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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. The implication MATH follows similarly by considering MATH and noting MATH. |
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 of any nonzero MATH is given by MATH. Thus by REF the NAME basis MATH satisfies MATH and therefore MATH. But with MATH and MATH (since MATH is a two-sided integral and MATH) REF implies MATH. CASE: If MATH is symmetric then it is unimodular and all modular automorphisms are inner. Hence MATH is inner. Conversely, if MATH is unimodular and MATH is inner pick MATH and MATH invertible such that MATH. Then MATH is a nondegenerate trace on MATH. |
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 MATH is normalized. Moreover, using again the identity MATH one easily verifies that MATH, that is, the modular automorphism of MATH is given by MATH. Hence, by REF , MATH is unimodular and MATH is the NAME integral in MATH. CASE: Following CITE, if MATH is a normalized left integral then MATH provides a separating idempotent in MATH, that is, MATH and MATH for all MATH. CASE: This is a standard textbook exercise, see for example, CITE. |
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 above calculation also gives MATH for all MATH. Hence MATH where we have used that by REF MATH and that MATH provides the NAME basis of MATH. |
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 last three factors of MATH to obtain MATH . Using the antipode REF , under the evaluation of MATH the third factor may be dropped and the first factor may be replaced by MATH. Hence we get MATH where we have used REF. Thus we finally arrive at MATH . By REF we have proved REF. Using REF and the definition of MATH given in REF, one obtains MATH . This finishes the proof of REF . |
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 MATH: MATH . Finally, if MATH is biinvariant we may use the identities REF to conclude MATH for all choices of MATH and MATH. Hence, MATH for all these choices and therefore MATH is independent of these choices. This proves the first statement of REF . To prove the second statement we compute MATH where in the last line we have used REF . On the other hand MATH . Thus, MATH is cocentral if and only if for all MATH . Clearly, REF holds, if MATH is a MATH-bimodule and a MATH-comodule map. Conversely, let MATH be a MATH-bimodule map satisfying REF. Then by REF MATH . Hence MATH is a MATH-comodule map. This proves the second statement of REF . |
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 MATH and MATH, respectively. This notation will also be used frequently below. We proceed as follows. Viewing MATH and MATH as MATH-modules, the semisimplicity of MATH implies the existence of a MATH-linear surjection MATH, satisfying REF. Denoting MATH, we now define the map MATH in terms of MATH by MATH where MATH is the normalized biinvariant cocentral form associated with MATH. To show REF note that MATH, since by assumption the embedding MATH is MATH-linear. Using MATH this implies MATH and therefore MATH where we have used REF, then the normalization condition MATH and finally the identity MATH. Thus we have proven REF. MATH-linearity REF follows from REF and biinvariance of MATH, since one computes MATH . We are left to show REF. For the l.h.s. we get MATH . Here we have used REF and the identity MATH - following from REF - and the biinvariance of MATH. A similar calculation yields for the right-hand side of REF MATH . Comparing REF, both expressions coincide since MATH is cocentral. This proves REF and therefore concludes the proof of REF . |
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 so there exist MATH, such that MATH . |
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 and MATH are elments of MATH. Equivalently, we have a sequence of REF and MATH. Now MATH, MATH for some MATH. Hence MATH . From MATH we get MATH. Hence MATH . By induction, we can show that MATH where each MATH lies in MATH. Consequently, MATH . Define MATH as follows. Set MATH. For MATH, assume that MATH have been defined such that MATH for MATH. Then it is straightforward to see that MATH, so MATH for some MATH. Finally, we have 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 MATH, we have MATH for some MATH. Now as above, we solve MATH by induction: first expand in power series to get a sequence of REF then inductively check the right hand side of each equation lies in MATH as above, hence one can find a solution MATH in MATH. |
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 that MATH has leading term MATH, we get a sequence of REF . Now MATH, and MATH . Therefore MATH, and so MATH, a contradiction to the assumption that MATH. CASE: MATH is surjective. By REF , any element of MATH can be represented by an element MATH. Then we have MATH, hence it can be extended to an element MATH. Consider now MATH, it can be written as MATH, where MATH. Hence by induction, MATH lies in the image of MATH. |
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 . Replacing MATH by MATH, we get MATH . Similarly, we have MATH and so MATH . Therefore, we have MATH . Here we use the observation that since both MATH and MATH has no zeroth order term, MATH has no first order term. |
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 and MATH in MATH such that MATH where MATH. (Let MATH. Then we choose MATH and MATH such that MATH and MATH. Further we choose MATH and MATH such that MATH, MATH, and, MATH. We just repeat this process.) Then the limit MATH exists and is a unitary such that MATH and MATH . Let MATH. Then MATH is a pure state with MATH and MATH is a pure state for every MATH. Put MATH. |
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 . Then MATH has a tensor product decomposition into pure states on the matrix subalgebras MATH, and MATH likewise on the subalgebras MATH. Thus we can define a pure state MATH by requiring that it decomposes under the tensor product decomposition MATH into states given by: MATH . |
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 disjoint from MATH. CASE: Let MATH be the support projection of MATH in MATH. It suffices to show that for any multi-indices MATH since the linear span of MATH is dense in MATH. If MATH, we have that MATH by using the fact that MATH is disjoint from MATH for MATH. If MATH, we have that MATH since MATH is pure. |
