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math/0003068
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The simply connected MATH-manifolds MATH and MATH are non-spin and have the same invariants MATH; NAME 's classification CITE therefore tells us that they are homeomorphic. On the other hand, because MATH, these complex surfaces have NAME dimension MATH. REF therefore tells us that MATH whereas MATH . Thus these manifolds have unequal invariants MATH, as claimed.
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math/0003068
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First recall that, in terms of the decomposition MATH the curvature operator of an oriented Riemannian MATH-manifold can be expressed in the form MATH where MATH and MATH are trace-free, and where MATH exactly corresponds to MATH. On the other hand, every REF-form MATH on MATH can be uniquely written as MATH, where MATH. Now a REF-form is expressible as a simple wedge product of REF iff MATH. Thus MATH is simple and has unit length iff MATH. The sectional curvature in the corresponding MATH-plane is then MATH . Notice, however, that the last term changes sign if MATH is replaced by MATH; geometrically, this means that the average of the sectional curvatures of an orthogonal pair of MATH-planes only depends on the scalar and NAME curvatures. Thus, there is always a MATH-plane MATH for which the sectional curvature is exactly MATH . Hence MATH where MATH is the lowest eigenvalue of MATH. In particular, MATH as claimed.
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math/0003068
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By REF we know that MATH . On the other hand, MATH at all points by virtue of MATH by REF . Hence MATH and so MATH . Taking the square root of this inequality then yields the claim.
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math/0003068
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Let MATH be the minimal model of MATH. Then for every metric MATH on MATH there is a spin-MATH structure with MATH and such that the corresponding NAME invariant is non-zero for the polarization determined by MATH; in particular, the scalar curvature MATH of MATH is negative somewhere. Thus, REF tells us that MATH satisfies MATH . Restricting our attention to metrics for which MATH, we thus get MATH . However CITE, MATH so it follows that MATH as claimed. If MATH is a complex-hyperbolic manifold, and if MATH is a multiple of the hyperbolic metric, normalized so that its sectional curvatures satisfy MATH, one then has MATH, and MATH, whereas MATH. Thus MATH, and MATH for any complex-hyperbolic MATH-manifold.
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math/0003068
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Suppose that MATH is a metric with MATH and total volume MATH. Thus, letting MATH denote the minimal model of MATH, and choosing a spin-MATH structure on MATH as in the proof of REF , MATH by virtue of REF . Since the spin-MATH structure is chosen so that MATH, equality must hold, it follows that MATH and MATH; in particular, MATH is minimal. Moreover, since REF is saturated, REF tells us that MATH is almost-Kähler, with MATH constant, MATH an eigenvector of MATH at each point, and MATH. Since MATH, the almost-complex structure satisfies MATH, so that our almost-Kähler manifold is monotonic in the terminology of CITE. Since equality holds in REF , we also have MATH and this, for starters, tells us that MATH by REF . Inspection of REF then shows that MATH, so that MATH is in fact an eigenvector of the full curvature operator MATH. An almost-Kähler manifold with this property is called weakly MATH-Einstein. Because we also know that MATH is monotonic, and that the sum MATH of the scalar and MATH-scalar curvatures is constant, a result of CITE asserts that MATH is NAME. But we also have also observed that MATH, so the entire curvature operator is parallel, and the universal cover of MATH is therefore the symmetric space MATH, equipped with the unique multiple of its standard metric for which MATH. Thus MATH admits a complex-hyperbolic metric; and this metric is moreover unique, up to rescalings and diffeomorphisms, by virtue of NAME rigidity.
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math/0003073
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The main idea of the proof relies on the identity MATH where MATH is the following coboundary operator: MATH . Observe that for MATH, MATH is a differential only for the MATH-reduction of the graded algebra of functionals. Using MATH is like working with a cosmological term, and, upon using the generalized NAME theorem, one gets MATH which proves the theorem. REF is a consequence of REF and of the following identities: MATH . The first identity follows from REF, from the fact that MATH is MATH-invariant and from NAME theorem. The second identity holds since MATH depends only on MATH and as a consequence of the already discussed property according to which the BV NAME operator contracts each field with the NAME dual (for some Riemannian metric on MATH) of the corresponding antifield at the same point in MATH. The last identity is ``formally" (that is, modulo regularization problems for MATH) true if MATH does not have transversal self-intersections, for the same reason as above. However, in order to rely upon this last identity confidently, we must then restrict ourselves to framed imbeddings and put each component of MATH on the imbedding and each component of MATH on its companion (as done in CITE).
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math/0003079
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Let MATH be a smooth function such that MATH for all MATH and define MATH . Then MATH and hence the tangent space of MATH at MATH is given by MATH . The tangent space to the MATH-orbit consists of all vector fields of the form MATH, where MATH. The map MATH identifies the quotient space MATH with the space of closed MATH-forms on MATH. If MATH are two smooth paths in MATH that satisfy MATH for some path MATH then the vector fields MATH and MATH are related by MATH where MATH generates the diffeomorphism MATH via MATH . Hence the MATH-forms MATH and MATH are related by MATH . Hence two closed MATH-forms MATH corresponding to two Lagrangian embeddings MATH and MATH represent the same tangent vector of MATH if and only if MATH or, equivalently, MATH. This proves the lemma. MATH .
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math/0003079
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Choose a smooth path MATH such that MATH for every MATH and let MATH be defined by REF . Then MATH if and only if MATH. It follows from the definitions that this is equivalent to MATH for all MATH and all MATH. This means that there exists a smooth family of vector fields MATH such that MATH . Equivalently, MATH where the isotopy MATH is generated by MATH via MATH and MATH . This proves the lemma. MATH .
