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math/0002243 | 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 R... |
math/0002243 | 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/0002243 | 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 tha... |
math/0002243 | 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. |
math/0002243 | 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. |
math/0002243 | 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... |
math/0002243 | The proof is the same as the one given by CITE. |
math/0002243 | 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 MAT... |
math/0002251 | 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: MA... |
math/0002251 | 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. |
math/0002251 | 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 t... |
math/0002251 | 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 f... |
math/0002251 | 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, u... |
math/0002251 | 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 r... |
math/0002251 | CASE: Follows from REF Follows from REF Follows from REF and CITE. CASE: Follows from REF and the fact that MATH. |
math/0002251 | REF follow from REF follows from REF. |
math/0002251 | 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. |
math/0002251 | 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. CA... |
math/0002251 | 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 a... |
math/0002251 | 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 . |
math/0002251 | Clearly, MATH is a composition series for MATH if and only if MATH is a composition series for MATH. |
math/0002251 | 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... |
math/0002251 | 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 isomo... |
math/0002253 | 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 inject... |
math/0002253 | 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 M... |
math/0002253 | 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... |
math/0002253 | 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 suitab... |
math/0002253 | 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... |
math/0002253 | 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 f... |
math/0002253 | 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 a... |
math/0002253 | 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 prim... |
math/0002253 | 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 n... |
math/0002253 | 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. |
math/0002253 | 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/0002253 | 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 ... |
quant-ph/0002057 | 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 a... |
quant-ph/0002057 | 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 . |
quant-ph/0002057 | 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 ... |
quant-ph/0002057 | 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 b... |
quant-ph/0002057 | 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. |
quant-ph/0002057 | 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... |
quant-ph/0002057 | 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 l... |
quant-ph/0002057 | 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 amp... |
quant-ph/0002057 | 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 g... |
cond-mat/0003123 | 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 trian... |
cs/0003009 | 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... |
cs/0003009 | 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 . |
cs/0003035 | MATH implies that an ordering MATH is compatible with MATH whenever it is compatible with MATH. We thus have MATH and therefore MATH. |
cs/0003035 | 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 ... |
cs/0003035 | 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 stric... |
cs/0003045 | 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 consecuti... |
cs/0003045 | 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 an... |
cs/0003045 | 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. |
cs/0003045 | 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. |
cs/0003045 | The first statement is trivial by definition. For the second statement, this is a corollary of the following REF with MATH and MATH. |
cs/0003045 | 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 pa... |
cs/0003045 | 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 ... |
cs/0003045 | 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 q... |
cs/0003045 | 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 b... |
cs/0003045 | 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 ... |
cs/0003045 | 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. Supp... |
cs/0003045 | 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... |
cs/0003045 | This is a direct corollary of REF (every recursive predicate in MATH is defined in MATH and hence tabled). |
cs/0003045 | Because MATH extends MATH and MATH extends MATH, MATH for MATH. The proposition follows then by definition of quasi-termination. |
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 . |
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 . |
cs/0003045 | This is a direct corollary of REF (every recursive predicate in MATH is defined in MATH and hence tabled). |
cs/0003045 | Because MATH extends MATH and MATH extends MATH, MATH, for MATH. The proposition follows then by definition of NAME. |
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 se... |
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 ca... |
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 MAT... |
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 MAT... |
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. |
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... |
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. |
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... |
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 t... |
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... |
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 s... |
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 allo... |
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 closel... |
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 d... |
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 se... |
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 ... |
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 o... |
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. CA... |
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. |
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 ... |
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 MAT... |
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 solut... |
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 a... |
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. |
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 bi... |
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 holomorph... |
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... |
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 symm... |
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