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 have a unitary MATH such that MATH; it follows that MATH. It is not automatical that MATH satisfies the condition that all MATH, MATH are mutually disjoint and are disjoint from MATH. But using the freedom in constructing MATH and MATH successively, we can certainly impose this condition. Thus we obtain three pure states MATH of MATH such that all MATH, MATH are mutually disjoint and MATH and MATH are spotwise asymptotically equal as specified in REF . It now suffices to prove the lemma for the pairs MATH and MATH. Thus replacing MATH by one of these pairs, we may assume the lemma satisfy the additional condition that there exists an increasing sequence MATH in MATH and an increasing sequence MATH of finite type I subfactors of MATH such that MATH . We shall construct a sequence MATH of unitaries in MATH such that MATH defines an automorphism of MATH with MATH. To ensure the existence of the limit we choose the unitaries such that they mutually commute and MATH. Since MATH commutes with the gauge action MATH, this will complete the proof. We fix a large MATH. We choose MATH so large that the support projections MATH are almost orthogonal and MATH. Let MATH be a partial isometry in MATH with MATH, MATH. By the polar decomposition of the approximate unitary MATH we obtain a unitary MATH such that MATH and MATH. We next choose MATH so large that MATH are almost orthogonal and MATH. (Though MATH is an endomorphism, MATH on MATH is well defined for MATH.) Let MATH be a partial isometry in MATH such that MATH and MATH and let MATH be a partial isometry in MATH such that MATH and MATH. Assume for the moment that MATH, MATH; MATH are all orthogonal and set MATH for MATH. Then MATH is a family of matrix units such that MATH when MATH. Let MATH as in CITE. Then MATH is a projection in MATH and satisfies MATH . Let MATH and MATH where MATH. By the orthogonality assumption on MATH, MATH is a unitary in MATH and satisfies MATH . Note also that MATH commutes with MATH and MATH. Now, the projections MATH, MATH, MATH are not actually orthogonal but choosing MATH so large that they are very close to being orthogonal, we may obtain a unitary MATH in MATH by polar decomposition of MATH such that MATH satisfies the same conditions as above, that is, MATH and MATH. Since MATH with MATH, and since the operators MATH, and MATH commute, we obtain that MATH . Here we have also used the fact that MATH commutes with MATH. We repeat this procedure. Thus we obtain an increasing sequence MATH in MATH and a sequence MATH of mutually commuting unitaries such that MATH where MATH and such that MATH maps MATH into MATH. Then the limit MATH defines the desired automorphism. |
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, then we mentioned in the introduction that MATH is isomorphic to the UHF algebra MATH and MATH corresponds to the one-sided shift on MATH. Thus MATH satisfies the NAME property, CITE, CITE. In particular for any MATH and MATH there is an orthogonal family MATH of projections in MATH such that MATH with MATH. The similar properties hold for MATH, that is, if MATH denotes the C*-subalgebra generated by MATH, then MATH corresponds to the one-sided shift on MATH. Hence for any MATH and MATH there is an orthogonal family MATH of projections in MATH such that MATH with MATH. Suppose we have chosen such projections MATH for the same MATH. Since MATH, we have that MATH in MATH and so obtain a partial isometry MATH such that MATH, MATH. We find unitaries MATH such that MATH, MATH, and MATH, MATH (depending on MATH). Let MATH where MATH is the right multiplication by MATH and MATH is the left multiplication by MATH. Since MATH is a partial isometry with initial projection MATH and final projection MATH, MATH is a unitary in MATH. Since MATH and MATH has real rank zero, we find a sequence MATH of unitaries in MATH such that MATH, MATH, MATH . Define a unitary MATH by MATH . Then since MATH it follows that MATH or MATH . This completes the proof of REF . Since MATH, MATH is dense. That MATH is a MATH set follows from MATH where MATH is a dense sequence in MATH. If MATH contains a non-empty open set, then it follows that MATH or MATH. Because for any unitaries MATH of MATH we find a unitary MATH such that MATH. (Apply the previous argument for the endomorphism MATH instead of MATH and the unitary MATH.) Since MATH for any unitary MATH, the above fact implies that MATH contains an arbitrary unitary. But we know that MATH. For example if MATH, then MATH and MATH. Thus we obtain that MATH is dense. |
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 further assume that MATH in MATH. Pick up a projection MATH such that MATH and MATH. Then MATH is a partial isometry with initial projection MATH and final projection MATH. Let MATH be a partial isometry such that MATH and MATH. Then MATH is a unitary in MATH such that MATH . Thus we have that MATH. To prove the last statement we shall modify MATH so that MATH is the unique state satisfying MATH . We have chosen MATH. We let MATH . Then MATH is self-adjoint with MATH and MATH is the only state satisfying MATH. Let MATH . Then MATH and the assertion follows. |
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 eigenvalues. Let the maximum of the absolute values of those eigenvalues be MATH and the maximum absolute value for an eigenvalue of their inverses be say MATH. Then MATH . After we conjugate back, the entries in the first summand contribute entries proportional to the fixed row and column eigenvectors, which are at least MATH. The entries in the REFnd part are at most MATH. Choose MATH large enough that MATH and the inequality will eventually hold. |
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. Nonnegativity follows from REF : the matrices are primitive, their NAME eigenvalue is REF and their left and right NAME eigenvectors are MATH and MATH, respectively. To verify integrality, we compute the determinants as both REF; then the row spaces of MATH each lie within the space of vectors MATH whose product with the column vector MATH is a positive integer multiple of MATH. But since MATH is their determinant, this is their exact row spaces, and the same holds for column spaces. It follows that each matrix is the product of the other matrix and a unimodular matrix, since the rows of the powers of each lie in the row spaces of the powers of the other. To show powers of MATH can never be conjugate over the rational numbers (which the last equation implies) we compute that their characteristic polynomials are MATH and the degree REF factors are irreducible in MATH. We restrict to the eigenspaces associated with the degree REF factors. The discriminants of the degree REF factors and of their algebraic number fields are REF. These are relatively prime, and the root field of MATH is cyclotomic (REFth roots of unity from its circulant form), its only nontrivial proper subfield is quadratic and can be determined also to have discriminant a multiple of REF. Therefore the only intersection of the fields is the rational numbers CITE. Suppose we have powers which are conjugate over the rational numbers, so have the same eigenvalues. These powers of eigenvalues all are in the intersection field, the rational numbers, and their product is REF by the determinant. So powers of the eigenvalues are rational units, MATH and all eigenvalues of both matrices (other than the NAME eigenvalues) must be roots of unity. But this is false. |
