paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0005234 | The simpler case is one where MATH is a genuine triangulation. Then, it is clear that a small perturbation of the vertices of MATH doesn't change the combinatorics of MATH, and so continuity follows from REF and the continuity of the cross-ratio. Things are very slightly more complicated when MATH has non-triangular fa... |
math/0005234 | Let MATH be a sequence of metrics on MATH converging to a metric MATH. Let MATH be such that MATH. We will show that there exists a MATH such that MATH. First choose a subsequence MATH such that all of the MATH have the same combinatorics. This is possible since the number of possible combinatorial structures is finite... |
math/0005234 | (also see REF ) It is easy to see that any minimal cutset as above is actually the set of internal edges of a triangulation MATH of an annulus MATH (possibly with one boundary component collapsed to a point MATH , if all of the MATH are incident to MATH . ) From now on all references will be to quantities in MATH . Let... |
math/0005234 | Let MATH be a complete finite-volume hyperbolic surface homeomorphic to MATH. Let MATH and MATH be two different embeddings of MATH into MATH as convex polyhedra. If MATH and MATH are combinatorially equivalent, REF implies that MATH and MATH are congruent. Assume that MATH and MATH are not combinatorially equivalent. ... |
math/0005235 | The basic approach is to use the fundamental relation REF. We will first show, using REF, that the characteristic function of MATH is bounded by MATH for each MATH. Mixing, this yields REF for MATH. Then we will use another consequence of REF, namely, the functional equation MATH or rather its consequence MATH and obta... |
math/0005235 | The case MATH is REF , and the case MATH follows by MATH. The remaining cases follows from these cases by induction on MATH and the following calculus lemma. |
math/0005235 | Fix MATH and let MATH. For MATH, MATH and thus, integrating from MATH to MATH, MATH . Consequently, MATH and the result follows. |
math/0005235 | For each MATH with MATH, the random variable MATH [with notation as in REF ] has the density function MATH where the integral converges for each MATH since, using REF , the integrand is bounded by MATH; dominated convergence using the continuity of MATH and the same bound shows further that MATH is continuous. This arg... |
math/0005235 | We again use the notation REF from the proof of REF . Fix MATH and MATH. Since MATH is almost everywhere positive CITE, the integrand in REF is positive almost everywhere. Therefore MATH. Now we integrate over MATH to conclude that MATH. |
math/0005236 | Setting MATH, REF implies MATH . Thus MATH is continuously differentiable on MATH and satisfies the differential equation MATH there. This is an easy differential equation to solve for MATH, and we find that MATH for some MATH. After rearrangement, the proposition is proved. |
math/0005236 | Let MATH . From REF, we see immediately that, for MATH, MATH . Therefore, for MATH, REF yields MATH where MATH . Consequently, again for MATH (but then trivially for all MATH), MATH . Fix MATH; later in the proof we shall see that MATH suffices for our purposes. Since MATH as MATH, we can choose MATH such that MATH. Th... |
math/0005236 | Combining REF - REF , we obtain MATH . Thus the integral MATH converges absolutely, and from REF we obtain the desired conclusion about MATH. |
math/0005236 | Combining REF - REF , we readily obtain MATH. Therefore, by REF , MATH with MATH. Since MATH for all MATH, we must have MATH. |
math/0005236 | Let MATH be independent random variables, with every MATH and every MATH. Then, using the definition of MATH repeatedly, MATH where we define the random variables MATH as follows. Using MATH in the obvious fashion, split the unit interval into intervals of lengths MATH and MATH. Now using MATH and MATH, split the first... |
math/0005236 | Extend MATH to a transformation MATH on the class MATH of probability measures on MATH by mapping the distribution MATH of MATH to the distribution MATH of MATH where MATH, MATH, and MATH are independent, with MATH, MATH, and MATH, and where MATH is given by REF. (Note that we use the same uniform MATH for the MATH-s a... |
