paper stringlengths 9 16 | proof stringlengths 0 131k |
|---|---|
math/9904164 | MATH-linearity of MATH follows straightforwardly from REF. Identifying MATH, quasi-coassociativity of MATH is equivalent to the identity MATH which follows from REF. |
math/9904164 | REF is equivalent to MATH . Thus, noting that in MATH the roles of MATH and MATH interchange, REF reduce to REF, respectively, in MATH. To prove REF we compute, using REF and denoting MATH . To prove REF we compute MATH . Now REF follows immediately from REF from REF. |
math/9904164 | The first statement follows from MATH and the fact that by REF MATH provides an isomorphism of quasi-Hopf MATH-bimodules. For the second statement see the remarks above. |
math/9904164 | By REF MATH, which is the space of right integrals MATH. Thus MATH and MATH induce the splittings MATH and MATH, respectively. |
math/9904164 | By the defining relation of MATH, MATH we conclude from REF MATH . This implies MATH and therefore MATH. |
math/9904164 | Clearly MATH is a right integral and for any MATH we have MATH, from which the statement follows by the nondegeneracy of the pairing MATH. |
math/9904164 | CASE: By REF MATH is an isomorphism of quasi-Hopf MATH-bimodules, implying MATH to be a NAME transformation. CASE: Holds by REF . CASE: Pick MATH and put MATH. Since MATH and MATH are right MATH-module maps, we get MATH. Moreover, REF implies MATH, and in particular MATH. Applying MATH gives MATH and therefore MATH. Th... |
math/9904164 | Since MATH, according to REF the inverse MATH is given by MATH . Here we have used MATH and REF applied to MATH. Hence we conclude MATH, and therefore MATH satifies MATH. |
math/9904164 | CASE: If there is a nonzero MATH-invariant integral in MATH, then it is two-sided, hence all integrals are two-sided and MATH-invariant (by the one-dimensionality of the space of left/right integrals), implying MATH. Conversely, if MATH is unimodular then all integrals are two-sided and by REF the modular automorphism ... |
math/9904164 | REF : see REF. CASE: Let MATH be a normalized right integral, then MATH. Moreover, to verify the above formula for MATH, we use REF and MATH to obtain MATH and therefore indeed MATH. Here we have used REF and the identity MATH. CASE: If MATH there exists a unique right integral MATH such that MATH yielding MATH. Thus M... |
math/9904164 | Using REF we have for all MATH and therefore MATH where we have used that according to REF the modular automorphism of MATH is given by MATH. We are left to show that MATH is a left inverse of MATH, implying by the above calculation MATH to be a right inverse and therefore MATH. First note that the second line of the a... |
math/9904164 | We first show the identities MATH . Noting that in MATH the roles of MATH and MATH as well as of MATH and MATH and of MATH and MATH interchange reduces REF to REF. For the left hand side of REF we obtain, using REF MATH where MATH denotes the map MATH and MATH is given by MATH . Now we use the pentagon equation for the... |
math/9904164 | For MATH we have by REF MATH and from REF we conclude for MATH and MATH . Thus, MATH is biinvariant and MATH independently of the choice of MATH and MATH. Conversely, let MATH be biinvariant, then for all choices MATH and MATH by the identities REF - REF. Hence, MATH is a MATH-bimodule map for all choices of MATH and M... |
math/9904164 | Pick MATH and put MATH. Then MATH is a nonzero morphism of quasi-Hopf MATH-bimodules, whence MATH for all MATH, where MATH. Moreover, MATH. Applying MATH gives MATH and therefore MATH. |
math/9904164 | We show that for any two MATH-modules MATH, where MATH is a submodule, there exists a MATH-linear surjection MATH. Denoting the canonical embedding MATH and MATH and MATH it therefore suffices to find a surjection MATH satisfying MATH where at the l.h.s. and right-hand side of REF we have used the shortcut notation MAT... |
math/9904168 | Since MATH, so MATH for some MATH. Now MATH, so MATH determines a class in MATH. Since MATH, there exist MATH, such that MATH. This completes the proof. |
