paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9905187 | If MATH and MATH so MATH for some MATH. Thus MATH so MATH. Thus MATH is well defined, and clearly it is a homomorphism. Let MATH and MATH be the generators of MATH and MATH respectively, and let MATH and MATH. Given MATH then MATH is a polynomial in MATH. This includes the functions MATH. Since MATH is a homomorphism, then for any polynomial MATH we have MATH . Given a point MATH, this has coordinates MATH. The point MATH has coordinates MATH. Also for each immersions equation MATH defining MATH, given in REF we have MATH from REF . So MATH. We define MATH via MATH for MATH. Since MATH is a polynomial in MATH we can calculate all the partial derivative. To show that the NAME structure is preserved, given MATH we have MATH . And since the NAME bracket is defined by a bi-vector then MATH must also respect the NAME structure. Let MATH be surjective and MATH such that MATH. Then there exists a function MATH such that MATH. Since MATH is surjective there exists a function MATH such that MATH. Thus MATH, giving MATH, and hence MATH. Thus MATH is injective. If MATH is bijective then we have MATH and hence MATH. Given MATH, MATH we have MATH . Since this is for all MATH, then MATH is bijective. If MATH is bijective then we can define MATH via MATH, and this satisfies MATH and MATH. If MATH is bijective then we first show that MATH is injective. Let MATH such that MATH. Then MATH. Since MATH is injective MATH. Thus MATH. So MATH. So MATH. Repeating this process shows MATH. If MATH is bijective then we show, by construction, that MATH is surjective. Choose any ordering MATH. Let MATH, MATH be defined inductively via MATH . Clearly MATH so MATH converge to MATH. Also MATH so MATH. |
math/9905187 | Since REF is linear, then we can extend MATH and MATH to MATH, thus REF . Now MATH . |
math/9905187 | Let us assume we have a representation MATH. Let MATH be the eigenvalues of MATH. Then MATH. This implies that there must exist a MATH such that MATH. By looking at the trace of MATH we can show that MATH. Hence MATH. Likewise MATH, for another integer MATH. This is impossible since MATH is not rational. |
math/9905187 | Choose some ordering on MATH. Define MATH as MATH. Since MATH is linear on MATH we can write MATH for some (distributional) weight function MATH on MATH. Let MATH be conjugate coordinates on a patch of MATH, and let MATH and MATH. Let MATH and MATH, for MATH where MATH is the set of permutations, and MATH is the signature of the permutation. However MATH since it is an odd permutation. Thus we have MATH for all (algebraic) functions MATH on MATH. Integration by parts gives MATH . By considering a sequence of MATH we can let MATH be the characteristic function on some subset MATH. This implies MATH . Repeat this process until we are left with MATH, for two points MATH. This implies MATH. Thus MATH is a constant, whose value is given by MATH. |
math/9905187 | First note MATH . These are given in CITE. Also MATH . Thus REF follows from induction on MATH, and REF is its inverse. REF follows from induction on MATH. For REF expand MATH and MATH. Then REF follows from induction on MATH and MATH. |
math/9905187 | To show this is true for MATH simply substitute MATH into REF and MATH into REF to obtain the corresponding product formulae. The results naturally extend for MATH. |
math/9905187 | The formula for the normal star product MATH extends naturally to the elements of MATH. Thus MATH . Hence result. |
math/9905187 | Trivial. |
math/9905187 | Simply go though all the axioms of a quantum group. |
math/9905187 | Simply go though all the axioms of a quantum group. |
math/9905187 | This is basic manipulation MATH . Inverting this gives MATH . Here MATH when written as a MATH matrix. MATH which gives REF |
math/9905187 | Take the expression for MATH and replace the metric with REF or REF , and replace the derivatives using MATH or MATH . |
math/9905187 | From REF we have MATH . Hence REF . Using REF shows that the star product must be NAME. We can also prove that the central ordering gives the NAME product directly. Let MATH and MATH then from the definition of the NAME product we have MATH where MATH and MATH refers to differentiating with respect to MATH and MATH refers to differentiating with respect to MATH. Thus MATH . |
math/9905187 | Trivial. |
math/9905188 | For REF-parameter subgroup MATH, MATH and left translations are isometries. |
math/9905188 | This follows from the complete integrability of the geodesic REF , or from the integration of them in REF . |
math/9905188 | MATH . |
math/9905188 | All the identities follow easily taking into account REF , and the definition of MATH -type. For the first, we have MATH . The second and the third identities follow from this one by polarization. For the second, we compute as follows: MATH on the other hand, MATH . Similarly for the third: MATH on the other hand, MATH . In order to prove the fourth identity, we polarize the formula in REF : MATH on the other hand, MATH . |
math/9905188 | Suppose MATH is of MATH -type. Consider the associated left-invariant Riemannian metric MATH given by MATH and let MATH and MATH be the respective MATH-maps. Then MATH for all MATH and MATH, noting that MATH is the same for both inner products by construction. Thus MATH for every MATH, so MATH and MATH is of MATH-type. The converse follows by choosing any MATH-orthonormal basis, then choosing an associated MATH and MATH, and then reversing the preceding argument. |
math/9905188 | If MATH, this follows immediately from the previous Corollary. If MATH and MATH, then from REF infra there are abelian subgroups of dimension greater than or equal to REF which give rise to flat submanifolds. Finally, the nonflat NAME group has nonconstant curvature. |
math/9905188 | To begin, observe that the first part of the hypothesis implies that MATH for all MATH. This eliminates most of the possible contributions to a nonzero curvature operator. With the second part included, one need only check REF , and this is readily seen to vanish as well. |
math/9905188 | This is mostly just straight-forward computation, but there are some details to be noted carefully. We give one case to illustrate this, in which we use the selfadjointness of MATH and the MATH-skewsymmetry of MATH extensively. From the definition, MATH . From REF and those properties of MATH and MATH, MATH . The basic trick in all these cases is to expand in an orthonormal basis and write MATH . Then we get MATH and summing on MATH yields MATH . This results in the formula for MATH, and the others are similar. |
math/9905188 | The second part follows from the observation at the beginning of the proof of REF . |