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 eigenvalues. Let the maximum of the absolute values of those eigenvalues be MATH and the maximum absolute value for an eigenvalue of their inverses be say MATH. Then by REF MATH . After we conjugate back, the entries in the first summand contribute entries proportional to the fixed row and column eigenvectors, which are at least MATH. The entries in the REFnd part and REFrd part are at most MATH. Choose MATH large enough that MATH and the inequality will eventually hold. The same holds true for the MATH term. |
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 sufficiently rapidly and MATH has determinant dividing some power of the determinants of MATH, these matrices exist over the integers, and positivity follows from REF (except for the first equation, which follows by a similar argument or can be checked step by step). |
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 group (see below). But the values of the two pairs of eigenvalues prevents any power of one matrix to be conjugate to a power of the other matrix over the rationals. |
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 of matrices whose column vectors come from the nonmaximal eigenspace of MATH. But this is impossible, since other nonmaximal eigenspaces of a positive matrix contain no nonnegative vectors (if they did, multiplication by powers of the matrix would increase them at a rate which is asymptotically the maximal eigenvalue, which means that they would have components in the maximal eigenspace). This proves REF and given REF we have MATH . If MATH projects to a negative multiple of MATH, then those terms will be dominant and make the entirety negative. If it projects to a zero multiple, then MATH which makes equality in REF impossible. This proves REF . Sufficiency of REF are proved in REF . For REF , it will suffice that MATH is eventually positive, looking at the dominant maximal eigenspaces. Conversely, if the vector is negative then the dominant part is negative, which is impossible. Suppose it is nonnegative but not positive. Then the indicated replacement continues to allow solution of the other equations with altered exponents, but it maps the nonnegative vector to a positive one. |
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 each power of MATH, MATH . Hence these matrices MATH have equal row spaces. Conversely, suppose that this condition holds. Then we can find MATH-adic MATH satisfying the last equations. Hence they satisfy them modulo each power of MATH. Consider a term like MATH and the problem of making it integral at the prime MATH for sufficiently large MATH. In order for it to be integral, it suffices that it be so modulo the power of MATH dividing the determinant of MATH. Modulo this power of MATH, increase MATH until we may replace MATH by MATH and use MATH and MATH . This guarantees the left hand side is divisible by MATH, so that fractions in MATH have no denominators MATH. Taking all these primes means we have no denominators at all. A special case is the first equation MATH . For this we allow a replacement similar to the above of MATH by MATH which won't affect solvability of the other equations. |
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 conditions are satisfied. The eigenvalues of MATH can be identified with the matrices themselves and their conjugates, under the map sending MATH to its maximal eigenvalue. If powers of MATH were shift equivalent, then the maximal eigenvalues would correspond to maximal eigenvalues up to powers. Hence we would have some equation MATH. But MATH are independent in the group of units of this degree REF field, so there can be no such equation. |
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 already implies that the two fields must be the same (they have the same set of nontrivial NAME actions under some finite NAME extension containing both). It follows essentially by REF that the prime factors of the two maximal eigenvalues must be the same over algebraic number fields. The intersection of all conjugates of MATH goes to the corresponding intersection for MATH by this isomorphism consistent with NAME action, so that the quotient of the rational dimension group by nonconjugate eigenvalues is mapped from the one to the other. The mapping MATH taken over the field MATH gives an isomorphism between the maximal eigenspaces. This will be multiplication by some element MATH. Let MATH be MATH. Multiplication by MATH will take MATH into MATH, by the effect of MATH, and its inverse will do the reverse, up to multiplication by powers of the primes in MATH which represent multiplication by MATH. Therefore REF MATH for some positive MATH. Recall that the dimension groups of MATH can be viewed as the direct limit of MATH sent to itself by MATH; if we embed them in the maximum eigenspace this direct limit is equivalent to making MATH or MATH, or all the primes in them invertible so that REF implies isomorphism. This implies that the dimension groups of the matrices MATH are isomorphic as modules over MATH for sufficiently large MATH. (This condition does not depend on MATH.) This proves necessity. Now assume the characteristic polynomials of MATH are irreducible and REF hold, hence REF . Choose such an isomorphism as in REF , and adjust its sign so that on the maximal eigenvector it is positive. Expanding out the coefficients gives a map MATH over the integers which preserves all the conjugates of the rational eigenspace. The effect of MATH on all vectors over MATH is isomorphically mirrored in its effect on vectors over MATH on the maximal eigenspace there. This fact together with REF ensures positivity of the other MATH. To get positivity of MATH we replace it by some product MATH and note that its effect on the maximum row eigenvector is positive and this dominates the product asymptotically. The assumption on the primes and REF , implies that for all MATH there exists a MATH such that MATH involves only nonnegative powers of all the primes and is divisible by any given power of each MATH. This implies that all the MATH exist over the integers. |
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 (negative) real parts. On the other hand, the characteristic speeds for the system REF , MATH, in the state MATH are given by MATH . A standard result (for example, NAME REF) tells us that, since MATH, MATH are symmetric and MATH is positive definite, MATH has the same number of positive, negative and zero eigenvalues as the set MATH. If, like NAME REF , we assume that MATH is symmetric as well as positive definite, then the theorem would follow immediately from REF . However, the following lemma shows that this is an unnecessary restriction. Let MATH be a non-singular symmetric matrix, MATH a positive definite matrix and MATH the solutions of MATH . Then the number of MATH with positive (negative) real part is the same as the number of positive (negative) eigenvalues of MATH. Define MATH where MATH and MATH . It easy to see that MATH is also positive definite. Now consider the eigenvalue problem MATH . The conclusion of the lemma is certainly true for MATH, since then MATH is symmetric. If we can show that the MATH are continuous functions of MATH and that MATH for MATH, then it will also be true for MATH. The MATH are the roots of a polynomial of degree MATH whose coefficients are polynomials in MATH. A root can therefore only change discontinuously by going to infinity, which can only occur if the coefficient, MATH, of the highest power of MATH vanishes. However, this cannot happen since MATH is positive definite for MATH. The MATH must therefore be continuous functions of MATH for MATH. In order to prove that the MATH cannot cross the imaginary axis, suppose that for some MATH, MATH, where MATH is real. If MATH is the corresponding eigenvector, we have MATH . Multiplying the first of these by MATH. the second by MATH and substracting gives MATH . Since MATH is positive definite this requires MATH and hence MATH, which cannot be true if the eigenvalues of MATH are non-zero. This completes the proof of the lemma. REF show that the theorem is true even if MATH is not symmetric. |