math/9904168 | The second statement is trivial. The first can be proved by a standard argument modeled on REF and CITE. Rewrite REF as a sequence of REF By REF , we can take MATH to represent MATH. Suppose now we have found MATH with MATH for MATH. By REF , MATH. Also we have the following standard calculation MATH . Hence MATH, and ... |
math/9904168 | We need to show MATH . We prove the second inclusion first. Assume that MATH, MATH. We get a sequence of REF where MATH. Now MATH, by REF , MATH, where MATH. Hence MATH . And from MATH we get MATH. Hence we have MATH . By induction, we find that MATH in other word, MATH, where MATH. Now assume MATH and MATH, where MATH... |
math/9904168 | When expanded into the formal power series in MATH, we can rewrite MATH as a sequence of REF To find MATH, notice that MATH that is, MATH, hence MATH can be found. Suppose now that we have found MATH, we have MATH . That is, MATH, hence we can find MATH. By REF , MATH, since MATH and MATH represents the same class in M... |
math/9904168 | We break the proof into two steps. CASE: MATH is injective. By REF , we can represent any class of MATH by an element MATH. Without loss of generality, we assume that MATH and hence MATH. In fact, if MATH for some MATH, we replace MATH by MATH and consider the MATH-th term. Now if MATH for some MATH, then by noticing t... |
math/9904168 | The first equality can be proved elementarily by induction. For the second equality, we have MATH . |
math/9904168 | It is easy to see that MATH . Then by induction, it is easy to show that MATH . Now MATH, so we can use REF to handle MATH as follows: MATH . Hence we have MATH . Replacing MATH by MATH then completes the proof. |
math/9904168 | It suffices to prove the following: if MATH, MATH, then there exist MATH, such that MATH. For MATH, we clear have MATH. Since MATH, there exists MATH such that MATH. Now modulo MATH, we have MATH . For MATH, we have MATH, and MATH hence there exists MATH such that MATH. Now modulo MATH, we have MATH . |
math/9904168 | Given any MATH, represent it by an element MATH and extend it to a power series MATH, such that MATH and MATH. Then MATH satisfies MATH, MATH and MATH. |
math/9904168 | The proof of the first statement is an easy modification of the proof of REF . To prove the second statement, we first linearize the gauge transformation: MATH . Then we have MATH where the subscript means the the first order term. Since MATH is odd, we have MATH . Notice that MATH . Now MATH hence MATH, and so MATH . ... |
math/9904168 | Just observe that a universal normalized solution is mapped to a universal normalized solution under quasi-isomorphism. |
math/9904174 | Since the automorphism group MATH of MATH acts transitively on the set of pure states of MATH, CITE, there exists an increasing sequence MATH of finite type I subfactors of MATH such that MATH and MATH is pure for every MATH. Then we can find sequences MATH and MATH of unitaries in MATH and increasing sequences MATH an... |
math/9904174 | It follows from the previous lemma that there exist pure states MATH, increasing sequences MATH of finite type I subfactors of MATH, and an increasing sequence MATH in MATH such that MATH . By passing to subsequences of MATH and MATH and setting MATH if MATH is odd and MATH if MATH is even, we may assume that MATH . Th... |
math/9904174 | CASE: Since MATH is pure, and gauge-invariant, it follows that MATH is pure. Let MATH be the support projection of MATH in MATH. Since MATH is minimal, and MATH is gauge-invariant, it follows that for any MATH and any multi-index MATH with MATH, MATH where MATH. Thus we obtain that MATH which implies that MATH is disjo... |
math/9904174 | By REF , MATH and MATH are pure states on MATH. Applying REF on MATH in lieu of MATH, with MATH, we obtain pure states MATH and MATH of MATH with the properties given there. Since MATH is equivalent to MATH, MATH is a pure state of MATH by REF and this state is equivalent to MATH. By NAME 's transitivity theorem we hav... |
math/9904174 | If MATH is disjoint from MATH, then it follows that MATH, MATH, MATH are mutually disjoint (by REF ); thus the assertion follows from REF . If MATH is equivalent to MATH, there is a unitary MATH such that MATH (by NAME 's transitivity). |