math/9905188 | For convenience, we shall also denote the corresponding subbundles of MATH by MATH and MATH. Then we note that MATH is parallel, integrable, and MATH-invariant. Also, MATH is an orthogonal direct sum and MATH and MATH are MATH-invariant, since we have assumed MATH is nondegenerate. First we show that MATH. Suppose not, with MATH. Then for all MATH, we have MATH. Since MATH and MATH are MATH-invariant, MATH, and MATH is parallel, we deduce that MATH for all MATH. To see that MATH for some MATH, choose MATH such that MATH. Then for MATH, we find MATH. Thus MATH whence MATH . Now, for any such MATH, we have MATH because MATH, MATH, and MATH is integrable. Moreover, for such a MATH, we would have MATH and MATH would be another nonzero element in MATH. But this would mean that a sectional curvature numerator for MATH, namely MATH, is nonvanishing. This contradicts MATH being flat, so we conclude that MATH as desired. Next we show that MATH. Let MATH and write MATH with MATH and MATH. To show that MATH it suffices by REF to show that MATH for all MATH. Let MATH be fixed but arbitrary. Then the parallel and MATH-invariant distribution MATH contains MATH. But from the first part of this proof it follows that MATH since MATH. Therefore MATH. We now claim that for all MATH, we have MATH. Indeed, for any MATH and MATH, since MATH is parallel and MATH-invariant, MATH. But MATH; similarly, MATH. Finally, we show that if MATH then MATH. For any such MATH set MATH, and observe that it suffices to prove that MATH is parallel and flat. Consider MATH and MATH and write MATH for some MATH and MATH, and MATH for some MATH and MATH. Then MATH so MATH is parallel. Now, from MATH it follows that MATH for all MATH, whence MATH is flat. |
math/9905188 | For any MATH, using REF and the third identity from REF , we have MATH . Thus MATH implies there is no component in the MATH direction. |
math/9905188 | The first two parts are immediate consequences of REF . Letting MATH denote the natural projection, the second part implies that the functions MATH and MATH for MATH are well defined and satisfy MATH for all MATH. From the first part, the MATH commute with MATH. The functions MATH are clearly continuous and nonconstant in MATH, but constant along MATH-orbits. Thus there is no dense orbit in any part of MATH. Finally, the last part follows from the definition of MATH and MATH. |
math/9905188 | The formulas follow from straightforward integrations of the geodesic REF - REF . We used the general fact about exponentials of matrices that MATH commutes with MATH for all MATH. Using this, it is routine to verify that MATH, MATH, and MATH satisfy the geodesic equations and initial conditions. |
math/9905188 | MATH is complete by REF , hence pseudoconvex. The preceding geodesic equations show that MATH is nonreturning. Thus the space of geodesics MATH is NAME by REF. Now REF yields geodesic connectedness of MATH. |
math/9905188 | Compare CITE. First note that on each MATH where the restriction MATH is invertible, MATH. Next, one verifies straightforwardly that MATH and, using the properties of MATH, that the derivative with respect to MATH of each MATH is zero. Hence MATH for all MATH and MATH. From this and the decomposition of MATH, we obtain MATH . Using the two facts about MATH again, another computation produces MATH . Combining this with REF and MATH, we get MATH . Finally, it follows from REF and this that MATH satisfies MATH, and from REF and this that MATH satisfies the initial conditions. |
math/9905188 | First use the decomposition of MATH. From the representation of MATH on MATH, we obtain in the first case MATH there, and the second case is similar. |
math/9905188 | It follows from REF that all planes containing any MATH are homaloidal. Thus the NAME equation trivializes and the result follows. |
math/9905188 | Consider an isometry of MATH which is also the inner automorphism determined by MATH. Then MATH is a linear isometry of MATH. The NAME group exponential map is surjective, so there exists MATH with MATH. Then MATH . Now MATH is nilpotent so all its eigenvalues are REF. Thus all the eigenvalues of MATH are REF, so it is unipotent. It now follows that MATH is the identity. Thus MATH lies in the center of MATH and the isometry of MATH is the identity. |
math/9905188 | Similarly to the observation by CITE, it is now easy to check that REF 's proof for Riemannian MATH of MATH-type CITE is readily adapted and extended to our setting, with the proviso that it is easier to just replace his expressions MATH with MATH instead of trying to convert to our MATH. |
math/9905188 | It is easy to see that MATH; compare REF . Then, on the one hand, MATH. On the other hand, MATH. It follows that MATH. The converse follows from REF . |
math/9905188 | We may assume without loss of generality that MATH. Then a straightforward calculation shows that MATH and the result follows. |
math/9905188 | We follow CITE. It is easy to check that MATH is a well-defined, smooth, injective homomorphism with image in MATH. Thus we need only show that MATH is surjective. Let MATH and let MATH be a path from MATH to MATH. The covering map MATH has the homotopy lifting property, so choose a lifting MATH as a path in MATH. Then for all MATH, it follows that MATH for all MATH. Hence for each MATH, there exists MATH such that MATH. Since MATH and MATH is a discrete group, it follows that MATH for every MATH, so MATH commutes with MATH for every MATH and MATH. From REF , there exist MATH and MATH such that MATH for all MATH. Now, every MATH commutes with every MATH, so MATH for all MATH and MATH. Extension from lattices is unique CITE, so MATH. By REF , MATH is the identity and MATH for all MATH. Thus MATH, so from the definition of MATH we obtain MATH. But this means MATH. |
math/9905188 | By REF , we may identify MATH as the group of the principal bundle MATH, so it acts freely on the total space. Since MATH is a local isometry and the MATH-orbits in MATH are complete, flat, and totally geodesic from REF , it follows (using the identification MATH supra) that the MATH-orbits are complete, flat, and totally geodesic. |
math/9905188 | We follow the proof of CITE. For each MATH, define MATH, and note that it is a local isometry. Clearly, MATH if and only if MATH for some MATH. Hence MATH induces an isometric embedding MATH. That the image is the MATH-orbit of MATH follows from the proof of REF . |
math/9905188 | We may assume MATH and MATH. From REF , MATH. |
math/9905188 | Let MATH and set MATH and MATH. Then MATH is equivalent to MATH, and the latter is routine to verify using REF . Now MATH is a geodesic if and only if MATH is, and it is easy to check directly that MATH is equivalent to MATH. |
math/9905188 | Because then MATH and MATH. |
