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math/0002243
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Observe that the condition of MATH having MATH is not much of a restriction at all, because the space of such MATH-forms is open and the set of generic perturbations is dense (see CITE). Suppose otherwise, there exists some MATH such that for every MATH we have MATH. This would imply that MATH since we have seen (see REF ) that MATH, but this is a contradiction because we have proven (see REF ), that MATH.
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math/0002243
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Assume that MATH. REF implies, for MATH there exist (at least) two different irreducible solutions MATH, MATH on MATH. By REF we would have an element of MATH at MATH the unique solution on MATH, obtained in REF . But this is a contradiction since MATH is a smooth point. The same kind of argument shows that MATH since MATH.
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math/0002243
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First we will prove that if MATH are fixed representatives for the generators for the free part of MATH then MATH generates MATH. It is enough to prove that for every MATH we can find MATH such that MATH. Consider the line bundle MATH, and observe that there is no obstruction to extend it to a line bundle MATH such that MATH. Let MATH be a MATH-connection on MATH and consider the map MATH . It is not difficult to see that MATH. After extending MATH to a MATH-connection on MATH for each MATH, we obtain (see remark below REF ) our desired maps MATH. To prove the last statement we proceed as follows: let MATH, MATH . Finally we have to show that if MATH then MATH generates the cohomology of the MATH factor. Since MATH acts freely on MATH, then it is easy to show that MATH, where MATH is the kernel of the homomorphism MATH given by evaluating on the fiber over MATH.
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math/0002243
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MATH is a cobordism invariant for every MATH. Consider the smooth cobordism induced by the family of metrics MATH on MATH as MATH and observe REF that MATH. This shows that MATH where MATH is a representative for the MATH factor of the connected sum. This, the definition of a MATH-class and REF complete the proof.
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math/0002243
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Theorem A shows that every time that we perform a connected sum with MATH we add a cycle to the moduli space, that lies entirely in the MATH part of MATH.
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math/0002243
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First observe that MATH. Since MATH is a NAME surface, we know that MATH is also a NAME surface, and its associated MATH structure MATH satisfies MATH, where MATH are generators for the pull-backs to MATH of the MATH copies of MATH so that MATH . Let MATH be the first NAME class of MATH which is a MATH-class by theorem A, and notice that MATH. One then has MATH .
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math/0002243
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The proof is the same as the one given by CITE.
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math/0002243
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Let MATH be an admissible pair and consider the pair of integers MATH. It is always possible to find (infinitely many) positive integers MATH and MATH such that MATH where MATH denotes the set of MATH that satisfy REF 's theorem. The reason for this last statement is that the region MATH determine by MATH such that MATH is open, connected and not bounded, where MATH is the same constant as in NAME 's theorem. If we denote by MATH the simply connected NAME surface with MATH and MATH, then MATH, is a manifold that realizes the pair MATH and does not admit any NAME metric. This last statement is a consequence of REF .
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math/0002251
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Since MATH, MATH is a MATH and MATH is a classifying map. CASE: We have to show that MATH, for MATH. Of course, MATH. Fix MATH, and assume that MATH, for MATH. By the NAME isomorphism theorem, MATH. By minimality of MATH and MATH, we have a commuting ladder between the (equivariant) chain complexes of MATH and MATH: MATH . The three vertical arrows are isomorphisms, since MATH, for MATH. It follows that MATH. But MATH is acyclic, and so MATH. CASE: Consider the commuting diagram MATH . By REF and NAME, MATH. A diagram chase yields isomorphisms MATH and we are done.
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math/0002251
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From presentation REF, we compute MATH. At the same time, MATH, since MATH is MATH-minimal, and MATH, since MATH lifts MATH. This proves the first claim. To finish the proof, note that MATH, by the definition of MATH.
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math/0002251
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The extension of MATH to group rings gives rise to an isomorphism between MATH and MATH, and thus maps MATH bijectively to MATH. Now identify MATH and MATH with MATH, and let MATH be the matrix of MATH under this identification. Let MATH be the corresponding monomial isomorphism, given by MATH. It is readily verified that MATH preserves the NAME varieties.
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math/0002251
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By REF , MATH is a finitely presented MATH-module, with presentation matrix MATH given in REF. Hence, MATH, where MATH is the evaluation of the matrix of NAME polynomials MATH at MATH. Let MATH be the inclusion. The lift to universal covers, MATH, gives rise to an exact sequence of MATH-equivariant chain complexes, a fragment of which is shown below: MATH . Now fix MATH. NAME REF over MATH with MATH, via the representation MATH, yields the commuting diagram MATH (Note also that MATH, by the definition of MATH.) Chasing this diagram, we see that MATH. From REF , we have MATH, and we are done.
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math/0002251
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Denote by MATH and MATH the complements of the associated projective arrangements in the projective spaces MATH and MATH, respectively. The triviality of NAME fibrations of arrangement complements readily gives homotopy equivalences, MATH and MATH, and implies that MATH. REF and MATH from CITE may be used to replace, up to homotopy, the inclusion MATH by a cellular map between minimal NAME, MATH, which restricts to the identity on MATH-skeleta; see the discussion preceding REF from CITE. Let MATH and MATH. As shown by CITE, all complements of complex hyperplane arrangements satisfy REF from REF. Thus, MATH and MATH satisfy that condition, too. Now set MATH. The claimed properties of the map MATH follow from the corresponding properties of MATH, and the fact that MATH (by REF ), which guarantees that MATH is a classifying map.
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math/0002251
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CASE: The supersolvable arrangement MATH is obtained from MATH by the deformation method introduced in CITE, and refined in CITE. REF follows from this deformation method, which proceeds inductively, using the given composition series of MATH. CASE: Up to homotopy, we may view each MATH as an arrangement in MATH, and replace each map MATH by a bundle map, MATH, with the specified fiber (more precisely, by a linear fibration, admitting a section, see CITE). Moreover, the defining polynomials for MATH may be written inductively as MATH, and MATH. Clearly, MATH is a completely solvable NAME polynomial over MATH. Thus, by CITE, the monodromy of the bundle map MATH factors through the pure braid group MATH, acting on the free group MATH via the NAME representation. CASE: The first assertion follows from CITE. The specified structure of MATH is provided by REF Since MATH is supersolvable, MATH, see CITE and CITE. Moreover, REF from CITE insures that MATH and MATH have the same collinearity relations, which implies that MATH. The canonical isomorphism in REF is then given by: MATH . The identification of MATH with MATH follows from the fact that the basis MATH of MATH is dual to the basis of MATH given by the meridians, and MATH preserves those meridians. CASE: The equality between the NAME polynomials of MATH and MATH follows from REF. The second equality follows from CITE, via REF This follows from REF This follows from REF .
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math/0002251
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CASE: Follows from REF Follows from REF Follows from REF and CITE. CASE: Follows from REF and the fact that MATH.
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math/0002251
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REF follow from REF follows from REF.
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math/0002251
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Recall from the proof of REF that MATH has defining polynomial of the form MATH, where MATH, and MATH is a completely solvable NAME polynomial over MATH. The decone MATH, obtained by setting MATH, has defining polynomial MATH. The result follows at once.
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math/0002251
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If MATH, then MATH, by REF. Up to homotopy, MATH, and we may use the NAME minimal structure from CITE. If MATH, we know from REF that MATH is an iterated generic hyperplane section of MATH, with MATH; see also the first paragraph of REF. The method of proof of REF from CITE provides the desired minimal complex MATH. CASE: The space MATH is aspherical if and only if MATH, by CITE (since MATH is fiber-type, if MATH). If MATH, then necessarily MATH (by REF ). It follows from REF that MATH, and so MATH must be aspherical. Conversely, if MATH, then MATH must be non-zero (use REF ). CASE: If MATH, we know from REF that MATH, where MATH. Everything then follows from REF .
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math/0002251
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CASE: From resolution REF, we see that MATH is isomorphic to MATH. If MATH, then MATH is a free MATH-module, with rank MATH given by REF. If MATH, then MATH is not projective, by the minimality of REF Note first that the MATH-adic filtration of the group algebra MATH is NAME, in the sense that MATH, where MATH is the augmentation ideal. This follows from the fact that MATH is an iterated semidirect product of free groups, where all homology monodromy actions are trivial (compare REF ); therefore, MATH is residually torsion-free nilpotent (see CITE), and so the MATH-adic filtration of MATH must be NAME (see CITE). It follows that the MATH-adic filtration of the free MATH-module MATH is also NAME. Assume now that either MATH is finitely generated as an abelian group, or nilpotent as a MATH-module. It follows that MATH, for some MATH, and thus MATH must be a nilpotent, non-trivial MATH-module. This implies that MATH, for some MATH, and MATH. On the other hand, MATH has no zero-divisors, since MATH is residually torsion-free nilpotent (see CITE). This gives the desired contradiction, proving REF.
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math/0002251
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Let MATH, MATH, MATH be three distinct edges of MATH. Notice that the corresponding defining equations, MATH, viewed as points in MATH, are collinear if and only if MATH are the edges of a MATH-cycle. Using this remark, it is a straightforward exercise to translate REF - REF from REF into REF from REF .
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math/0002251
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Clearly, MATH is a composition series for MATH if and only if MATH is a composition series for MATH.
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math/0002251
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There are no collinearity relations among the defining equations of MATH, since MATH has no MATH-cycles. REF then follows from REF from REF , and REF from the definition of the quadratic NAME. Now set MATH. If MATH, then REF is trivially verified. If MATH, we may take a generic MATH-plane MATH in MATH with the property that MATH (by CITE), and such that MATH and MATH have the same collinearity relations. The decone of the arrangement MATH is thus generic, and so MATH (by CITE).