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math/0003079
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The ``if" part was proved in REF . Suppose that the path MATH is tangent to MATH. Choose a smooth function MATH such that MATH for every MATH and let MATH be defined by REF . By assumption, MATH is exact for every MATH. Fix a smooth path MATH and, for every MATH, choose MATH such that MATH . Then the function MATH is smooth. We construct a smooth function MATH such that MATH . Choose an almost complex structure MATH on MATH that is compatible with MATH. Let MATH be so small that, for every MATH, the map MATH restricts to a diffeomorphism from the MATH-neighbourhood of the zero section in MATH onto the open neighbourhood MATH of MATH in MATH. Choose a cutoff function MATH such that MATH for MATH and MATH for MATH. Define MATH by MATH for MATH and MATH with MATH, and by MATH for MATH. Then MATH satisfies REF and hence MATH . By REF , the Hamiltonian isotopy MATH generated by MATH satisfies MATH for every MATH. This proves the lemma. MATH .
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math/0003079
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Choose a lift MATH of MATH and denote MATH . Then MATH and hence MATH as claimed. MATH .
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math/0003079
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The parallel transport of MATH along a curve MATH is determined by the Hamiltonian functions MATH via REF . The functions MATH in REF correspond to the path MATH. By REF , the Hamiltonian isotopy determined by MATH preserves MATH if and only if MATH for every MATH. This shows that REF is equivalent to REF . To prove the equivalence of REF note that MATH where MATH denotes the Hamiltonian vector field of MATH as in REF . The right hand side vanishes if and only if MATH and the left hand side vanishes if and only if MATH. This proves the lemma. MATH .
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math/0003079
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Let MATH be defined by MATH . Since MATH is connected there exists a smooth path MATH such that, for every MATH, MATH . For example choose MATH in the interval MATH such that MATH for MATH and MATH for MATH. Then define MATH for MATH such that REF is satisfied. Let MATH be defined by MATH . By REF , the function MATH is MATH-periodic in MATH and the proof of REF shows that the MATH-forms MATH are MATH-periodic in MATH. Hence the functions MATH are MATH-periodic in MATH and hence, so are the functions MATH defined in the proof of REF . Now define MATH . Let MATH be a smooth cutoff function such that MATH for MATH and MATH for MATH and define MATH by MATH where MATH is given by MATH . Explicitly, MATH has the form REF where MATH are given by MATH for MATH. These functions have mean value zero and satisfy REF with MATH replaced by MATH. Since MATH it follows as in the proof of REF that MATH . By REF , the parallel transport of MATH along the boundary preserves MATH. Hence MATH is an element of MATH. This proves REF . We prove REF . Assume first that there exists a fibrewise Hamiltonian symplectomorphism of the form MATH such that MATH for every MATH. Define MATH for MATH and MATH. Then MATH is an exact Lagrangian loop for every MATH and MATH is exact for all MATH and MATH. Hence the Lagrangian loop MATH is Hamiltonian isotopic to MATH. Since MATH is a Hamiltonian symplectomorphism, the loop MATH is Hamiltonian isotopic to MATH. Conversely, suppose that MATH and MATH are two exact Lagrangian loops that are Hamiltonian isotopic. Choose an exact isotopy MATH such that MATH and MATH for MATH. As in the proof of REF , one can construct a smooth family of Hamiltonian functions MATH such that MATH . Define the Hamiltonian symplectomorphisms MATH by MATH . Then MATH for MATH and the required fibrewise Hamiltonian symplectomorphism is given by MATH. We prove REF . Let MATH be given by REF and suppose that MATH is a fibrewise Hamiltonian symplectomorphism. Choose smooth functions MATH such that the functions MATH and MATH have mean value zero and the Hamiltonian vector fields MATH and MATH satisfy MATH . Then MATH where MATH . Hence MATH . The last equality follows from the definition of MATH and MATH in REF . This proves the lemma. MATH .
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math/0003079
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Let MATH be an exact Lagrangian loop. Let MATH be the connection MATH-form defined by REF in the proof of REF , where the cutoff function MATH is chosen to be nondecreasing. Then MATH where MATH are given by REF and MATH. Taking the differential of this MATH-form on MATH we find MATH . Since MATH and MATH we obtain MATH . Moreover, MATH and hence MATH . This implies MATH . If MATH and MATH are Hamiltonian isotopic then, by REF , there exists a fibrewise Hamiltonian symplectomorphism MATH of MATH such that MATH. Hence MATH if and only if MATH. By REF , MATH . Hence MATH . We prove that MATH. Let MATH. We shall construct an exact Lagrangian loop MATH that is Hamiltonian isotopic to MATH and satisfies MATH . Suppose that MATH has the form REF . Since the function MATH in REF has no effect on the curvature we may assume, without loss of generality, that MATH. Define MATH and MATH by the formula MATH . Explicitly, MATH . Define the Hamiltonian symplectomorphisms MATH by MATH . Then the loop MATH is evidently Hamiltonian isotopic to MATH. We shall prove that it satisfies REF . To see this, denote by MATH the fibrewise Hamiltonian symplectomorphism of MATH given by MATH . Then, as in the proof of REF , we obtain MATH where MATH and MATH is defined by MATH. These functions satisfy MATH . Moreover, by REF , we have MATH . Hence the length of MATH is given by MATH . Thus we have proved that for every MATH there exists an exact Lagrangian loop MATH that is Hamiltonian isotopic to MATH and satisfies MATH. Hence MATH and this proves the theorem. MATH .