math/9905188 | As before, we may assume without loss of generality that MATH. REF are equivalent by REF . The following lemma shows that REF implies REF . As in the preamble to REF , write MATH as an orthogonal direct sum MATH, where MATH, and use MATH, MATH, MATH, and MATH as given there. Then MATH and MATH fixes MATH. By REF , we have MATH for every positive integer MATH. By induction, from REF in the statement of the proposition we obtain MATH for every MATH. Decomposing MATH, this yields MATH, MATH, and MATH for every MATH. Now, MATH is an element of the identity component of the pseudorthogonal group of isometries of MATH, and as such can be decomposed into a product of reflections, ordinary rotations, and boosts. With respect to appropriate coordinates, which may be different from our standard choice, a boost will have a matrix of the form MATH on some pair of basis vectors, for some MATH. If MATH is composed only of reflections and ordinary rotations, then the right-hand side of REF is uniformly bounded in MATH (say, with respect to the positive definite MATH) while the left-hand side is unbounded, so MATH. On the other hand, if MATH is a pure boost, then the right-hand side grows exponentially in MATH while the left-hand side grows but linearly, and again we obtain MATH. The Lemma now follows from this and REF for MATH. The proof that REF implies REF is the same as the relevant part of the proof of REF. |
math/9905188 | Note that under this hypothesis, MATH. Now REF implies that MATH, and this is now equivalent to MATH. Thus the relevant parts of the proof of REF apply mutatis mutandis. |
math/9905188 | Let MATH translate a unit-speed geodesic MATH by MATH. As usual, we may as well assume that MATH. Write MATH and MATH. From REF we get MATH, MATH, and MATH. Note that MATH. Substituting and rearranging, we obtain MATH where MATH. |
math/9905188 | As usual, we may assume that MATH. Note that this replaces MATH as given in the statement with MATH and MATH with MATH. For the first part of REF , since MATH there exists an orthonormal basis of MATH having MATH as a member. (This may well be a different basis from our usual one.) Fix one such basis, and consider the positive-definite inner product with matrix MATH on this basis. Let MATH denote the norm associated to this positive-definite inner product. By REF , MATH. Then MATH so MATH. For the rest of REF , we begin with REF and get MATH . By REF , MATH and MATH. Inspecting the formula for MATH in REF , we find MATH. Thus MATH . By REF , MATH. Then MATH . Recall that MATH is skewadjoint with respect to MATH (whence MATH is an isometry of MATH for all MATH), that MATH commutes with every MATH (whence so does MATH), and that MATH. We compute MATH . Therefore, MATH . Now MATH so MATH, where MATH as usual. Substituting in REF for MATH, we obtain MATH so MATH . Adding MATH to both sides, we get MATH . There are several cases: MATH is REF or MATH and MATH is timelike, spacelike, or null. If MATH is null, then MATH and MATH. If MATH and MATH is timelike, or if MATH and MATH is spacelike, then MATH whence MATH. If MATH and MATH is spacelike, or if MATH and MATH is timelike, then it follows similarly that MATH. This completes the proof of REF . Now we prove REF . If MATH is as given there, then MATH whence MATH. Conversely, assume MATH and consider the associated positive-definite inner product MATH. Changing the basis of MATH if necessary, we may assume that MATH, MATH, and MATH are mutually orthogonal with respect to both MATH and MATH. Let MATH now denote the norm for MATH. Then MATH so MATH if and only if MATH. But now MATH has the same initial data as MATH, so by uniqueness they must coincide. Finally, we prove the last part of REF ; the first part is immediate from the last part of the proof of REF above. So assume MATH or MATH. Continue with the immediately previous positive-definite norm MATH and basis of MATH. Substituting in REF we get MATH whence MATH if and only if MATH. |
math/9905188 | We compute: MATH . |
math/9905197 | If MATH, MATH, MATH, MATH realize the weak equivalence of MATH and MATH, let MATH and MATH be given by MATH and MATH. It is immediate that MATH and MATH. |
math/9905197 | The approximation theorem of Mort NAME CITE asserts that there are MATH so that if MATH is uniformly within MATH of MATH (fix some metric on MATH) for each MATH then MATH. Clearly, such MATH can be chosen to be strictly piecewise monotone, in MATH, and follow the pattern of MATH. |
math/9905197 | Let MATH be as in REF with MATH and MATH, all following the pattern of MATH. Let MATH be the identity. Suppose that, for MATH, homeomorphisms MATH have been defined in such a way that MATH for each MATH and the diagram commutes. Since MATH are strictly piecewise monotone, and they follow the same pattern, for each MATH, MATH and MATH where the MATH and MATH are pairwise disjoint open arcs, MATH and MATH are dense in MATH, MATH and MATH (the order given by the positive orientation on MATH), MATH and MATH are both orientation preserving or both orientation reversing homeomorphisms, and MATH for some MATH. Let MATH be defined by MATH. MATH then extends from MATH to a homeomorphism of MATH onto itself so that MATH. The preceding commuting diagram then induces a homeomorphism MATH by MATH. |
math/9905197 | This follows immediately from REF by taking, for instance, MATH to be any of the MATH of REF. |
math/9905197 | By REF , we may assume that MATH and MATH are strictly piecewise monotone. Suppose that MATH, MATH and let MATH, MATH, MATH, MATH be such that, after extending the substitutions by concatenation, MATH, MATH. We will construct continuous surjections MATH and MATH such that MATH and MATH. Let MATH be any continuous strictly piecewise monotone surjection that takes MATH to MATH and follows the pattern MATH. Suppose that MATH, MATH, and MATH, MATH, are all continuous strictly piecewise monotone surjections taking MATH to MATH following the patterns MATH and MATH that have been defined so that the following diagram commutes. Now MATH breaks each MATH into a certain number of intervals; we define the same number of intervals in MATH in two steps, with the aid of the substitutions MATH and MATH. Fix MATH and suppose that MATH. Then MATH with MATH mapping MATH homeomorphically onto MATH in an orientation preserving (respectively, reversing) way iff MATH (respectively, MATH). Let MATH. Since MATH, there are integers MATH such that MATH for MATH. Now fix MATH and let MATH be such that MATH. Suppose that MATH. Since MATH follows the pattern MATH, MATH and MATH maps MATH homeomorphically onto MATH where MATH. Define MATH on MATH by MATH. Extend MATH to a continuous surjection from MATH onto MATH. Carry out the above for each MATH and MATH. That there is no ambiguity in the definition of MATH is guaranteed by the construction and the fact that MATH. It is clear that MATH is a continuous strictly piecewise monotone surjection that follows the pattern MATH and that MATH. Now define MATH using MATH and MATH, etc. The commuting diagram induces a homeomorphism between MATH and MATH. |
math/9905197 | By REF , we may assume that MATH is strictly piecewise monotone (the homeomorphisms of REF can be chosen to take MATH to MATH). Moreover, we may assume that MATH, since MATH, and that MATH is the identity for some MATH where MATH is a closed arc in MATH and MATH or MATH. Then, if MATH is an open arc in MATH containing MATH, we may assume that MATH for all MATH. But then MATH, and this is clearly not possible since MATH is an endpoint of the closed arc MATH. |
math/9905197 | Replacing MATH by MATH, if necessary, we may suppose that there are distinct edge germs MATH such that MATH, for MATH. Say, without loss of generality, that MATH, MATH. We may also assume that MATH are such that MATH is the identity for MATH. Let MATH. |