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: Sample MATH bits. If the fraction MATH of REF is MATH, then output REF and stop. CASE: If MATH, then increase MATH by REF and repeat REF . CASE: If MATH, then count MATH exactly using MATH queries and output the correct answer. It is easy to see that this is a bounded-error algorithm for MATH. Let us bound its average-case complexity under the uniform distribution. If MATH, the expected number of queries for REF is MATH . The probability that REF is needed (given MATH) is at most MATH. This adds MATH to the expected number of queries. Under the uniform distribution, the probability of the event MATH is at most MATH for some constant MATH. This case contributes at most MATH to the expected number of queries. Thus in total the algorithm uses MATH queries on average, hence MATH. Since MATH, we also have MATH. Since a deterministic classical algorithm for MATH must be correct on every input MATH, it is easy to see that it must make at least MATH queries on every input, hence MATH. |
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 orthogonal to this subspace. We estimate the probability that this happens. Consider some fixed non-zero vector MATH. The probability that MATH and MATH are orthogonal is at most MATH, as follows. With probability at least REF/REF, the measurement of the second register gives MATH such that MATH for a unique MATH. In this case, the measurement of the final superposition REF gives a uniformly random MATH. The probability that a uniformly random MATH has MATH is REF/REF. Therefore, the probability that MATH is at most MATH. The vectors MATH are chosen independently. Therefore, the probability that MATH is orthogonal to each of them is at most MATH. There are MATH possible non-zero MATH, so the probability that there is a MATH which is orthogonal to each of MATH, is MATH. |
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 MATH (MATH is the maximum probability of error on an input with MATH and MATH is the probability of getting an input with MATH), so the probability that MATH outputs REF is at least MATH. The average probability (under MATH) that MATH does not stop before MATH queries is at most MATH, for otherwise the average number of queries would be more than MATH. Therefore the probability under MATH that MATH outputs REF after at most MATH queries, is at least MATH. In contrast, the MATH-probability that MATH outputs REF is MATH because MATH for any input MATH from MATH. This shows that we can distinguish MATH from MATH. |
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 is MATH CASE: since MATH queries only MATH of the MATH, it queries only MATH distinct pairs MATH . If no pair MATH, MATH is queried, the probability that all answers are different is MATH . It is easy to see that all sequences of MATH different answers are equally likely. Therefore, for both distributions MATH and MATH, we get a uniformly random sequence of MATH different values with probability MATH and something else with probability MATH. Thus MATH cannot ``see" the difference between MATH and MATH with sufficient probability to distinguish between them. |
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 a quantum algorithm MATH queries are necessary and sufficient on an input MATH CITE, so MATH . |
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 value of inputs from MATH is probably detected around the MATH-th counting stage. We will use MATH queries in the MATH-th counting stage. Our NAME is the following: For MATH to MATH do: quantum count MATH using MATH queries (call the estimate MATH) if MATH, then output whether MATH and stop. Classically count MATH using MATH queries and output its majority. Let us analyze the behavior of the algorithm on an input MATH. For MATH, we have MATH. By REF , with probability MATH we have MATH, so with probability MATH we have MATH for all MATH. This ensures that the algorithm outputs the correct value with high probability. We now bound the expected number of queries the algorithm needs on input MATH. Consider the MATH-nd counting stage. With probability MATH we will have MATH. In this case the algorithm will terminate, because MATH . Thus with high probability the algorithm needs no more than MATH counting stages on input MATH. Later counting stages take exponentially more queries (MATH), but are needed only with exponentially decreasing probability MATH: the probability that MATH goes down exponentially with MATH precisely because the number of queries goes up exponentially. Similarly, the last step of the algorithm (classical counting) is needed only with negligible probability. Now the expected number of queries on input MATH can be upper bounded by MATH . Therefore under the uniform distribution the average expected number of queries can be upper bounded by MATH . |
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 MATH. The algorithm uses MATH queries and should output REF with probability at least MATH. Thus the probability of output REF on MATH must be at most MATH, in particular MATH . Since the MATH queries of the algorithm can be viewed as sampling without replacement from a set containing MATH REFs and MATH REFs, this error probability is given by the hypergeometric distribution MATH . We can approximate the hypergeometric distribution using the normal distribution, see for example, CITE. Let MATH and MATH, then the above probability approaches MATH . Note that MATH and that MATH if MATH. Thus we can only avoid having an error probability close to REF/REF by using MATH queries on MATH with MATH. A similar argument shows that we must also use MATH queries if MATH. |
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 it uses an expected number of MATH queries on every input. Using standard techniques, we can turn this into an algorithm for PARITY with worst-case MATH queries. Since the worst-case lower bound for PARITY is MATH CITE, the theorem follows. |
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 that MATH (compare NAME 's inequality CITE). The lemma now follows from REF . |
