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math/0005234
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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 faces. Then, MATH is combinatorially unstable: a small perturbation in the vertices changes the combinatorial structure, but only in the following simple fashion: For a sufficiently small MATH, a perturbation MATH of MATH is combinatorially equivalent to MATH with some diagonals added to the non-triangular faces. There exists a MATH, such that if a point MATH is closer than MATH to MATH then MATH, MATH, MATH, and MATH are co-circular, for any triangle MATH and vertex MATH of MATH. Every way of adding diagonals to MATH until we get a triangulation corresponds to a different coordinate system of MATH (where MATH is the number of vertices of MATH), and REF shows that every sufficiently small perturbation of MATH is close to MATH in at least one of the coordinate systems. Since the transition maps between the various coordinate systems are continuous REF follows.
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math/0005234
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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. As before, the vertices and faces of MATH are labelled in such a way that MATH, MATH, MATH, and MATH lies above the real axis. By compactness of the sphere MATH, there exists a limiting tessellation MATH. If MATH is non-degenerate (that is, no two vertices of a triangle have coalesced into one), then by REF it follows that MATH. We will show that MATH is always non-degenerate. If this is not the case, let MATH be a collapsing face of MATH which is abutting a non-collapsing face MATH. Such a pair of faces must exist, since at least one face REF is not collapsing. By relabelling and a hyperbolic isometry, send MATH to the triangle MATH, and MATH to MATH. Since MATH, and MATH is a non-degenerate metric it follows that MATH stays away from MATH and MATH, and so MATH is not collapsing after all.
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math/0005234
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(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 the inner and outer boundary components of MATH be MATH and MATH, respectively. The edges of MATH naturally fall into three categories - outer boundary edges, inner boundary edges and internal edges. Similarly, divide the angles of the triangles of MATH into the three sets - MATH (angles opposite outer boundary), MATH (angles opposite inner boundary) and MATH (angles opposite inner edges). Obviously, MATH (where MATH is the number of triangles and the cardinality of the cutset). Furthermore, by REF , MATH . Now, note that the sum of the angles incident (not opposite) to MATH is MATH, and further note that this sum is equal to MATH . That is true since if MATH is opposite to MATH, then the other angles of the triangle containing MATH are incident to MATH . Now, since MATH it follows that MATH . Since MATH is greater than zero precisely when the inner boundary of MATH is non-degenerate, it follows that MATH whenever the inner boundary of MATH is non-degenerate and MATH otherwise. The statement of the theorem then follows from REF , and REF.
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math/0005234
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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. Then MATH and MATH induce two different cell decompositions MATH and MATH of MATH, where the edges of MATH are preimages of corresponding edges of MATH. Produce a new cell decomposition MATH of MATH by superimposing MATH and MATH. The vertex set of MATH is the union of MATH (the cusps of MATH) with the set MATH of intersections of edges of MATH with those of MATH. The image of MATH under MATH will be MATH with some extra edges and vertices drawn on it. Then we can treat MATH and MATH as being of the same type (that of MATH), and REF may be applied. Thus MATH and MATH are congruent.
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math/0005235
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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 obtain successive improvements in the exponent MATH. We give the details as a series of lemmas, beginning with a standard calculus estimate CITE. Note that it suffices to consider MATH in the proofs because MATH and thus MATH. Note also that the best constants satisfy MATH (although we do not know in advance of proving REF that these are finite), and thus MATH, which clearly satisfies REF because MATH. Suppose that a function MATH is twice continuously differentiable on an open interval MATH with MATH . Then MATH . By considering subintervals MATH and letting MATH, we may without loss of generality assume that MATH is defined and twice differentiable at the endpoints, too. Then, using integration by parts, we calculate MATH . So MATH . For any real numbers MATH and MATH, the random variable MATH defined by REF satisfies MATH . We will apply REF , taking MATH to be MATH. Observe that MATH and that MATH . Let MATH and MATH. If in REF we take MATH and MATH, and assume that MATH, then note MATH . So, by REF , MATH . Trivially, MATH so we can conclude MATH . This result is trivially also true when MATH so it holds for all MATH. The optimal choice of MATH here is MATH, which yields MATH . Similarly, for example by considering MATH, MATH and we conclude that the lemma holds for all MATH, and thus for all real MATH. For any real MATH, MATH. REF shows that MATH and thus MATH . The preceding lemma is the case MATH of REF . We now improve the exponent. Let MATH. Then MATH . By REF and the definition of MATH, MATH and the result follows by evaluating the beta integral. In particular, recalling MATH, REF yield MATH . This proves REF for MATH, with MATH, and thus by REF for every MATH with MATH; applying REF again, we obtain the finiteness of MATH in REF for all MATH. Somewhat better numerical bounds are obtained for MATH by taking a geometric average between the cases MATH and MATH: the inequality MATH shows that MATH, MATH. In particular, we have MATH, and thus, by REF , MATH. Let MATH. Then MATH . Assume that MATH. Then, again using REF , MATH . We have derived the desired bound for all MATH. But also, for all MATH, we have MATH so the estimate holds for all MATH. REF completes the proof of finiteness of every MATH in REF (by induction), and of the estimate REF. The bound for MATH obtained above now shows (using Maple) that MATH, which then gives MATH. We can rewrite REF as MATH . Hence, by induction, if MATH for a nonnegative integer MATH, then MATH where MATH, using the above estimate of MATH. Consequently, MATH when MATH. For general MATH we now use REF with MATH and MATH, obtaining MATH; the case MATH follows from REF and the estimate MATH. This completes the proof of REF and hence of REF .
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math/0005235
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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.
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math/0005235
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Fix MATH and let MATH. For MATH, MATH and thus, integrating from MATH to MATH, MATH . Consequently, MATH and the result follows.
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math/0005235
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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 argument yields the bound MATH, and since MATH by symmetry in REF, we have MATH. This uniform bound, REF, and dominated convergence again imply that MATH is a continuous density for MATH, and thus equals MATH for every MATH.
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math/0005235
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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.
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math/0005236
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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.
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math/0005236
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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. Then, for MATH, MATH and thus MATH . Since MATH is bounded, this is trivially true also for MATH. Summarizing, for some constant MATH we have, with MATH, MATH . Now fix the value of MATH to be any number in MATH, say MATH. Then a straightforward induction [substituting REF into REF for the induction step] shows that for any nonnegative integer MATH we have, for all MATH, MATH . Recalling that MATH is bounded and letting MATH, we obtain the desired conclusion, with MATH.
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math/0005236
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Combining REF - REF , we obtain MATH . Thus the integral MATH converges absolutely, and from REF we obtain the desired conclusion about MATH.
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math/0005236
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Combining REF - REF , we readily obtain MATH. Therefore, by REF , MATH with MATH. Since MATH for all MATH, we must have MATH.
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math/0005236
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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 interval into subintervals of lengths MATH and MATH and the second interval into subintervals of lengths MATH and MATH. Continue in this way (using MATH) until the unit interval has been divided overall into MATH subintervals. Call their lengths, from left to right, MATH. Let MATH. We show that MATH converges in probability to MATH as MATH. Luckily, the complicated dependence structure of the variables MATH does not come into play; the only observation we need is that that each MATH marginally has the same distribution as MATH. Indeed, abbreviate MATH as MATH; briefly put, we derive a NAME 's bound for MATH and then simply use subadditivity. To spell things out, let MATH be fixed and let MATH. Then MATH . Choosing the optimal MATH (valid for MATH), this yields MATH and thus MATH as MATH. Since MATH converges in probability to MATH, we can therefore choose MATH so that MATH. To prove the theorem, it then suffices to prove MATH . For this, we note that the characteristic function MATH of MATH is given for MATH by MATH . We will show that MATH converges to MATH for each fixed MATH, and [since, further, MATH] this will complete the proof of the lemma. Indeed, we need only consider the second term in REF. For that, the calculus estimates outlined in the proof of the lemma preceding REF demonstrate that, when MATH, MATH for complex random variables MATH (depending on our fixed choice of MATH) satisfying MATH for a deterministic sequence MATH [with MATH as MATH]. [Leaving out the error estimates, the argument is MATH . It now follows easily that MATH, as desired.