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math/0003079
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Let MATH be given by REF with MATH replaced by MATH for MATH. Denote MATH and let MATH be defined by REF . Since MATH it follows from REF that there exists a function MATH such that MATH for every MATH. We shall prove that the required identity holds with MATH . To see this note that, by REF , MATH where MATH denotes the pushforward of the MATH-form MATH under the diffeomorphisms MATH. Let MATH be a compact oriented NAME surface and MATH be a smooth function such that MATH. Denote MATH. Then MATH . The penultimate equality follows from REF and the last from the identities MATH and MATH . Here REF is the degree theorem for maps between compact MATH-manifolds and REF is the degree theorem for maps between MATH-manifolds with boundary. More precisely, if the function MATH has mean value zero then there exists a MATH-form MATH such that MATH and MATH . This implies that the left hand side of REF vanishes. Hence it suffices to establish REF for constant functions MATH and this reduces to REF . This proves the lemma. MATH .
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math/0003079
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Let MATH be an exact Lagrangian loop and MATH be smooth functions such that the functions MATH defined by REF satisfy MATH for every MATH. For every smooth function MATH let MATH be given by REF . In particular, MATH is given by REF with MATH. We shall prove that MATH . To see this, note that MATH and hence MATH . This shows that MATH is nondegenerate if and only if MATH for all MATH. Fix a number MATH . Choose a smooth function MATH such that MATH for all MATH and MATH . Then MATH is nondegenerate and represents the class MATH . This proves REF follows from a similar argument. It follows from REF that MATH . Since the curvature of MATH is equal to the curvature of MATH for every MATH it follows that MATH for every MATH and hence MATH. This proves the theorem. MATH .
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math/0003079
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We prove that MATH. To see this, choose smooth functions MATH such that MATH and the map MATH defined by MATH is an embedding with MATH. Then MATH . Hence MATH where MATH. Hence MATH and this implies MATH . By REF in the appendix, two loops MATH and MATH, associated to two embedded discs MATH via REF , are Hamiltonian isotopic if and only if MATH and MATH have the same area. Now for every MATH there exists an embedded disc MATH (as illustrated in REF ) such that MATH where MATH are chosen such that the map MATH defines an embedding MATH whose image is MATH. Hence the length of the loop MATH is bounded above by MATH. Thus we have proved that MATH . To show the reverse inequality let MATH be an exact Lagrangian loop that is Hamiltonian isotopic to MATH. Then MATH where MATH is a smoothly embedded closed disc of the same area as MATH. Let MATH be a Hamiltonian isotopy such that MATH . We shall prove that MATH . To see this, choose an embedded closed discs MATH such that MATH and let MATH be a lift of MATH. Since MATH is Hamiltonian isotopic to MATH we have MATH and hence MATH . Let MATH be the Hamiltonian functions that generate MATH and have mean value zero over the fundamental domain MATH. Choose MATH such that MATH for every MATH and let MATH be a compactly supported cutoff function such that MATH. Then the functions MATH generate a compactly supported Hamiltonian isotopy MATH of MATH that satisfies MATH . Now it follows from the energy-capacity inequality in CITE that the displacement energy of MATH is bounded below by the area. Hence MATH . Since MATH this proves REF . It follows from REF that MATH for every exact Lagrangian loop MATH that is Hamiltonian isotopic to MATH. Hence MATH. MATH .
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math/0003079
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We compute MATH . This proves the lemma. MATH .
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math/0003079
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Let MATH and MATH. Let MATH be given by REF . Then MATH is a MATH-holomorphic curve. By REF , MATH is tamed by MATH. Hence MATH . The infimum of the numbers on the right is MATH. This proves the lemma. MATH .
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math/0003079
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Let MATH and MATH. Let MATH be given by REF . Then MATH is a MATH-holomorphic curve. By REF , MATH is tamed by MATH. Hence MATH . The supremum of the numbers on the right is MATH. This proves the lemma. MATH .
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math/0003079
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REF . MATH .
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math/0003079
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REF . MATH .
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math/0003079
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If MATH for every MATH then MATH (with reversed orientation) and MATH form a sphere and the difference MATH is equal twice the first NAME number of this sphere. Hence the difference of the NAME numbers is an even multiple of MATH. This continues to hold whenever MATH is homotopic to MATH as a section of the bundle MATH. For any two maps MATH there exists a smooth function MATH such that MATH and the connected sum MATH is homotopic to MATH along the boundary. Hence, by what we have just proved, MATH . Since MATH is an integer multiple of MATH, the lemma is proved. MATH .
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math/0003079
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In the case of MATH, consider the constant function MATH . Then a trivialization of the pullback tangent bundle MATH is determined by the coordinate chart MATH . In these coordinates the Hamiltonian flow is MATH . Since MATH we see that the NAME index of the loop MATH is equal to MATH. This proves the second equation in REF and the first follows from a similar argument. MATH .
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math/0003079
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Let MATH be given by REF with MATH and MATH where MATH and MATH is given by REF . As in REF , MATH is a smooth nondecreasing cutoff function such that MATH for MATH near MATH and MATH for MATH near MATH. The formula MATH for MATH shows that MATH for MATH and MATH for MATH. By REF with MATH and MATH, MATH . The explicit formulae for MATH and MATH show that MATH consist entirely of critical points of MATH and MATH for all MATH. This shows that the constant functions MATH with values in MATH are horizontal for the symplectic connection determined by MATH. In explicit terms MATH and MATH. Hence these constant functions satisfy both REF for every MATH. The constant functions with values in MATH satisfy in addition the boundary REF . REF shows that the constant solutions with values in MATH and those with values in MATH represent different homology classes. We prove that, for every MATH, MATH . To see this, let MATH. Then, by REF , MATH . Hence every MATH satisfies MATH and MATH . The latter implies that MATH for some point MATH and the former implies that MATH is a horizontal section of MATH with respect to MATH. Now let MATH, choose a path MATH that connects MATH to MATH, and define MATH by MATH . Then MATH and MATH . Since MATH consists of critical points of MATH and MATH for all MATH it follows that MATH for all MATH. Hence MATH is constant. The boundary condition shows that this constant lies in MATH. This proves REF . Hence MATH is diffeomorphic to MATH for every MATH and, in particular, for every MATH. The evaluation map MATH is obviously an embedding of MATH into MATH. A similar assertion holds for MATH and this proves the theorem. MATH .
