, where MATH is computed as follows: On input MATH, guess MATH with MATH, let MATH, and accept if and only if REF MATH, and REF MATH, where MATH, and REF MATH. Since MATH, MATH, and thus our algorithm is in MATH. Note that REF has no analog in the proof of REF . We need this extra condition here as otherwise the different MATH might latch onto different strings MATH and this would cause unpredictable behavior (as different MATH-s would create different MATH-s). We say that MATH is hard for length MATH if MATH and MATH is not easy for length MATH, that is, if MATH and for all MATH with MATH, MATH, where MATH. If MATH is a hard string for length MATH, then MATH induces a many-one reduction from MATH to MATH, namely, MATH, where MATH. Note that MATH is computable in time polynomial in MATH. We can use hard strings to obtain a MATH algorithm for MATH, and thus (since MATH) certainly a MATH algorithm for MATH. Again, let MATH be languages in MATH such that MATH. For all MATH, let MATH be a MATH machine such that MATH, where MATH. Let the run-time of all MATH-s be bounded by polynomial MATH, which without loss of generality satisfies MATH (as promised above, we have now specified MATH). Then for all MATH, MATH where MATH. If there exists a hard string for length MATH, then this hard string induces a reduction from MATH to MATH. Thus, with any hard string for length MATH in hand, call it MATH, MATH with oracle MATH recognizes MATH for strings of length MATH, where MATH is the machine that simulates MATH but replaces each query to MATH by the first component of MATH. It follows that if there exists a hard string for length MATH, then this string induces a MATH algorithm for MATH, and therefore certainly a MATH algorithm for MATH. It follows that there exist MATH sets, say, MATH for MATH, such that the following holds: For all MATH, if MATH (functioning as MATH above) is a hard string for length MATH, then MATH if and only if MATH. However, now we have an outright MATH algorithm for MATH: For MATH define a MATH machine MATH as follows: On input MATH, the NP base machine of MATH executes the following algorithm: CASE: Using its MATH oracle, it deterministically determines whether the input MATH is an easy string for length MATH. This can be done, as checking whether the input is an easy string for length MATH can be done by one query to MATH, and MATH by our MATH hypothesis. CASE: If the previous step determined that the input is not an easy string, then the input must be a hard string for length MATH. So simulate the MATH algorithm for MATH induced by this hard string (that is, the input MATH itself) on input MATH (via our NP machine itself simulating the base level of the MATH algorithm and using the NP machine's oracle to simulate the oracle queries made by the base level NP machine of the MATH algorithm being simulated). CASE: If the first step determined that the input MATH is easy for length MATH, then our NP machine simulates (using itself and its oracle) the MATH algorithm for MATH on input MATH. It follows that for all MATH, MATH if and only if MATH. Since MATH is complete for MATH, it follows that MATH. |
cs/9910010 | Suppose we have some exact protocol for MATH that uses MATH qubits of communication and MATH prior NAME. We will build a clean qubit protocol without prior entanglement for MATH. First NAME makes MATH NAME and sends one half of each pair to NAME (at a cost of MATH qubits of communication). Now they run the protocol to compute the first instance of MATH (MATH qubits of communication). NAME copies the answer to a safe place which we will call the `answer bit' and they reverse the protocol (again MATH qubits of communication). This gives them back the MATH NAME, which they can reuse. Now they compute the second instance of MATH, NAME ANDs the answer into the answer bit (which can be done cleanly), and they reverse the protocol, etc. After all MATH instances of MATH have been computed, NAME and NAME have the answer MATH left and the MATH NAME, which they uncompute using another MATH qubits of communication. This gives a clean protocol for MATH that uses MATH qubits and no prior entanglement. By REF : MATH hence MATH . Since this must hold for every MATH, the theorem follows. |
cs/9910010 | Since a qubit and an NAME can be used to send REF classical bits CITE, we can devise a qubit protocol for MATH using MATH qubits (compute the two copies of MATH in parallel using the classical bit protocol). Hence by the previous theorem MATH. |
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