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math/0005236
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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 as for the MATH-s!) Of course, MATH maps the marginal distributions MATH of MATH and MATH of MATH into MATH and MATH, respectively; more importantly for our purposes, it maps the distribution, call it MATH, of MATH into the distribution MATH, with MATH defined at REF. Now let MATH have marginals MATH, MATH. Then MATH has constant marginals MATH as MATH varies and so is a tight sequence. We then can find a weakly convergent subsequence, say, MATH of course, the limit MATH again has marginals MATH, MATH. Moreover, MATH . But, by supposition, the characteristic function of MATH satisfies REF, so REF implies that MATH is Cauchy-MATH. Thus MATH supplies the desired coupling.
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math/0005236
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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 characteristic functions corresponding respectively to MATH, MATH. By REF (or see REF ), MATH satisfies REF [for some MATH]. Of course, MATH satisfies REF with MATH taken to be MATH, so the characteristic function MATH of MATH satisfies REF for the same MATH as for MATH. Applying REF , there exists a coupling MATH of MATH and MATH such that MATH . Cauchy-MATH. Without loss of generality (by building a suitable product space), we may assume the existence of a random variable MATH on the same probability space as MATH and MATH such that MATH and MATH are independent. We know that the distribution MATH of MATH is a fixed point of MATH. But so is the distribution MATH of MATH. By REF applied to MATH, MATH, as desired.
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math/0005243
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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 with orbits MATH . Here and in the sequel we denote by MATH the MATH-th iteration of MATH and MATH, MATH, the MATH-th coordinate of MATH. Let MATH be a MATH-representation of MATH on a NAME space MATH by bounded operators, let MATH be the resolution of the identity for the commutative family MATH of the positive operators MATH, MATH, MATH and let MATH be the joint spectrum of the family MATH. Next step is to show that any irreducible representation is concentrated on an orbit of this dynamical system. If MATH is an irreducible representation of MATH then the spectral measure MATH is ergodic with respect to the action of the dynamical system generated by MATH, MATH, MATH and there exists an orbit MATH such that MATH. From REF and the spectral theorem it follows that MATH for any MATH. Hence any subset MATH such that MATH, MATH, MATH, MATH, MATH defines a subspace MATH which is invariant with respect to the operators MATH, MATH for any MATH as above. Moreover, such subspace is invariant with respect to any operator of the representation MATH. In fact, the following relations hold in MATH, MATH, MATH, which gives MATH . Therefore if MATH is invariant with respect to all MATH and MATH we obtain MATH and hence MATH for any MATH. Taking MATH gives MATH, that is, MATH. Similarly, MATH. The ergodicity of the measure MATH follows immediately, that is, MATH or MATH for any NAME MATH which is invariant with respect to MATH, MATH. The simplest invariant sets are the orbits of the dynamical system. The next step is to show that only atomic measures concentrated on an orbit give rise to irreducible representation of the MATH-algebra. It is easily seen that the dynamical system generated by MATH is one-to-one and possesses a measurable section, that is, a set MATH which intersects any orbit in a single point. This implies that any ergodic measure is concentrated on a single orbit of the dynamical system and therefore MATH for some orbit MATH. We now clarify which orbits MATH give rise to bounded irreducible representation MATH, that is, MATH, and classify all such representations up to unitary equivalence. We claim first that there is no bounded representations MATH with MATH if MATH. From REF we have MATH where MATH is the eigenspace for MATH corresponding to the eigenvalue MATH. Since MATH, where MATH, implies MATH we conclude that MATH and MATH. This clearly forces MATH for any MATH. However, the set MATH is unbounded which contradicts the boundness of the representation MATH. Similar arguments show that there is no bounded representation MATH with MATH, MATH or MATH. In this case MATH. The only possibility is MATH and in this case we obtain MATH, MATH. It follows now from REF - REF that MATH, MATH satisfy the relations MATH . This implies that MATH commutes with all images of the generators in the algebra under th e representation MATH and therefore MATH is a multiple of the identity operator if MATH is irreducible. By REF we have also MATH, MATH. Irreducible representations of the relation MATH are well-known and can be easily calculated using the method of dynamical systems (see CITE). Any such representation is either one-dimensional: MATH, MATH, or infinite-dimensional which is unitary equivalent to the following one MATH. The corresponding irreducible representations of MATH are MATH and MATH. Since MATH and MATH as MATH, it follows from REF that MATH, MATH and the corresponding orbit contains a point MATH. We have MATH. Similar arguments show that MATH, where MATH or MATH, is impossible if the representation MATH is bounded. From the positiveness of MATH we obtain also that the only orbits corresponding to irreducible representation of the MATH-algebra are MATH, MATH, MATH, MATH and MATH . The last one was treated above. We consider now the case MATH, MATH. Let MATH, MATH be the projection onto the eigenspace corresponding to the eigenvalue MATH. Using REF we get MATH (``+" for MATH and ``-" for MATH) MATH. By REF we have MATH and MATH . Setting MATH the projection onto an eigenspace which corresponds to the eigenvalue MATH we obtain MATH that is, MATH . Moreover, MATH. The operator MATH can be written as a sum of its diagonal part MATH, and the operator MATH. Let now MATH, where MATH or MATH. It follows from REF by direct computation that MATH . The only bounded operator MATH satisfying this relation is the zero-operator. Therefore MATH and MATH is irreducible iff so is the family MATH. Let MATH be the polar decomposition of MATH. Using easy arguments one can show that MATH, MATH and MATH . Here MATH, MATH. Moreover, if MATH REF we have MATH (MATH respectively) commutes with any operators from the family MATH and therefore with any operator of the representation. This clearly forces MATH, MATH (MATH, MATH respectively). Let MATH. Consider MATH, MATH, MATH. Then MATH is an orthonormal system which defines an invariant subspace. The corresponding irreducible representation is MATH. Analogously MATH, MATH, MATH, build an orthonormal basis of an irreducible representation space if MATH, the corresponding action is given by formulae REF . If MATH we have that MATH, MATH is invariant with the corresponding action given by REF . We now turn to the case MATH. From REF - REF we have MATH . Note that MATH, MATH. Moreover, it follows from the above relation that if MATH is irreducible then the family MATH restricted to the subspace MATH is irreducible for any MATH. We have MATH where MATH . Any irreducible family (MATH, MATH) is either one-dimensional and given by MATH, MATH, or infinite dimensional defined on MATH by MATH. These representations give rise to irreducible representations of the MATH-algebra MATH. Namely, in the first case we have that MATH, where MATH, MATH, define an orthonormal basis of the space where the irreducible representation MATH acts, and for the second irreducible family we have that MATH, MATH, define an orthonormal basis of the space where the irreducible representation MATH acts. This finishes the proof.
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math/0005244
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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 next section) for the field MATH and MATH, we have a commutative diagramm with exact lines MATH . Since MATH is smooth MATH and by a previous exact sequence the map MATH is injective. By snake lemma the map MATH is also injective and the proof is done.
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math/0005244
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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 groups could be deduced from the previous remarks. We may calculate MATH in a different way. The exact sequence MATH gives MATH . We already know MATH, MATH. Let MATH denote the kernel of the degree map defined on MATH, then we have an exact sequence MATH . It is well known that MATH is a divisible group, hence the previous exact sequence splits since a divisible group is an injective MATH-module. Moreover MATH and MATH. Using MATH and MATH one gets MATH and MATH. Counting dimensions in the previous exact sequence MATH we obtain MATH.
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math/0005244
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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 group and MATH. Since MATH is a divisible group, using the split exact sequence MATH and REF , we have MATH. Since MATH and MATH, counting dimensions in MATH, one obtains MATH . Now, using the exact sequence MATH one gets MATH if MATH, and MATH if MATH.
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math/0005244
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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 MATH, we obtain MATH. Thus MATH . Then counting dimensions in MATH we have MATH . The result is compatible with the fact MATH and MATH are isomorphic via MATH REF .
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math/0005244
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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 . Let MATH and MATH denote the class of MATH in MATH, then MATH. We may prove that MATH is surjective using the classes of the MATH in MATH. Moreover the following diagramm is commutative MATH . Since MATH is an isomorphism, one gets the result.
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math/0005244
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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 sequence, MATH which gives MATH we find MATH. NAME formula and comparison theorem give the result.