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math/0003079
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The loop MATH, given by REF , is Hamiltonian isotopic to MATH and hence MATH . By REF , MATH and MATH . Hence, by REF , MATH . Here MATH denotes the connection MATH-form introduced in the proof of REF . Hence MATH . Since MATH the result follows from REF . MATH .
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math/0003079
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Let MATH be a smooth isotopy such that MATH and MATH. Define MATH for MATH. Then MATH and MATH for all MATH and MATH. Fix a Riemannian metric on MATH with volume form MATH and let MATH be defined by MATH . Choose MATH such that MATH and define MATH by MATH . Then MATH and MATH. Hence MATH for all MATH and MATH. Moreover, MATH and MATH for all MATH. Hence MATH is the required symplectic isotopy from MATH to MATH. MATH .
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math/0003079
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The result follows again from NAME isotopy. We prove that there exists a MATH-form MATH such that MATH . To see this, choose any MATH-form MATH such that MATH. Then the integral of MATH over MATH vanishes and so MATH is exact. Hence there exists a smooth function MATH such that MATH and the MATH-form MATH satisfies REF . Now let MATH and define MATH and MATH by MATH . Then MATH is tangent to MATH for every MATH. Hence MATH preserves MATH and MATH for every MATH. This proves the lemma. MATH .
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math/0003079
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Choose orientation preserving embeddings MATH such that MATH and MATH. We prove the result in four steps. CASE: There exists a diffeomorphism MATH that is isotopic to the identity and satisfies MATH. Choose a path MATH such that MATH and MATH. Next choose a smooth family of vector fields MATH such that MATH for every MATH. Then the diffeomorphisms MATH, defined by MATH and MATH, satisfy MATH for every MATH. Hence MATH satisfies the requirements of REF : MATH can be chosen such that MATH. We prove that, for every matrix MATH such that MATH, there exists a diffeomorphism MATH such that MATH. To see this, let MATH and choose MATH and MATH such that MATH. Next choose smooth functions MATH such that MATH and MATH for all MATH and MATH . Then the diffeomorphism MATH defined by MATH satisfies MATH. Hence the function MATH is a diffeomorphism of MATH and satisfies MATH. To prove REF , let MATH be defined by MATH choose a diffeomorphism MATH such that MATH, and replace MATH by MATH. CASE: MATH can be chosen such that MATH for MATH sufficiently small. By REF , we may assume that MATH. Choose MATH such that MATH and consider the function MATH . This function is an embedding and satisfies MATH. Choose a smooth cutoff function MATH such that MATH for MATH and MATH for MATH. For MATH define MATH by MATH . Then MATH is a diffeomorphism for MATH sufficiently small and MATH for MATH. Hence the embedding MATH satisfies the requirements of REF for MATH sufficiently small. CASE: We prove the lemma. By REF , there exist embeddings MATH, a constant MATH, and a diffeomorphism MATH such that MATH is isotopic to the identity and MATH . Choose MATH such that MATH and MATH extend to embeddings of MATH into MATH . Choose a smooth function MATH such that MATH for every MATH and MATH . Let MATH be given by MATH for MATH and by MATH in MATH. Then MATH is isotopic to the identity and MATH. Similarly, there exists a diffeomorphism MATH that is isotopic to the identity and satisfies MATH. The diffeomorphism MATH is isotopic to the identity and maps MATH to MATH. This proves the lemma. MATH .
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math/0003080
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It is clear from the definitions that the equivalence relation MATH is contained in MATH. For the converse, suppose MATH. Then there exist MATH and MATH, such that MATH. By splitting MATH and MATH into their component terms for MATH we obtain MATH for some MATH, MATH. It follows immediately from this that MATH.
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math/0003080
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All that remains to be verified is that NAME resulting from matches found in MATH can be added to MATH without altering MATH. We assume all polynomials in MATH to be monic (possible since MATH is a field). Now NAME result from two types of overlap. For the first case let MATH be polynomials in MATH such that MATH for some MATH. Then the NAME is MATH where MATH for MATH. Now MATH therefore MATH, and hence the congruence generated by MATH coincides with MATH. For the second case let MATH be polynomials in MATH such that MATH for some MATH. Then the NAME is MATH. Now MATH therefore MATH, and hence the congruence generated by MATH coincides with MATH.
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math/0003081
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Assume for the block MATH the notations given in REF . Since MATH and MATH belong to different components of MATH, the MATH-edge MATH, together with its endpoints MATH, is a dipole of type REF in MATH. The cancellation of this dipole produces a dipole of type REF involving the MATH-coloured edge MATH. The sequence of cancellations of this dipole and of the resulting dipoles of type REF successively involving MATH leads to MATH.
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math/0003081
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CASE: By deleting all dipoles involving the MATH- and MATH-edges connecting MATH with MATH we obtain the ``normal" crystallization of the lens space MATH (see CITE). CASE: MATH.