math/0005236 | As discussed in REF, it is simple to check that MATH (and that the elements of MATH are all distinct). Conversely, given MATH, let MATH and MATH be independent random variables (on some probability space); recall that MATH is the limiting NAME measure, with zero mean and finite variance. Write MATH, MATH, for the chara... |
math/0005243 | Let us consider a MATH-subalgebra MATH of MATH which is generated by MATH, MATH, MATH and MATH, MATH, MATH. Direct computation shows that MATH, MATH, MATH generate a commutative MATH-subalgebra of MATH and satisfy the following relations: MATH where MATH . The functions MATH, MATH, MATH define an action of MATH on MATH... |
math/0005244 | The first part of the lemma is CITE, the statement follows from the fact that MATH is finite and étale. If MATH is not connected, since MATH is an isomorphism, obviously MATH. If MATH then MATH is a product of two fields and MATH is not connected. We prove now the last part of the lemma. Using NAME exact sequence (see ... |
math/0005244 | By REF MATH and the restrictions of MATH to MATH and MATH are isomorphisms. Thus MATH is a smooth complete connected curve over MATH of genus MATH. Then MATH. The group MATH is trivial since MATH is an isomorphism and MATH is complete over MATH. Using the exact sequence REF we get MATH. The other étale cohomology group... |
math/0005244 | By CITE MATH with MATH a divisible subgroup, thus MATH. Using NAME Theorem on NAME groups and REF , we may calculate MATH. For MATH. Hence we are reduced to calculate MATH. We first calculate MATH. There is an exact sequence MATH which is part of the NAME sequence CITE. We have MATH using REF since MATH is a divisible ... |
math/0005244 | For MATH, MATH, the proof of the previous proposition works. We have a decomposition MATH as in REF . The closed subset MATH of MATH consists of MATH closed points. Again by REF , MATH with MATH complete over MATH. Obviously MATH consists of MATH closed points of MATH. Using the exact sequence MATH for MATH, MATH and M... |
math/0005244 | We may assume that MATH. Let MATH denote the semi-algebraic connected components of MATH. By REF, there exist MATH such that the signature of MATH, denoted by MATH, is MATH on MATH and MATH outside. Clearly one gets MATH and MATH since its signature is not divisible by MATH. In CITE the author has defined homomorphisms... |
math/0005244 | With the previous notations MATH has a structure of a compact REF-manifold, more precisely a sphere with MATH handles. By a comparison theorem CITE, for any finite abelian group MATH we have MATH and MATH. We write MATH. Let MATH where MATH is a small closed ball of MATH centered at MATH. Then MATH. Using NAME exact se... |
math/0005244 | Assume MATH. We have MATH since MATH is the kernel of the total signature homomorphism MATH. Since MATH, MATH is a group of exponent MATH. By REF , MATH. Since MATH is surjective REF and MATH, the proof is done. Assume MATH. Thus MATH and MATH. Hence MATH is a group of exponent MATH and we have to determine MATH and MA... |
math/0005244 | Let MATH. We have a short exact sequence MATH which induces a long exact sequence in NAME cohomology MATH . Since MATH we get MATH. Moreover, MATH induces a flat pull-back (see CITE) MATH which is an injection and respects rational equivalence. We obtain an injection MATH. The statement follows now easily. |
math/0005244 | We set MATH. We fix a basis MATH of MATH and we get the associated MATH for MATH. We first claim that any MATH can be written uniquely as a product MATH with MATH and MATH. We look at MATH as a rational function on MATH, then MATH is an integral combination of points in MATH and the degree of MATH is zero. The divisor ... |
math/0005244 | We have MATH and for any MATH an exact sequence MATH . By CITE MATH, if MATH is odd one has to drop all summands MATH (For MATH it gives the result of REF ). For MATH, we denote by MATH the MATH-torsion subgroup of MATH. Using REF and the exact sequence MATH, the sequence MATH is exact if MATH is even, and MATH is exac... |
math/0005244 | By a topological computation, we may prove that MATH and then MATH for any MATH. By REF and since MATH is algebraically closed, MATH for any MATH. Arguing as for real curves, for any MATH, MATH, thus we get an exact sequence MATH . If MATH is an abelian group, MATH is the NAME dual of MATH. Suppose that MATH is finite,... |