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math/0005244
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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 MATH. By REF , the rank mod REF homomorphism MATH is an isomorphism and also the discriminant homomorphism MATH. Consequently MATH. To determine MATH we look at the following exact sequence: MATH . Since MATH, the only non-zero element is the class of MATH. If MATH then MATH in MATH. Since we have injections MATH, MATH and MATH, we get MATH that is, MATH. Conversely, if MATH then MATH. So MATH. Consequently MATH and MATH . If MATH that is, MATH then MATH and MATH .
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math/0005244
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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.
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math/0005244
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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 is in fact in MATH and clearly MATH. Hence MATH can be written uniquely as an integral combination of MATH. Thus we have the claim. The classes of MATH (and the constant function MATH if MATH is even) in MATH generate this group. Since MATH is a basis of MATH, we may prove that the order of MATH is exactly MATH in MATH and MATH (respectively, MATH) is the the minimal number of generators in MATH if MATH is odd (respectively, MATH is even). The statement follows now easily.
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math/0005244
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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 exact if MATH is odd. Then for any prime number MATH we get MATH . By a structure theorem on divisible groups MATH is a direct sum of some quasicyclic MATH-groups MATH for some primes MATH. The group MATH could be seen as the MATH-primary component of MATH. The result on the MATH-torsion part of MATH implies that, for any prime MATH, we have exactly MATH copies of MATH in the decomposition of MATH as a direct sum of quasicyclic groups. Thus the proof is done.
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math/0005244
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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, then MATH. According to the above remark, we get another exact sequence MATH which is a split exact sequence of MATH-modules. Hence MATH for any MATH. Since MATH is a divisible group, using a structure theorem on divisible groups, the statement follows.
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math/0005244
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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.
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math/0005244
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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 assume that it is MATH or MATH. We have shown that there is a one-to-one mapping between MATH and the set of functions MATH such that MATH modulo MATH. Hence the level of MATH is MATH if and only if there exists MATH such that MATH. Since MATH by NAME 's REF , MATH if only if MATH for MATH if only if MATH. We assume now that the level of MATH is MATH and MATH. Since MATH, it means that every MATH is invariant by MATH since it can be written as an integral combination of elements in MATH. But then MATH is trivial and MATH, contradiction.
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math/0005244
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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 corresponds to the class of MATH modulo MATH. We could see that MATH is well defined. By a well known Theorem of NAME, the NAME form MATH is universal over MATH, hence MATH and MATH is an isomorphism. We claim now that MATH is the cokernel of MATH (we don't have to assume that MATH), and the proof will be done. The short exact sequence of MATH-modules MATH gives a long exact sequence MATH . Since MATH, the statement follows easily.
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math/0005244
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The proof is straightforward using the exact sequence MATH since every positive function MATH on MATH is a sum of two squares in MATH.
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math/0005244
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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.
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math/0005248
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The function MATH factors as follows. MATH for MATH, with the interpretation that the function MATH is MATH when MATH in MATH.
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math/0005248
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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 to higher dimensions as follows: MATH with MATH, and MATH . Of course, then there are the obvious modifications of those cases resulting from permutation of the MATH coordinates; but the assertion is that, when MATH, these configurations do not suffice for cataloguing all the possible spectra MATH which turn MATH into a MATH-spectral pair. We now turn to the non-trivial spectral-theoretic content of the conclusion of the theorem. We claim that the two REF - REF suffice when MATH. Note that the sequence MATH is completely arbitrary. We will show in REF below that, up to a single translation in the plane, the possibilities for the coordinates of points in a spectrum MATH for MATH are given by two sequences MATH, MATH satisfying the following two cocycle relations: MATH as identities in MATH, and MATH. Note that the respective sequences are determined from this only up to MATH at each coordinate place. Simple algebra shows that the two identities REF - REF imply the following single identity MATH again for all MATH and MATH. But it follows from REF that at least one of the two sequences, MATH or MATH, must then vanish identically. This yields the connection to the two cases for MATH stated in REF - REF. Hence the result giving two classes for MATH in REF may be derived from our more general result in REF. The proof sketch of REF is completed for now, but details will be resumed in REF below.
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math/0005248
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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 spaces MATH have the same dimensions. In that case, the corresponding extensions are given by partial isometries between the respective defect spaces MATH and MATH (see CITE, CITE, or CITE). For convenience, we have chosen a slightly different ``normalization" in our treatment of the NAME transform REF and its inverse REF . We did not normalize the functions MATH and MATH in the defect spaces. They have MATH-norm equal to MATH. The fact that MATH in REF then defines a partial isometry as claimed amounts to the identities: MATH . This means that the vectors in the domain REF are given by the boundary REF which in turn determine the unitary one-parameter group MATH . This group is defined from REF by using translation modulo MATH in the MATH-variable. Then the operator MATH in REF is used in defining the representation MATH via induction from MATH. If MATH is a partial isometry, then the domain of the corresponding extension MATH REF is MATH and MATH . It follows that the lemma amounts to an identification of the defect spaces MATH when the symmetric operator is as specified. When the variables in MATH are separated as MATH, MATH, MATH, then vectors MATH are precisely the solutions to MATH . This amounts to solving MATH in the sense of distributions, but with the restrictions MATH. The result of the lemma then follows from NAME 's characterization. If the minimal operator is not closed at the outset, then the resulting self-adjoint extension comes from passing to the operator closure in REF .
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math/0005248
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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 the unitary one-parameter group MATH which is generated by the extension operator MATH.
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math/0005248
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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 some sequence MATH, MATH . But, according to REF , this means that MATH as an induced representation decomposes accordingly, which is to say that the basis vectors MATH simultaneously diagonalize each operator MATH as stated in REF .
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math/0005248
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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. We have MATH and MATH. Then from REF we get MATH . The assumed commutativity REF then takes the form: MATH . If MATH is not a scalar times MATH then two terms on either side are independent when evaluated on MATH. Hence both MATH and MATH hold. Addition of these two identities yields MATH which is the commutativity of MATH with periodic translation. If on the other hand MATH is a scalar, then it follows from the argument in REF that REF must hold. The two possibilities for the other boundary operator MATH lead to REF by symmetry.
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math/0005248
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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 reflect the following two obvious identities, MATH as functions on the unit interval. When the resulting REF - REF are substituted into MATH the equivalence to REF - REF results.
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math/0005248
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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.
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math/0005248
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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 proof of REF . For MATH, we may introduce the leg-notation: MATH, MATH, MATH. When evaluated at a general point in MATH of the form MATH, the respective eigenvalues are: MATH . Specifically, MATH where MATH are the fixed phase angles from the quasi-commutativity. Then the three pairs of cocycle identities from the theorem are as follows: REF - REF - REF , and REF - REF below. The argument for the equivalence of commutativity and the cocycle identities is essentially the same as the one used in the proof of REF above. The cocycle identities for MATH are: MATH and MATH .
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math/0005254
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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, MATH are basic vector fields, then MATH is constant along the fibre MATH. Therefore, MATH along the fibre MATH for every basic vector field MATH. Hence MATH along MATH. Since MATH is injective, it follows that MATH along the fibre MATH.
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math/0005254
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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, which makes the tangent bundle trivial.