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math/0002251
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Set MATH, MATH and MATH. Recall from REF the construction of the NAME of MATH, together with the graphic counterpart from REF. From REF , we know that MATH hypersolvable, MATH, and MATH. If MATH, then MATH, by REF, and thus MATH, by REF. If MATH, the same argument shows that MATH. From REF, we deduce that MATH is isomorphic to MATH, which is generated by MATH. A quick inspection of the construction of the nbc-basis in degree MATH reveals that MATH is free abelian, with rank equal to the number of broken MATH-circuits. Now, in the case of graphic arrangements, the associated broken circuit uniquely determines a cycle in the graph. Hence, the map MATH is actually an isomorphism.
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math/0002253
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Assume REF . By REF, the natural homomorphism MATH is surjective. Now REF follows from NAME 's Lemma. Clearly, REF implies REF . Assume REF . Then the natural homomorphism MATH factors through a surjective homomorphism MATH, that is, MATH is a MATH-module. Since MATH is a simple algebra and MATH, the map MATH is injective and therefore surjective. Thus the composition MATH is surjective, and REF follows. If REF holds, then the natural map MATH is surjective by dimension arguments, and REF follows. REF is an immediate corollary of REF. The form MATH in REF induces a MATH-invariant pairing MATH. The left and right kernels of MATH are MATH-submodules of the simple MATH-module MATH, and thus are MATH, since MATH. It follows that MATH is perfect.
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math/0002253
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Let MATH and MATH. Let MATH denote the image of MATH. By REF iii, MATH. Suppose MATH is an (MATH)-stable lattice in MATH. Then MATH is a MATH-module and thus a MATH-module. Since every finitely generated MATH-module is a direct sum of copies of MATH REF with the standard action, therefore MATH, as MATH-modules. Since MATH commutes with MATH, it commutes with the image of MATH in MATH, that is, with MATH. Therefore the image of MATH in MATH is contained in MATH, and the isomorphism MATH is also an isomorphism of MATH-modules. Therefore MATH as MATH-modules, so MATH as MATH-modules. Since MATH is a simple MATH-module, therefore MATH and MATH are semisimple. It follows that MATH as MATH-modules, and we can identify MATH with a MATH-stable lattice MATH in the MATH-module MATH. If MATH is an (MATH)-invariant MATH-valued pairing on MATH, then MATH is self-dual with respect to MATH, since MATH is simple as a MATH-module. Therefore MATH is self-dual, that is, admits a non-degenerate MATH-invariant MATH-valued pairing MATH. Since MATH is absolutely simple as a MATH-module, every MATH-invariant MATH-valued pairing on MATH is a multiple of MATH. Similarly, since MATH, thus every (MATH)-invariant MATH-valued pairing on MATH is of the form MATH, where for some MATH we have MATH for every MATH. Since MATH is MATH-invariant, so is MATH.
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math/0002253
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Replacing MATH by a suitable multiple, we may suppose that MATH. By REF , we have MATH where MATH is a MATH-stable lattice in MATH, and MATH where MATH is a non-degenerate MATH-invariant MATH-valued pairing on MATH. Since MATH is alternating, MATH is symmetric. Since MATH, we have MATH. Since MATH is not perfect, there exists MATH such that MATH. If MATH, then MATH but MATH . Therefore MATH is not perfect, so MATH. The NAME factors of the MATH-module MATH are all isomorphic to MATH. Since MATH for some positive integer MATH, the NAME factors of the MATH-module MATH are all isomorphic to MATH. Therefore MATH is a power of MATH as desired.
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math/0002253
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Since MATH, we have REF . The group isomorphism MATH defined by MATH gives REF . For REF , multiplying MATH by a suitable power of MATH, we may assume that MATH contains MATH but does not contain MATH. Assume that MATH is not contained in MATH. Then there exist MATH and MATH such that MATH. Multiplying MATH by a suitable MATH-adic integer, we may assume that MATH. Since MATH and MATH, therefore MATH. Since the orbit MATH contains all the roots and MATH is MATH-stable, thus MATH. This contradiction gives REF . The MATH-module MATH is absolutely simple, so every MATH-invariant MATH-valued pairing on MATH is of the form MATH for some MATH. We have MATH, since MATH. Further, MATH if and only if MATH. Suppose that MATH is divisible by MATH. Since the dual of MATH with respect to MATH is MATH, thus MATH is never perfect. Let MATH be the largest power of MATH dividing MATH. Since MATH, we have MATH. Further, MATH if and only if MATH. The dual of MATH with respect to MATH is MATH. If MATH, then MATH so MATH is perfect for no MATH, and we have REF . REF follows from REF , and REF . Suppose that MATH is not divisible by MATH. Let bars denote reduction mod MATH. By REF , MATH. By REF , every MATH-stable MATH-lattice in MATH is of the form MATH. Therefore the MATH-module MATH is simple, by REF. It thus suffices to show MATH. Let MATH denote the set of elements of MATH invariant under MATH. Then MATH generates the one-dimensional MATH-vector space MATH. Suppose MATH. Then MATH is MATH-stable, so there is a MATH such that MATH. Since MATH commutes with MATH and the orbit MATH contains all the MATH's, we have MATH. Since MATH is generated by the MATH's, we have MATH.
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math/0002253
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The commutator subgroup MATH lies in MATH. First suppose MATH. Then MATH contains MATH, and we are done by REF vii. Suppose MATH. We have MATH. Let MATH. By REF vii, we may suppose that MATH is a proper subgroup of MATH. If MATH, then MATH. Thus, we may assume MATH. Then MATH contains a primitive cube root of unity and therefore MATH is a pro-cyclic MATH-group. Then MATH is a cyclic MATH-group. Since MATH, MATH has order MATH which is prime to MATH and therefore the MATH-module MATH is semisimple. Since MATH is non-commutative and MATH, it follows that the MATH-module MATH is absolutely simple.
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math/0002253
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Write MATH and MATH. The mod MATH representation MATH is surjective, since MATH is surjective. Let MATH. Since MATH is a pro-MATH-group, the index of MATH in MATH is a power of MATH. Thus either MATH and we are done, or MATH. Assume the latter. We have MATH . Let MATH be the composition MATH and let MATH be the fixed field of MATH. Then MATH is a quadratic extension of MATH. By REF , MATH. Since MATH, MATH is the only quadratic extension of MATH in MATH. Therefore MATH is not contained in MATH, so MATH. This contradicts the assumption that MATH.
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math/0002253
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As shown on pp. REF, there is an elliptic curve MATH over the function field MATH, with MATH-invariant MATH, such that MATH and MATH is algebraically closed in MATH. Thus MATH and we have MATH . By REF on p. REF and NAME 's Irreducibility Theorem (p. REF), there are infinitely many specializations MATH such that MATH and therefore MATH.
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math/0002253
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Choose MATH such that MATH (such a MATH exists since MATH is unbounded as MATH). Let MATH be the product of the prime divisors of MATH. Let MATH if MATH is odd and let MATH if MATH is even. By REF , there are infinitely many elliptic curves MATH over MATH, non-isomorphic over MATH, for which MATH. Thus for all odd prime divisors MATH of MATH we have MATH, so MATH is MATH-full by REF . If MATH is even, then MATH is MATH-full by REF .
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math/0002253
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Let MATH, a MATH-adic NAME group. By the Proposition and REF, there exists an open normal subgroup MATH of finite index such that whenever MATH is a closed normal subgroup in MATH with MATH, then MATH. Let MATH be the fixed field of MATH. Suppose MATH is a finite extension of MATH, let MATH, and view MATH as a closed normal subgroup of MATH. If MATH and MATH are linearly disjoint, then the restriction to MATH of MATH is surjective, that is, MATH. Therefore MATH, so MATH and MATH are linearly disjoint.
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math/0002253
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The kernel of MATH is MATH. Since MATH, we have MATH and REF follows. From the natural injections MATH we have REF . Suppose MATH and MATH are linearly disjoint over MATH. Then the natural injection MATH is surjective, and REF follows. Now REF follows from REF , by applying REF with MATH in place of MATH.
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math/0002253
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If MATH is not an isogeny then MATH and we are done. Suppose now that MATH is an isogeny. Let MATH denote the image of MATH in MATH. By REF , the image of MATH in MATH is MATH. Therefore the MATH-stable MATH-lattice MATH is also (MATH)-stable. Since MATH is MATH-invariant, it is also (MATH)-invariant. Applying REF with MATH, MATH, and MATH, then MATH for some positive integer MATH, and the result follows.
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math/0002253
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We may assume that MATH. Let MATH be a positive integer relatively prime to MATH and such that for every prime divisor MATH of MATH, MATH does not divide MATH. Let MATH if MATH is a power of MATH, and otherwise let MATH be the product of the prime divisors of MATH. Identify the symmetric group MATH in the standard way as an irreducible subgroup of MATH. By REF , for all prime divisors MATH of MATH the MATH-module MATH is well-rounded. Note that MATH and MATH, so for MATH and MATH we have MATH. By REF v, no MATH-stable MATH-lattice in MATH has a perfect MATH-invariant MATH-valued pairing. By REF , there exists an elliptic curve MATH over MATH that is MATH-full for all prime divisors MATH of MATH. By REF , for every prime MATH there exists a finite NAME extension MATH such that every finite extension of MATH linearly disjoint from MATH is linearly disjoint from MATH. There exists a finite NAME extension MATH of MATH with MATH which is linearly disjoint from MATH over MATH for all prime divisors MATH of MATH (see CITE, especially REF ). We can write MATH with MATH and MATH. Then MATH and MATH are linearly disjoint over MATH, and MATH (respectively, MATH) and MATH are linearly disjoint over MATH. Let MATH be the MATH-form of MATH attached to MATH and let MATH be the MATH-form of MATH attached to MATH . By REF b, MATH, so MATH and MATH are linearly disjoint over MATH. By REF d, MATH is MATH-full. REF now follows from REF with MATH. REF follows from REF for all prime divisors MATH of MATH.