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 the observable MATH. We can now prove the claim by induction. For MATH, the state of the automaton is pure, so MATH. Now assume that MATH. After the MATH-th random input symbol is read, the state of MATH becomes MATH . By the definition of MATH, measuring MATH according to MATH yields MATH with probability at least MATH. So by REF , we have MATH . But the entropy of a mixed state is preserved by unitary transformations REF , so MATH . REF now gives us the claimed bound. |
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 measurement MATH distinguishes them correctly with probability MATH. Thus, by REF , we get MATH which gives us the claimed bound. |
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 basis for the entire decoding NAME space. Now, MATH . The last inequality follows because the length of the projection of any vector onto a space MATH is at least the length of its projection onto a subspace MATH of MATH. Observe that MATH. So, MATH since the orthonormal basis MATH for MATH has size at most MATH, which is a bound on the dimension of MATH. |
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 REF is satisfied if for any MATH, MATH, that is, for MATH by choosing MATH and MATH as in the proof of REF . Under REF, one knows, compare CITE, that MATH. We use finally MATH when MATH to get REF. We let MATH be the excitation corresponding to the interface MATH, MATH, where MATH is given by the height MATH for MATH and MATH otherwise. That is MATH is the boundary of the rectangle MATH, see REF . Its energy is: MATH. Denote by MATH the part of the wall between the points MATH and MATH. Then, MATH . Indeed the first term of the Right-hand side of REF corresponds to the excitation MATH and the second terms run over the other clusters containing MATH the length of them being at least MATH. By REF, this term is bounded as follows: MATH . Next we observe that for all excitations whose intersection with the wall is MATH satisfy MATH. Hence the bound REF is improved as follows: MATH when MATH. All the associated clusters have length MATH. Therefore MATH . Thus, MATH . The first term inside the parenthesis comes from the summation over MATH, MATH and the second term from the summation over MATH, MATH. Using that MATH the proof follows from REF . |
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: MATH . For the excitations included in the strips MATH and MATH, one has MATH. Therefore MATH and the associated clusters satisfy: MATH . We let MATH be the excitation corresponding to the interface MATH, MATH, where MATH is given by the height MATH for MATH, and MATH otherwise REF . Its energy is MATH and MATH. We let MATH denote the excitations of width MATH, height MATH (and length MATH) whose intersection with the wall is the segment MATH (of length MATH). Their energy when MATH or when MATH and they do not intersect the vertical part of the wall are: MATH. CASE: The excitations MATH, MATH, and MATH translated by MATH. Then we the decompose the sum involved in REF as follows: MATH where MATH . Let us compare the differences MATH, MATH. When MATH is even, we have MATH where MATH . Indeed, there is a one - to - one correspondence between the clusters MATH of base of size MATH occurring in MATH and MATH till MATH reach some value. This value, when MATH is precisely MATH, because in that case there are clusters (of base of size MATH) which belong to MATH bot neither to MATH nor to MATH. There is precisely one excitation MATH of base of size MATH (and length MATH) which belong to MATH bot neither to MATH nor to MATH. Its energy is MATH and gives the corresponding term in REF. The other clusters have length MATH. This gives the first bound on the reminder MATH. The second bound in REF follows from REF. When MATH the argument works in the opposite direction. The value MATH is MATH and there is a corresponding MATH of length MATH which belong to MATH but nor to MATH nor to MATH. Its energy is MATH and provides the corresponding term in REF. The other clusters have length MATH. This gives the bound on the reminder MATH, the second inequality in REF following from REF. When MATH is odd: MATH where MATH must be understood as MATH with MATH as large as we wish. Indeed the same reasoning as for MATH even applies. In addition there is the particular case MATH where the width of the cylinder MATH equals the one of MATH and the width of MATH equals the width of MATH. For MATH, we have: MATH . Indeed the clusters of minimal energy containing MATH and for which the correspondence is not one - to - one are the excitations MATH of length MATH and the excitation MATH of length MATH. All other clusters have length greater or equal than either MATH or than MATH. To bound the first sum we used REF and to bound the second sum we used REF. Finally, by REF MATH where MATH . Assume now that MATH if MATH is even or MATH if MATH is odd, then from REF - REF MATH where MATH . Therefore for MATH large the sign of REF will be given by the sign of the difference MATH . More precisely MATH where MATH and MATH. On the other hand when MATH if MATH is even or MATH if MATH is odd, we get from REF - REF MATH . Therefore, for MATH odd, REF proves REF of the theorem while REF gives MATH that is, REF for MATH the result for the other value following from REF, all this provided MATH is large enough as stated in the hypotheses of the theorem. When MATH is even we can conclude only if MATH. In this case, REF follows from REF while REF follows from REF. To find the lower bound on MATH, we let MATH and MATH be respectively the upper bounds on MATH and MATH obtained by replacing MATH by MATH and MATH by MATH in REF. We use also MATH. Then, we take MATH. The value of MATH giving the lower bound stated in the theorem ensures that MATH, MATH and MATH. The above analysis leads also to: MATH . The second term in the Right-hand side of REF comes from the energy of the excitation MATH for MATH and the second term in the Right-hand side of REF comes from the energy of the excitation MATH. Relations REF give REF and end the proof of REF . QED . |
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 program still cannot evaluate correctly. In last analysis the perfect evaluator for chess would have to be an extremely long program, and for more complex systems of this kind, an infinite one. To see it is not possible to program a static evaluator for chess that works correctly on all positions, notice that it would have to evaluate on the basis of the state itself without recourse to the tree. The evaluation of the state has to be done using heuristics, that is, using rules that say how good a state is on the basis of the positions of the pieces and not calculating the tree. But this is not possible if chess is chaotic because then we know the smallest difference between two states leads to completely diverging paths in configuration space, that is, to wholly differing states a few plies later. Therefore the heuristic rules of the static evaluator have to take into account the smallest differences between states, and the evaluators have to be long or infinite routines. Static evaluators, on the other hand, should be short programs, since they have to evaluate the states at the end of each branch of the tree. |