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math/0005254
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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 MATH. By NAME 's equations, we have MATH for a horizontal vector field MATH and for a vertical vector field MATH. Along the fibre MATH we obtain for every MATH . Thus MATH is an orthonormal system. Hence MATH. Let MATH. For every integer MATH such that MATH, let MATH be a horizontal vector field along the fibre MATH such that MATH is the horizontal lifting of some unit vector (that is, MATH), that MATH is orthogonal to MATH and that MATH. Then, by REF , MATH belongs to MATH for every MATH. Therefore, for MATH and MATH, we get MATH along the fibre MATH. By NAME 's equations, we obtain MATH for basic vector fields MATH, MATH and for vertical vector fields MATH, MATH. Thus, along the fibre MATH we get for every MATH and MATH . Since MATH along the fibre MATH, it follows that MATH . We proved that for some positive integer MATH, MATH is an orthonormal basis of MATH along the fibre MATH. Thus MATH for some positive integer MATH. Counting the timelike vectors in MATH, we get MATH for some nonnegative integers MATH, MATH with MATH. CASE: Let MATH and MATH, MATH such that MATH. We shall construct an isometry MATH such that MATH and MATH. Note that we may assume that MATH. Let MATH, MATH be the horizontal liftings along the fibre MATH of MATH and MATH, respectively. Take MATH. Let MATH be two orthonormal bases constructed as above such that MATH, MATH, MATH for MATH, and that MATH and MATH are orthonormal bases of the tangent bundle of the fibre MATH, where MATH and MATH are the horizontal liftings along MATH of the vectors MATH and MATH, respectively (as in REF ), for which MATH for MATH. Let MATH be the linear map given by MATH, MATH, MATH for every MATH and MATH. Since both MATH, MATH are orthonormal bases, we see that MATH is a linear isometry. We shall apply REF . Thus we need to prove that MATH for every MATH, MATH. Indeed, we obtain for MATH and MATH, MATH . Hence MATH. MATH is a basic vector field along the fibre MATH for every MATH and MATH. We have MATH. For every basic vector field MATH along the fibre MATH we know that MATH is constant along the fibre MATH. Hence MATH is a basic vector field along the fibre MATH. Now we assume MATH. Since MATH, it follows that MATH. Hence MATH is a basic vector field along the fibre MATH if and only if the following conditions are satisfied: MATH is constant along MATH for every MATH, which is a basic vector field along MATH, and MATH is constant along the fibre MATH for every MATH, which is a basic vector field along MATH. If MATH, then MATH along MATH. So MATH along MATH. If MATH, then MATH along MATH. By NAME 's equations, we get along the fibre MATH since MATH and MATH. Hence MATH is constant along MATH for every MATH, which is a basic vector field along MATH. We proved that MATH is a basic vector field along MATH for every MATH and MATH. We denote by MATH the induced NAME connection on the fibre MATH. MATH . By the relation REF together with REF , we obtain for MATH and MATH that MATH . In the last equality we used the fact that MATH is a Killing vector field along the fibre MATH (see CITE or CITE). Thus MATH . The following assertions are true-MATH CASE: MATH. CASE: If MATH, then MATH. CASE: If MATH and if we set MATH, then MATH and MATH where MATH is the signature of the permutation MATH. Since MATH are Killing vector fields along MATH and MATH for every MATH, we get MATH for every MATH. CASE: The case MATH is not possible. Indeed, if MATH, then the relation MATH implies MATH. On the other hand, MATH since MATH and each fibre has constant curvature MATH. So we get a contradiction. CASE: If MATH, then MATH for every MATH and MATH, because MATH implies MATH. CASE: In the case MATH we shall prove MATH is constant along the fibre MATH. Since NAME 's integrability tensor MATH is skew-symmetric, it follows that MATH. Then MATH is a Killing vector field along MATH. We then obtain MATH . Analogously, we get MATH. We also obtain MATH since MATH and MATH is a Killing vector field along MATH. It is easy to see that MATH . Thus MATH is constant along the fibre MATH for each MATH. Therefore MATH is constant along MATH. Also, we compute for MATH . Hence MATH is a basic vector field for each MATH. We choose MATH. Since MATH is a basic vector field along MATH, we get the horizontal lifting along MATH of MATH is MATH. On the other hand, MATH is, by definition, the horizontal lifting of MATH along MATH. It follows that MATH along MATH. Thus MATH along the fibre MATH. For MATH, we choose MATH. If we repeat the argument above for the basis MATH, by REF , we get MATH along the fibre MATH. It follows that MATH for each MATH. Returning to the computation of MATH, in both cases MATH and MATH, we get for every MATH and MATH . Hence MATH and MATH . By REF we see that MATH extends to an isometry on MATH, denoted by MATH, such that MATH and MATH. Hence MATH and MATH. Since MATH for every MATH, we see, by REF , that there is an isometry MATH such that MATH. Thus MATH and MATH. Therefore MATH is an isotropic semi-Riemannian manifold. This completes the proof of REF .
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math/0005254
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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.
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math/0005254
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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 that MATH is a fibre bundle. So we obtain an exact homotopy sequence: MATH . Thus MATH.
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math/0005254
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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 explicit formulas for curvature tensors, in all cases we obtain REF .
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math/0005254
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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 vector MATH. It follows that MATH . Thus MATH for every MATH. By NAME 's equations, we have MATH for every horizontal vector field MATH along MATH. Hence MATH has constant curvature is a contradiction. We established that MATH . So we can find a horizontal vector field MATH along the fibre MATH such that MATH, MATH, MATH and MATH is the horizontal lifting of some MATH. We then have MATH . Since MATH is a simply connected isotropic semi-Riemannian manifold with nonconstant curvature, we see that MATH is isometric to one of the following semi-Riemannian manifolds: CASE: MATH, MATH, MATH, or CASE: MATH, MATH, MATH. We shall prove that only REF is possible. First, we suppose that MATH is isometric to one of the following semi-Riemannian manifolds: MATH, MATH, MATH. By REF , we get MATH . Therefore MATH for every vertical vector MATH. Since MATH and MATH are basic vector fields along MATH with MATH and MATH along MATH, it follows from the relation REF that MATH. On the other hand, by REF , we get MATH . Hence MATH and MATH. Thus MATH . So for any vertical vector MATH we get MATH . Since the induced metrics on fibres are nondegenerate, it is not possible to have both REF. So we obtain the required contradiction. It follows that MATH is isometric to one of the following semi-Riemannian manifolds: MATH, MATH, MATH. We shall now prove that MATH. Suppose MATH. By REF , we get MATH . Hence MATH from which follows that MATH. Therefore MATH for every vertical vector field MATH. In particular, we have MATH, which implies MATH for every horizontal vectors MATH and MATH. We have the following cases: CASE: MATH. We can choose vector fields MATH, MATH on MATH such that MATH and that one of the following conditions is satisfied: CASE: MATH if MATH, where MATH is the natural complex structure on MATH, CASE: MATH if MATH, where MATH are local almost complex structures which give rise to the quaternionic structure on MATH, or Let MATH, MATH be the horizontal liftings of MATH, MATH. REF then implies MATH . Hence MATH. Therefore MATH. On the other hand, we can choose horizontal vector fields MATH, MATH such that MATH, MATH and MATH, because MATH. Then REF becomes MATH. So we get a contradiction. CASE: MATH. Similarly, we can choose vector fields MATH, MATH on MATH such that MATH and MATH. REF then implies MATH. By the hypothesis of REF , we get MATH. On the other hand, REF becomes MATH. So we get a contradiction. We have proved MATH. REF then becomes MATH for tangent vector fields MATH, MATH on MATH. Then we have MATH for a vertical vector field MATH and for a horizontal vector field MATH. Hence MATH . Therefore the induced metrics on fibres are negative definite.
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math/0005254
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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 equations, we have MATH for horizontal vectors MATH, MATH. Since MATH is surjective and since the induced metrics on fibres are nondegenerate, we get MATH if and only if MATH. Thus MATH . We have the following possibilities: CASE: MATH is isometric to MATH. So MATH, MATH. From the geometry of the complex pseudo-hyperbolic space (see relation REF), we get MATH. It follows that MATH. CASE: MATH is isometric to MATH. So MATH, MATH. From the geometry of the quaternionic pseudo-hyperbolic space (see relation REF), we get MATH. It follows that MATH. CASE: MATH is isometric to the NAME pseudo-hyperbolic plane MATH. So MATH, MATH. From the geometry of the NAME pseudo-hyperbolic plane, we obtain MATH. Hence MATH. Now, we discuss the remaining case MATH. From MATH, we have either CASE: MATH, MATH, or CASE: MATH, MATH. If MATH, MATH, then MATH is a semi-Riemannian submersion with totally geodesic fibres from an anti-NAME space onto a Riemannian manifold. In this case, investigated by NAME in CITE, it follows that MATH is isometric to the complex hyperbolic space MATH and MATH. For MATH, MATH, we get, by REF , MATH with MATH. Thus MATH. It follows that MATH. Hence MATH is bijective. Since MATH, we see that MATH has constant curvature MATH, which contradicts our assumption of nonconstant curvature of the base space.
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math/0005254
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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 in CITE and CITE is to see that the tangent bundle of any fibre is trivial and that fibres are diffeomorphic to spheres, and then to apply a well-known result of NAME which claims that the spheres of dimensions REF are the only spheres with trivial tangent bundle.