math/0005244 | The first assertion is clear. If MATH then MATH since MATH is not complete. Since MATH injects into MATH, the class of a divisor MATH is zero in MATH if only if its class is zero in MATH. It is the case, in particular, if MATH is contained in MATH, which gives the proof. |
math/0005244 | The level of MATH is MATH if and only there exists MATH such that MATH. Assume such MATH exists and let MATH. Clearly MATH and since MATH one gets MATH. This proves that MATH. Conversely assume MATH lies in MATH. It corresponds to the divisor of a non trivial element MATH in MATH. Moreover MATH, so MATH and we may assu... |
math/0005244 | Consider the short exact sequence MATH which induces a long exact sequence MATH . Now MATH by NAME 's REF and one gets an exact sequence MATH with MATH denoting the norm. We have to understand the boundary map MATH. An element of MATH could be seen as the class of MATH (denoted by MATH) with MATH and MATH. Then MATH co... |
math/0005244 | The proof is straightforward using the exact sequence MATH since every positive function MATH on MATH is a sum of two squares in MATH. |
math/0005244 | We have MATH by REF . The curves MATH and MATH are isomorphic over MATH by MATH. Hence MATH. Since they have the same number of points at infinity, REF shows that MATH. |
math/0005248 | The function MATH factors as follows. MATH for MATH, with the interpretation that the function MATH is MATH when MATH in MATH. |
math/0005248 | See CITE for details. The following are some remarks of relevance to the general extension problem for operators. The assertion in the theorem about MATH-translations tiling the plane with MATH is also clear from REF - REF , and it is illustrated graphically in REF . It is clear that the pattern REF for MATH continues ... |
math/0005248 | The proof is based on NAME 's deficiency-space analysis of self-adjoint extensions, and we refer to CITE, CITE, and CITE for background material on the theory of operator extensions. If MATH is a symmetric operator with dense domain MATH in a NAME space MATH, then it has self-adjoint extensions if and only if the two s... |
math/0005248 | The realization on the space MATH is the interpretation of MATH as a unitary representation of the group MATH which is induced from the subgroup MATH via REF . The advantage of this viewpoint is that the spectral resolution of the unitary operator MATH leads directly to an associated direct integral decomposition for t... |
math/0005248 | Recall that some fixed unitary one-parameter group MATH on MATH is determined uniquely by the corresponding boundary operator MATH. But it follows from REF that MATH satisfies the commutativity property of the theorem if and only if it is diagonalized by the basis functions MATH in MATH for some MATH, that is, if, for ... |
math/0005248 | When the two one-parameter groups MATH and MATH are written in the form REF from REF , then the alternatives in REF may be expanded as follows. Let MATH denote periodic translation in MATH, and let MATH denote the projection of MATH onto MATH, with MATH denoting then the projection onto the complement MATH, for MATH. W... |
math/0005248 | Recall from REF - REF that the two one-parameter groups are expressed in terms of multiplication operators on MATH with the respective indicator functions MATH and MATH. The sequences REF - REF are the NAME coefficients of these indicator functions, acting by multiplication in MATH, and the relations REF - REF simply r... |
math/0005248 | Consider the formula MATH and expand the inside function, MATH according to the quasi-periodicity assumption on MATH: specifically, MATH . Setting MATH and using MATH together with NAME summation (also in the second variable) we arrive at the desired formula. |
math/0005248 | Since the commutativity for the one-parameter groups of unitary operators MATH may be stated for pairs, that is, MATH, MATH, MATH, MATH, the argument for the general case MATH is the same as for MATH. To see this, just use the formulas for the respective one-parameter groups which are analogues to REF - REF in the proo... |
math/0005254 | CASE: By NAME 's equations, we get MATH for a horizontal vector field MATH and for vertical vector fields MATH and MATH. Thus MATH for every vertical vector field MATH. Therefore MATH is injective and MATH is surjective. CASE: By NAME 's equations, we have MATH for horizontal vector fields MATH, MATH and MATH. If MATH,... |