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quant-ph/0002057
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We apply the operators MATH, MATH, and MATH in that order to the state MATH, and check that the result has the same effect as MATH. The operator MATH simply applies MATH to each of the MATH qudigits of MATH, which yields, MATH where MATH is a compact notation for MATH, and MATH denotes MATH. Then applying MATH to the above state yields, MATH . By a change of variable, the above can be re-written as, MATH . Finally, applying MATH to the above undoes the NAME transform and puts the coefficient of MATH in the exponent into the last slot of the state. The result is, MATH which is exactly what MATH would yield.
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quant-ph/0002057
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By CITE, any fixed dimension unitary matrix can be computed in fixed depth using one-qubit gates and controlled nots. Hence MATH can be computed in MATH, as can MATH. The result now follows immediately from REF .
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quant-ph/0002057
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First note that MATH and MATH are equivalent, since a MATH gate can be simulated by a MATH gate with MATH extra inputs set to the constant REF. Hence we can freely use MATH gates in place of MATH gates. It is easy to see that, given a MATH gate, we can simulate a MATH gate. Applying MATH to MATH digits (represented as bits, but each digit only taking on the values REF or REF) transforms, MATH . Now send the bits of the last block REF to a NAME gate with all inputs negated and control bit MATH. The resulting output is exactly MATH. The bits in the last block can be erased by re-negating them and reversing the MATH gate. This leaves only MATH, MATH auxiliary bits, and the output MATH. The converse (simulating MATH given MATH) requires some more work. The first step is to show that MATH can also determine if a sum of digits is divisible by MATH. Let MATH be a set of digits represented as MATH bits each. For each MATH, let MATH REF denote the bits of MATH. Since the numerical value of MATH is MATH, it follows that MATH . The idea is to express this last sum in terms of a set of Boolean inputs that are fed into a MATH gate. To account for the factors MATH, each MATH is fanned out MATH times before plugging it into the MATH gate. Since MATH, this requires only constant depth and MATH auxiliary bits (which of course are set back to REF in the end by reversing the fanout). Thus, just using MATH and constant fanout, we can determine if MATH. More generally, we can determine if MATH using just a MATH gate and constant fanout. Let MATH denote the resulting circuit, that determines if a sum of digits is congruent to MATH mod MATH. The construction of MATH is illustrated in the figure below for the case of MATH. In the figure, MATH denotes MATH. The notation on the right will be used as a shorthand for this circuit: We can get the bits in the value of the sum MATH using MATH circuits. This is done, essentially, by implementing the relation MATH. For each MATH, MATH, we compute MATH (where now the MATH's are digits). This can be done by applying the MATH circuits in series (for each MATH) to the same inputs, introducing an auxiliary REF-bit for each application, as illustrated here. Let MATH denote the MATH bit of MATH. For each MATH and for each MATH, we take the AND of the output of the MATH with MATH (again by applying the AND's in series, which is still constant depth, but introduces MATH extra auxiliary inputs). Let MATH denote the output of one of these AND's. For each MATH, we OR together all the MATH's, that is, compute MATH, again introducing a constant number of auxiliary bits. Since only one of the MATH's will give a non-zero output from MATH, this collection of OR gates outputs exactly the bits in the value of MATH. Call the resulting circuit MATH, and the sum it outputs MATH. Finally, to simulate MATH, we need to include the input digit MATH. To do this, we apply a unitary transformation MATH to MATH that transforms it to MATH. By NAME, et al. CITE (as in the proof of REF ), MATH can be computed in fixed depth using one-qubit gates and controlled NOT gates. Now using MATH and all the other auxiliary inputs, we reverse the computation of the circuit MATH, thus clearing the auxiliary inputs. This is illustrated in this figure: The result is an output consisting of MATH, MATH auxiliary bits, and MATH, which is the output of a MATH gate.
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quant-ph/0002057
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By the preceeding lemmas, MATH and MATH are MATH-equivalent. By NAME 's result, MATH is MATH-reducible to MATH. Hence MATH is MATH-reducible to MATH. Conversely, arrange each block of MATH input bits to a MATH gate as follows. For the control-bit block (which contains the bit we want to fan out), set all but the last bit to zero, and call the last bit MATH. Set all bits in the MATH input-bit block to REF. Now the MATH output of the MATH circuit is MATH, represented as MATH bits with only one possibly nonzero bit. Send this last output bit MATH and the input bit MATH to a controlled-NOT gate. The outputs of that gate are MATH and MATH. Now apply MATH to the bits that were the outputs of the MATH gate (which are all left unchanged by the controlled-not's). This returns all the MATH's to REF except for the control bit which is always unchanged. The outputs of the controlled-not's give the desired MATH. Thus the resulting circuit simulates MATH, with MATH auxiliary bits.
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quant-ph/0002057
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By the preceeding lemmas, fanout of bits is equivalent to the MATH function. By NAME 's result, we can do MATH if we can do fanout in constant depth. By our result, we can do fanout, and hence MATH, if we can do MATH. Hence QACC MATH QACCCITE MATH QACCCITE.
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quant-ph/0002057
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We will abuse notation in this proof and identify the encoding MATH with its value MATH. So MATH and MATH will mean the encoding of MATH and MATH respectively. CASE: To do sums, the first thing we do is form the list MATH. Then we create a flattened list MATH from this with elements which are the MATH's from the MATH's. MATH is in MATH using our definition of sequence from the preliminaries, and closure under sums and MATH to find the length of the longest MATH. To flatten MATH we use MATH to find the length MATH of the longest MATH for MATH. Then using max twice we can find the length of the longest MATH. This will be the second coordinate in the pair used to define sequence MATH. We then do a sum of size MATH over the subentries of MATH to get the first coordinate of the pair used to define MATH. Given MATH, we make a list MATH of the distinct MATH's that appear as MATH in some MATH for some MATH. This list can be made from MATH using sums, MATH and MATH. We sum over the MATH and check if there is some MATH such that the MATH-th element of MATH has same MATH as MATH and if not add the MATH-th elements MATH times REF raised to the appropriate power. We know what power by computing the sum of the number of smaller MATH that passed this test. Using MATH and closure under sums we can compute in MATH a function which takes a list like MATH and a MATH and returns the sum of all the MATH's in this list. So using this function and the lists MATH and MATH we can compute the desired encoding. For products, since the MATH's of MATH are algebraically independent, MATH is isomorphic to the polynomial ring MATH under the natural map which takes MATH to MATH. We view our encodings MATH as MATH-variate polynomials in MATH. We describe for any MATH a circuit that works for any MATH computable MATH such that MATH is of degree less than MATH viewed as a MATH-variate polynomial. In MATH we define MATH to consist of the sequence of polynomially many integer values which result from evaluating the polynomial encoded by MATH at the points MATH where MATH and MATH. To compute MATH at a point involves computing a polynomial sum of a polynomial product of integers, and so will be in MATH. Using closure under polynomial integer products we compute MATH where MATH is the sequence projection function from the preliminaries. Our choice of points is what is called by CITE the MATH-th order principal lattice of the MATH-simplex given by the origin and the points MATH from the origin in each coordinate axis. By REF of that paper (proved earlier by a harder argument in CITE) the multivariate NAME Interpolant of degree MATH through the points MATH is unique. This interpolant is of the form MATH where the MATH's are polynomials which do not depend on the function MATH. An explicit formula for these MATH's is given in REF as a polynomial product of linear factors. Since these polynomials are all of degree less than MATH, they have only polynomial in MATH many coefficients and in PTIME these coefficients can be computed by iteratively multiplying the linear factors together. We can then hard code these MATH's (since they don't depend on MATH) into our circuit and with these MATH's, MATH, and closure under sums we can compute the polynomial of the desired product in MATH. CASE: We do sums first. Assume MATH. One immediate problem is that the MATH and MATH might use different MATH's for their denominators. Since MATH is closed under poly-sized maximum, it can find the maximum value MATH to which MATH is raised. Then it can define a function MATH which encodes the same element of MATH as MATH but where the denominators of the MATH's are now MATH. If MATH was MATH we need to compute the encoding MATH. This is straightforward from REF . Now MATH where MATH's are the numerators of the MATH's in MATH. From REF we can compute the encoding MATH of MATH in MATH. So the desired answer MATH is in MATH. For products MATH, we play the same trick as the in the MATH product case. We view our encodings of elements of MATH as d-variate polynomials in MATH under the map MATH goes to MATH. (Note that this map is not necessarily an isomorphism.) We then create a function MATH which consists of the sequence of values obtained by evaluating MATH at polynomially many points in a lattice as in the first part of this lemma. Evaluating MATH at a point can easily be done using the first part of this lemma. We then use REF of this lemma to compute the products MATH. We then get the interpolant MATH. We non-uniformly obtain the encoding of MATH expressed as an element of MATH. that is, in the form MATH. Thus, the product MATH is MATH . The encoding of the products is the d-tuple given by MATH . Each of its components is a polynomial sum of a product of two things in MATH and can be computed using the first part of the lemma.