math/9905197 | We may assume that MATH fixes each edge germ in MATH (by replacing MATH by MATH). For the sake of argument, assume that MATH and that MATH, MATH occur consecutively in MATH. Further, assume that MATH is strictly piecewise monotone and that MATH is the identity for some MATH. Let MATH be such that MATH, MATH, MATH. Let MATH, MATH, and choose intervals MATH such that MATH takes MATH homeomorphically onto MATH for MATH. Let MATH. Then every neighborhood of MATH in MATH contains a neighborhood of MATH in MATH, which in turn contains a homeomorphic copy of MATH. Note that MATH is homeomorphic with a neighborhood of the point MATH in the ``topologist's sine curve" MATH. No such MATH is contained in the product of a zero-dimensional set with an arc. |
math/9905197 | Let MATH be an orientation preserving homeomorphism onto MATH (the orientation of MATH agreeing with the orientation of the MATH containing it). Let MATH be the arc components of MATH and MATH be the product space MATH where MATH has the discrete topology. Then MATH is a closed subset of the NAME set MATH. Let MATH be the homeomorphism defined by MATH. |
math/9905197 | Choose MATH to be strictly piecewise monotone and let MATH be such that MATH. Let MATH be a closed arc in MATH with MATH. Then MATH is a matchbox with MATH. |
math/9905197 | This follows immediately from REF: MATH is the only point which does not admit a matchbox neighborhood. |
math/9905197 | For each MATH, so in particular for each MATH, there is a MATH such that MATH for all MATH. Then, by compactness of MATH there is a MATH such that MATH for all MATH and all MATH. That is, MATH for all MATH. |
math/9905197 | Let MATH be a parametrization of MATH. By REF , there exists a MATH such that MATH for all MATH. Given MATH and an open set MATH, where MATH, there exists a MATH such that MATH. The matchbox MATH inherits an orientation from MATH. For MATH and MATH, MATH. Also for MATH and MATH, MATH and MATH are either disjoint or equal. So given MATH, there exist MATH such that MATH is a collection of pairwise disjoint matchboxes whose union is MATH. For each MATH (cross-section of MATH), choose MATH such that MATH. So there exist MATH such that MATH is a minimal cover of MATH (that is, MATH for each MATH). Let MATH be the matchbox MATH and MATH. Choose MATH such that, for all MATH, MATH and MATH and MATH are oriented coherently or anticoherently where MATH is such that MATH contains MATH. Let MATH. Then there exist MATH such that MATH is a collection of pairwise disjoint matchboxes whose union is MATH. |
math/9905197 | Choose MATH such that MATH for each MATH in a cross-section of MATH and MATH. For purposes of contradiction, assume there exist MATH and MATH such that MATH. If MATH for some MATH and MATH then MATH for each MATH covered by MATH where MATH. So there exist MATH such that MATH and MATH for MATH . There exists a MATH such that MATH for MATH . Choose MATH so that MATH and MATH lie on the same arc component of MATH and let MATH be the arc between MATH and MATH. There is then a MATH such that MATH. Thus MATH. But this contradicts MATH. |
math/9905197 | Let MATH be a homeomorphism from MATH to MATH. We inductively choose integers MATH in the following way. For a given MATH, MATH are pairwise disjoint matchboxes in MATH, with orientation given by REF , and MATH intersects each arc component of MATH. So MATH are pairwise disjoint oriented matchboxes in MATH and MATH intersects each arc component of MATH. By REF , we can choose a MATH such that, for each MATH, MATH and, for each MATH, MATH is the union of a finite collection of pairwise disjoint matchboxes satisfying conditions MATH of REF . Similarly, for a given MATH, MATH are pairwise disjoint oriented matchboxes in MATH and MATH intersects each arc component of MATH. By REF , we can choose a MATH such that, for MATH, MATH and, for each MATH, MATH is the union of a finite collection of pairwise disjoint matchboxes satisfying conditions MATH of REF . For MATH we define a substitution MATH in the following way. Choose a match MATH of MATH. Let MATH be the arc components of MATH, with the ordering given by the positive orientation of MATH. Let MATH where MATH (respectively, MATH) if MATH and MATH and MATH are oriented coherently (respectively, anticoherently). Let MATH. REF ensure that this substitution is independent of the chosen match. We define MATH similarly. Let MATH and MATH. If MATH is a match of MATH, then the substitution MATH describes the way in which MATH passes through MATH, but MATH also describes the way in which MATH passes through MATH. So MATH. Similarly MATH. Thus MATH. |
math/9905197 | By REF , if MATH then MATH. Conversely, if MATH and MATH, then MATH by REF . Let MATH, MATH be as above. Then MATH and MATH. So by REF , MATH. |
math/9905197 | See CITE. |
math/9905197 | If MATH, this follows from REF . If MATH, this follows from REF . If MATH, CITE proves that MATH, MATH, so the corollary follows from REF . There are no other possibilities for MATH. |
math/9905197 | By construction of MATH, there is a nonnegative integer matrix MATH of rank MATH such that MATH. Let MATH be a positive eigenvector of MATH corresponding to the eigenvalue MATH. Then MATH. Thus MATH is a positive eigenvector for MATH with eigenvalue MATH. Thus MATH by the NAME Theorem. |
math/9905197 | NAME and NAME CITE prove that there are branched one-manifolds MATH and MATH and expanding, orientation preserving immersions MATH and MATH such that MATH and MATH. Furthermore, the NAME matrices associated with MATH and MATH are MATH and MATH, respectively. Using the NAME moves to split the branched manifolds into wedges of circles, there are MATH and MATH with MATH and MATH (see CITE and CITE). Moreover, MATH and the associated matrices MATH and MATH are shift equivalent to some powers of MATH and MATH, respectively. In particular MATH and MATH. Assuming MATH, it follows from REF that MATH. |
math/9905197 | This follows from REF . |
math/9905197 | Let MATH be orientation preserving homeomorphisms for MATH. Let MATH by MATH, MATH and MATH as appropriate. Extend MATH to a map in MATH that follows the pattern of MATH and so that for MATH there is a MATH such that MATH. Let MATH and MATH be defined by MATH. Then the diagram commutes. Given MATH, let MATH. Then MATH and MATH for MATH . Claim: MATH is a homeomorphism. Proof: Suppose that MATH for some MATH. Then MATH for all MATH. Then MATH for each MATH . Given MATH there is a MATH such that diam-MATH for all MATH and MATH. So there is a MATH such that if the NAME distance between MATH and MATH is less than MATH then diam-MATH. Since there are MATH with MATH and MATH, for sufficiently large MATH, the NAME distance between MATH and MATH and between MATH and MATH is less than MATH (MATH such that MATH). It follows that the distance between MATH and MATH goes to zero as MATH. Thus MATH. Since MATH and MATH hits each MATH in a unique point, MATH. MATH is surjective because MATH follows the pattern MATH so given MATH pick MATH; then MATH. Note that for MATH, MATH. Extend MATH to all of MATH by conjugation; that is, for MATH let MATH be such that MATH. Put MATH. Then (by the above note) MATH is well defined on MATH and, by REF , MATH is continuous. Clearly MATH. If MATH is a sequence in MATH without a limit point in MATH then the (minimal) MATH for which MATH diverge to MATH. Thus the sequence MATH as MATH and it follows MATH is continuous. |