math/0005254 | Let MATH and MATH. Let MATH be an orthonormal basis in MATH. Let MATH be the horizontal liftings along the fibre MATH of MATH, MATH,, MATH, respectively. Let MATH for each MATH. Since MATH is constant along the fibre MATH and MATH we see that MATH is a global orthonormal basis of the tangent bundle of the fibre MATH, w... |
math/0005254 | Normalizing the metric on MATH, we can suppose MATH. Let MATH. Since the tangent bundle of the fibre MATH is trivial, we can choose a global orthonormal frame MATH for the tangent bundle of MATH. We have MATH, MATH, and card-MATH. CASE: Let MATH be the horizontal lifting along the fibre MATH of a vector MATH, so that M... |
math/0005254 | By REF , we have MATH for some nonnegative integers MATH and MATH. Hence MATH. Since the right hand side is the sum of two non-negative numbers, it follows that MATH and MATH. Therefore MATH. This implies MATH. |
math/0005254 | If MATH is geodesically complete, then so is MATH (see CITE or CITE). Since MATH is a complete semi-Riemannian manifold and the fibres are totally geodesic, any fibre is also geodesically complete. By a theorem in CITE, it follows that the horizontal distribution MATH is an NAME connection. Therefore, by CITE, we see t... |
math/0005254 | For each tangent vectors MATH, MATH of MATH, we have the following formulas for the curvature tensors: CASE: If MATH and MATH is the natural complex structure on MATH, then MATH . CASE: If MATH and MATH are local almost complex structures which give rise to the quaternionic structure on MATH, then MATH . By these expli... |
math/0005254 | Since MATH for some positive integer MATH, we get MATH. Let MATH be a horizontal vector field along a fibre MATH such that MATH and MATH is the horizontal lifting of some tangent vector of MATH. First, we shall prove that MATH . Suppose that MATH. Then MATH is bijective. For every MATH we get MATH for some vertical vec... |
math/0005254 | First, we shall discuss the case MATH. By REF , MATH is isometric to one of the semi-Riemannian manifolds MATH, MATH, MATH for some MATH. Let MATH and let MATH such that MATH, and let MATH be the subspace in MATH given by MATH . Let MATH and let MATH be the horizontal lifting vector at MATH of MATH. By NAME 's equation... |
math/0005254 | Since MATH has constant curvature, the curvature of MATH is MATH and MATH. By REF , MATH and MATH. Then either MATH or MATH. If MATH, then MATH. If MATH then MATH. Summarizing, we have MATH. If MATH, then, by REF , we obtain MATH. Hence, by CITE, we have REF . If MATH, then, by CITE, we have REF . The idea of the proof... |
math/0005254 | Let MATH. Let MATH be two orthonormal bases of MATH along MATH and of MATH along MATH constructed as in the proof of REF such that MATH for MATH, MATH for MATH and for MATH, MATH and MATH. Let MATH be the linear map given by MATH, MATH, MATH for every MATH and MATH. In a manner similar to the proof of REF , we obtain M... |
math/0005254 | Let MATH be the canonical semi-Riemannian submersion. By REF, we see that MATH and MATH are semi-Riemannian submersions with totally geodesic fibres. We denote by MATH, MATH, MATH, MATH, MATH NAME 's integrability tensors of MATH, MATH, MATH, MATH, MATH, respectively. In order to reduce the proof of the equivalence the... |
math/0005254 | If MATH, then MATH is simply connected and hence, by REF , MATH is an isotropic semi-Riemannian manifold and MATH. By REF , we see that the base space of the semi-Riemannian submersion is isometric to a complex pseudo-hyperbolic space if the dimension of fibres is one, or to a quaternionic pseudo-hyperbolic space if th... |
math/0005254 | If the dimension of the fibres is less than or equal to MATH, then, by REF , MATH is equivalent to the canonical semi-Riemannian submersions: CASE: MATH, MATH or CASE: MATH, MATH . Now we assume that the dimension of the fibres is greater than or equal to REF. CASE: If we assume that the dimension of the fibres is grea... |
math/0005254 | Let MATH be the canonical semi-Riemannian submersion. By REF, one obtains that MATH is a semi-Riemannian submersion with connected totally geodesic fibres. CASE: If the dimension of the fibres of MATH is MATH and MATH, then the dimension of the fibres of the semi-Riemannian submersion MATH is less than or equal to MATH... |
math/0005254 | We suppose that there are such semi-Riemannian submersions. It is well-known that any quaternionic submanifold in MATH is totally geodesic. Let MATH, MATH, be the canonical semi-Riemannian submersions. By REF, we see that MATH is a semi-Riemannian submersion with connected totally geodesic fibres. We remark that the di... |