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quant-ph/0002057
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The proof is by induction on MATH. In the base case, MATH, we do not multiply any layers, and we can easily represent this as a tensor graph of width REF. Assume for MATH that MATH can be written as color consistent tensor graph of width MATH and polynomial size. There are two cases to consider: In the first case the layer is a tensor product of matrices MATH where the MATH's are NAME gates, one qubit gates, or fan-out gates (since MATH-QACC); in the second case the layer is a controlled-not layer. For the first case we ``multiply" MATH against our current graph by ``multiplying" each MATH in parallel against the terms in our sum corresponding to MATH's domain, say MATH. If MATH with domain MATH is a one-qubit gate, then we multiply the two amplitudes in each vertical edge of height MATH in our tensor graph by MATH. This does not effect the width, size, or number of paths through the graph. If MATH is a NAME gate, then for each term MATH in MATH in our tensor graph we add one new term to the resulting graph. This term is added by adding a horizontal edge going out from the source node of MATH followed by the new MATH-term followed by a horizontal edge into the terminal node of MATH. The new term is obtained from the old one by setting to MATH the left hand amplitudes of all edges in MATH of height between MATH and MATH and then if MATH is the amplitude of an edge of height MATH in the new term we change it to MATH. This new term adjusts the amplitude for the case of a MATH vector in MATH tensored with either a MATH or MATH. This operation increases the width of the new tensor graph by the width of the MATH-term for each MATH-term in the graph. Since the original graph has width MATH there are at most this many starting and ending vertices for such terms. So there at most MATH such terms. Each of these terms has width at most MATH. Thus, the new width is at most MATH . Notice this action adds one new path through the MATH part of the graph for every existing one. Now suppose MATH is a fan-out gate, let MATH be a MATH-term in our tensor graph and let MATH be any vertical edge in MATH in MATH. Suppose MATH has amplitude MATH for MATH and amplitude MATH for MATH. In the new graph we change the amplitude of MATH to MATH. We then add a horizontal edge out of the source node of MATH followed by a new MATH-term followed by a horizontal edge into the terminal node of MATH. The new term is obtained from MATH by changing the amplitude for edges in MATH with amplitudes MATH in MATH to MATH. The amplitudes of the non-MATH edges in this term are the reverse of the corresponding edge in MATH, that is, if the edge in MATH had amplitude MATH then the new term edge would have amplitude MATH. The same argument as in the NAME case shows the new width is bounded by MATH and that this action adds one new path through the MATH part of the graph for every existing one. For the case of a controlled-not layer, suppose we have a controlled-not going from line MATH onto line MATH. Let MATH be a new color, anti-color pair not yet appearing in the graph. Let MATH be a vertical edge of height MATH in the graph and let MATH be respectively its color product and two amplitudes. Similarly, let MATH be a vertical edge of height MATH in the graph and MATH be its color product and two amplitudes. In the new graph we multiply MATH times the color product of MATH and MATH and change the amplitude of MATH to MATH. We then add a horizontal edge going out from the starting node of MATH, followed by a vertical edge with values MATH followed by a horizontal edge into the terminal node of MATH. In turn, we add a horizontal edge going out of the starting node of MATH, followed by a vertical edge with values MATH followed by a horizontal edge into the terminal node of MATH. We handle all other controlled gates in this layer in a similar fashion (recall they must go to disjoint lines). We add at most a new vertex of a given height for every existing vertex of a given height. So the total width is at most doubled by this operation and MATH. In the MATH case, simulating a layer which is a NAME product of spaced controlled-not gates and identity matrices, notice we would at most add one to the color depth at any place. So if a controlled-not layer is a composition of MATH many such layers it will increase the color depth by MATH. In the MATH case, notice that simulating a single controlled-not we add one new path for each existing path through the graph at each of the two heights affected. This gives three new paths on the whole subspace for each old one. Since we have handled the two possible layer cases and the changes we needed to make only increase the resulting tensor graph polynomially, we thus have established the induction step and REF. For REF , observe for each multi-line gate we handle in adding a layer we at most quadruple the number of paths through the subspace where that gate applies. Since there are at most logarithmically many such gates, the number of paths through the graph increases polynomially.
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quant-ph/0002057
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Let MATH be a particular graph in the family and let MATH be the vector whose amplitude we want to compute. Assume that all graphs in our family have fewer than MATH colors in any color product and have a width bounded by MATH. We will proceed from the source to the terminal node one height at a time to compute the amplitude. Since the width is MATH the number of MATH-terms is at most MATH and each of these must have width at most MATH. Let MATH (some of which may be zero) denote the amplitudes in MATH of MATH in each of these terms. The MATH are each sums of at most MATH amplitudes times the color products of at most MATH colors and anticolors, so the encoding of these MATH amplitudes is MATH computable. Because of the restriction on the width of MATH there are at most MATH many MATH-terms, MATH many MATH-terms, and MATH many MATH-terms. Fixing some ordering on the nodes of height MATH and MATH let MATH be the amplitude of MATH in the MATH-term with source the MATH-th node of height MATH and with terminal node the MATH-th node of height MATH. The amplitude is zero if there is no such MATH-term. Then the amplitudes MATH of the MATH-terms can be computed from the amplitudes MATH of the MATH-terms using the formula MATH . Thus MATH can be computed from the MATH using a polynomial sized circuit to do these adds and multiplies. Similarly, each MATH can be computed by polynomial sized circuits from the MATH's and so on. Since we have log-color depth the number of terms consisting of elements in our field times color products in a MATH will be polynomial. So the size of the MATH's MATH, MATH will be polynomial in the input MATH. So the size of the circuits for each MATH where MATH and MATH will be polynomial size. There is only one MATH-term in MATH and its amplitude is that of MATH, so this shows it has polynomial sized circuits. For the MATH result, if the number of paths is polynomially bounded, then the amplitude can be written as the polynomial sum of the amplitudes in each path. The amplitude in a path can in turn be calculated as a polynomial product of the amplitudes times the colors on the vertical edges in the path. Our condition on every color appearing at exactly two heights guarantees the color product along the whole path will be REF or REF, and will be zero iff we get a color and its anticolor on the path. This is straightforward to check in MATH, so this sum of products can thus be computed in MATH using REF .
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quant-ph/0002057
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Given a a family MATH of MATH operators and a family MATH of states we can use REF to get a family MATH of log color depth, color-consistent tensor graphs representing the amplitudes of MATH. Note MATH is also a family of MATH operators since NAME and fan-out gates are their own inverses, the inverse of any one qubit gate is also a one qubit gate (albeit usually a different one), and finally a controlled-not layer is its own inverse. REF shows there is a P/poly circuit computing the amplitude of any vector MATH in this graph. This amounts to calculating MATH . If this is nonzero, then MATH, and we know MATH is in the language. In the MATH case everything is a rational so P/poly can explicitly compute the magnitude of the amplitude and check if it is greater than MATH. The MATH result follows similarly from the MATH part of REF .
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cond-mat/0003123
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Let MATH be such that MATH for every MATH. The existence of MATH bonds like that in the set MATH is ensured by the hypotheses that MATH and that MATH forms a complete coupon term structure. Let MATH be a MATH matrix such that MATH for all MATH. Finally let MATH be an arbitrary bond of MATH. Since MATH is an upper triangular matrix with no vanishing element on its principal diagonal, it is non-singular. Now let MATH be such that MATH . Then MATH . By REF , this implies REF with MATH . In order to conclude the proof of the first part of the theorem, we have to show that the MATH's are independent of the choice of MATH such that MATH for every MATH if several such choices are possible. To do this, let MATH and let MATH be such that MATH. Let MATH be a MATH matrix such that MATH for all MATH, and let MATH . We want to show that MATH for every MATH. Since both MATH and MATH has been arbitrarily chosen, this will conclude the proof of the first part of the theorem. Observe that MATH for all MATH and that by REF MATH and MATH . Then for every MATH . In order to prove the second part of the theorem, let MATH be such that MATH for every MATH and let MATH be a portfolio of these bonds such that MATH for some MATH and some MATH. By REF, the price of this portfolio is MATH . If MATH and MATH, then by REF , MATH, which implies that MATH and, since MATH has been chosen arbitrarily, that MATH . Now suppose, by contradiction, that MATH and let MATH be such that MATH . Since MATH, we can assume, without loss of generality, that MATH and that MATH. However REF implies that either MATH or MATH. This is the wanted contradiction.
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cs/0003009
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According to REF , the equivalence relation REF provides a rough classification of the worlds in MATH with respect to the conditionals in MATH. Obtaining a representation of the form REF then amounts to checking the solvability of a linear equational system. The proof of this theorem is very similar to the proof of the analogous theorem for probabilistic representation of knowledge given in CITE.
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cs/0003009
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MATH is a c-representation of MATH iff MATH and MATH is indifferent with respect to MATH. From REF , we obtain representation REF . Due to REF , MATH iff MATH, that is, iff MATH which is equivalent to MATH . This shows REF .
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cs/0003035
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MATH implies that an ordering MATH is compatible with MATH whenever it is compatible with MATH. We thus have MATH and therefore MATH.
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cs/0003035
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If MATH has no preferred extension the proposition is trivially true. So assume MATH possesses preferred extension REF . A simple induction shows that each preferred extension is among the extensions compatible with the formulas computed in each step of the iteration of MATH. Therefore each preferred extension is also an accepted extension.
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cs/0003035
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We show by induction that, for arbitrary MATH, the set of formulas obtained after MATH applications of MATH is consistent. For MATH this is trivial. Assume the set of REF obtained after MATH iterations is consistent. Since MATH is consistent and MATH formalizes a strict partial ordering there must be at least one strict partial ordering MATH compatible with MATH, so the set of all extensions compatible with MATH is nonempty. Since each extension is by definition consistent the intersection of an arbitrary nonempty set of extensions must also be consistent.
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cs/0003045
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By definition of MATH, for every arc from MATH to MATH in MATH, there exists a sequence of consecutive NAME steps, starting from MATH and having a variant of MATH as its selected atom at the end. Because (a variant of) MATH is selected at the end-point, any two such derivation-step sequences, corresponding to consecutive arcs in MATH, can be composed to form a new sequence of NAME steps. In this sequence, all REF nodes of the consecutive arcs remain selected atoms in the new sequence of derivation steps. Transitively exploiting the above argument yields the result.
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cs/0003045
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Let MATH be an atom such that MATH. Let MATH be the NAME with respect to MATH for MATH. If MATH NAME with respect to MATH, it is easy to see that MATH is finite. Hence, MATH is finite, and MATH consists of a finite number of NAME. Now we prove that no tree in MATH has an infinite branch. Suppose this is not the case and there is a tree in MATH with an infinite branch. Let MATH be the leftmost atom of a query labeling a node in this infinite branch. Then, MATH has an infinite NAME (just plug in, for each tabled atom MATH in the infinite branch which is resolved with an answer, the branch of the tree with root MATH which leads to this answer). This gives a contradiction.
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cs/0003045
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Since MATH, an NAME cannot have infinite branches. The equivalence then follows from the fact that for every MATH such that MATH, MATH is the root of an NAME in the NAME of MATH iff MATH.