math/9905197 | If MATH then MATH so that, with MATH, MATH as in REF , MATH. From REF , MATH so that MATH. Conversely, if MATH then MATH so that MATH, by REF . Since MATH, a homeomorphism of MATH to MATH must take MATH to MATH REF . Thus MATH. |
math/9905197 | Given MATH with MATH of period MATH under MATH, let MATH take MATH (see notation introduced before this theorem) homeomorphically, and in an orientation preserving manner, onto MATH for each MATH, and let MATH for MATH. Then there is a MATH such that MATH. Moreover, MATH. Let MATH denote the point in MATH with MATH for MATH and let MATH be its orbit. Then the map MATH defined by MATH is a surjection and is one-to-one except on MATH where MATH. For MATH, the endpoints of the continuum MATH are exactly the points of MATH (see CITE). For MATH, any homeomorphism from MATH to MATH must take endpoints to endpoints so such a homeomorphism would descend to a homeomorphism of MATH with MATH by way of MATH, MATH. Moreover, the arc components of the endpoints of MATH are in one-to-one correspondence with elements of MATH, the eventual range of the action of MATH on edge germs. Since MATH is not periodic of period REF for any MATH, MATH for any MATH. That is MATH for MATH. Thus a homeomorphism of MATH with MATH would necessarily take MATH to MATH and would then lift to a homeomorphism of MATH with MATH. Thus for MATH, MATH. |
math/9905217 | If MATH, we claim first that MATH . REF was proved in the rational case in CITE; the proof is sketched here in the general case. The height is functorial in the sense that it respects field extensions. Thus we may assume MATH corresponds to an embedding of MATH into MATH. Take MATH in the elliptic NAME REF . The points of MATH-torsion are dense and uniformly distributed in MATH as MATH, so the limit sum over the torsion points will tend to the integral when the integrand is continuous. Note that the torsion points occur in pairs usually. Working with MATH they only occur with multiplicity REF, hence the formula differs from the usual elliptic NAME formula in this respect. The only potential problem arises from torsion points close to MATH: by CITE, for MATH with MATH, MATH for some MATH which depends on MATH and MATH only. This inequality is enough to imply that the NAME sum given by the MATH-torsion points for MATH converges, which gives REF , and the explicit error term gives the estimate in REF . Assume now that MATH is non-archimedean, corresponding to a prime of singular reduction. Let MATH denote any complete, algebraically closed field containing MATH. Assume MATH is integral, MATH. Now use the parametrisation of the curve described before. The points of order dividing MATH on the NAME curve are precisely those of the form MATH, MATH, where MATH denotes a fixed, primitive MATH-th root of unity in MATH. We claim that MATH . Let MATH denote the closure of the torsion points: MATH is not compact, so the MATH-adic elliptic NAME formula cannot be used. Instead we use a variant of the NAME integral: for MATH define the elliptic NAME integral to be MATH whenever the limit exists. We claim firstly that for any MATH, the NAME integral MATH . First assume that MATH is the identity. Using the explicit formula for the local height gives MATH . The first sum is bounded by MATH, which vanishes in the limit; the second sum converges to MATH. For the general case, let MATH correspond to the point MATH on the multiplicative NAME curve. If for some large MATH no MATH has MATH then the analogous sum to REF is close to MATH by the same argument. Assume therefore that there is a MATH with this property. Then the first sum in REF is replaced by MATH where MATH only depends on MATH. By MATH-adic elliptic transcendence theory (see CITE), there is a lower bound for MATH of the form MATH, where MATH depends on MATH and MATH only. It follows that the first sum vanishes in the limit as before. The second sum in REF is simply rearranged under rotation by MATH, so converges to MATH as before. This proves REF . The claimed limit REF now follows by taking the elliptic NAME integral of both sides of the parallelogram law REF and noting that we count torsion points in pairs. REF shows that three terms cancel to leave the required limit. The error term in REF comes from the lower bound used above. |
math/9905217 | It will be convenient to use normalized heights, so define MATH . Then MATH is invariant under isomorphism (see CITE). By the product formula, MATH . Also, by REF , MATH . For any MATH, MATH . Therefore, using the product formula again, MATH . The reason for introducing the factor MATH is to take account of the possibility that REF is not in minimal form at some non-archimedean MATH corresponding to a prime of singular reduction. The change of coordinates to put the equation into minimal form is an isomorphism, so it leaves the local height MATH invariant. Now REF follows directly from REF . |
math/9905218 | See NAME, REF p REF . |
math/9905218 | MATH CASE: The subgroup MATH of MATH is from hypothesis a cyclic group. It has one and only one subgroup MATH of order MATH. Let MATH be an integral ideal of MATH not principal and such that MATH is principal: note that such an ideal exists. We have MATH. From MATH, we get MATH and hence MATH. Therefore, there exists MATH such that MATH. In the same way, we have MATH because MATH is a principal ideal, so MATH is principal and so MATH. But MATH and finally MATH. Then, we have for MATH the relation MATH. CASE: In an other part MATH is a principal ideal of MATH. Therefore MATH is a principal ideal for some MATH. CASE: MATH . Then MATH is a principal ideal and MATH. CASE: MATH and MATH. We have MATH. From classical properties of NAME form MATH, we deduce immediatly that there exists at least one divisor MATH of MATH such that MATH. CASE: MATH and MATH. Then MATH, then same result as second case with the form MATH. CASE: MATH and MATH. CASE: Show at first that the ideal MATH is principal: Then MATH and thus MATH is a principal ideal for all MATH-isomorphisms which generate the class group MATH of MATH. CASE: Let us consider the ideal MATH. In the field MATH, we have from above MATH. Therefore, we get MATH. The case MATH has been treated in the first part of the proof, so we can suppose a fortiori that MATH, thus the ideal MATH of MATH is not principal. Therefore, the ideal MATH of MATH is not principal in the quadratic extension MATH of MATH between MATH and MATH, and MATH divides its class number MATH, which achieves the proof. |