math/0005260 | MATH : MATH is obviously contained in the sum of all ideals MATH where MATH runs through the generators of MATH. On the other hand, let MATH be a (standard) bitableau contained in MATH, that is, MATH. If even MATH, then we can apply NAME expansion and write MATH as a linear combination of bitableaux of the the same sha... |
math/0005260 | CASE: It follows immediately from the definition of str-monotonicity that MATH is the MATH-vector space generated by the bitableaux MATH with MATH, and that the standard bitableaux MATH form a MATH-basis. CASE: We have to show that MATH for every standard bitableau MATH. First note that MATH. Set MATH. Since by REF is ... |
math/0005260 | Let MATH be a (standard) bitableau, and suppose that MATH is the biggest index such that MATH. Then obviously MATH. The verification of the inclusion MATH is likewise simple. |
math/0005260 | Let MATH be a (standard) bitableau. Suppose that MATH with MATH, and MATH (equality can only occur if MATH). Set MATH and MATH. Then MATH is the smallest ideal of type MATH containing MATH. Among the shapes of the elements in the standard basis of MATH there exists a unique element that is minimal with respect to the p... |
math/0005260 | Let MATH be the tableau obtained from MATH by the NAME insertion algorithm. We have already discussed REF 's theorem, namely that the sum MATH of the lengths of the first MATH rows of MATH is the length of the longest subsequence of MATH that has a decomposition into MATH increasing subsequences. But the theorem contai... |
math/0005262 | Let MATH and MATH. Denote with MATH the composition of an element in MATH and an element in MATH. Then by definition of the operator valued weight MATH we get MATH . By the right invariant version of CITE we get that MATH . From this it follows that MATH. So we get that MATH. If MATH and MATH we have MATH . So we get t... |
math/0005262 | We have MATH . Choose MATH. Define MATH by MATH for all MATH. Then MATH by invariance of MATH. So we may conclude that MATH . Then the result of the lemma follows immediately. |
math/0005262 | Choose MATH and MATH. Let MATH be an orthonormal basis for MATH. Choose MATH. Because MATH we have MATH . Hence applying MATH gives MATH in the MATH-strong-MATH topology. Choose now MATH. For every finite subset MATH we have by REF that the element MATH where MATH is the projection on MATH. So we get that the net MATH ... |
math/0005262 | Choose MATH and MATH. Then MATH and hence MATH with MATH . Because MATH is a MATH-strong-MATH - norm core for MATH we may conclude that for every MATH we have MATH and MATH . Then the lemma follows immediately. |
math/0005262 | Because MATH it follows from the previous lemma that for every MATH we have MATH and MATH . Taking the adjoint we may replace MATH by MATH in the formula above. Let now MATH. Then we have MATH and MATH . Because MATH we also have MATH and MATH . So we get MATH and so MATH with MATH. It now follows from the results of C... |
math/0005262 | Denote with MATH the NAME algebra of MATH-matrices over MATH. Denote with MATH the matrix units. Define MATH . Then MATH is an action of MATH on MATH. Denote with MATH the balanced weight on MATH (see for example, CITE) given by MATH . It is immediately clear that MATH is MATH-invariant for the action MATH. Let MATH. D... |
math/0005262 | Recall that the dual action MATH is an action of MATH on MATH. We claim that the weight MATH is MATH-invariant. Observe that MATH is the modular element of MATH and that is the reason to have MATH-invariance rather than MATH-invariance. To prove our claim, choose MATH, MATH, MATH and MATH. Then define MATH . It follows... |
math/0005262 | Let MATH. Because MATH implements the automorphism MATH on MATH we get that MATH will also leave MATH invariant. So we can define the automorphism group MATH on MATH by MATH . So, for every MATH we have MATH. Further we have MATH for all MATH and MATH. Here we used the formula MATH stated in the beginning of the paper.... |
math/0005262 | Let MATH and MATH. Let MATH. Then, by CITE, MATH and MATH . So MATH and MATH . Because the elements MATH span a core for MATH and because MATH is closed (both in the MATH-strong-MATH - norm topology), we have for all MATH that MATH and MATH . Denoting with MATH and MATH the relative modular apparatus of the weights MAT... |