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cs/0003045
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The implication follows from the fact that for every MATH such that MATH, MATH is the root of an NAME in the NAME with respect to MATH of MATH iff MATH.
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cs/0003045
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The first statement is trivial by definition. For the second statement, this is a corollary of the following REF with MATH and MATH.
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cs/0003045
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Let MATH be such that MATH quasi-terminates with respect to MATH and MATH and MATH does not NAME with respect to MATH and MATH. Then, there exists a predicate MATH such that there is a MATH-atom in MATH which has infinitely may different (nonvariant) computed answers. Since tabling does not influence the set of call patterns nor the set of computed answer substitutions (see for example, CITE), there cannot exist a tabling such that MATH NAME with respect to that tabling and the set MATH.
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cs/0003045
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Let MATH be an atom such that MATH. Let MATH be the NAME with respect to MATH of MATH and let MATH be the NAME with respect to MATH of MATH. We know that MATH consists of a finite number of finite NAME. So, MATH, hence, MATH and MATH consists of a finite number of NAME. We prove that the NAME of MATH are finite. Since each NAME in MATH can be extended to obtain an NAME in MATH, this follows from the finiteness of the NAME in MATH.
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cs/0003045
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MATH . Suppose MATH NAME with respect to MATH and MATH. Then MATH quasi-terminates with respect to MATH and MATH. It is easy to see that, since for every MATH such that MATH the NAME for MATH consists of a finite number of finite trees, the set of computed answers for atoms in MATH is finite. MATH . Suppose that MATH quasi-terminates with respect to MATH and MATH and for all MATH such that MATH the set of NAME answers for atoms in MATH is finite. We prove that MATH NAME with respect to MATH and MATH. Let MATH be an atom such that MATH. We already know that the NAME MATH of MATH consists of a finite number of NAME without infinite branches. We prove by contradiction that these NAME are finitely branching. Suppose there is an NAME in MATH which is infinitely branching. Then, there is an NAME in MATH with an infinitely branching node, which contains a query which has a tabled, recursive atom at the leftmost position. That is, there is an atom in MATH which has infinitely many computed answers. This gives a contradiction.
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cs/0003045
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MATH . Suppose MATH is quasi-terminating with respect to MATH and MATH. Let MATH be an atom such that MATH. Let MATH be the NAME with respect to MATH of MATH. We prove that MATH consists of a finite number of finite NAME. We know that the NAME MATH with respect toMATH of MATH is a finite set of NAME, without infinite branches. It is easy to see that hence, MATH consists also of a finite number of trees without infinite branches. We prove that the NAME in MATH are finitely branching. Suppose this is not the case, that is, there is an NAME in MATH which is infinitely branching. Then, there is an NAME MATH in MATH which is infinitely branching in a non-root node, which is a query with leftmost atom MATH, with MATH, which is directly descending from an atom MATH, with MATH, via a recursive clause MATH. Let MATH be the NAME in MATH corresponding to MATH. Note that the clause MATH instead of MATH is used in MATH. Because of this, the atom to the right of MATH in the infinitely branching node is MATH. Thus, MATH consists of a infinite number of NAME (there are an infinite number of NAME with predicate MATH in the root). This gives a contradiction. MATH . Suppose MATH is NAME with respect to MATH and MATH. Let MATH be an atom such that MATH. Let MATH be the NAME with respect to MATH of MATH. Then, MATH consists of a finite number of finite NAME. Let MATH be the NAME with respect to MATH of MATH. By definition of the MATH-transformation, we see that MATH also consists of a finite number of finite NAME. Hence, MATH quasi-terminates (and even NAME) with respect to MATH and MATH.
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cs/0003045
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MATH . Suppose that MATH is quasi-acceptable with respect to MATH, MATH and a level mapping MATH. We prove that MATH quasi-terminates with respect to MATH and MATH. Let MATH be an atom such that MATH, let MATH be the NAME with respect to MATH of MATH. CASE: MATH consists of a finite number of NAME, that is . MATH. Due to the quasi-acceptability condition, any call in MATH directly descending from MATH, say MATH, is such that MATH. The same holds recursively for the atoms descending from MATH. Thus, the level mapping of any call, recursively descending from MATH, is smaller than or equal to MATH. Since MATH is finitely partitioning on MATH, we have that: MATH. Hence, MATH, that is . MATH consists of a finite number of trees. CASE: The NAME in MATH have finite branches. Suppose there is a tree in MATH with an infinite branch. This infinite branch contains an infinite directed subsequence MATH. It is easy to see that the leftmost atoms in the nodes of this infinite directed subsequence all are MATH-atoms (because MATH-atoms are resolved using answers). There is a MATH, such that each MATH, MATH, has as leftmost atom MATH and for all MATH, MATH and MATH does not hold. Because of the quasi-acceptability condition, MATH, for all MATH. This gives a contradiction. MATH . Suppose that the tabling MATH is well-chosen with respect to MATH and suppose that MATH quasi-terminates with respect to MATH. We have to construct a level mapping MATH such that MATH is quasi-acceptable with respect to MATH, MATH and this level mapping MATH. We will only define MATH on elements of MATH. On elements of the complement of MATH in MATH, MATH can be assigned any value, as these elements do not turn up in the quasi-acceptability condition. In order to define MATH on MATH, consider the MATH-graph REF . Consider a strongly connected component MATH in MATH. Then, there is at least one MATH-atom in MATH. To see this, suppose this is not the case. Consider a cyclic path MATH in MATH. This consists only of MATH-atoms. But then, because of REF , there is an infinite branch in a tree of the NAME of an element of MATH. This gives a contradiction. Also, there is only a finite number of MATH-atoms in MATH. To see this, suppose this is not the case. Then there is an infinitely long path MATH through infinitely many MATH-atoms of MATH. Because of REF , there is an infinite number of MATH-atoms selected in a derivation of an element of MATH, that is, there are infinitely many trees in the NAME of that element of MATH. This gives a contradiction. For every two non-tabled atoms, say MATH and MATH, in MATH (note that thus MATH), MATH does not hold (since there is at least one MATH-atom in MATH). Thus, since the tabling is well-chosen, MATH holds. Define MATH as the graph obtained from MATH by replacing any strongly connected component by a single contracting node and replacing any arc from MATH pointing to (respectively, from) any node in that strongly connected component by an arc to (respectively, from) that contracting node. MATH does not have any (non-trivial) strongly connected components. Moreover, any strongly connected component from MATH that was collapsed into a contracting node of MATH necessarily contains at least one and at most a finite number of MATH-atoms. Note now that each path in MATH which is not cyclic (there are only trivial cycles in MATH) is finite. This also follows directly from REF . Note also that it is possible that MATH has an infinitely branching (possibly contracting) node. Let MATH be an atom in that infinitely branching node. It follows from REF that, because MATH quasi-terminates with respect to MATH, MATH is a descendant of MATH in MATH. We now construct MATH from MATH starting from the top nodes MATH downwards as follows: CASE: replace all direct descendants of MATH in MATH different from MATH, by a single contracting node MATH; CASE: replace any arc from MATH pointing to (respectively, from) any node in that (possibly infinite) set of direct descendants by an arc to (respectively, from) that contracting node MATH; CASE: repeat this for the nodes MATH. This process stops because, as we already noted, each path in MATH which is not cyclic is finite. It is easy to see that MATH is a graph in which each node has at most one direct descendant different from itself. Also, each node in MATH consists of a (possibly infinite) set of nodes of MATH which contains only finitely many MATH-atoms. We define the level mapping MATH as follows. Consider the layers of MATH (there are only a finite number of layers). Let layer-REF be the set of leaves in MATH. We assign to these nodes a number in MATH, such that all nodes get a different number. Then, we move up to the next layer in MATH. This layer, layer-REF, consists of all nodes MATH such that the path starting from MATH has length REF. We assign to each such node MATH a natural number, such that the number assigned to MATH is strictly larger than the number assigned to its descendant (in the previous step). We continue this process layer by layer. The value of the level mapping MATH on elements of MATH is defined as follows: all calls contained in the node MATH receive the number assigned to the node MATH. We prove that MATH is quasi-acceptable with respect to MATH, MATH and this level mapping MATH. CASE: for every MATH, MATH is finitely partitioning on MATH. Note that MATH is even finitely partitioning on MATH. This is because each (contracting) node of MATH contains only a finite number of MATH-atoms and because of the construction of MATH. CASE: Let MATH be an atom such that MATH, let MATH be a clause in MATH, such that MATH exists, let MATH be a MATH for MATH: CASE: then MATH. This is because there is a directed arc from MATH to MATH in MATH and because of the construction of MATH. CASE: then MATH if MATH and MATH does not hold (that is, MATH holds). There is a directed arc in MATH from MATH to MATH. Note that MATH and MATH do not belong to the same strongly connected component of MATH. This is because MATH holds. So, MATH and MATH belong to a different layer and MATH is a direct descendant of MATH. Hence, because of the construction of MATH, MATH.