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math/0003081
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With the previous assumptions and notation, it suffices to show that the MATH-coloured graph MATH obtained by cancelling MATH in MATH is c.p.-isomorphic to MATH. First of all, exchange the names of the two cycles MATH and MATH, together with the first components in the labelling of their vertices. After this relabelling, MATH becomes the unique gluing subgraph of MATH connecting MATH with MATH by colour MATH; denote by MATH the length of MATH. REF sketches, with the usual conventions, the graphs MATH and MATH; here we also point out the labelling of some ``strategic" vertices of MATH. The computation of the integers MATH, depending on the components of MATH, is described in the following table: MATH. Note that: CASE: MATH if and only if MATH; CASE: MATH if and only if MATH . Now, we are going to look into the shape of MATH and MATH. It is clear that, by cancelling MATH in MATH, the two MATH-cycles MATH and MATH of MATH give rise to a unique MATH-cycle MATH of MATH; moreover, since the length of MATH (respectively, MATH) is MATH (respectively, MATH), the length of MATH is MATH. Since the gluing subgraph MATH is coherent with the orientations on MATH and MATH, the cycle MATH inherits an orientation in a natural way. On the other hand, the length of MATH (respectively, MATH) in MATH is still MATH (respectively, MATH). Hence, the MATH-coloured graphs MATH and MATH can be sketched as in REF respectively. By REF , it is easy to check that, in all four cases of REF , the graph MATH is planar and has the shape of REF , where the numbers inside the strips can be computed by REF . The graph MATH satisfies the assumptions of REF and hence MATH is c.p.-isomorphic to MATH, where MATH. We are now going to compute MATH and MATH. If MATH (respectively, MATH) denotes the key-vertex of MATH (respectively, MATH) belonging to MATH, MATH, then MATH is the distance from MATH to MATH according to the orientation of MATH. Now, MATH (respectively, MATH) is the vertex of MATH which is MATH-adjacent with MATH (respectively, with MATH) in MATH. On the other hand, the vertex which is MATH-adjacent with MATH (respectively, with MATH) in MATH is MATH (respectively, MATH); hence, the distance from MATH to MATH (respectively, from MATH to MATH), according to the orientation of MATH (respectively, MATH), equals the distance from MATH to MATH (respectively, from MATH to MATH), according to the orientation of MATH (respectively, MATH). Since MATH (respectively, MATH) is the vertex MATH (respectively, MATH) of MATH, we obtain: MATH . Furthermore, MATH is the vertex of MATH which is MATH-adjacent in MATH with the vertex preceding MATH in MATH; hence, MATH is the vertex MATH in MATH. In the same way, MATH is the vertex of MATH which is MATH-adjacent with MATH in MATH. Therefore, we have the following two possibilities: MATH . By recalling that MATH if and only if MATH, we can conclude that, in both cases MATH . The graph MATH is c.p.-isomorphic to MATH, where: MATH . Substituting in this expression the values of MATH and MATH of REF we obtain REF ; moreover, MATH satisfies REF and, by REF . So, MATH is an admissible MATH-tuple and this completes the proof.
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math/0003081
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See REF .
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math/0003081
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The relations MATH, MATH, MATH and MATH hold. Therefore, we get for MATH the classical presentation of the NAME four group: MATH. To obtain the second result, define MATH and observe that both MATH and MATH commute with MATH. By NAME transformations, we have: MATH, MATH, MATH, MATH, which is a classical presentation of MATH. Finally, the last sentence holds since MATH and MATH.
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math/0003081
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By MATH and MATH we can permute MATH in all possible way and therefore REF can be achieved. REF follow by a suitable application of MATH. REF follow by a combined application of the three maps. The unicity of such a MATH is straightforward.
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math/0003081
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By REF there exists MATH and MATH such that MATH. Hence, by REF : MATH.
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math/0003081
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MATH is MATH-equivalent to a MATH-tuple MATH verifying condition (MATH). By REF , MATH is MATH-equivalent to either MATH or MATH or MATH. Since MATH if and only if MATH, it is easy to see that the admissible MATH-tuples MATH and MATH both verify condition MATH and therefore the statement is achieved.
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math/0003081
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CASE: There is a finite number of traps of a fixed type. CASE: Trivial.
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math/0003081
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In one direction REF the statement is trivial since MATH are MATH-equivalent to MATH. To prove the converse, denote by MATH the graph whose vertex-set is the set MATH of all canonical MATH-tuples and whose edge-set is defined by the following rule: join two different vertices MATH and MATH by an edge iff there exist two admissible MATH-tuples MATH, such that MATH. The graph MATH is well-defined because MATH; moreover, it is an infinite graph without loops or multiple edges. Each connected components of MATH corresponds to a MATH-orbit and each vertex of MATH has degree MATH by REF : in fact, the vertices which are adjacent to a given vertex MATH are the canonical representatives of the MATH-orbits MATH distinct from MATH. We are now going to prove that if MATH is adjacent to MATH and MATH, then the other vertices which are adjacent to MATH have complexity MATH. First of all, if MATH is not MATH-equivalent to MATH then MATH and therefore MATH by REF . Moreover, from REF we get MATH. Suppose now MATH, then MATH by REF and both MATH. As a consequence, any path in MATH whose sequence of vertices is MATH has the following property: if MATH (respectively, MATH), then MATH (respectively, MATH), for each MATH. Now, if MATH (respectively, MATH whenever MATH), for MATH, then all vertices which are adjacent to MATH have not lower (respectively, have greater) complexity; hence, each path of positive length starting from MATH ends in a vertex MATH such that MATH (respectively, MATH) and therefore MATH is minimal (respectively, is a root).