math/9905218 | The NAME group MATH is of order MATH. From MATH, we get MATH where MATH is a cyclic group of order MATH and MATH is a cyclic group of order MATH. We have MATH and therefore MATH is cyclic of order MATH. Let MATH be a MATH-isomorphism of MATH. Then the proof is similar to REF proof, with the cyclic extension MATH instead of the cyclic extension MATH. Here we have, for the ideal MATH of MATH which generates a cyclic class of order MATH . The ideal MATH of the field MATH generates a cyclic group of order MATH or is principal. It cannot generate a cyclic group of order MATH because MATH cannot divide the order of the class of the ideal MATH of MATH : in fact from hypothesis and REF , MATH is greater than the class number of MATH. The end of proof is then similar to REF proof, with MATH in place of MATH. In this situation, the case MATH cannot occur, because roughly MATH and the case MATH cannot occur because roughly MATH. |
math/9905218 | We apply REF for MATH observing that, for MATH, we have MATH, therefore MATH and then the result. |
math/9905218 | MATH . Immediate consequence of previous REF . |
math/9905218 | MATH . Immediate consequence of previous REF . |
math/9905218 | MATH . Immediate consequence of previous REF . |
math/9905218 | MATH CASE: From hypothesis on MATH, we get MATH. Therefore, there exists at least one prime MATH dividing MATH. CASE: From MATH, we get classically MATH. CASE: Decomposition of MATH in MATH : There is one prime MATH above MATH in MATH with inertial degree REF, ramification indice MATH. CASE: Decomposition of MATH in MATH : there are MATH ideals MATH with inertial degree MATH and ramification indice MATH. CASE: Decomposition of MATH in MATH : CASE: The ramification indices and inertial degrees are multiplicative in tower and MATH; CASE: therefore there is only on possible configuration : there are MATH ideals MATH of inertial degree MATH and ramification indice MATH above MATH. CASE: Then MATH where MATH. |
math/9905218 | MATH CASE: Let MATH be a prime ideal of MATH above MATH. From previous lemma, MATH, where MATH is a prime ideal of MATH. Then, from MATH we get MATH where MATH is an integral ideal of MATH. CASE: We can extend this result to all prime MATH which divides MATH. We note that MATH for all prime MATH dividing MATH, which then leads to the result. |
math/9905218 | We shall suppose that MATH to obtain a contradiction. CASE: From previous theorem, we have MATH, where MATH is an integral ideal of MATH. From MATH, then MATH is a principal integral ideal of MATH. Then we can write MATH where MATH and where MATH. CASE: The cyclotomic MATH-field MATH is a complex normal extension. From a generalization of NAME Lemma, see REF p REF , we have MATH where MATH and where MATH. Therefore, MATH CASE: By conjugation, we obtain MATH . Then MATH which can be written MATH . CASE: Let us define the map MATH . MATH . CASE: From this relation, by division, we get MATH . Therefore, there exists MATH such that MATH . Therefore, at this stage of the proof, we have simultaneously MATH and MATH. CASE: We have MATH where the prime MATH verify MATH. Then MATH leads to MATH. CASE: There is one and only one NAME extension MATH with MATH and with MATH, given by MATH. CASE: There are two possibilities MATH or MATH. If MATH then, from MATH, we deduce that MATH is a NAME extension with MATH. Therefore, from MATH, we get MATH and hence MATH where MATH. CASE: Therefore we have MATH with MATH . CASE: Then, we have MATH . Classically MATH. Then MATH hence MATH . CASE: Then, with hypothesis MATH, we conclude, clearly, that this congruence is not possible. CASE: Then MATH would lead to a contradiction, which achieves the proof. |
quant-ph/9905042 | REF follows immediately from REF, and the comments following the proof of that proposition. REF follows immediately from REF. |
quant-ph/9905042 | ` REF ' If MATH, then MATH is a normal state of MATH (compare CITE). Since MATH is (by hypothesis) dispersion-free on MATH, MATH is a MATH-normal homomorphism of MATH onto MATH, and the conclusion that MATH satisfies NAME on MATH follows immediately from CITE. ` REF ' Suppose that MATH satisfies NAME on MATH. Let MATH be any family of mutually orthogonal projections in MATH. Since MATH is separable, MATH must be countable, and we may assume that MATH. Let MATH. Since MATH is NAME, MATH. Further, MATH, where MATH is the characteristic function of the (singleton) set MATH. Let MATH be the characteristic function of the (entire) set MATH. Now, MATH in the sense of pointwise convergence of partial sums. Since the map MATH from NAME functions on MATH into MATH is a MATH-normal homomorphism CITE, MATH. Using the previous two facts, we may compute: MATH where we used NAME in the third and fifth equalities. Since MATH is an arbitrary family of orthogonal projections in MATH, MATH is totally additive on MATH. By CITE, there is a sequence MATH of unit vectors in MATH, and sequence MATH of non-negative real numbers with sum REF, such that MATH. We may assume that MATH, and the last equation can then be written as MATH, with MATH a state of MATH. Since MATH is dispersion-free on MATH, it is pure on MATH REF . Thus, MATH, and MATH is a vector state. |
quant-ph/9905042 | CASE: Suppose that MATH and that MATH for all MATH. Fix MATH. Clearly MATH itself leaves MATH invariant (since MATH is a MATH-algebra and MATH is continuous). Thus, MATH. Since MATH was arbitrary, it follows by the linearity and continuity of MATH that MATH leaves MATH invariant. CASE: Let MATH, and let MATH. Since MATH is linear and continuous, it is sufficient to show that MATH for any MATH. But, this is immediate from the fact that MATH and MATH. |
quant-ph/9905042 | We prove REF and then REF . `` REF " Suppose that MATH is beable for MATH. Then, there is a measure MATH on the set MATH of dispersion-free states of MATH such that REF holds. Fix arbitrary MATH. Since each MATH in MATH is a MATH-homomorphism of MATH into MATH, and states are hermitian, MATH for each MATH, and thus MATH by REF . `` REF " Suppose that MATH for all MATH. In order to show that MATH is abelian, let MATH. Thus, for any MATH, MATH . Thus, MATH. Now, since MATH we may apply REF , with MATH-algebra MATH and vector MATH, to conclude that MATH. Therefore, MATH, and MATH is abelian. `` REF " If MATH is abelian, we may identify it with the set of continuous, complex-valued functions, MATH, on some compact NAME space MATH CITE. Consider the vector state MATH on MATH induced by MATH. By the NAME Representation Theorem CITE, there is a probability measure MATH on MATH such that MATH . For each MATH, define MATH by MATH. The reader may verify without difficulty that each MATH defines a dispersion-free state on MATH (using the fact that MATH is a MATH-homomorphism, and the definition of multiplication on MATH). Finally, for each MATH, MATH . Therefore, MATH is beable for MATH. `` REF " MATH is a MATH-subalgebra of MATH, which is in turn a MATH-subalgebra of MATH. Thus, the mapping MATH is a representation of MATH on MATH with cyclic vector MATH (compare CITE), and the composition map MATH is a cyclic representation of MATH on MATH. We now show that MATH is unitarily equivalent to MATH. Recall that MATH is the completion of the pre-hilbert space MATH, where MATH. For elements of this latter set, there is a natural isometric mapping MATH into MATH; namely the mapping that takes MATH to MATH. It is not difficult to verify that MATH extends uniquely to a unitary operator MATH from MATH onto MATH, and that MATH for all MATH in MATH. Thus, MATH is unitarily equivalent to MATH. The equivalence of REF now follows from the fact that MATH is MATH-isomorphic to MATH. |