math/0005262 | Because MATH we have MATH . So for every MATH and with MATH an orthonormal basis of MATH we have, by applying MATH in the MATH-strong-MATH topology. From this it follows that MATH for all MATH. But MATH for all MATH. So MATH is a well-defined MATH-homomorphism. By symmetry MATH will be surjective and hence it is a MATH... |
math/0005262 | Recall that MATH for all MATH and MATH. Because the positive operators MATH and MATH strongly commute, we can define the closure MATH of the product MATH. Denoting with MATH the characteristic function of a subset MATH we consider the following subspace of MATH. MATH . Let now MATH, MATH, MATH and MATH. Put again MATH ... |
math/0005262 | Consider the bidual action MATH of MATH on MATH. Let MATH be a n.s.f. weight on MATH and denote with MATH the bidual weight on MATH. It follows from REF that MATH is a MATH-invariant weight for the action MATH. With the notation of REF we define MATH. Then MATH will be a n.s.f. and MATH-invariant weight on MATH for the... |
math/0005262 | Because MATH we get easily that MATH . But MATH, so that we have MATH . Because MATH is a corepresentation the MATH-strong-MATH closure of MATH is self-adjoint and then the result follows. |
math/0005262 | Let us first suppose the first statement is valid. Because MATH is a MATH-strong-MATH closed, two-sided ideal of MATH we can take a central projection MATH such that MATH. Let MATH be the restriction of MATH to MATH. Then MATH is a MATH-isomorphism onto MATH. When MATH is a n.s.f. weight on MATH we have MATH for all MA... |
math/0005262 | Suppose the first statement is true. Because MATH is represented on the NAME of MATH, there exists a unitary MATH on MATH such that MATH for all MATH and MATH. Define MATH from MATH to MATH by MATH for all MATH. Then MATH for all MATH. Further MATH . Because MATH, we get MATH, which leads to MATH. So we may suppose fro... |
math/0005262 | When MATH is integrable, one can use REF and then observe that the MATH-homomorphism MATH is faithful because MATH is a factor. Next suppose that the inclusions stated above are isomorphic. By REF there exists an integrable action MATH which is cocycle equivalent with MATH and satisfies MATH. Let MATH be a MATH-cocycle... |
math/0005262 | For clarity we stress that MATH is the operator valued weight MATH from MATH to MATH, that MATH is obtained out of MATH by modular theory and the basic construction, and it goes from MATH to MATH. Finally MATH is the canonical operator valued weight MATH from MATH to MATH, giving the dual weights by the formula MATH fo... |
math/0005262 | Using the notations introduced above we will identify the inclusions MATH and MATH. Then we get that MATH. Now it is obvious that MATH and MATH. So the restriction of MATH to MATH is semifinite. Next observe that MATH. Applying the first part of the proof to the dual action MATH, which is integrable and for which the M... |
math/0005262 | Choose a n.s.f. weight MATH on MATH and let MATH be the dual weight on MATH. Represent MATH on the NAME of MATH such that MATH is a NAME for MATH. Let MATH be the canonical NAME for MATH and denote with MATH the modular conjugation of MATH. Then it follows from REF that MATH is the unitary implementation of MATH. The b... |
math/0005262 | Because MATH is a factor the MATH-homomorphism MATH from REF is faithful. Then we apply REF to obtain the regularity of MATH and REF to get that MATH has depth REF. It is clear that MATH is irreducible, because MATH . |
math/0005262 | Let us first prove the first statement. Denote with MATH the dual action, which is an action of MATH on MATH. Because MATH is the right NAME weight of MATH, the role of MATH is played by MATH. So we have to find a unitary MATH satisfying MATH. Then it is clear that we can take MATH and so MATH is semidual. To prove the... |
math/0005262 | Let MATH be minimal. Let MATH. Then certainly MATH and hence MATH by minimality. We now claim that for MATH we have MATH if and only if MATH. Suppose MATH. It is clear that for every MATH we have MATH. So we get MATH. From this it follows that MATH. So we may conclude that MATH, where MATH. Because MATH we get that MAT... |