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math/0003081
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From REF we always get MATH and MATH. Moreover, MATH when either MATH or MATH or MATH, by REF . Now, if MATH then MATH, for MATH. Therefore MATH whenever MATH. On the other hand, it is easy to check that REF (respectively, REF ) includes all the minimal MATH-tuples MATH such that MATH and MATH (respectively, such that MATH and MATH).
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math/0003082
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Let MATH and MATH be a standard solution. If MATH and MATH is another solution of the conjugate equation, then MATH and MATH for some invertible MATH CITE, hence the conjugate of MATH with respect to MATH and MATH is given by MATH where MATH is the modular group of MATH , so that MATH because the minimal left inverse MATH of MATH is tracial on MATH.
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math/0003082
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We may suppose that MATH. Set MATH, where MATH is the minimal left inverse of MATH as before. Since the NAME cocycle MATH is a covariance cocycle for MATH CITE, there exists a one-dimensional character MATH of MATH such that MATH (both cocycles intertwine MATH and MATH). As shown in CITE, MATH . Moreover the NAME cocycle is holomorphic and MATH CITE, therefore MATH . If MATH is continuous, then MATH is continuous. So MATH extends to an entire function, hence MATH is holomorphic, and the second formula in the statement follows from the first one.
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math/0003082
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We have MATH CITE (the conjugate map is the one associated with MATH), hence the above REF applies, provided MATH is irreducible in End-MATH. If MATH is reducible in End-MATH we have MATH with MATH. By using the tracial property of MATH it is easy to check that MATH is a one-parameter group of unitaries in the finite-dimensional algebra MATH, and therefore can be diagonalized. If MATH is a decomposition of MATH into irreducibles (with eigen-projections of MATH) in End-MATH, we have a corresponding decomposition of MATH as direct sum of the MATH thus MATH with MATH, hence MATH .
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math/0003082
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See REF .
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math/0003082
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Let MATH be a cyclic separating vector such that MATH, MATH, and let MATH be the isometry of MATH given by MATH . The final projection of MATH is given by MATH thus MATH is a homomorphism of MATH onto MATH and MATH . Now the central support of MATH in MATH is MATH as MATH, hence if MATH there exists a unique MATH such that MATH, providing an extension of MATH to MATH. As MATH is separating, MATH is an isomorphism.
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math/0003082
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The map MATH is a functor of MATH tensor categories from MATH to a sub-tensor category of End-MATH, hence MATH. As MATH has a unitary braiding, every real object MATH is amenable REF , thus MATH, where MATH is the MATH norm of the fusion matrix MATH associated with MATH, therefore MATH, thus MATH. For any MATH, the object MATH is real hence, by the multiplicativity of the dimension, MATH.
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math/0003082
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Let MATH a holomorphic unitary covariance MATH-cocycle. As MATH is continuous, the map MATH is strongly continuous. To check this, note that by cocycle property it is enough to verify the continuity at MATH because then the strong limit MATH due to the normality of MATH. Let then MATH be a weak limit point of MATH as MATH. Then MATH and MATH for some sequence MATH. Thus MATH by the limit case of the NAME inequality, thus MATH because MATH is separating. Therefore by NAME 's theorem CITE there exists a normal faithful semifinite weight MATH on MATH with MATH and, by the finite holomorphic dimension assumption, MATH so that MATH is indeed a positive linear functional. Then MATH is a KMS functional for its modular group MATH (we are setting MATH here). The functional on MATH is KMS with respect to MATH . Hence, by the uniqueness of the KMS state, MATH on MATH, for some MATH. Thus MATH is a normal faithful functional and MATH is normal too. Therefore the proof is completed by REF .
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math/0003082
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By assumption and REF both MATH and MATH extend to MATH. As MATH and MATH are also intertwiners on MATH by weak continuity and the conjugate equation for MATH and MATH is obviously satisfied on MATH, the extension of MATH to MATH has finite index.
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math/0003082
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By REF MATH extends to MATH and has finite index. We are then in the case covered by REF .
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math/0003082
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As usual MATH is identified with MATH. If MATH, there exist two non-zero projections MATH with sum MATH. Thus MATH are different graded KMS functionals on MATH. This can be checked since both the extensions of MATH and MATH act trivially on MATH, and by usual approximation arguments. By the uniqueness assumption there exists a constant MATH with MATH, then by continuity the same equality holds for MATH, thus MATH. Analogously MATH, thus MATH.
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math/0003082
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The holomorphic property of MATH is a direct consequence of the holomorphic property of the NAME cocycle because MATH is indeed a NAME cocycle, up to phase, with respect to two bounded positive normal functionals of MATH (see CITE and the previous section). Note now that, since the extension of MATH to MATH (still denoted by MATH) commutes with MATH, we have MATH, thus MATH because MATH is self-adjoint, hence MATH, where MATH is the (unique) left inverse of MATH. To check REF , recall that, by REF cocycles, we have (see CITE): MATH . As above we have the polar decomposition MATH. Then MATH namely the holomorphic dimension with respect to MATH coincides with the holomorphic dimension with respect to MATH, up to a sign.
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math/0003082
|
Because of additivity and wedge duality one can write MATH (intersection over all wedges containing MATH). If MATH is spacelike separated from MATH there is a wedge MATH with MATH and MATH, hence MATH is local. As MATH extends MATH and is local, wedge duality must hold for MATH too, therefore MATH for all wedges MATH. In particular the global NAME algebra associated with MATH coincides with the one associated with MATH: MATH . Analogously, it follows from REF that MATH, namely MATH, that is to say NAME duality holds for MATH.