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cs/0003045
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MATH . Suppose that MATH is NAME with respect to MATH and MATH. We prove that MATH NAME with respect to MATH and MATH. Let MATH be an atom such that MATH. Let MATH be the NAME with respect to MATH of MATH. We prove that MATH consists of a finite number of finite NAME. CASE: The NAME in MATH are finitely branching. Suppose this is not the case, that is, there is an NAME in MATH which is infinitely branching. Then, there is an NAME MATH in MATH which is infinitely branching in a non-root node, which is a query with leftmost atom MATH, with MATH, which is directly descending from an atom MATH, with MATH, via a recursive clause MATH. Now, consider the NAME MATH of MATH. Let MATH be the NAME in MATH corresponding to MATH. Note that the clause MATH instead of MATH is used in MATH. Because of this, the atom on the right of MATH in the infinitely branching node is MATH. Thus, MATH consists of a infinite number of NAME (there are an infinite number of NAME with predicate MATH in the root). But, all these MATH-atoms directly descend from the node MATH via the clause MATH in MATH and hence, because of the NAME condition, their value under the level mapping MATH is smaller or equal to MATH. Because MATH is finitely partitioning on MATH, this gives a contradiction. CASE: MATH consists of a finite number of NAME, that is, MATH. Suppose this is not the case. A first possible reason for an infinite number of NAME in MATH is an infinitely branching NAME in MATH. But we already proved that this does not occur. The other possibility is that there exists an infinite NAME of MATH in MATH which contains an infinite directed subsequence, such that this infinite directed subsequence has a tail MATH with MATH, MATH, such that MATH is an infinite set and MATH for all MATH. So, MATH. Since MATH, and since MATH is finitely partitioning on this set and MATH for all MATH (by the NAME condition), this gives a contradiction. CASE: The NAME in MATH have finite branches. The same argumentation as in the proof of REF can be applied here. MATH . Suppose that the tabling MATH is well-chosen with respect to MATH and suppose that MATH NAME with respect to MATH and MATH. We prove that there exists a level mapping MATH such that MATH is NAME with respect to MATH, MATH and this level mapping MATH. Since MATH NAME with respect to MATH and MATH, we know by REF that MATH quasi-terminates with respect to MATH and MATH. Note that, since MATH is well-chosen with respect to MATH, MATH is well-chosen with respect to MATH. By REF , there exists a level mapping MATH such that MATH is quasi-acceptable with respect to MATH, MATH and this level mapping MATH. It is straightforward to verify that MATH is NAME with respect to MATH, MATH and this level mapping MATH. (Note that, as we already discussed in the beginning of this subsection, the level mapping obtained in this way satisfies more conditions than required by the notion of NAME.)
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cs/0003045
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Let MATH be an atom such that MATH. Let MATH be the NAME with respect to MATH of MATH. We prove that MATH consists of a finite number of NAME without infinite branches. If MATH is defined in MATH, this follows directly from the fact that MATH extends MATH and that MATH quasi-terminates with respect to MATH and MATH. So, suppose MATH is defined in MATH. Because of the second condition in the proposition statement, every call directly descending from MATH, say MATH, is such that MATH. This holds recursively for atoms descending from MATH using clauses of MATH. Because MATH is finitely partitioning on MATH, the set of tabled atoms, descending from MATH, using clauses of MATH, is finite. For atoms MATH, defined in MATH and descending from MATH using clauses of MATH, we know that MATH. So, MATH. We now prove that there is no tree in MATH with an infinite branch. Suppose this is not the case, and there is a tree in MATH with an infinite branch. Because, MATH quasi-terminates with respect to MATH and MATH, and because MATH extends MATH, this infinite branch contains an infinite directed subsequence MATH, with leftmost atoms MATH, belonging to MATH. This infinite directed subsequence has a tail, such that for all MATH such that MATH belongs to this tail, MATH and MATH does not hold. But because of the condition in the proposition statement, MATH and this gives a contradiction.
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cs/0003045
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This is a direct corollary of REF (every recursive predicate in MATH is defined in MATH and hence tabled).
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cs/0003045
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Because MATH extends MATH and MATH extends MATH, MATH for MATH. The proposition follows then by definition of quasi-termination.
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cs/0003045
|
The proof is a simple adaptation of the proof of the if-direction of REF ; the adaptation is similar to the adaptation needed to transform the proof of the if-direction of REF into a proof of REF .
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cs/0003045
|
Because no defined predicate in MATH is tabled, MATH. Also, for all MATH with MATH, MATH holds. The proposition follows then from REF .
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cs/0003045
|
This is a direct corollary of REF (every recursive predicate in MATH is defined in MATH and hence tabled).
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cs/0003045
|
Because MATH extends MATH and MATH extends MATH, MATH, for MATH. The proposition follows then by definition of NAME.
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cs/0003045
|
CASE: Let MATH. We prove that MATH. MATH and this last set is finite. CASE: Let MATH. We prove that MATH. MATH . It is obvious that the first two sets in the union are finite (MATH, respectively, MATH, is finitely partitioning on MATH, respectively, MATH). The set MATH is finite, because it is a subset of the finite set MATH.
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cs/0003045
|
Because of REF , the level mapping MATH is finitely partitioning on MATH. We prove that MATH is quasi-acceptable with respect to MATH, MATH and the level mapping MATH (see REF ). Let MATH be an atom such that MATH. Let MATH be a clause of MATH such that MATH exists. Let MATH be a MATH in MATH for MATH. There are two cases to consider: CASE: MATH is defined in MATH, MATH. Then, because of REF in the proposition statement, MATH (note that because MATH extends MATH, for a clause MATH in MATH and MATH, a MATH for MATH in MATH is the same as a MATH for MATH in MATH only). Since MATH, MATH. In case MATH and MATH does not hold, MATH. CASE: MATH is defined in MATH, MATH. CASE: MATH, MATH. Because of REF in the proposition statement, MATH. Hence, MATH. Note that in this case we always have that MATH (because MATH extends MATH). CASE: MATH, MATH. Because of REF in the proposition statement, MATH. Also, because of REF , MATH. Hence, MATH. In case MATH and MATH does not hold, we have that MATH, hence MATH. In each case, we conclude that MATH and that, in case MATH and MATH does not hold, MATH.
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cs/0003045
|
Note that because MATH extends MATH and vice versa, MATH, MATH. By REF , MATH is finitely partitioning on MATH. Also, if MATH is a clause in MATH and MATH, then a MATH in MATH for MATH is a MATH in MATH REF for MATH and vice versa. Then it directly follows that MATH is quasi-acceptable with respect toMATH, MATH and MATH.
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cs/0003045
|
Suppose the above condition is satisfied for MATH. We prove that MATH is quasi-acceptable with respect to MATH, MATH and the level mapping MATH. Let MATH be an atom such that MATH. Let MATH be a clause in MATH such that MATH exists. Let MATH be an LD- computed answer substitution for MATH. Then, MATH. We prove that MATH. By the condition in the proposition, we know that MATH. Because MATH, MATH. Now, since MATH and MATH is rigid on MATH, MATH. Thus, MATH, and therefore MATH. The proof that MATH in case MATH and MATH does not hold, is analogous.
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cs/0003045
|
The union of the relations MATH, define an interpretation of MATH on the domain MATH. The condition expresses that for this interpretation, MATH holds. Thus, the interpretation is a model and therefore each MATH is a valid interargument relation.
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cs/0003045
|
This symbolic condition for quasi-termination is derived from the rigid quasi-acceptability condition in a way analogous to the derivation of the symbolic condition for NAME from the rigid acceptability condition (see CITE). In order for this article to be self-contained, we include the proof. Suppose that there exists a symbol mapping MATH satisfying the above condition. We prove that MATH is rigid quasi-acceptable with respect to MATH and MATH, and hence that MATH quasi-terminates with respect to MATH and MATH. We propose as a level mapping the level mapping MATH induced by MATH (based on the norm MATH induced by MATH). Because MATH measures only input positions in MATH, we have by REF (and by the fact that MATH and MATH are well-moded) that MATH is rigid on MATH. Also, because MATH measures all input positions in MATH, we have by REF (and by the fact that MATH and MATH are simply moded) that MATH is finitely partitioning on MATH. Take any clause MATH in MATH, and any body atom MATH, MATH. Let MATH be a substitution such that MATH. We prove that MATH (the proof that MATH in case MATH and MATH does not hold, is analogous). REF of this proposition holds for any instantiation of it, so MATH holds. Now, we prove that for any MATH, MATH holds. By REF of this proposition, we have that, for all MATH, MATH is valid with respect to MATH. Then, since for all MATH, MATH holds, we have that MATH holds. So, by MATH we conclude that MATH holds, which implies that MATH holds.
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cs/0003062
|
The proofs of CITE regarding cut-elimination for definitions do not appear to extend to our setting where induction is included. A complete proof of this theorem appears in CITE and CITE and is modeled on proofs by NAME and NAME that use the technical notions of normalizability and reducibility.
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cs/0003062
|
This rule expresses the following idea: we want to show that MATH follows from MATH and the fact that MATH is a natural number. Since MATH is a natural number, it must be either zero or the successor of another natural number. Thus if we can show that MATH holds for zero and for the successor of any natural number (the first two premises), then we know that MATH holds for MATH. It then remains to show that MATH follows from MATH and MATH (the third premise). To derive this rule, we assume that we have derivations of the premises and proceed to prove the conclusion. That is, we construct in MATH a partial derivation of the sequent MATH, leaving unproved premises of the form MATH, MATH, and MATH. This corresponds to working under the assumption that MATH holds both for zero and for the successor of any number and that MATH and MATH imply MATH. We proceed by induction on MATH, using MATH as our induction predicate. As a result, we must establish three things: CASE: the base case: zero is a natural number and MATH holds for it; CASE: the induction step: if MATH is a natural number and MATH holds for it, then the same is true for MATH; CASE: the relevance of the induction predicate: if MATH is a natural number and MATH holds for it, then MATH implies MATH. This staging of the problem is represented in MATH by applying the MATH rule: MATH . The three premises to the MATH rule correspond to the three proof obligations enumerated above. Let us first consider the relevance of the induction predicate. This is clear, since we are working under the assumption that MATH follows from MATH and MATH. This is formally represented by the partial derivation MATH . The base case is also simple: zero is obviously a natural number, and we are working under the assumption that MATH holds for zero. This is expressed in MATH by the partial derivation MATH . It remains to prove the induction step. Since MATH is a natural number, MATH is as well. In addition, MATH holds for MATH by our working assumption. The formal representation of this reasoning is MATH .