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math/0005254
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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 MATH for every MATH, MATH. By REF, MATH extends to an isometry on MATH, denoted by MATH, satisfying MATH and MATH. From REF it follows that MATH induces an isometry MATH on MATH, such that MATH. Hence MATH and MATH are equivalent.
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math/0005254
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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 theorem of semi-Riemannian submersions from a complex pseudo-hyperbolic space to that from a pseudo-hyperbolic space, we need to establish relations among the integrability tensors MATH, MATH, MATH. First, we prove that MATH for MATH-basic vector fields MATH and MATH. Let MATH. Let MATH, MATH be two orthonormal MATH-vertical vectors in MATH and let MATH, MATH be the MATH-horizontal liftings at MATH of MATH, MATH, respectively. Let MATH be a unit MATH-vertical vector in MATH. Then MATH gives an orthonormal basis of MATH. Since the induced metrics on the fibres of MATH are negative definite, we have MATH . Thus MATH for MATH-basic vector fields MATH and MATH, where MATH denotes the metric on MATH and MATH is the NAME connection of MATH. Let MATH be the MATH-horizontal lifting along the fibre MATH of some unit vector in MATH. Let MATH, MATH, MATH be the MATH-horizontal liftings along the fibre MATH of MATH, MATH, MATH, respectively. Let MATH for MATH. As in REF , we choose MATH, which implies that MATH (see REF ). We remark that MATH is a MATH-vertical vector field along the fibre MATH. Indeed, we have MATH for any MATH. Since MATH, MATH are orthogonal to the vertical vector field MATH along MATH, we see that MATH, MATH are MATH-horizontal. Since MATH for MATH and for MATH, we obtain that MATH, MATH are MATH-basic vector fields along MATH. Thus MATH along MATH. Here MATH and MATH denote the MATH-horizontal and MATH-vertical projections, respectively. We also obtain that MATH. Therefore, MATH along MATH. We shall prove that MATH along MATH for every MATH-basic vector field MATH along MATH. We first obtain along MATH that MATH for a MATH-basic vector field MATH along MATH. Analogously, we get MATH. Thus MATH along MATH for every MATH-basic vector field MATH along MATH. Let MATH be an orthonormal basis of MATH along the fibre MATH constructed as in REF , for the semi-Riemannian submersion MATH. From the proof of REF , we have MATH for MATH, and MATH for MATH. We then obtain along MATH that MATH from which follows MATH. Hence MATH, because MATH is MATH-basic. We also have MATH. Let MATH. Summarizing all the above, we obtain that MATH is an orthonormal basis of the MATH-horizontal space MATH along the fibre MATH and MATH satisfies all conditions imposed in the construction of the basis MATH in the proof of REF . We notice that MATH along MATH, and that along MATH, MATH is equal to the MATH-horizontal lifting of MATH. Let MATH. Let MATH be an orthonormal basis of MATH along MATH constructed in the same way as MATH, but for the semi-Riemannian submersion MATH (see the proof of REF ), in such a way that MATH for MATH, MATH for MATH, and MATH. Let MATH be the linear map given by MATH, MATH for MATH and for MATH. By REF, MATH extends to an isometry MATH such that MATH and MATH. By the proof of REF , we have MATH for every MATH, MATH. By the proof of REF and by REF , MATH induces an isometry on MATH, denoted by MATH, such that MATH. Since the MATH-vertical space at MATH is spanned by MATH, since the MATH-vertical space at MATH is spanned by MATH, and since MATH, for MATH, we see that MATH maps the MATH-vertical space at MATH into the MATH-vertical space at MATH. For MATH-horizontal vectors MATH and MATH we obtain MATH . Therefore, by REF , we see that MATH and MATH are equivalent.
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math/0005254
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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 the dimension of fibres is MATH. In REF we solved the equivalence problem. The existence problem is solved by the explicit construction given in the preliminaries (see REF ). If MATH, then either REF MATH, MATH, or REF MATH, MATH. Since the fibres are assumed to be negative definite, REF cannot occur. CASE: If MATH, MATH, then MATH is a semi-Riemannian submersion from an anti-NAME space onto a Riemannian manifold. By CITE, MATH is equivalent to the canonical submersion MATH. This falls in REF .
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math/0005254
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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 greater than or equal to MATH and MATH is an isotropic semi-Riemannian manifold with non-constant curvature, then, by REF , MATH is isometric to MATH, MATH, and the dimension of the fibres is MATH. By REF , there are no such semi-Riemannian submersions with base space MATH. Therefore, REF and MATH imply that MATH has constant curvature, and hence, by REF , we obtain MATH. CASE: If MATH and MATH, then, by CITE, the semi-Riemannian submersion MATH is equivalent to the canonical semi-Riemannian submersion MATH. If MATH, then, by REF , we get MATH. By changing the signs of the metrics on the base and on the total space, MATH becomes a Riemannian submersion with connected totally geodesic fibres from a sphere onto a Riemannian manifold. So, by CITE and CITE, one obtains the conclusion.
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math/0005254
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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 and greater than or equal to MATH. By REF , MATH is isometric to MATH and MATH, MATH. Then MATH, MATH. By REF , we see that MATH is equivalent to the canonical semi-Riemannian submersion. REF If MATH is an isotropic semi-Riemannian manifold or if MATH, then, by REF , MATH is equivalent to one of the following canonical semi-Riemannian submersions: CASE: MATH, MATH; CASE: MATH, MATH; CASE: MATH, MATH. If the dimension of the fibres of MATH is greater than or equal to MATH, then the dimension of the fibres of MATH is greater than or equal to MATH. Hence, in this case, MATH is equivalent to MATH, MATH. For MATH, the semi-Riemannian submersion MATH is, after a change of signs of the metrics on the total space and on the base space, of type MATH. For MATH, MATH is of type MATH. In CITE (for case t=REF) and CITE (for case t=REF), it is proved that there are no such semi-Riemannian submersions with totally geodesic fibres. We proved that the dimension of fibres of MATH is less than or equal to MATH.
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math/0005254
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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 dimension of the fibres of MATH is greater than or equal to MATH. Thus, by REF , we see that MATH is equivalent to the canonical semi-Riemannian submersion MATH . It follows that MATH is one of the following types: CASE: MATH, or CASE: MATH. In CITE, NAME proved that there are no Riemannian submersions with fibres MATH from MATH onto MATH. Therefore, REF is not possible. The fibres of semi-Riemannian submersion MATH are totally geodesic by REF, and complex submanifolds, since the horizontal lifting of the tangent space of the quaternionic line MATH is invariant under the canonical complex structure on MATH. By CITE, there are no semi-Riemannian submersions with complex totally geodesic fibres from MATH onto MATH. Thus REF is impossible.
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math/0005260
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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 shape as MATH. Suppose that MATH, and let MATH be the smallest index such that MATH. Then there must be an index MATH such that MATH. Now one applies the balancing lemma above, increasing the MATH-th row at the expense of the MATH-th. After finitely many balancing steps we have written MATH as a MATH-linear combination of bitableaux MATH such that MATH. MATH : This is evident, as well as MATH . MATH : We have to show that MATH for all standard bitableaux MATH such that MATH for some standard bitableau MATH. Set MATH. Suppose first that MATH. By CITE there is a decomposition of MATH stable subspaces MATH where MATH is generated by all standard bitableaux MATH such that MATH; in this decomposition MATH is irreducible and the unique MATH-stable complement of MATH in MATH. Thus we can write MATH where MATH, MATH. Since MATH, we deduce that MATH. For MATH it follows that MATH. In fact, MATH has a unique decomposition as a direct sum of irreducible MATH-modules, and these are exactly the MATH; since the projection to MATH is non-trivial, it must appear in the decomposition of MATH. For general MATH it now follows that MATH, and therefore all the standard bitableaux in the standard representation of MATH must belong to MATH. However, since MATH, MATH must appear in this standard representation. Suppose now that MATH. If we find some standard bitableau MATH of shape MATH such that MATH, then the argument just given shows that MATH. Since we can connect MATH and MATH by a chain of shapes with respect to the partial order MATH, it is enough to consider the case in which MATH is an upper neighbor of MATH. One obtains the upper neighbors by either inserting a new box below the bottom of the diagram (and thereby increasing MATH by MATH) or by removing an ``outer corner" box of the NAME diagram of shape MATH and inserting it at an ``inner corner" in such a way that the box travels one row up or one column to the right. (At the end of the first row is also an inner corner, provided its length is MATH). REF illustrates the three cases. The case in which a new box is added, is trivial. In fact, we fill it with MATH, and MATH is standard of the right shape. We now assume that a box travels one row up, say from row MATH to row MATH. With MATH let MATH and MATH. One forms the standard bitableau MATH where MATH of shape MATH. Since MATH is a standard bitableau of shape MATH, it belongs to MATH, as was shown above. Note that MATH (where MATH if MATH and MATH if MATH). Now we apply the cyclic permutation MATH to both the rows and columns of the matrix MATH; the transformation MATH belongs to MATH. Therefore MATH. All the factors of MATH except MATH and MATH are invariant under MATH, whereas MATH . If we straighten this product and multiply its standard presentation with the remaining factors of MATH, then we obtain the standard representation of MATH. Therefore it is enough that a standard bitableau of shape MATH appears in the standard representation of MATH. Whereas the proof of ( MATH is based on ``balancing", we now need the ``unbalancing" effect of straightening.) Mapping all the indeterminates MATH with MATH, MATH or MATH, to MATH and MATH to MATH, one reduces the claim to the assertion that in the standard representation of MATH with MATH the minor MATH shows up, and that is immediate from NAME expansion. The case in which the box travels one column to the right is similar and left to the reader. Essentially it is the case in which MATH consists of a single column.