math/0005262 | Let MATH be the free group with a countably infinite number of generators MATH. It is well known that the free group factor MATH is a MATH-factor. Let MATH be the automorphism of MATH satisfying MATH for all MATH. Let MATH be the automorphism of MATH satisfying MATH for all MATH. Define the automorphism group MATH in t... |
math/0005262 | Consider the action MATH of MATH on MATH given by MATH for MATH. Here we used the notations of REF : MATH, MATH and MATH. Let us define now MATH . Using matrix notation and referring to REF , it is then clear that MATH if and only if MATH, MATH and MATH. Choose a n.s.f. weight MATH on MATH and represent MATH on the NAM... |
math/0005262 | The first statement follows easily from the definition of MATH. Let MATH, then MATH . So we get the first statement. To prove the second one we define MATH . It is clear that for MATH such a MATH is necessarily unique. We denote it with MATH. Then for every MATH we have MATH and MATH. Define MATH. We claim that MATH is... |
math/0005262 | Let MATH be the sequence of operators defined in the proof of CITE. Because MATH it is clear that MATH . Because MATH we have MATH. We know that MATH, so that MATH and MATH . Because MATH is MATH-strong-MATH - norm closed, the conclusion follows. |
math/0005262 | Let MATH and MATH. Then we have for all MATH that MATH . So we get that MATH and MATH. Take now a net MATH in MATH such that MATH in the MATH-strong-MATH topology for all MATH. Then we have for all MATH . Hence MATH for all MATH. Because MATH is MATH-strong-MATH - norm closed we get MATH and MATH. |
math/0005262 | Because MATH is an isometry, MATH is injective. Define MATH. Then MATH is a dense subspace of MATH. We make MATH into a MATH-algebra by using MATH and the MATH-algebra structure on MATH. We claim that MATH is a left NAME algebra. The only non-trivial point is to prove that the map MATH for MATH is closable. But, suppos... |
math/0005265 | Choose MATH. Take MATH. Using REF , we get for all MATH, MATH implying that MATH and hence MATH. We conclude that MATH. Therefore MATH. |
math/0005265 | By definition of MATH, we have that MATH . So we get for all MATH and MATH that MATH implying that MATH. Arguing as in the proof of REF , the lemma follows. |
math/0005265 | Choose MATH and MATH. Then MATH . Define MATH such that MATH for all MATH. It is clear that MATH. Now define MATH by setting MATH for MATH, then it is clear that MATH. Thus, MATH . |
math/0005265 | For MATH, MATH and MATH, we have MATH . This implies the existence of an isometric linear map MATH such that MATH for all MATH, MATH and MATH. Choose MATH, MATH and MATH. Then MATH where we used the normality of MATH and REF . Now we get for all MATH, MATH and MATH that MATH . Hence MATH. From this all we can also conc... |
math/0005265 | CASE: Since MATH strongly, the net MATH is a bounded net in MATH that converges to REF in the weak operator topology. Therefore the normality of MATH implies that the net MATH also converges to REF in the weak operator topology. Thus MATH converges to REF in the weak operator topology. It follows that MATH converges st... |
math/0005265 | CASE: Choose MATH. We have for all MATH and MATH that MATH from which it follows that MATH. CASE: Choose MATH, MATH and MATH, then MATH implying that MATH for all MATH. |
math/0005265 | Referring to REF we get for all MATH, MATH and MATH that MATH . From this chain of equalities the existence of MATH follows in the usual way. Choose MATH. Take MATH, MATH and MATH. Then, applying result REF twice and remembering that MATH and MATH, we get that MATH . Hence MATH. We conclude from this that MATH belongs ... |
math/0005265 | Define MATH. For all MATH, we have that MATH implying that MATH belongs to MATH. Let MATH denote the multiplicative unitary of MATH in the NAME MATH. We know that MATH for all MATH. Choose MATH, MATH and MATH. Fix also an orthonormal basis MATH for MATH. CASE: We have that MATH . Choose MATH. Using REF in the first and... |
math/0005265 | For MATH and MATH, we have MATH . This implies the existence of an isometry MATH such that MATH for all MATH and MATH. It is then immediately clear that MATH has dense range and is therefore unitary. Let MATH be any NAME space and MATH an element in MATH. Choose MATH, MATH. Take also an orthonormal basis MATH for MATH,... |
math/0005265 | Choose MATH, MATH, MATH. Then the previous proposition implies that MATH and the proposition follows. |
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