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 MATH. Let MATH denote the subspace of MATH given by the constant functions, and let MATH denote the open submanifold of MATH consisting of negative functions: MATH . It is easy to see that MATH. Let MATH denote the quotient space MATH, which is naturally identified with the set of functions with null average in MATH. Let MATH be the map: MATH given by the quotient map on the first factor and the identity on the second factor. To prove the Proposition we use the Inverse Mapping Theorem (see CITE). According to this Theorem, MATH is a smooth submanifold of MATH provided that the map MATH be transversal over MATH, that is, if for all MATH the composite map: MATH is surjective. This amounts to saying that, for all MATH there exists a constant MATH such that the system of differential equations: MATH has at least one solution MATH. Using the fact that MATH, we can rewrite REF as: MATH where MATH is in MATH. Let MATH be a vector field along MATH satisfying MATH . To prove the existence of such a vector field MATH, consider first the vector field along MATH given by MATH, which is the orthogonal projection of MATH onto the distribution MATH orthogonal to MATH. Formally, we have: MATH . Obviously, we have: MATH because MATH in MATH. Observe that MATH, and it does not have the required MATH-regularity. Now, let MATH be any section of class MATH of MATH which is uniformly close to MATH, in such a way that MATH as well. Observe in particular that, since MATH is continuous, then MATH is in MATH. In order to solve REF , we set MATH where MATH are to be determined. Observe that such a MATH belongs to MATH provided that MATH and MATH satisfy the boundary conditions: MATH . Since MATH, REF are translated into: MATH . We solve for MATH REF obtaining: MATH substituting REF gives: MATH where MATH and MATH . Observe that MATH, MATH and MATH are in MATH. Thus, the unique solution MATH of REF satisfying MATH, given by: MATH is in MATH. Observe that MATH in MATH, and so MATH. In particular, there exists MATH such that MATH. Finally, MATH can be chosen as the unique solution of REF satisfying MATH. Observe that the right hand side of REF is in MATH, so MATH and MATH is transversal over MATH. Hence, MATH is a smooth submanifold of MATH. By the Inverse Mapping Theorem, for MATH, the tangent space MATH is identified with the kernel of the map MATH, which consists of the vector fields MATH such that MATH. Recalling REF , we have that MATH belongs to MATH if and only if there exists MATH such that MATH satisfies the equations: MATH . From REF we easily obtain REF and we are done. |
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 other hand, by REF we have: MATH which proves the claim. For REF , observe that MATH is a well defined element in the dual of MATH. The linearity of the map MATH is trivial; the continuity depends on the fact that convergence in MATH implies uniform convergence. Finally, the injectivity is simply the Fundamental Theorem of Calculus of Variations. REF is proven analogously. Namely, using an orthonormal frame along MATH, one reduces the problem to the case MATH. In this case the proof of REF can be repeated verbatim for each component of MATH. Using REF , we will identify each MATH with MATH and every MATH with MATH. Suppressing the symbol MATH, for all MATH, MATH, MATH and MATH we will write concisely: MATH . To prove REF , observe that the map MATH from MATH to MATH is continuous and surjective. Hence, if MATH, then the map MATH, where MATH is the unique primitive of MATH such that MATH, gives a continuous linear functional on MATH, and the conclusion follows by NAME Theorem. The second half is proven similarly. The formulas in REF are the product rules for the MATH-derivative. We prove REF as follows. For MATH, we have: MATH . On the other hand, we compute: MATH . REF is proven similarly. We omit the proof of REF and we prove REF . For MATH, using REF , we have: MATH . This concludes the proof of REF . |
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 MATH for some MATH and MATH with MATH clearly has dimension MATH and it is contained in the annihilator of MATH. Thus, MATH and we are done. |
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 the element MATH would satisfy MATH. Finally, suppose that MATH. Then, MATH, and by REF , MATH. |
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 REF of MATH, we obtain that MATH; multiplying by MATH we have: MATH . Since MATH (because MATH has image in MATH), it follows that MATH and the proof is concluded. |
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 . Since MATH, then, the uniqueness of the solution for REF implies MATH. Now, if we multiply REF by MATH, knowing that MATH is constant and MATH, we obtain: MATH that can be written as: MATH . Again, since MATH, REF implies MATH and we are done. |
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 MATH is horizontal. Therefore, the critical points of MATH on MATH belong to MATH, and clearly they are critical points of the restriction of MATH to MATH. Conversely, if MATH is a critical point of the restriction of MATH to MATH, then the NAME derivative MATH vanishes for all MATH. Let's define: MATH . Since MATH is Killing in the metric MATH, an easy calculation shows that for all MATH, the NAME derivative MATH vanishes for all MATH. Moreover, for all MATH it is (see CITE): MATH which implies MATH for all MATH. This concludes the proof. |
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 computes easily: MATH . Here we have used the facts that MATH is constant along the flow lines of MATH, that MATH is an isometry for all MATH and the conservation law of the energy of the Riemannian geodesics. Observe that, since MATH is Killing in the metric MATH, then a critical point of MATH in MATH is indeed a geodesic with respect to MATH (see CITE). It follows that the quantity MATH is constant along each horizontal geodesic MATH. Recalling REF , integrating REF yields: MATH . From REF it follows that MATH is a local minimizer for the travel time if and only if MATH is a local minimizer for the energy functional MATH in MATH, that is, if and only if MATH is a horizontal geodesic between MATH and MATH with respect to MATH. The last statement of the thesis follows easily by integrating REF over MATH. |
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 . Using REF and the commutation relations REF , we have: MATH . Keeping in mind that MATH and arguing as in the proof of REF (see REF ), the boundary term in REF can be computed as: MATH . Finally, we have: MATH . REF follows from REF . |