math/9904174 | NAME proved REF and the other statements are more or less known. We shall give a proof of REF . We again denote by MATH the canonical endomorphism of MATH, MATH. Since the unitary corresponding to MATH is MATH, it suffices to show that MATH, MATH, is dense in MATH. If MATH denotes the C*-subalgebra generated by MATH, t... |
math/9904174 | It suffices to show that if MATH is a pure state there is a unitary MATH such that MATH . Since MATH has real rank zero, there is a decreasing sequence MATH of projections in MATH such that MATH is the unique state satisfying MATH for MATH, that is, MATH converges to the support projection of MATH in MATH. We may furth... |
math/9904177 | The parts at the largest eigenvalue will multiply, be positive, and will swamp all the others, since they grow at an exponential rate corresponding to this eigenvalue. More precisely, we can conjugate and then write MATH corresponding to the eigenspace for the maximal eigenvalue MATH, and the eigenspaces for all other ... |
math/9904177 | The MATH-equivalence is the existence of an infinite sequence MATH of nonnegative matrices, and suitable powers, such that we have MATH, MATH. Let MATH. We can solve recursively MATH and so on. It follows that if for any MATH there exists MATH such that MATH are nonnegative integer matrices, the results hold. Nonnegati... |
math/9904177 | The parts at the largest eigenvalue will multiply, be positive, and will swamp all the others, since they grow at an exponential rate corresponding to this eigenvalue. More precisely, we can conjugate and then write MATH corresponding to the eigenspace for the maximal eigenvalue MATH, and the eigenspaces for all other ... |
math/9904177 | We write out the equations for a general MATH which is assumed to be a nonnegative unit MATH . The nilpotence modulo MATH guarantees that some powers of MATH are divisible by MATH, hence any sufficiently large power are divisible by the determinant of a given power of the other matrix, so that if the powers increase su... |
math/9904177 | We check that the condition in REF holds with MATH and that the matrices have eigenvalues respectively MATH; MATH. At the negative eigenvalues both column eigenvectors are MATH so the identity maps one to the other. One can alternatively check by the recipe REF - REF that the two matrices define the same dimension grou... |
math/9904177 | If REF does not hold true, then MATH will map vectors from MATH into the maximal column eigenspace of MATH nontrivially. At the largest eigenvalue where this occurs, these terms will become dominant in MATH and give the asymptotic value of the entire matrix. This will make the limit of MATH as MATH in any way, a limit ... |
math/9904177 | If MATH gives an isomorphism then for arbitrary large powers of MATH, the matrices MATH are integral. Modulo any fixed power of MATH, we can arrange by increasing these powers and altering MATH to other integral matrices that the powers are in each case those giving rise to the idempotent limits MATH. Therefore modulo ... |
math/9904177 | These matrices come from REF and CITE, main theorem. It can be checked they are unimodular and are units in the field generated by a root of the characteristic polynomial of MATH. By diagonalizing the field, it follows that multiplication by MATH sends MATH to MATH in the notation of REF , so that the isomorphism condi... |
math/9904177 | First we argue for the necessity. Let the maximal and nonmaximal eigenspaces of MATH be MATH as in REF . The matrix MATH must map MATH to MATH nontrivially, and send MATH to MATH nontrivially by REF . Since MATH is rational, it commutes with NAME actions among the different conjugates of the maximal eigenvalue. This al... |
physics/9904035 | Suppose that MATH (the proof for MATH is identical). Then linearizing REF in the neighbourhood of MATH gives MATH where MATH, MATH and MATH. If this equilibrium point is hyperbolic, then the dimension of its stable (unstable) manifold are given by the numbers of eigenvalues, MATH, satisfying MATH with positive (negativ... |