quant-ph/9905042 | Since MATH is faithful, MATH is an isomorphism of MATH onto MATH CITE. However, MATH is abelian REF . |
quant-ph/9905042 | CASE: The ``if" implication is trivial. Suppose then that MATH is beable for MATH, for all MATH. Consider the closed subspace of MATH given by MATH . Clearly, MATH is precisely the set of all MATH such that MATH is beable for MATH (see REF ). By supposition, MATH; thus, MATH will also contain MATH's closed, linear span MATH. CASE: The ``if" implication is trivial, since MATH contains the identity. Conversely, suppose that MATH is beable for MATH. Fix MATH. Then, for any MATH, MATH, and moreover MATH. Thus, we may apply REF to conclude that MATH for any MATH and for any MATH. Since MATH was an arbitrary element of MATH, it follows REF that MATH is beable for MATH whenever MATH. CASE: Recall, first, that as a positive, trace-MATH operator, MATH has a pure-point spectrum CITE. By the spectral theorem (and the fact that MATH leaves MATH invariant), MATH is the closed span of the eigenvectors of MATH in its range. Thus, there is a countable set MATH, such that MATH for all MATH, MATH, where MATH, and MATH. ` REF ' Suppose that MATH is beable for MATH. Recall from REF that MATH is beable for MATH if and only if MATH for all MATH. Given any eigenvector MATH in MATH, we may write MATH for some positive, trace-MATH operator MATH, and MATH. Thus, by the linearity of the trace, MATH where the inequality in REF follows since MATH and MATH. Thus, MATH for any eigenvector MATH of MATH in its range. Since the closed linear span of these eigenvectors is just MATH, the conclusion follows by REF . ` REF ' Let MATH. Then, by hypothesis, MATH whenever MATH. In particular, MATH, for each MATH, where MATH. Therefore, MATH. Since MATH were arbitrary, MATH is beable for MATH. |
quant-ph/9905042 | The ``if" implication follows trivially from REF . Conversely, suppose MATH is beable for MATH. By REF , MATH is beable for MATH, for all MATH. Fix MATH. By REF , MATH is beable for MATH, for all MATH. Finally, MATH, so by REF , MATH is beable for MATH, for all MATH. |
quant-ph/9905042 | REF ` REF ' Suppose that MATH is beable for MATH. Clearly, we have defined MATH in such a way that MATH reduces MATH. Thus, each element of MATH will decompose uniquely into the direct sum of an operator on MATH and an operator on MATH. We must show that whenever MATH and MATH are in MATH, then MATH. In other words, we must show that elements of the set MATH commute with each other. Let MATH. Thus, MATH for some MATH and MATH for some MATH. Let MATH be arbitrary. Then, MATH for (unique) MATH and MATH. Since MATH, MATH is beable for MATH REF . Thus, MATH where the last two equalities hold because both MATH and MATH leave MATH invariant. By symmetry, MATH. But, MATH since MATH, MATH, and MATH is beable for MATH. Thus, MATH, and since MATH was arbitrary, MATH. Since MATH were arbitrary, any two elements of MATH commute. ` REF ' Suppose that MATH consists of mutually commuting operators. Let MATH. Then, since MATH contains the identity, MATH. Let MATH. Then, we may write MATH and MATH. Hence, MATH. By symmetry, MATH. But, since elements of MATH commute, MATH. Thus, MATH for any MATH; that is, MATH is beable for MATH. Furthermore, since MATH was an arbitrary element of MATH, we see that MATH is beable for every state defined by a (unit) vector in MATH. By REF , MATH is beable for MATH. CASE: We have proved in REF that any algebra MATH which is beable for MATH will be commutative in its action on MATH, and that any algebra MATH which is commutative in its action on MATH will be beable for MATH. To complete the proof, then, it will suffice to show that if MATH, where MATH is maximal abelian, then MATH is not properly contained in any beable algebra for MATH. Suppose then that MATH, and that MATH is beable for MATH. (We show that MATH.) Since MATH is beable for MATH, MATH is beable for MATH whenever MATH REF . Furthermore, MATH. Thus, MATH is beable for MATH whenever MATH. Now, MATH since the latter is maximal abelian and since the former is the identity on MATH. Thus, MATH. Let MATH be a self-adjoint element of MATH. Then, for all MATH, MATH, since MATH is beable for MATH. That is, MATH leaves MATH invariant. However, since MATH is self-adjoint, it also leaves MATH invariant, and therefore MATH. Finally, let MATH. For any MATH, MATH . The first, second, and fifth equalities hold since MATH and MATH leave MATH invariant. The third equality holds since MATH, and MATH is beable for MATH. Hence, MATH and thus MATH. We have shown that MATH, from which it follows that MATH and MATH is maximal beable for MATH. |
quant-ph/9905042 | CASE: Let MATH be a positive trace-MATH operator that induces the state MATH on MATH. If MATH is maximal beable for MATH, then MATH, where MATH is a maximal abelian subalgebra of MATH. Since MATH is a maximal abelian subalgebra of MATH, it follows that MATH is a NAME algebra. Therefore, MATH is a NAME algebra. CASE: Now suppose that MATH is beable for MATH. Then, MATH is contained in some maximal beable algebra MATH for MATH. By REF of this Corollary, MATH. Thus, MATH, and since beable status is hereditary, the conclusion follows. |
quant-ph/9905042 | CASE: Suppose that MATH, and consider MATH, the characteristic function of MATH. Define MATH as in REF , so MATH if MATH, MATH otherwise. Let MATH be the unique function in MATH that agrees with MATH on the complement of a meager set. Thus, MATH and MATH. Suppose, for reductio ad absurdum, that MATH is not defined at MATH, so that MATH. Since MATH is self-adjoint, MATH. Since MATH is bounded, there is an open neighborhood MATH of MATH such that MATH, for all MATH. Thus, MATH. However, since MATH is a continuous map from MATH into MATH and MATH, there is an open neighborhood MATH of MATH such that MATH. But then MATH and MATH disagree on the non-empty open set MATH, which is impossible. Therefore, MATH is defined at MATH. Again, suppose for reductio that MATH is defined at MATH but that MATH. Since MATH is defined at MATH, it is continuous at MATH. Hence MATH and there is an open neighborhood MATH of MATH such that MATH. Thus, MATH and MATH. Since MATH is continuous, there is an open neighborhood MATH of MATH such that MATH. But then MATH and MATH disagree on the non-empty open set MATH, which again is impossible. Therefore MATH. CASE: Suppose that MATH is open and MATH. Let MATH. We must show that MATH. Recall from REF that MATH agrees on the complement of a meager set with MATH. By assumption, then, MATH. Suppose, for reductio, that MATH. Since MATH is continuous, there is an open neighborhood MATH of MATH such that MATH. Since MATH is continuous on MATH, MATH is open in MATH (and thus open in MATH, since MATH is open in MATH). Then, MATH and MATH (which contains MATH) is a non-empty open set on which MATH and MATH disagree - a contradiction. Therefore, MATH. |