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math/0005260
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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 G-KRS we have that MATH for some MATH and consequently MATH. It follows that for every MATH one has MATH and hence MATH . Now let MATH be a monomial and set MATH. Then MATH is in MATH which proves that MATH.
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math/0005260
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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.
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math/0005260
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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 partial order defined by the functions MATH (see REF). In fact let MATH and MATH. Then the smallest element is MATH where MATH appears MATH times, MATH appears MATH times, and MATH as often as in MATH. We have MATH. Therefore, if MATH, the ideal MATH is contained in MATH, as follows from REF . Now suppose that MATH. Then it is enough to show that there exists a standard bitableau MATH of shape MATH such that MATH whenever MATH. Since we have MATH from the KRS invariance of the functions MATH (or MATH) it really suffices to find MATH of shape MATH such that MATH for all MATH with MATH. Instead of constructing MATH directly, we find a monomial MATH such that MATH has the desired shape, but MATH cannot be written as MATH where MATH. The shape of MATH can be controlled via the MATH or MATH functions. Let us first consider MATH; we set MATH, MATH. Then MATH and MATH. It is harmless to assume MATH. With MATH let MATH . For MATH and MATH this monomial has the following ``picture": MATH . The reader can check that MATH indeed as shape MATH. However, it is not possible to decompose MATH into a product MATH where MATH, MATH. In the general case one must multiply MATH with suitable factors. These are not hard to find.
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math/0005260
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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 contains a second (dual) assertion: the sum MATH of the lengths of the first MATH columns of MATH is the length of the longest subsequence of MATH that can be decomposed into MATH decreasing subsequences. It follows that a sequence MATH has no increasing subsequence of length MATH if and only if it can be decomposed into MATH decreasing subsequences. Then MATH is the equal to the maximal length of a subsequence of MATH which can be decomposed into MATH decreasing subsequences. Therefore MATH. On the other hand, by REF we know that MATH is equal to MATH which is the sum of the length of the columns of MATH of index MATH. Therefore MATH is equal to the number of entries of MATH which is the length of MATH.
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math/0005262
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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 that MATH is indeed an operator valued weight. Because both MATH and MATH are faithful and normal, also MATH is faithful and normal.
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math/0005262
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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.
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math/0005262
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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 converges MATH-strong-MATH to the element MATH . Then we may conclude that MATH and MATH . Because the considered elements MATH form a MATH-strong-MATH - norm core for MATH we conclude that for every MATH and MATH we have MATH and MATH. It is easy to prove that for every MATH and MATH we have MATH and MATH. From this follows the lemma.
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math/0005262
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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.
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math/0005262
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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 CITE that MATH for every MATH. But then it follows that for all MATH we have MATH. Combining this with the formula above we get MATH . By the density of such elements MATH we get that MATH for all MATH. From the definition of MATH follows immediately the final formula we had to prove.
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math/0005262
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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. Denote with MATH the automorphism of MATH defined by MATH. Here MATH denotes the modular automorphism group of MATH. Then MATH is implemented by MATH. It follows from REF that MATH for all MATH. In particular we have MATH . So we get MATH for all MATH.
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math/0005262
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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 from REF of the appendix that MATH . So we may conclude that MATH and MATH . Because MATH is a MATH-strong-MATH - norm core for MATH we conclude that MATH for all MATH and MATH . Because MATH is unitary we immediately get that MATH is MATH-invariant. From REF it follows that MATH is MATH-invariant. Then we conclude from REF that MATH for all MATH. So by REF we can take unitaries MATH such that MATH for all MATH. From the theory of operator valued weights we know that MATH. Because MATH is a MATH-cocycle, we get that MATH is a MATH-cocycle. By CITE we can take a (uniquely determined) n.s.f. weight MATH on MATH such that MATH for all MATH. With MATH we can define the n.s.f. weight MATH on MATH in the sense of REF . Then it follows from the theory of operator valued weights that MATH for all MATH. So MATH. Because MATH is a restriction of MATH we get that MATH is a restriction of MATH. Fix MATH with MATH. Choose MATH. Then it follows from REF that MATH and MATH . Because MATH is a restriction of MATH we get that MATH and MATH . Then it follows from REF that MATH. This means that MATH is a restriction of MATH. Further we have, using the theory of operator valued weights in the first equality and REF in the last one, MATH . So MATH for all MATH. Because MATH is a restriction of MATH we may conclude that MATH and then MATH.
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math/0005262
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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. Let now MATH, MATH and MATH, where MATH. Then MATH . Now MATH and MATH and MATH commute. So MATH and MATH. Then we get MATH . Hence we may conclude that MATH and MATH . Because of the previous lemma we get MATH for all MATH. By taking the adjoint we get MATH for all MATH. So for all MATH . Denoting with MATH the automorphism MATH of MATH we get that for every MATH we have MATH and MATH. Then the results of CITE allow us to conclude that MATH for every MATH and MATH. This gives MATH for all MATH. Combining this with REF we get for every MATH . So we get MATH for every MATH, and hence for every MATH. Rewriting this we get MATH for every MATH. This gives MATH.
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math/0005262
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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 MATH and MATH, it follows from CITE that MATH and MATH . Because MATH for every MATH we see that MATH. So we have MATH. Combining this with the equation above we get MATH . The last formula is valid for all MATH. Because MATH implements MATH we may then conclude that MATH. Then we get MATH and so MATH. Now MATH and MATH is the unitary intertwining the two standard representations of MATH. This proves the first claim of the proposition. In particular we get MATH . This proves the proposition.
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math/0005262
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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-isomorphism. Consider now the dual weights MATH and MATH on MATH and MATH, with canonical NAME MATH and MATH. Take MATH, MATH and MATH. Then MATH in the MATH-strong-MATH topology. For every finite subset MATH of MATH we define MATH . By REF of the appendix we get that MATH belongs to MATH and MATH where MATH denotes the projection onto MATH. Now define MATH. Then we see that MATH-strong-MATH and MATH . So we get that MATH and MATH . Because the elements MATH span a core for MATH we have MATH for every MATH and MATH in that case. By symmetry MATH if and only if MATH. But then it is clear that MATH and so MATH . Now suppose that MATH is a corepresentation, meaning that MATH. Then MATH where MATH is the MATH-anti-isomorphism from MATH to MATH defined by MATH for all MATH. Then we can compute MATH . So, when MATH is a corepresentation then MATH is a corepresentation. By symmetry also the converse implication holds.
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math/0005262
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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 and MATH. Then MATH . Now MATH . So we may conclude that MATH for all MATH, MATH and MATH. Because MATH is closed we can conclude that MATH and MATH for all MATH and MATH. Because MATH is a core for MATH and MATH for MATH we get MATH . We now claim that MATH for every MATH. Together with the fact that MATH this leads to the conclusion that MATH is a core for MATH. Then we get that the inclusion in REF is in fact an equality. Uniqueness of the polar decomposition gives us MATH and so MATH. So we only have to prove our claim. For this choose MATH and MATH. Then using REF we get MATH because MATH and MATH commute. Now observe that MATH and MATH commute, so that MATH and MATH . From this immediately follows our claim, and then the proof of the proposition is complete.