quant-ph/9904079 | Suppose we randomly sample MATH bits of the input. Let MATH denote the fraction of REF in the input and MATH the fraction of REF in the sample. The NAME bound (see for example, CITE) implies that there is a constant MATH such that MATH . Now consider the following randomized algorithm for MATH: CASE: Let MATH. CASE: Sa... |
quant-ph/9904079 | It can be shown by a small modification of CITE that with probability at least MATH (MATH), there are at least MATH values MATH such that MATH for exactly one MATH (and hence MATH). We assume that this is the case in the following. If MATH generate a proper subspace of MATH, then there is a non-zero MATH that is orthog... |
quant-ph/9904079 | Without loss of generality we assume MATH has error probability MATH. To distinguish MATH and MATH, we run MATH until it stops or makes MATH queries. If it stops, we output the result of MATH. If it makes MATH queries and has not stopped yet, we output REF. Under MATH, the probability that MATH outputs REF is at most M... |
quant-ph/9904079 | For a random input from MATH, the probability that all answers to MATH queries are different is MATH . For a random input from MATH, the probability that there is a MATH such that MATH queries both MATH and MATH (MATH is the hidden vector) is MATH, since: CASE: for every pair of distinct MATH, the probability that MATH... |
quant-ph/9904079 | Any classical algorithm for OR requires MATH queries on an input MATH. The upper bound follows from random sampling, the lower bound from a block-sensitivity argument CITE. Hence (omitting the intermediate MATH-s): MATH where the last step can be shown by approximating the sum over MATH with an integral. Similarly, for... |
quant-ph/9904079 | By NAME 's inequality, if MATH is concave then MATH, hence MATH . |
quant-ph/9904079 | For all MATH, define MATH. The probability under the uniform distribution of getting an input MATH is MATH, since the number of inputs MATH with MATH REFs is MATH for all MATH. The idea of our algorithm is to have MATH runs of the quantum counting algorithm, with increasing numbers of queries, such that the majority va... |
quant-ph/9904079 | Let MATH be a bounded-error quantum algorithm for MAJORITY. It follows from the worst-case results of CITE that MATH uses MATH queries on the hardest inputs, which are the MATH with MATH. Since the uniform distribution puts MATH probability on the set of such MATH, the average-case complexity of MATH is at least MATH. |
quant-ph/9904079 | We will prove the lemma for MATH, which is the hardest case. We assume without loss of generality that the algorithm queries its input MATH at MATH random positions, and outputs REF if the fraction of REF in its sample is at least MATH. We do not care what the algorithm outputs otherwise. Consider an input MATH with MA... |
quant-ph/9904079 | The previous lemma shows that any algorithm for MAJORITY needs MATH queries on inputs MATH with MATH. Since the uniform distribution puts MATH probability on the set of such MATH, the theorem follows. |
quant-ph/9904079 | Let MATH be a bounded-error quantum algorithm for PARITY. Let MATH be an algorithm that flips each bit of its input MATH with probability MATH, records the number MATH of actual bitflips, runs MATH on the changed input MATH, and outputs MATH. It is easy to see that MATH is a bounded-error algorithm for PARITY and that ... |
quant-ph/9904093 | Consider MATH to be an encoding of the bit MATH. If MATH is an unbiased boolean random variable, then MATH represents the encoding of MATH. Let MATH be the outcome of the measurement of this encoding according to MATH. By the hypothesis of the lemma, MATH. It is easy to see from the concavity of the entropy function th... |
quant-ph/9904093 | Let MATH be the unitary operator of MATH corresponding to the symbol MATH. Let MATH be the span of the accepting basis states of MATH and let MATH be the subspace orthogonal to it. Define the measurement MATH as applying the transformation MATH (recall that `' is the right end-marker) and then measuring with respect to... |