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math/0003082
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Let MATH denote the sub-MATH-algebra of all elements with pointwise norm continuous orbit under the action of MATH of MATH, and set MATH. We have MATH where MATH is the MATH-weak closure of MATH. Indeed if MATH and MATH is an approximation of the identity in MATH by continuous functions with support in a ball of radius MATH, then MATH (MATH-weak convergence) and MATH has pointwise MATH-orbit and, for large MATH, belongs to MATH (the MATH-continuity is checked similarly). Set MATH, where MATH and MATH, where MATH is the expectation of MATH onto MATH as above. Clearly MATH is a MATH-invariant locally normal state of MATH and so is its restriction MATH to MATH. Let MATH be an extremal MATH-invariant state of MATH extending MATH. By the AHKT theorem CITE MATH is KMS with respect to a one-parameter automorphism group MATH of MATH. Since MATH is locally normal and dominates MATH, also MATH is locally normal, thus it extends to a locally normal state MATH of MATH. By usual arguments, MATH is a KMS state on MATH with respect to MATH.
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math/0003082
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Let MATH be a NAME space of isometries implementing MATH on MATH and let MATH be an orthonormal basis of MATH, thus the MATH's are isometries in MATH with orthogonal final projections summing up to the identity and MATH, MATH. Then MATH where MATH and MATH. Therefore MATH. On the other hand MATH is a separating vector for MATH; indeed elements of MATH have the form MATH, MATH, thus MATH iff MATH, thus iff MATH because MATH is separating, which is equivalent to MATH because MATH.
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math/0003082
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Let MATH a KMS state of MATH extending MATH. As MATH is a separating state of MATH, also MATH is also separating as in the proof of REF , where MATH is the inner endomorphism of MATH implemented by MATH. As MATH extends MATH we have by REF that MATH, where the symbol MATH denotes quasi-equivalence.
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math/0003082
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The proof now follows by REF .
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math/0003082
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With MATH irreducible, we have to show that MATH is irreducible too. Let MATH, namely MATH and MATH for all MATH in MATH. As MATH is localized in a double cone MATH, MATH acts identically on MATH, hence MATH, thus MATH as desired.
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math/0003082
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As above noticed, MATH is a two-variable cocycle for the action of MATH on MATH, where MATH is the tensor category with conjugates generated by MATH (see CITE). Hence by CITE MATH . On the other hand, by the results in the previous section, we may extend MATH to the weak closure of MATH in the GNS representation of MATH, and if then compare with the NAME cocycle, we have MATH . But MATH, hence MATH namely MATH. The last point is a consequence of REF .
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math/0003082
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Assume first that MATH is a half-line and let MATH be orthogonal to MATH; we have to show that MATH. Indeed if MATH is a half-line and MATH, then for all MATH for all MATH such that MATH. But, because of the KMS property, the function MATH is the boundary value of a function analytic in the strip MATH (as MATH and Dom-MATH), hence it must vanish everywhere. It follows that MATH is orthogonal to MATH for all MATH, hence MATH is orthogonal to MATH. Assume now that MATH and MATH is additive. Set MATH. We shall show that MATH is cyclic for MATH, hence for MATH. By the same argument as above MATH namely the orthogonal projection MATH onto MATH is independent of MATH and thus belongs to MATH (by additivity). As MATH is separating for MATH and MATH it follows that MATH, namely MATH.
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math/0003082
|
Clearly MATH for positive MATH. Setting MATH we have MATH therefore, by using the relation MATH, it follows that MATH . On the other hand MATH hence MATH showing the last part of the statement.
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math/0003082
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We first show that the triple MATH is a +hsm factorization with respect to MATH in the sense of CITE, namely these three algebras mutually commute and MATH, MATH and MATH are +half-sided modular inclusions. Now MATH and MATH commute by the isotony of MATH; for the same reason MATH commute with MATH, indeed MATH is MATH-invariant where, as above, MATH. Again MATH, hence MATH because MATH if MATH by translation-dilation covariance of MATH on positive half-lines REF . Concerning the hsm properties, the only non-trivial verification is that MATH is a +hsm inclusion, namely that MATH . We thus need to show that for any fixed MATH we have MATH . Indeed if MATH and MATH, there exist MATH and MATH such that MATH, as follows immediately by the corresponding relation in the MATH group. Therefore MATH as desired. By a result in CITE there exists a conformal net MATH on MATH such that the local NAME algebras associated to MATH, MATH and MATH are respectively MATH, MATH and MATH and having MATH and MATH as translation and dilation unitary groups. By translation-dilation covariance, MATH is then conformal thermal completion of MATH.
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math/0003082
|
If MATH is strongly additive, then MATH is strongly additive (on the intervals of MATH), hence MATH is strongly additive and the above comment applies.
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math/0003082
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MATH follows by the above comments. On the other hand MATH because they are equivalent to the relative commutant property MATH for MATH and either MATH or MATH, which are indeed equivalent conditions in the conformal case CITE.
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math/0003082
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If MATH satisfies essential duality then, since MATH by MATH of the above proposition, we have MATH for MATH. On the other hand, by essential duality, we have MATH, hence MATH as desired. The case of arbitrary MATH is obtained by translation covariance.
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math/0003082
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By translation covariance there exists a unitary MATH such that MATH is localized in an interval contained in MATH, thus MATH on MATH for all MATH, MATH. It follows that MATH is a normal extension of MATH to MATH.
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math/0003082
|
Setting MATH, MATH, we have MATH for large MATH. Moreover MATH and MATH converge increasingly respectively to MATH and MATH. Thus REF applies and gives MATH .