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math/0005262
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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 action MATH of MATH on MATH. Combining REF the unitary implementation of MATH constructed with the weight MATH is a corepresentation. By REF the unitary implementation of MATH constructed with MATH is a corepresentation as well. Then it follows from REF that the unitary implementation MATH of MATH constructed with the n.s.f. weight MATH on MATH will be a corepresentation. Here MATH denotes the usual trace on MATH. Represent MATH on the NAME of MATH such that MATH is a NAME for MATH. Let MATH be a NAME for MATH. Then we have a canonical NAME MATH for MATH. With this we construct the NAME MATH of the dual weight MATH on MATH. Denote with MATH the modular operator of this dual weight. As before we denote with MATH the modular operator of the weight MATH on MATH with NAME MATH. It is an easy exercise to check that MATH where MATH flips the first two legs of MATH. By uniqueness of the polar decomposition we get MATH and hence MATH. Because MATH is a corepresentation, also MATH will be a corepresentation.
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math/0005262
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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.
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math/0005262
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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 MATH, because MATH is central. So the restriction MATH of MATH to MATH is a n.s.f. weight and MATH is the restriction of MATH to MATH for all MATH. For every n.s.f. weight MATH on MATH we define the n.s.f. weight MATH on MATH by MATH. Here MATH denotes as before the dual weight on MATH. For every MATH we have MATH . When MATH and MATH are both n.s.f. weights on MATH we have MATH for all MATH. So, by CITE, there exists a unique n.s.f. operator valued weight MATH from MATH to MATH such that MATH for all n.s.f. weights MATH on MATH. So MATH for all n.s.f. weights MATH on MATH. When MATH is a n.s.f. weight on either MATH or MATH we denote again with MATH the n.s.f. weight on either MATH or MATH given by MATH for all positive MATH in either MATH or MATH. By CITE there exists a unique n.s.f. operator valued weight MATH from MATH to MATH such that MATH for all n.s.f. weights MATH on MATH and MATH on MATH. Choose now a n.s.f. weight MATH on MATH. Put MATH and MATH. When we change the weight MATH which was chosen on MATH in the beginning of the story, the tower MATH will be transformed into a unitarily equivalent tower. The unitary implementing this transformation is the unique unitary intertwining the two standard representations of MATH. This unitary also intertwines the two implementations of MATH by REF . Hence also MATH can be transformed. So we may suppose that MATH. From REF follows that MATH . So we also have MATH. But MATH because MATH is the NAME of MATH. To compute MATH we need a NAME for the weight MATH. But MATH. So we put MATH and as before we denote with MATH the NAME of MATH. For every MATH we define MATH. Then MATH and it is easy to check that MATH is a NAME for MATH. Also observe that for all MATH and MATH we have MATH and MATH . Now choose MATH. Then MATH . Choose now a family MATH of vectors in MATH such that MATH for all MATH. Fix MATH. Then MATH and so MATH. Further MATH . We will compute the final expression. For this we choose MATH and MATH. Recall that the subset MATH was introduced in the introduction. Observe that MATH . So we have MATH . By continuity we get that MATH for all MATH, MATH and MATH. By REF of the appendix, it follows that MATH and MATH for all MATH and MATH. Fix an orthonormal basis MATH for MATH. Then we may conclude that MATH . Combining this with REF we get that MATH for all MATH and MATH. Summing over MATH we get MATH for all MATH. Using REF we get that MATH for all MATH. Hence the normal faithful weight MATH is semifinite. From CITE it follows that MATH is semifinite. So MATH is integrable. Proof of the second implication. The second implication can be proved along the same lines as in the case of an Abelian group action, see CITE. So let us suppose that MATH is integrable. Choose a n.s.f. weight MATH on MATH and put MATH. Then MATH is a n.s.f. weight on MATH. Represent MATH on the NAME MATH of MATH such that MATH is a NAME for MATH. Choose a family of vectors MATH in MATH such that MATH . Define MATH and let MATH be the MATH-fold direct sum of the identical representation MATH of MATH on MATH. Recall that for any operator valued weight MATH we define MATH as the left ideal of elements MATH for which MATH is bounded. Also recall that we introduced the canonical NAME MATH for MATH in the introduction. When MATH we define MATH by MATH for all MATH, where MATH denotes the canonical NAME of MATH. One can check easily that MATH. For this see for example, CITE. Put MATH. For all MATH we define MATH . Observe that MATH is well-defined: MATH . Because MATH is a MATH-strong-MATH - norm core for MATH we get that MATH is dense in MATH. So MATH can be extended uniquely to an isometry from MATH to MATH. We now want to prove that the range of MATH is invariant under MATH. So we first choose MATH. Then for every MATH we have MATH . Next we will look at the invariance under MATH. Analogously as in REF of the appendix we have that for every MATH, MATH and MATH, MATH and MATH . Then it follows easily that for all MATH we have MATH . So for all MATH, MATH and MATH we have MATH by REF . So the range of MATH is invariant under MATH. Then we can define a MATH-homomorphism MATH . By the computations above we get that MATH for all MATH and MATH. Then it follows from REF that MATH and so the theorem is proved.
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math/0005262
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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 from the beginning that MATH for all MATH. Define the unitary MATH by MATH . Put MATH. Clearly MATH and MATH . For every MATH we have MATH . So we get MATH and hence MATH. In the next computation we denote again with MATH the MATH-anti-automorphism of MATH given by MATH for all MATH. Then we have MATH . So MATH is a MATH-cocycle. Define the action MATH of MATH on MATH given by MATH for all MATH. Then MATH for all MATH. Because the MATH-strong-MATH closure of MATH equals MATH we get that MATH. To conclude the proof of the first implication we have to show that MATH is integrable. For this we will use the previous theorem. From REF it follows that the unitary implementation MATH of MATH is given by MATH . From the proof of REF we also get that MATH gives an isomorphism from MATH onto MATH. So we can define MATH . Then MATH is a surjective MATH-homomorphism onto MATH and clearly MATH for all MATH. Because MATH is a MATH-cocycle we get that MATH . From this it follows that MATH . By the previous theorem we get that MATH is integrable. Proof of the second implication. Conversely suppose that the second statement is valid and take such an action MATH. Let MATH be a MATH-cocycle such that MATH for all MATH. It follows from the proof of REF that MATH is an isomorphism and MATH for all MATH. By the previous theorem we can find a surjective MATH-homomorphism MATH from MATH onto MATH satisfying MATH for all MATH. Putting MATH and observing that MATH we get the first statement.
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math/0005262
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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 such that MATH for all MATH. Then for all MATH we have MATH . Hence we get MATH. From our assumption and the fact that MATH is outer it follows that MATH. But then also MATH . So we can take MATH such that MATH. Because MATH is a MATH-cocycle we get that MATH. By the unicity of right invariant weights on MATH there exists a number MATH such that MATH for all MATH. Then we get that for all MATH we have MATH. Because MATH is integrable it follows that MATH is integrable.
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math/0005262
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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 for all n.s.f. weights MATH on MATH. Choose a n.s.f. weight MATH on MATH. Put MATH and let MATH. We will prove that MATH. As in the proof of REF we may suppose that MATH is represented on the NAME of MATH such that MATH is a NAME for MATH. Let MATH be the canonical NAME for MATH and put MATH. We now make a kind of converse reasoning of the proof of REF . Denote again with MATH the n.s.f. weight on MATH given by MATH for all positive MATH. We claim that for all MATH . So choose MATH. Take a family of vectors MATH in MATH such that MATH for all MATH. Because MATH we have MATH for all MATH. Fix MATH. Then we conclude from the previous formula that MATH . So, when MATH is an orthonormal basis for MATH we can define the element MATH by MATH . It is easy to check that for all MATH we have MATH and MATH . Using the notation MATH introduced in the introduction, we get for all MATH and MATH that MATH . From this we get that MATH for all MATH. Hence MATH and MATH . This means that MATH . Summing over MATH we get our claim stated in REF . But now MATH and so MATH for all MATH. Next we claim that MATH and MATH commute strongly. Then we will be able to conclude that MATH. But then MATH, and so we will get MATH . So we may conclude that MATH. By definition of MATH we have MATH and then MATH. By CITE we get that MATH. So we only have to prove our claim. Hence we want to prove that MATH and MATH commute for every MATH. For this it is sufficient to prove that MATH leaves both MATH and MATH invariant and MATH for all MATH. When MATH we have MATH . Then it is immediately clear that MATH. Because MATH we have that MATH leaves MATH invariant. Recall that we denoted with MATH the modular group of MATH on MATH. Then we have, for all MATH . Finally, for all MATH we have by REF that MATH where we used that MATH. From the proof of REF it follows that MATH and so we see that MATH for all MATH and MATH. Combining this with REF we get that MATH for all MATH and MATH. Then we get immediately that MATH for all MATH. This proves our claim and ends the proof of the proposition.