quant-ph/9904093 | The proof is by downward induction on MATH. The base case MATH is satisfied easily: MATH for all MATH-bit strings MATH. Suppose the claim is true for MATH. We have MATH . By hypothesis, MATH for MATH. Moreover, since the two density matrices are mixtures arising from strings that differ in the MATH-th bit, the measurem... |
quant-ph/9904093 | The probability of correct decoding is equal to MATH . Now, MATH. So MATH. Taking expectation over MATH, and noting that MATH is a convex function, we have MATH which gives us the claimed lower bound on the decoding probability. |
quant-ph/9904093 | Let MATH be the subspace spanned by the codewords MATH, and let MATH be the projection onto MATH. Since the codes are over MATH qubits, MATH has dimension at most MATH. Let MATH be an orthonormal basis for MATH. Let MATH be an orthonormal basis for the range of MATH. The union of all these bases MATH is an orthonormal ... |
quant-ph/9904107 | MATH . |
quant-ph/9904107 | ( MATH.) MATH . Where MATH. Now let's bound MATH by MATH, and we think of MATH, with the standard interpretation of MATH as MATH in the original MATH, and MATH as MATH. MATH . Note that MATH, therefore, MATH . |
quant-ph/9904107 | MATH . |
quant-ph/9904107 | MATH . |
cond-mat/9905213 | We first observe that all excitations satisfy MATH, that is, MATH, so that REF reads MATH . Under the condition, MATH, MATH, the cluster expansion converges and moreover MATH where for a cluster MATH we use MATH to denote its length. To prove REF, we write for a cluster MATH of length at least m, MATH. Then we use that... |
cond-mat/9905213 | CASE: The lower bounds REF on the energy can be improved for some excitations. Let MATH denote the infinite cylinder between the vertical lines MATH and MATH. For the excitations included in the strips MATH and MATH, one has MATH, and thus by arguing as in the proof of REF MATH . The associated clusters satisfy thus: M... |
cs/9905010 | MATH . |
cs/9905010 | MATH . The equality MATH holds iff MATH is a fixed point of MATH, that is, MATH with MATH. Furthermore, MATH is a fixed point of MATH iff MATH, MATH, MATH, MATH, by REF MATH . |
cs/9905010 | Let MATH be a subsequence of MATH converging to MATH. Then for all MATH: MATH and in the limit as MATH, for continuous MATH and MATH: MATH. Thus MATH is a maximum of MATH, using REF , and MATH is a fixed point of MATH. Furthermore, MATH, using REF , and MATH is a critical point of MATH. |
cs/9905016 | The point of the proposition is that a program with a good static evaluator is always going to have shortcomings: it will always evaluate incorrectly at least some positions. If one programs another static evaluator that evaluates correctly these positions, one will notice soon that there are others that the new progra... |
math-ph/9905007 | For MATH, the maps MATH, MATH and MATH are in MATH. Let MATH be a fixed constant. We consider the following map: MATH given by: MATH . It is not difficult to prove that MATH is a smooth map and that, for MATH and MATH, the NAME derivative MATH is given by: MATH . Here we have used the fact that MATH is Killing, thus MA... |
math-ph/9905007 | Since MATH is a smooth submanifold of MATH, then the restriction of the action functional MATH to MATH is smooth. For MATH, we have: MATH hence MATH is also smooth. Equality REF follows easily by differentiating the expression MATH and using the equality MATH. |
math-ph/9905007 | Convergence in the space MATH clearly implies the convergence in MATH, which implies that the inclusion MATH is continuous. For the second part of the thesis, it suffices to adapt the proof of REF , and the details will be omitted. |
math-ph/9905007 | For REF , it is MATH. By REF , it is MATH and by REF MATH. REF is simply the fact that elements in the dual space of MATH are characterized by the property of vanishing on constant functions. For REF , it suffices to observe that the map MATH is surjective from MATH to MATH. For REF , using REF , we have: MATH . On the... |
math-ph/9905007 | Under the hypotheses, it is MATH for all MATH such that MATH and such that MATH is a multiple of the covector MATH. The subspace MATH of such MATH's has codimension equal to MATH in MATH. Then, the annihilator MATH of MATH in MATH has dimension MATH. The subspace MATH of MATH consisting of elements MATH of the form MAT... |