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cs/0003062
|
The proof is a simple case analysis on MATH. To represent this in MATH, we apply the MATH and MATH rules to get MATH and then use the derived rule of REF , which yields the three sequents MATH . In this case, the third premise is immediate: MATH . If MATH is zero, then it is immediate that zero is equal to itself and thus less than or equal to itself: MATH . If MATH is the successor of some number MATH, then MATH by definition, and so MATH also by definition. This is represented formally by the derivation MATH .
|
cs/0003062
|
To derive this rule, we construct a partial derivation of the sequent MATH, leaving unproved premises of the form MATH, MATH, and MATH. This corresponds to proving that MATH follows from MATH and the fact that MATH is a list under the assumptions CASE: MATH holds for MATH; CASE: for any MATH and MATH, if MATH holds for MATH, then it also holds for MATH; CASE: MATH and MATH imply MATH. The proof is by induction on the length of the list MATH. Since MATH holds, by REF has a length which is a natural number: MATH . We now claim that MATH holds for lists of any length, and wish to prove this claim by induction on the length of the list. Thus we must prove CASE: the base case: MATH holds for lists of length zero; CASE: the induction step: if MATH holds for lists of length MATH, it holds for lists of length MATH; CASE: the relevance of the claim: MATH follows from MATH, the fact that MATH has length MATH, and the fact that MATH holds for lists of length MATH. This is represented in MATH by applying the MATH rule with the induction predicate MATH, which yields the three sequents MATH . Once we have proved that MATH holds for lists of length MATH, then we know it holds for MATH. Thus we know that MATH follows from MATH, since our third working assumption says that MATH follows from MATH and MATH. This is represented formally by the partial derivation of the third premise of the MATH rule: MATH . The unproved premise of this partial derivation is actually a weakening of the third premise of the induction rule we are deriving. We do not have an explicit weakening rule in MATH, but it suffices here to use the cut rule: MATH . The first premise of the cut rule is derivable for any MATH and MATH, since the consequent MATH also occurs as an antecedent. The second premise is the desired premise of the rule we are deriving. In the base case of the induction, we must show that MATH holds for lists of length zero. Since the only list of length zero is MATH, this follows from the first working assumption, which says that MATH holds. This case is formalized in the following partial derivation of the first premise of the MATH rule: MATH . The induction step requires us to prove that MATH holds for all lists of length MATH, given that it holds for all lists of length MATH. Since a list of length MATH is constructed by adding an element to the front of a list of length MATH, this step follows from the second working assumption, which says that if MATH holds for a list MATH, then for any MATH, MATH holds for MATH. This reasoning is represented in the partial derivation of the second premise of the MATH rule: MATH . In this use of the MATH rule, the complete set of unifiers for the atomic formula MATH and the head of the clause MATH is the singleton set MATH. The unproved premise of the partial derivation above is a weakening of the second premise of the induction rule we are deriving. We can achieve this weakening using the cut rule in the same manner as we did for the third premise: MATH .
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cs/0003062
|
We prove this by induction on MATH; using the right rules for MATH and MATH and the derived rule of REF with the induction predicate MATH, we get the three sequents MATH . Since the induction predicate applied to MATH is the same as the consequent, the relevance of the induction predicate is immediate. Thus the third sequent follows from the MATH rule. The base case follows immediately from the definition of split, and so the first sequent is derivable using the MATH and MATH rules. The induction step also follows easily from the definition of split: MATH .
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cs/0003062
|
The reverse direction follows easily from the definition MATH. For the forward direction, the use of the MATH rule with MATH will cause the structure of the MATH derivation to closely follow that of the corresponding derivation in intuitionistic logic. However, we need to be sure that the MATH and MATH rules don't allow us to derive anything that we can't derive in intuitionistic logic. In fact, we can show that a cut-free derivation of MATH will consist only of sequents with empty antecedents CITE. Thus the MATH and MATH rules are not used, since they both require a formula in the antecedent.
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cs/0003062
|
We can restrict our attention to uniform derivations in linear logic, since they are complete for this fragment of linear logic CITE. As before a cut-free derivation of MATH will consist only of sequents with empty antecedents. Thus the definition of seq will ensure that the structure of the MATH derivation will closely follow that of the corresponding derivation in linear logic. The proof of the forward direction goes by induction on the structure of the MATH derivation, and the reverse direction by induction on the structure of the linear logic derivation. In general each case follows easily from the induction hypothesis. A more detailed proof of this theorem, including a definition of the MATH translations, can be found in CITE.
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cs/0003062
|
We prove this theorem by induction on the height of the derivation of MATH. Since MATH is atomic, its derivation must end with the use of one of the formulas encoding evaluation. If the MATH formula for abs is used, then MATH and MATH are both equal to MATH, for some MATH, and the consequent is immediate. If MATH was derived using the MATH formula for app, then MATH is of the form MATH, and for some MATH there are shorter derivations of MATH and MATH. Since MATH is MATH, MATH must have been derived using the formula encoding the typing rule for app. Hence, there is a MATH such that MATH and MATH. Applying the inductive hypothesis to the evaluation and typing judgements for MATH, we have MATH. This atomic formula must have been derived using the typeof formula for abs, and, hence, MATH. Since our specification logic is intuitionistic logic, we can instantiate this quantifier with MATH and use cut and cut-elimination to conclude that MATH. Applying the inductive hypothesis to the judgements for MATH yields MATH.
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cs/0003062
|
We show the derivation of the first subject reduction property, which is a formalization of REF . We wish to show that evaluation preserves types: MATH . (We have changed the names of the quantified variables to agree with those in the informal proof.) Applying the MATH, MATH, MATH, MATH, and MATH rules to the above sequent yields MATH . (Recall that MATH is an abbreviation for MATH.) As in the informal proof, we proceed with an induction on the height of the derivation of MATH, which is represented here by MATH. We will use the derived rule for complete induction REF and our induction predicate will be MATH which we will denote by IP. The derivation of the conclusion from the induction predicate applied to MATH is trivial, so it only remains to derive the induction step MATH . We use the MATH and MATH rules to obtain MATH . In the informal proof we use the fact that the derivation of the atomic formula MATH must end with the use of a clause from the specification of evaluation. We deduce this formally by applying the MATH rule to MATH, which yields MATH . We next apply the MATH, MATH, and MATH rules, and then apply the MATH rule to MATH which yields the two sequents MATH . This use of the MATH rule corresponds to the case analysis of the formula used to derive MATH. As in the informal case, the abs case (represented here by the first sequent) is immediate. The derivation of the second sequent, representing the app case, begins with the use of the MATH, MATH, and MATH, bringing us to the sequent MATH . (We use the term MATH as an abbreviation for MATH.) The informal proof continues with an analysis of the derivation of MATH . Again we accomplish this through two uses of the MATH rule, the first to indicate that the derivation must end with the use of a specification clause, and the second to determine the applicable clauses. In this case there is only one applicable clause, so we are left to derive the sequent MATH . Additional uses of the MATH, MATH and MATH rules bring us to the sequent MATH . In the informal proof we now apply the induction hypothesis to the evaluation and typing judgments for MATH. We accomplish this here by applying the appropriate left rules to the elided induction hypothesis MATH. This requires the derivation of the five sequents MATH . The first two of these represent the fact that the measure of the evaluation derivation for MATH is a natural number that is smaller than the measure of the original evaluation derivation for MATH. By REF these are derivable in MATH from MATH. The third sequent is immediate, and the fourth also follows easily from REF . The derivation of the fifth sequent proceeds with another two applications of the MATH rule, corresponding to the analysis of the proof of MATH in the informal proof. This yields the sequent MATH . This is followed by applications of the MATH and MATH rules to give us MATH . The informal proof proceeds with a use of the cut rule, and here we use the derived object-level cut rule REF with the elided assumption MATH to obtain MATH . The first two of these follow easily from REF . The informal proof concludes by applying the induction hypothesis to the evaluation and typing judgments for MATH. Again we accomplish this by applying the appropriate left rules to the induction hypothesis MATH, which requires the derivation of the five sequents MATH . The first two sequents follow from REF , and the last three are all immediate.
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cs/0003062
|
The derivation of the unicity of typing is by complete induction on the height of the first typing derivation MATH. Let MATH be the predicate MATH and MATH the predicate MATH . These predicates encode the requirements that the list of assumptions contains only typing assignments for variables and assigns only one type to any one variable. Our induction predicate IP is then MATH . The details of the proof are presented in CITE.
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cs/0003076
|
We already noted in REF that all such computations are finite. Suppose that MATH. We consider now the following partial ordering MATH. The elements of MATH are the sequences MATH such that MATH for MATH, ordered componentwise with respect to the reversed subset ordering MATH. So MATH is the least element MATH in this ordering and MATH . We replace in each rule each premise atom MATH by MATH and MATH by MATH. Since for all MATH we have MATH iff MATH and MATH iff MATH, it follows that the applications of the original and of the resulting rules coincide. This allows us to confine our attention to the rule each premise of which is of the form MATH. Consider now a membership rule MATH associated with a constraint MATH from MATH defined on a set of variables MATH. We interpret this rule as a function on the just defined set D as follows. First, denote by MATH the extension ``by padding" of MATH to all the variables MATH, that is, MATH and MATH iff MATH. Next, given a constraint MATH and its variable MATH denote the set MATH by MATH. Finally, assume for simplicity that MATH is MATH. The function MATH that corresponds to the rule MATH is defined as follows: MATH . Denote the set of so defined functions by MATH. By definition each function MATH is inflationary and monotonic with respect to the componentwise reversed subset ordering MATH. Now, there is a one-one correspondence between the common fixpoints of the functions from MATH at which the iterations of MATH eventually stabilize and the outcomes of the computations by means of MATH starting at MATH. In this correspondence a common fixpoint MATH is related to the CSP MATH closed under the rules of MATH, where MATH are the constraints from MATH restricted to the domains MATH. The conclusion now follows by REF .