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math/0005262
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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 MATH-homomorphism MATH is faithful by REF , we get that the restriction of MATH to MATH is semifinite.
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math/0005262
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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 basic construction from MATH is then given by MATH . To prove that MATH has depth REF, we have to show that MATH is the basic construction. But it is immediately clear that the restriction of MATH to MATH is an action MATH of MATH on MATH. So by the first part of the proof it is sufficient to prove that MATH . Now it follows from REF that MATH . From this we can immediately deduce REF , and that concludes the proof.
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math/0005262
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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 .
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math/0005262
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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 second part suppose that MATH is unitary and MATH. Define the isomorphism MATH by MATH. Using the notation of REF we get that MATH for all MATH. So the action MATH of MATH on MATH is isomorphic with the action MATH, which is integrable because it is isomorphic with the bidual action MATH. Hence MATH is integrable, and so MATH is integrable. Fix now a n.s.f. weight MATH on MATH and represent MATH on the NAME of MATH such that MATH is a NAME. Let MATH be the basic construction from MATH and let MATH be the MATH-homomorphism from REF . Then define MATH and define MATH . Because MATH we have MATH for all MATH. Further we have MATH and so MATH. Putting MATH around this equation and using that MATH (see CITE), we get MATH . Flipping the first two legs of this equation and rewriting it we get MATH . From this it follows that MATH . Then we get for all MATH that MATH . Hence we may conclude that MATH and so MATH is faithful.
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math/0005262
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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 MATH for all MATH. By minimality we get MATH. But then MATH and so MATH. Hence MATH is outer. Let now MATH be outer and integrable. Choose a n.s.f. weight MATH on MATH and represent MATH on the NAME of MATH. Let MATH denote the modular conjugation of MATH and let MATH be the basic construction from MATH. Let MATH be the MATH-homomorphism given in REF . Then MATH is faithful because MATH is a factor. Because MATH is an isomorphism we get MATH and so MATH . Next we claim that MATH. Because, by REF , MATH, we get MATH . When MATH denotes the modular conjugation of the dual weight MATH, we already observed in the proof of REF that MATH. Then the outerness of MATH implies that MATH and so MATH . Then we may conclude from the previous computation that MATH where we have used that MATH, MATH and MATH. Then our claim follows and hence it is clear that MATH . So MATH is minimal.
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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 the usual way by MATH for all MATH. It is easy to verify that MATH is a free action and hence MATH is outer (see CITE). Further it is easy to check that MATH.
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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 NAME of MATH such that MATH is a NAME. Then we fix MATH and MATH and we claim that the element MATH defined by MATH belongs to MATH. Here we used the notation MATH introduced in the proof of REF . To prove our claim we observe that for all MATH, MATH and MATH . We can conclude that MATH . So MATH and for all MATH and MATH we have MATH . Next we observe that for all MATH . Inserting this in the computation above we get that MATH . Then it follows that MATH. So we see that MATH. Because MATH is minimal we also have that MATH is outer by REF . In particular MATH is a factor. Also MATH is a factor. Because MATH we then get immediately that MATH is a factor. Because MATH is supposed to be MATH-finite, the NAME space MATH is separable. So MATH is MATH-finite. Denoting with MATH the matrix units in MATH we see that the projections MATH and MATH both belong to MATH. Because MATH both projections are infinite. Hence they are equivalent in the MATH-finite factor MATH. Take MATH such that MATH and MATH. Then there exists a unitary MATH such that MATH. Now we can consider the isomorphism MATH . It is easy to check that MATH for all MATH. So the actions MATH and MATH are isomorphic. Because MATH is isomorphic to the bidual action MATH by REF , we get that MATH is a dual action. Because MATH is properly infinite and because MATH is a separable NAME space we get that the action MATH on MATH is isomorphic with the action MATH on MATH. So MATH is a dual action.
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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 a MATH-strong-MATH - norm core for MATH. Denote with MATH the domain of the MATH-strong-MATH - norm closure of the restriction of MATH to MATH. Let MATH and MATH. Define MATH. Then MATH and for all MATH we have MATH where we used the notation of CITE. So MATH and MATH. Hence MATH . So we get that MATH is invariant under MATH. Then it is easy to conclude that MATH is invariant under MATH for all MATH. Let now MATH and suppose that there exists a MATH such that MATH for all MATH. Let MATH. Define MATH. Then MATH and for all MATH we have MATH . So we see that MATH and MATH. Then we may conclude that MATH . Because such elements MATH form a MATH-strong-MATH dense subset of MATH it is easy to conclude that MATH is a left ideal in MATH. Because MATH is a MATH-strong-MATH dense left ideal of MATH, invariant under MATH and because MATH, we may conclude that MATH is a MATH-strong-MATH - norm core for MATH. But then MATH and we have proven our claim. Then it follows easily that also MATH is a MATH-strong-MATH - norm core for MATH. This last space equals MATH and so the proposition is proven.
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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.
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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.
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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, suppose that MATH is a sequence in MATH such that MATH and MATH. Applying MATH we get MATH and MATH. Because MATH is a n.s.f. weight we get that MATH and so MATH. This gives our claim. It is clear that the NAME algebra generated by the left NAME algebra MATH is MATH. Because MATH is injective we have that MATH is injective. So the n.s.f. weight on MATH which is canonically associated with MATH can be composed with MATH to obtain a n.s.f. weight MATH on MATH. Let MATH be the canonically associated NAME. Then, by definition of MATH, every MATH will belong to MATH and MATH. Let now MATH. Take a net MATH in MATH such that MATH-strong-MATH. Then we have MATH-strong-MATH, MATH and MATH . Because MATH is MATH-strong-MATH - norm closed we get MATH and MATH. Conversely, suppose MATH. By the main theorem of CITE there exists a net MATH in MATH such that MATH for all MATH, MATH-strong-MATH and MATH in norm. But MATH for every MATH. Because MATH is MATH-strong-MATH - norm closed we get that MATH. This concludes our proof.
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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.
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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.
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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 .
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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 conclude that MATH has dense range implying that it is a unitary transformation.
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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 strongly to MATH. CASE: Choose MATH. Since MATH strongly, the net MATH is a bounded net that converges to REF in the weak operator topology. Therefore MATH is a bounded net in MATH that converges to REF in the weak operator topology. Thus the normality of MATH implies that the net MATH also converges to REF in the weak operator topology. In other words, the net MATH converges to REF in the weak operator topology, implying that MATH converges strongly to MATH. Hence MATH converges to MATH for all MATH. Because the net MATH is bounded and MATH, we conclude from this all that MATH converges strongly to MATH.
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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.
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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 to MATH.
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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 last step of the next chain of equalities, we get for all MATH and MATH that MATH . Hence MATH . This implies that MATH . Combining this with REF , we find that MATH where we used REF in the last step. CASE: Let MATH denote the flip map on MATH. Then we have that MATH . Choose MATH, MATH. Then REF implies that MATH . Now, CASE: For all MATH, MATH implying that the linear map MATH is bounded. CASE: For MATH, MATH, we get, by applying REF twice, MATH . Combining these fact, we get that MATH for all MATH. Using this in combination with the chain of equalities in REF , we see that MATH . Hence, by REF , MATH . Comparing the above equation with REF , we conclude that MATH .
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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, then MATH thus MATH.
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math/0005265
|
Choose MATH, MATH, MATH. Then the previous proposition implies that MATH and the proposition follows.
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