math-ph/9905007 | First of all, observe that there exists no MATH such that MATH. Namely, if MATH and MATH, then it would be MATH, and so MATH. On the other hand, for all constants MATH, it would be MATH, and MATH, which is a contradiction. It follows that there exists no MATH such that MATH with MATH. Indeed, if such MATH existed, then... |
math-ph/9905007 | If MATH, then MATH for all smooth function MATH with support contained in MATH. This implies MATH. The proof of the second part of the Lemma is analogous. |
math-ph/9905007 | From REF , we have MATH for all MATH such that MATH and MATH is parallel to MATH. From REF it follows that MATH is a linear combination of delta's, and in particular, MATH is in MATH. Hence, MATH is in MATH, and, by REF , MATH. Since MATH, then MATH is in MATH; computing explicitly, we have: MATH hence MATH. Then, from... |
math-ph/9905007 | From REF , all we need to prove is that any smooth curve MATH that satisfies the differential REF and whose initial velocity MATH satisfies REF is in MATH. To this aim, it suffices to show that the functions MATH and MATH are constant. If we multiply REF by MATH, we obtain: MATH that can be written as: MATH with MATH .... |
math-ph/9905007 | The critical points of MATH in MATH are precisely the geodesics in MATH with respect to MATH that join MATH and MATH and that are orthogonal to MATH, that is, MATH. Since MATH is MATH-invariant, then MATH is Killing in the metric MATH, thus, for every such geodesic MATH, the quantity MATH is constant. Hence MATH and MA... |
math-ph/9905007 | The smooth dependence on MATH of the solution MATH of REF proves that MATH is a smooth map. REF are easily obtained by differentiating REF using REF , and keeping in mind that MATH. In particular, REF follows immediately from REF . |
math-ph/9905007 | The equivalence of REF follows from the fact that the brachistochrones of energy MATH between MATH and MATH and the local minimizers for the travel time are characterized by the same differential equation (see REF and Ref. CITE). The equivalence of REF is based on the fact that, for MATH and MATH, using REF , one compu... |
math-ph/9905007 | Since both sides of REF are symmetric, it suffices to prove the equality in the case MATH. Let MATH, MATH be a smooth curve in MATH such that MATH and MATH. Then, clearly, MATH is a smooth curve in MATH such that MATH and MATH. Using REF , we have: MATH which concludes the proof. |
math-ph/9905007 | The geodesic equation for the metric MATH is easily computed as the NAME - NAME equation for the functional MATH, and it is given by: MATH . In analogy with the proof of REF , let MATH be a fixed vector field in MATH and let MATH denote a variation of MATH in MATH such that MATH. Then, we compute as follows: MATH . Usi... |
math-ph/9905007 | The condition that the Killing vector field MATH is never vanishing is needed to prove that the space MATH is a smooth submanifold of MATH (see for instance Ref. CITE). We start proving the second equality in REF ; we denote by MATH and MATH respectively the covariant derivative and the curvature tensor of the NAME - N... |
math-ph/9905007 | REF is obtained by patiently linearizing the brachistochrone differential REF , using the following dictionary: CASE: MATH; CASE: MATH; CASE: MATH; CASE: MATH; CASE: MATH; CASE: MATH; CASE: MATH. The formulas above are obtained by considering the basic properties of the NAME - NAME connection and the curvature tensor o... |
math-ph/9905007 | We use a sort of brachistochrone exponential map, as follows. Given a vector MATH such that MATH then there exists a unique brachistochrone MATH and such that MATH. This is obtained by solving the differential REF with initial conditions MATH and MATH. Moreover, the map MATH is MATH, due to the regular dependence on th... |
math-ph/9905007 | Following the proof of REF , MATH is the variational vector field corresponding to a variation MATH of MATH. |