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 - NAME connection of MATH; moreover, let MATH denote the index form in MATH with respect to the metric MATH, defined as in REF . Moreover, let MATH be the geometric index of the geodesic MATH in the metric MATH, defined as in REF . Let MATH be defined by: MATH . Clearly, MATH. Observe that, since MATH is orthogonal to MATH, then MATH for all MATH. Let MATH and MATH be fixed; using the fact that MATH, MATH and that MATH, it is easy to see that MATH. This implies that MATH and MATH are orthogonal with respect to the bilinear form MATH; in particular, it is: MATH . It is easy to see that MATH; indeed, for MATH, since MATH, from REF we get: MATH and since MATH and MATH, the above inequality implies that MATH is positive definite in MATH, and so MATH. This proves the second equality in REF . To prove the first equality, we prove that MATH and the conclusion will follow directly from REF . To this goal, we will use also some abstract arguments in functional analysis on NAME spaces, and we introduce the following notation. For all MATH, let MATH be a real NAME space with relative inner product, defined by: MATH . Observe that MATH is non degenerate on MATH, because of the condition MATH. Let MATH be the relative norm. We also define a continuous symmetric bilinear form MATH on MATH, by: MATH observe that for MATH, the NAME space MATH coincide with MATH and the bilinear form MATH is precisely the Hessian MATH. The symmetry of MATH is easily obtained using the symmetry of the curvature tensor and of the second fundamental form of MATH. Observe also that, since MATH and MATH are multiples of the Killing field MATH, then we have: MATH . Using the NAME representation theorem, we can write MATH as: MATH where MATH is a self-adjoint linear operator on MATH. Comparing REF , we see that we can write: MATH where MATH is the identity on MATH and MATH is the self-adjoint operator on MATH defined by: MATH . Since the inclusions of MATH into MATH and into MATH are compact (see CITE) and keeping in mind REF tells us that MATH is a compact operator for every MATH. For all MATH, let MATH be the sequence of all the eigenvalues of MATH; they can be characterized by the following minimax property: MATH where the first maximum is taken over all possible subspaces MATH of MATH having dimension equal to MATH. By standard arguments (see for instance CITE) using the above characterization of the MATH's one proves that the map MATH is continuous. We now prove the following claims: CASE: for MATH small enough, MATH is positive definite in MATH; CASE: for all MATH, the kernel of MATH consists precisely of all MATH-Jacobi fields along MATH that vanish at MATH; CASE: for all MATH, the map MATH is increasing on MATH; moreover, if for some MATH it is MATH, then MATH for all MATH. Observe that the proof will be concluded once the above claims are proven. Indeed, by definition, a point MATH is a MATH-focal point along MATH with multiplicity MATH if and only if there exists MATH such that MATH. From REF , the NAME index of MATH on MATH is given by the sum of the dimensions of the eigenspaces of MATH corresponding to eigenvalues MATH which are strictly larger than one. By REF above, such number is given by the sum of the dimensions of the kernels of MATH, the sum being taken over all MATH. By definition, this number is equal to the geometric index MATH of MATH. Let's prove REF ; observe that another way of stating this claim is that, for all MATH and for MATH small enough, we have: MATH . For MATH, since MATH, we have: MATH hence, using NAME 's inequality we have: MATH . Integrating REF on MATH we obtain: MATH from which we get: MATH . Moreover, another application of NAME 's inequality gives us: MATH from which we obtain the inequality: MATH . The proof of REF follows immediately from REF . For REF , we need to show that MATH and the equality MATH holds for all MATH if and only if MATH is a MATH-Jacobi field along MATH, that is, if and only if MATH satisfies the four conditions: MATH . For the first part of the claim, it suffices to show that if MATH is a vector field along MATH such that REF holds, then MATH. Indeed, for any vector field MATH that satisfies REF , the equality MATH is easily verified using integration by parts. Since MATH is Killing and MATH is a geodesic in the metric MATH, then the quantity MATH is constant along MATH, hence REF implies that MATH. Conversely, let's assume that MATH for all MATH. Let MATH be an arbitrary smooth vector field along MATH such that MATH. Let us set: MATH and MATH where MATH . From REF of MATH it is easily checked that MATH; we compute as follows: MATH . We now show that MATH. Since MATH is Killing, then its restriction to MATH is a NAME field (see CITE), and so it satisfies: MATH . Integration by parts and REF yield: MATH where in the last equality we have used the anti-symmetry of the map MATH. By the symmetry of the curvature tensor, we have: MATH hence, we have: MATH because MATH (see REF ). If we use the equality MATH we get: MATH . Since REF holds for all smooth vector field MATH along MATH vanishing at the endpoints, the fundamental lemma of Calculus of Variations tells us that: MATH which is the first condition in REF . The other three conditions of REF are satisfied by any vector field in MATH, hence REF is proven. Let's go now to the proof of REF . Let's fix MATH in MATH; we prove first that, for all MATH, we have: MATH . To this goal, let MATH be fixed and let MATH be a MATH-dimensional subspace of MATH such that: MATH . We define a linear and continuous map MATH given by: MATH . We observe that, with the above definition, MATH does indeed belong to MATH (see REF ); observe also that MATH is an isometry, and in particular injective. Moreover, the following equality holds trivially: MATH . Let MATH be the MATH-dimensional subspace of MATH defined by: MATH . Then, by REF , we have: MATH which proves REF . To prove the second part of REF , it suffices to observe that if MATH then MATH is a MATH-focal point along MATH. Since the set of MATH-focal points along MATH is discrete (see REF ), it follows that, if MATH, then MATH in a neighborhood of MATH. Finally, by the monotonicity of MATH, we conclude that MATH in MATH, and we are done. |
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 of MATH. In particular, in the sixth formula we have used the fact that MATH, by the anti-symmetry in the last two variables. The initial REF is obtained by linearizing the first equation of REF . |
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 the data of the solution of the differential REF . Let MATH be the set of vectors MATH satisfying REF ; MATH is a submanifold of MATH. Indeed, the condition MATH is open; moreover, the gradient of the smooth map MATH is easily computed as: MATH . Multiplying by MATH we obtain: MATH where the last inequality depends on the fact that both MATH and MATH are timelike, hence MATH, and MATH in MATH. This implies that MATH, hence MATH is a smooth submanifold of MATH. Clearly, MATH. Let MATH be a smooth map such that MATH and MATH. Observe that MATH belongs to MATH, because, from REF , we have: MATH . Since MATH, then MATH, so we have: MATH where the last equality follows immediately from REF . Hence, MATH and the curve MATH is well defined. Now, for all MATH, let MATH be the unique brachistochrone in MATH satisfying MATH; clearly, MATH, and MATH is a smooth variation of MATH. Observe that, since MATH is defined on the closed interval MATH, then we can assume that also MATH is defined on MATH for all MATH. In order to conclude the proof, we need to show that the variational field MATH coincides with MATH. By REF , MATH satisfies the second order differential REF , while MATH satisfies REF by assumption, and MATH, because we are fixing the initial point MATH. By uniqueness, in order to prove that MATH along MATH it suffices to show that MATH. This is easily established by the following calculation, that concludes the proof: MATH . |