quant-ph/9905042 | ` REF ' Immediate from REF . ` REF ' Suppose that MATH is defined at MATH. Then, since there is an open convex neighborhood MATH of MATH such that MATH is compact (and convex), MATH (by REF ). Moreover, since a projection-valued measure is monotone and states are order-preserving, MATH which entails MATH. Suppose now that the the above equivalent conditions hold for MATH and MATH, and let MATH denote the intersection in the statement of this proposition. By assumption, there is at least one compact convex set MATH such that MATH, so that MATH is nonempty. Let MATH be any other such set where MATH. Then, by REF , MATH. Therefore, MATH. Finally, to see that MATH is the unique point in MATH, suppose that MATH yet MATH. Since MATH is NAME, there is an open convex neighborhood MATH in MATH such that MATH but MATH. (We choose MATH such that its closure is compact.) Since MATH and MATH is open, MATH (by REF ). Therefore, MATH, contradicting our assumption. It follows that MATH is the unique element of MATH. |
quant-ph/9905042 | Note that MATH is continuous, being the composition of two continuous functions, MATH and MATH. Moreover, since the normal function MATH agrees with MATH on the complement of a meager set, MATH must agree with MATH throughout MATH. Thus, MATH. |
quant-ph/9905042 | Let MATH be defined as MATH if MATH and MATH otherwise. Then, MATH where MATH is the unique function in MATH which agrees with MATH on the complement of a meager set. Clearly then, MATH converges pointwise to MATH, the characteristic function of MATH, as MATH. Thus, MATH, and MATH where the second equality follows from the Monotone Convergence Theorem CITE. Hence, MATH. Since MATH is a self-adjoint function on MATH, we may decompose MATH as MATH where MATH is the set of points MATH of MATH such that MATH and MATH is the set of points MATH of MATH such that MATH CITE. Now, let MATH be the positive part of MATH and let MATH the negative part. Then, MATH . Thus, MATH, and if MATH then either REF or REF is infinite and either MATH is undefined or MATH. Therefore, MATH has a finite value only if MATH. |
quant-ph/9905042 | Since MATH is beable for MATH, we have a probability measure MATH on the set of dispersion-free states MATH of MATH such that MATH . We will show that MATH, where MATH is the subset of MATH consisting of those states that are well-defined for each MATH. Fix MATH. Let MATH be the NAME algebra generated by MATH, let MATH be the projection-valued measure on MATH induced by MATH, and let MATH. Let MATH . It is not difficult to verify that MATH is a measurable subset of MATH. Suppose, for reductio, that MATH so that MATH. Choose any MATH. Then, MATH . REF follows by the definition of MATH, and the inequality in REF follows since MATH for all MATH. Since MATH was arbitrary MATH. By REF , MATH does not correspond to a convergent measure, contradicting our assumption that MATH is well-defined for MATH. Thus, MATH. Since MATH is countably additive, MATH. Let MATH. Then, for MATH, MATH where the penultimate equality follows since MATH. Finally, suppose that MATH; that is, for each MATH there is a MATH such that MATH. But MATH since MATH is dispersion-free on MATH and MATH is a projection. Thus, for each MATH there is a MATH such that MATH, and by REF , MATH is well-defined for each MATH. Therefore, MATH is a mixture of dispersion-free states of MATH, all of which are well-defined for each MATH. |
quant-ph/9905042 | We show first that MATH for all MATH. For this, let MATH and let MATH. Then, invoking the NAME form of MATH (taking either sign), we have MATH for all MATH. Thus, MATH. Moreover, since MATH, MATH is dispersion-free on MATH and MATH. Again, since MATH is dispersion-free on MATH, MATH for all MATH, and MATH for all MATH. Let MATH. Then we may choose MATH such that MATH, and hence MATH. But MATH was an arbitrary non-zero number; thus, MATH for all MATH. Moreover, MATH, from which it follows that MATH for all MATH. Recall that MATH . Let MATH. From REF we may deduce the operator identity: MATH . Thus, from the linearity of MATH, in combination with the result of the previous paragraph, we may conclude that MATH whenever MATH. And, using NAME 's approximation for the factorial, MATH for large MATH, whence MATH. Now let MATH be a compact interval in MATH. Then, MATH, for all MATH and all MATH. Consider the extended function representation MATH of the abelian NAME algebra MATH. Let MATH. (MATH is clopen since it is the support of the continuous idempotent function MATH.) Fix MATH and let MATH for each MATH. We claim that MATH converges pointwise to MATH as MATH. Note first that MATH is defined at all points of MATH and MATH by REF . Now, for any MATH, we may choose MATH small enough that MATH for all MATH (since MATH is compact and MATH is fixed). Thus, for any MATH (and using NAME in the fourth step), MATH which is what we needed to show. Since MATH converges pointwise to MATH we may apply the Dominated Convergence Theorem CITE to conclude that MATH . However, since MATH, for all MATH, it follows that MATH. Since this is true for all MATH, and MATH, it follows that MATH. |
quant-ph/9905042 | Let MATH. Let MATH. Then, from the preceding Proposition, MATH, and thus MATH, for all MATH. However, MATH, and it follows from the countable additivity of MATH that MATH. |
quant-ph/9905042 | (MATH-Priv) MATH. CASE: See REF . (Def) Suppose that MATH is a unitary element of MATH such that MATH and MATH. Then, since MATH satisfies (Def), MATH. Since the spatial automorphism MATH of MATH induced by MATH is a NAME from MATH to MATH, it follows that MATH. |
quant-ph/9905042 | To show that MATH, it will suffice to show that MATH for any self-adjoint MATH (since MATH is a MATH-algebra). If MATH, then MATH is unitary for all MATH. By hypothesis, then, MATH, for all MATH. Since MATH is bounded, MATH is a compact subset of MATH. Consider the one-parameter family MATH of (complex valued) continuous functions on MATH. Clearly, this family converges uniformly to the constant MATH function as MATH. Employing the continuous function calculus CITE, it follows that MATH converges uniformly to MATH as MATH. Thus, MATH. Since MATH and MATH commute, we may write MATH, MATH, where MATH and MATH are pairwise orthogonal projections. Then, MATH. Choose MATH such that MATH for all MATH. Suppose that MATH for some MATH. Then MATH and we may choose a unit vector MATH. But then MATH, which contradicts the fact that MATH. Thus, MATH for all MATH, and by symmetry MATH for all MATH. Hence, for all MATH, MATH, that is, MATH. Employing the functional calculus for MATH again, we see that MATH converges uniformly to MATH as MATH; thus, MATH uniformly as MATH. We may then compute, MATH . The second (and fourth) equalities follow since right (and left) multiplication by MATH is norm continuous. The third equality follows since there is a MATH such that MATH for all MATH. Therefore, MATH. |
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