math-ph/9905007 | By REF , MATH; the first part of the statement follows immediately by observing that a vector field MATH belongs to MATH if and only if MATH is parallel to MATH (see REF ). To prove the second part of the thesis, we need to show that MATH for all MATH. By REF , we have: MATH hence, to conclude the proof it suffices to ... |
math-ph/9905007 | Let MATH and MATH be a smooth variation of MATH consisting of horizontal geodesics and such that MATH. Let MATH; since MATH is smooth, then MATH is a smooth variation of MATH. Moreover, MATH, and since MATH is a horizontal geodesic, by REF , MATH is a brachistochrone in MATH for all MATH. By REF , MATH is a b-Jacobi fi... |
math-ph/9905007 | It suffices to prove that MATH has a left inverse, that is, that there exists a linear bounded operator MATH such that MATH is the identity on MATH. Such a map MATH is given by the differential of the map MATH defined by REF . Indeed, by REF , MATH is the identity on MATH,and by differentiating we have that MATH is the... |
math-ph/9905007 | We first show that MATH. To this end, let MATH be fixed; by REF , it satisfies: MATH . Since MATH, then clearly MATH. Moreover, let MATH. For the inclusion MATH we need to show that REF is satisfied. Using REF , we compute easily: MATH . For the opposite inclusion, we argue as follows. Let MATH be fixed in MATH and let... |
math-ph/9905007 | The proofs of REF can be repeated verbatim, by replacing the initial point MATH with the point MATH. The only technical subtlety to worry about is that, when replacing the initial point, it will not hold, in general, that MATH. Nevertheless, this fact is not essential, because one can always reduce to this case by cons... |
math-ph/9905007 | REF proves that any b-Jacobi field along MATH is in the kernel of MATH. Conversely, let MATH be a fixed brachistochrone and MATH. From REF , it suffices to prove that the vector field MATH is a MATH-Jacobi field with respect to the Riemannian metric MATH along the geodesic MATH. Moreover, since MATH, from REF it suffic... |
math-ph/9905007 | By REF , since isomorphisms preserve dimensions, for all MATH we have: MATH . This implies that MATH is a b-focal point along MATH if and only if MATH is a MATH-focal point; moreover, summing over all MATH, we obtain REF . From REF , we have: MATH from REF we obtain: MATH finally, from REF we have the equality: MATH . ... |
math-ph/9905007 | By REF we see that MATH if and only if MATH, where MATH is the deformation map of REF and MATH is a travel time brachistochrone of energy MATH between MATH and MATH. By the first part of REF implies that every MATH is a nondegenerate critical point of MATH in MATH. Moreover, by REF , we have MATH, while, by REF , it is... |
math-ph/9905007 | Let MATH and MATH be a smooth curve such that MATH and MATH. By REF , we have: MATH since MATH, then MATH which concludes the proof. |
math-ph/9905007 | The computation is done by brute force, as follows. Let MATH be a fixed brachistochrone. Observe that MATH is smooth. Hence, by a density argument, it suffices to restrict our attention to smooth variations. Let MATH be a fixed smooth variational vector field and let MATH, MATH, be a smooth variation of MATH in MATH co... |
math/9905001 | For any MATH let MATH be the unique point that is proximate to MATH and to MATH, and consider MATH and MATH. By performing successive unloading steps on MATH we see that MATH is equivalent to MATH and since MATH results from MATH by adding two simple points, MATH . The hypothesis on MATH implies that the inequality is ... |
math/9905001 | The total transform of MATH in MATH is MATH where MATH is the strict transform. Let MATH be the equation of a second germ MATH. If MATH the total transform of MATH is MATH with MATH effective, so clearly MATH. Conversely, if MATH, then the total transform of MATH is MATH with MATH effective, so the total transform of M... |
math/9905001 | Let MATH. It will be enough to see that there are open neighbourhoods MATH of MATH and MATH containing both MATH and MATH, and isomorphisms MATH commuting with the blowing-ups, such that MATH and MATH. This is equivalent to prove that there are isomorphic unibranched germs of curve MATH and MATH going sharphly through ... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.