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 show that MATH is in the kernel of the Hessian MATH. By REF , this amounts to proving that MATH is the variational vector field corresponding to a smooth variation MATH of MATH consisting of horizontal geodesics in the metric MATH between MATH and some integral curve MATH of MATH lying in MATH (recall that a vector field along a geodesic is NAME if and only if it is the variational vector field corresponding to a variation by geodesics). To see this, let MATH be a smooth variation of MATH consisting of brachistochrones in MATH between MATH and some curve MATH in MATH, and with variational vector field MATH. Such a variation exists by REF . Then, if we consider the curves MATH, by REF , each MATH is a horizontal geodesic between MATH and MATH; by REF , MATH is a smooth variation of MATH. Finally, we have: MATH which concludes the proof. |
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 field in MATH. Note that MATH because MATH for all MATH. It is easily computed: MATH which concludes the proof. |
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 identity on MATH for all MATH. |
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 be a variation of MATH with variational vector field MATH such that MATH and MATH (constant). Such a variation exists, provided that we do not require the condition MATH. For all MATH, define MATH where MATH is the map defined in REF . Then, MATH is a variation of MATH in MATH; if MATH is the corresponding variational vector field, then clearly MATH. To conclude the proof, we need to show that MATH, that is, that MATH is parallel to MATH. Recalling REF , his follows easily from the fact that MATH is a multiple of MATH and from REF . This concludes the proof. |
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 considering a suitable isometry of MATH given by MATH. |
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 suffices to show that MATH is a NAME field along MATH, that is, that it satisfies REF . Observe that, by REF , MATH is in MATH, hence it satisfies the two equations: MATH . To prove that MATH is NAME, let MATH be any smooth vector field along MATH vanishing at the endpoints. We set: MATH where MATH and MATH are functions in MATH to be determined in such a way that the resulting vector field MATH belongs to MATH. Straightforward computations show this condition is satisfied by setting: MATH . Observe that, with the definitions above, since MATH is a geodesic with respect to MATH one has: MATH . Arguing as in the proof of REF since MATH is Killing in the metric MATH, then its restriction to MATH is a NAME field (see also REF ): MATH . Recalling REF , keeping in mind REF and the fact that MATH, we have: MATH . From REF , the second equation of REF and the anti-symmetry of the curvature tensor MATH, the term MATH is easily seen to vanish: MATH . From REF , integrating by parts and using REF and the symmetry of the curvature tensor MATH, we have: MATH . Finally, from REF , we have proven the equality: MATH . Since MATH, then MATH is in the image of MATH, say MATH for some MATH. Since MATH and MATH, then, by REF , it is MATH, and, in particular, MATH. Hence, we have that MATH for all smooth vector field along MATH vanishing at the endpoints, and by REF this implies that MATH is a NAME field, concluding the proof. |
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 . If MATH is not a MATH-focal point along MATH, or equivalently if MATH is not a b-focal point along MATH, then, REF implies: MATH the equality REF follows at once from REF - REF . This concludes the proof. |
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. Then, REF can be written as: MATH . Finally, it is well known CITE that MATH has the same homotopy type of MATH, which concludes the proof. |
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 corresponding to MATH. This means that MATH for all MATH, MATH, the map MATH is smooth, and MATH. We differentiate the expression MATH with respect to MATH twice, and we evaluate at MATH. From REF , we have: MATH . Since MATH is a brachistochrone, by REF the second integral in REF vanishes: MATH . By REF , the last term in REF can be written as: MATH . We now consider the three terms in REF that contain two derivatives with respect to MATH, and, using REF , we write them as follows: MATH . Integration by parts in the last two integrals of REF gives: MATH because, by REF , we have: MATH and, since MATH, MATH . Here, we have used the equalities: MATH . Let now MATH be defined by: MATH . We have: MATH hence, multiplying by MATH, we obtain MATH . Moreover, from REF we easily get: MATH . Substitution of REF into REF gives: MATH . In conclusion, we have proven the equality: MATH . Observe that, since MATH, with MATH then the boundary term in REF vanishes: MATH . Since MATH, then: MATH hence MATH . Finally, by the anti-symmetry of the curvature tensor MATH, we have: MATH . REF now follows from REF . |
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 in fact an equality and MATH so MATH for some MATH. MATH is the equation of a germ of curve MATH which goes through MATH but not through MATH, therefore its virtual transform at MATH relative to the system of multiplicities MATH is smooth, so it has a unique point MATH. It is clear that MATH and MATH, so MATH. |
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 MATH must be MATH and MATH. |
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 MATH and MATH respectively, because then there is a neighbourhood MATH of MATH where both MATH and MATH have a representative and an automorphism MATH of MATH sending one to the other and which therefore lifts to the desired MATH. Let MATH and MATH be arbitrary unibranched germs of curve going sharphly through MATH and MATH. They are equisingular with a single characteristic exponent MATH and their intersection multiplicity at MATH is MATH, so there is an automorphism of the completion MATH with MATH (compare CITE). Let MATH be a system of parameters; we have MATH, and MATH can be described by MATH . Choose MATH such that MATH. Then MATH define an automorphism of MATH such that MATH is the equation of an irreducible germ MATH going sharply through MATH. Indeed, MATH so MATH and we know also that MATH. Therefore MATH is the equation of a germ MATH going through MATH. As MATH is an automorphism, MATH is irreducible and goes sharply through MATH. |
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