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cs/0003076
|
Suppose that MATH is closed under all minimal valid rules in MATH for MATH. Take a rule MATH from MATH that is valid for MATH. CASE: MATH is feasible for MATH. Then, because MATH is finite, MATH extends some minimal valid rule MATH in MATH for MATH. But MATH is closed under MATH, so it is closed under MATH, as well. CASE: MATH is not feasible for MATH. Then MATH is not feasible for MATH either since MATH. Consequently, since MATH is non-empty, MATH is closed under MATH.
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cs/0003076
|
Assume that the rule MATH can be applied to MATH, that is, that for all MATH we have MATH. Suppose now that the rule MATH to MATH does not maintain equivalence. Then for some MATH we have MATH. MATH is based on MATH, so MATH. By the validity of the rule for MATH we get MATH. This yields a contradiction.
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cs/0003076
|
First note that in the algorithm all possible feasible equality rules are considered and in the list L only the valid equality rules are retained. Additionally, a valid equality rule is retained only if it does not extend a rule already present in L. Finally, the equality rules are considered in the order according to which those that use less variables are considered first. By virtue of Note REF this implies that if a rule MATH extends a rule MATH, then MATH is considered first. As a consequence precisely all minimal valid equality rules are retained in L.
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cs/0003076
|
Assume that MATH is arc consistent. Choose a constraint MATH of MATH and consider an equality rule MATH that is valid for MATH, where MATH and MATH are as in REF . Suppose by contradiction that MATH is not closed under this rule. So for MATH and MATH the domain of each variable MATH in MATH equals MATH and moreover MATH, where MATH is the domain of the variable MATH in MATH. By the arc consistency of MATH there exists MATH such that MATH. Because of the form of the domains of the variables in MATH, also MATH holds. Additionally, because MATH is based on MATH, we have MATH. But by assumption the equality rule MATH is valid for MATH, so MATH. A contradiction.
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cs/0003076
|
REF implication is the contents of REF . To prove the reverse implication suppose that some constraint MATH of MATH is not arc consistent. We prove that then MATH is not rule consistent. The constraint MATH is on some variables MATH with respective domains MATH. For some MATH some MATH does not participate in any solution to MATH. Let MATH be the sequence of all domains in MATH that are singletons. Suppose that MATH for MATH and let MATH and MATH. Consider now the equality rule MATH and take MATH, where MATH and MATH are as in REF . For appropriate domains MATH of MATH we have MATH. Next, take some MATH such that MATH. We show that MATH. Since MATH it suffices to prove that MATH. For each variable MATH lying inside of MATH we have MATH. In turn, for each variable MATH lying outside of MATH its domain MATH has two elements, so, by the assumption on MATH, MATH is the same as the corresponding domain MATH of MATH and consequently MATH, since MATH. So indeed MATH and hence MATH by the choice of MATH. This proves validity of the equality rule MATH for MATH. But MATH is not closed under this rule since MATH, so MATH is not rule consistent.
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cs/0003076
|
CASE: This part of the proof is a simple modification of the proof of REF . Assume that MATH is arc consistent. Choose a constraint MATH of MATH and consider a membership rule MATH that is valid for MATH, where MATH and MATH are as in REF . Suppose by contradiction that MATH is not closed under this rule. So for MATH and MATH the domain of each variable MATH is included in MATH and moreover MATH, where MATH is the domain of the variable MATH. By the arc consistency of MATH there exists MATH such that MATH. Because of the form of the domains of the variables in MATH, also MATH for MATH holds. Additionally, because MATH is based on MATH we have MATH. But by assumption the rule MATH is valid for MATH, so MATH. A contradiction. CASE: This part of the proof is a modification of the proof of REF . Suppose that some constraint MATH of MATH is not arc consistent. We prove that then MATH is not membership rule consistent. The constraint MATH is on some variables MATH with respective domains MATH. For some MATH some MATH does not participate in any solution to MATH. Take MATH, where MATH and MATH are as in REF . For appropriate domains MATH of MATH we have MATH. Let MATH be the sequence of domains in MATH that are respectively different than MATH. Further, let MATH and MATH. Consider now the membership rule MATH. Take some MATH such that MATH for MATH. We show that MATH. Since MATH it suffices to prove that MATH. For each variable MATH lying inside of MATH we have MATH. In turn, for each variable MATH lying outside of MATH its domain MATH is the same as the corresponding domain MATH of MATH in MATH and consequently MATH, since MATH. So indeed MATH and hence MATH by the choice of MATH. This proves validity of the rule MATH for MATH. But MATH is not closed under this membership rule since MATH, so MATH is not membership rule consistent.
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cs/0003076
|
The proof is analogous to that of REF . We only need to check that the membership rules are considered in such an order that if a rule MATH extends a rule MATH, then MATH is considered first. This follows from directly from Note REF
|
hep-th/0003018
|
MATH due to invariance. Inserting this into REF gives the desired result.
|
hep-th/0003018
|
First observe that the mentioned property of MATH is automatically satisfied by MATH as well. Then use invariance in the form MATH and apply MATH.
|
hep-th/0003034
|
The only possible extra poles can appear when MATH does not define a local coordinate, that is, at the points MATH such that MATH. Assuming for simplicity that MATH has a simple pole at MATH, let MATH, and choose MATH to be a local coordinate near MATH. Then we have MATH which is holomorphic at MATH.
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hep-th/0003034
|
This proposition is proved by a direct calculation. We illustrate it by considering the following cases. First, let MATH and MATH. Then we have to prove the following identity: MATH . But MATH . Taking real parts and observing that MATH we obtain the desired identity. Let MATH, MATH. Then MATH is just the usual NAME bilinear identity for a canonical basis of real-normalized holomorphic differentials and it is proved in the regular way: MATH . The most difficult identities to establish correspond to MATH, MATH and the like, since MATH, MATH have simple poles at MATH. In this case, let MATH be a small circle around MATH, MATH, and let MATH be a cut from MATH to MATH. We consider the integral MATH along the contour MATH . First, we compute that MATH . Note that MATH since the only non-trivial imaginary contribution for periods of MATH comes from MATH-periods, and in order for MATH-contribution not to be canceled by MATH-contribution, we need MATH, in which case MATH . On the cut MATH we have MATH . To evaluate MATH, we rewrite everything in terms of a local coordinate MATH in the neighborhood of MATH: MATH . Then MATH . Collecting all of the above together, we obtain MATH which proves our identity. All other cases are somewhat intermediate in difficulty to the cases considered above and are proved along the same lines.
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hep-th/0003034
|
First we prove the following formula. For any two NAME differentials MATH, MATH we have MATH . To establish this identity, note that since MATH is holomorphic outside of MATH, the right hand side of our equation is a sum of residues at MATH and at poles of MATH. Using MATH as a coordinate, we see that MATH is holomorphic. So the only extra poles can appear at MATH such that MATH. Assuming that MATH is a simple zero of MATH and using MATH as a local coordinate near MATH, we have MATH . At the same time, MATH . Therefore, MATH and we have MATH . The rest of the proof is a direct calculation, which we illustrate by three different cases. MATH since MATH is holomorphic at MATH and MATH . MATH since MATH . MATH since MATH and MATH is well-defined on MATH.
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hep-th/0003034
|
Let us rewrite the formula for the third derivative with MATH: MATH . First let us see what is happening near the puncture MATH. Since MATH near MATH, MATH has a zero of order MATH at MATH. It is clear that for the truncated hierarchy the only differentials that can contribute to the MATH-term are MATH and MATH. Assume that near MATH, MATH . Then MATH . Now let us consider the boundary term MATH. For it to be non-zero, at least one of MATH should be MATH or MATH. For example, let MATH. Then, if MATH, MATH and if MATH or MATH, MATH . All the remaining cases are similar to the ones considered, and this concludes the proof of the theorem.
|
hep-th/0003046
|
The proof is a simple modification of the proof of the NAME maximum principle given in CITE. We omit the details.
|
hep-th/0003114
|
Let us look for the solution to REF in the form of the explicit power series expansion in the variables MATH. MATH . It turns out that it is sufficient to consider functions MATH which do not depend on the momenta MATH: MATH . Functions MATH can be considered as the coefficients of the tensor field on MATH that is symmetric with respect to all indices except the first one. In the zeroth and first order we respectively have MATH with MATH standing for antisymmetrisation in MATH. There is a particular solution to these equations: MATH . Taking MATH one can in fact consider more general solutions for MATH. In this case, the second equation of REF implies that MATH are the coefficients of a symmetric symplectic connection on MATH. This arbitrariness in the solution of REF can be absorbed by the redefinition of the symmetric symplectic connection entering the NAME bracket REF. The ambiguity in MATH might be able to reflect additional geometrical structures on MATH. As we will see below, standard NAME 's construction of the star-product on MATH corresponds to the ``minimal" solution REF. However we consider here a general solution to REF. Taking any fixed solution to REF for MATH one sees that in the r-th MATH order in MATH . REF implies MATH where the quantities MATH are given by MATH . Now relations REF are to be considered as the equations determining MATH. We need the following Let the quantity MATH be such that MATH and MATH then there exist MATH such that MATH . The statement is an obvious generalisation of the standard NAME Lemma. In the case where MATH, it is precisely the NAME Lemma. It follows from the lemma that REF has a solution iff MATH satisfies MATH . To show that it takes place let us introduce the partial sum MATH and consider expression MATH where MATH denote terms of order higher than MATH in MATH. Assume that REF hold for MATH. Excluding the contribution of order MATH from the NAME identity MATH one arrives at MATH . It follows from the lemma that for MATH REF considered as that on MATH admits a solution. The induction implies that REF also admits solution, at least locally. To show that solution exists globally we construct the particular solution to REF for MATH: MATH . This solution satisfies the condition MATH which does not depend on the choice of the local coordinates on MATH. Given a fixed first term MATH, REF can be considered as the condition on the solution to REF . It is easy to see that solution to REF is unique provided REF is imposed. Indeed, general solution to REF is given by MATH where MATH is the particular solution REF and MATH is an arbitrary function. One can check that REF implies that second term in REF vanishes. Choosing MATH to satisfy REF in each domain MATH one gets the global solution to REF .
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