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math/0003082
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If MATH is KMS state, then MATH is a separating state. If MATH is a non-zero multiple of an isometry, then also MATH is a separating positive functional, which is quasi-equivalent to MATH because MATH is a separating vector for MATH. Take MATH, the element of MATH that implements MATH. If MATH we have MATH where MATH, thus MATH. Thus MATH, hence MATH and this implies that MATH is normal with respect to MATH. Conversely, let us assume that each MATH is normal with respect to MATH and still denote by MATH the extension of MATH to the NAME algebra MATH. By rescaling the parameter we may set the inverse temperature MATH equal to MATH. We may then define the *-algebra MATH as in the appendix with MATH replaced by its weak closure MATH (but the MATH are still defined with respect to MATH). Let MATH, thus MATH is a (not necessarily standard) non normalized left inverse of MATH. With MATH, define a one-parameter automorphism group MATH of MATH extending MATH as was done for MATH, but using MATH instead of MATH, thus, with obvious notations, MATH. By using the holomorphic properties of the NAME cocycles and the corresponding two-variable cocycle property (that can be checked similarly as in CITE), it can be seen by elementary calculations that MATH is KMS with respect to MATH. Clearly both MATH and MATH extend to MATH, indeed the extension of MATH is the modular group with respect to MATH, thus MATH and MATH commute and MATH is a one-parameter group of MATH leaving MATH pointwise fixed.
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math/0003082
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Let MATH be such that MATH if MATH. If MATH, write MATH with MATH. By the cocycle equation MATH we see that MATH.
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math/0003082
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We shall associate a covariant localized endomorphism MATH to a given MATH. Similarly as in REF , we set MATH where MATH is a vector in the time-zero hyperplane and MATH, with MATH the radius of the double cone in REF . We have to show that, for a fixed MATH, the above definition is independent of the choice of MATH, namely MATH if also MATH. As MATH is connected, by iterating the procedure, it is enough to check this if MATH with MATH. Then, by local commutativity, MATH because MATH belongs to MATH and MATH.
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math/0003082
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Setting MATH, MATH, we trivially obtain a local net on MATH and MATH is a KMS with respect to translations. The conformal extension of MATH is then the conformal completion of MATH given by REF ; the additivity assumption is here unnecessary since MATH is independent of MATH. A detailed proof can be found in REF.
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math/0003082
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Note first that there is a conditional expectation MATH, MATH, given by MATH, MATH, where MATH is the orthogonal projection onto MATH. The case MATH is clearly a consequence of NAME 's theorem since MATH is globally invariant under the modular group of MATH. Now the translations on MATH (constructed as in REF ) restrict to the translations on MATH and commute with MATH due to the cyclicity of MATH for MATH on MATH. Hence MATH maps MATH onto MATH, MATH. We then have MATH hence MATH is strongly additive, thus it satisfies NAME duality on MATH because MATH is conformal.
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math/0003082
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If MATH is localized in the interval MATH REF of MATH, then clearly MATH restricts to the NAME algebra MATH for all MATH as it acts trivially on MATH and by duality for MATH. Hence MATH restricts to the MATH-algebra MATH, and then it extends to MATH by REF . Let MATH be also localized in the interval MATH. If MATH are a standard solutions for the conjugate equation of MATH and MATH, then MATH and commute with MATH, hence they belong to MATH by strong additivity CITE. Conversely, if MATH are a standard solution for the conjugate equation of the restriction of MATH and MATH, then MATH and MATH belongs to MATH by the NAME duality for MATH, hence the conjugate equation is valid for all elements of MATH. This implies the dimension is the same for MATH and its restriction to MATH.
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math/0003082
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REF states that MATH . By the asymmetry of the chemical potential MATH, thus MATH . Summing up REF and setting MATH we obtain MATH .
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math/0003082
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Immediate.
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math/0003082
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Indeed MATH . Because of REF , one checks easily that MATH, MATH, where MATH and MATH is the one-parameter automorphism group. Since MATH commutes with MATH, it follows that MATH is an invariant subspace for MATH, thus the latter restricts to MATH on MATH and REF implies the statement.
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math/0003082
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To simplify notations we shall consider the tensor product MATH of the same net MATH by itself. The decomposition REF of the NAME space MATH of MATH gives a decomposition of MATH and we can define the operators MATH . The tensor product Hamiltonian MATH is equal to MATH.
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math/0003082
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Immediate by REF and the fact that inner automorphisms give the identity map in cyclic cohomology CITE.
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math/0003082
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Since MATH is the canonical endomorphism of MATH into MATH (see the appendix), the result is immediate.
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math/0003082
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Setting MATH we have (see REF and CITE) MATH, thus MATH which is a super-KMS functional with respect to the evolution Ad-MATH and the superderivation MATH. A direct verification shows that MATH is the JLO cocycle on MATH associated with MATH.
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math/0003082
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The proof could be given similarly to the one given in REF . However, it is easier to observe that one can consider sectors and conjugate sectors between different MATH-algebras. Then the canonical endomorphism MATH, as a map of MATH into MATH, is just the conjugate sector for the embedding MATH of MATH into MATH and thus the uniqueness of MATH modulo inner automorphisms of MATH is just a consequence of the uniqueness of the conjugate in a tensor REF-MATH-category, see CITE.
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math/0003082
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Let MATH be the operators in the conjugate equation for MATH; setting MATH and MATH REF holds true.
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math/0003087
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CASE: Let MATH the polar decomposition of MATH and MATH the spectral resolution of MATH. Then MATH and MATH, that is, MATH . Set now MATH with a unitary MATH. Since MATH is a trace vector we have MATH that is, MATH. Since every element of MATH is the linear combination of at most REF unitaries, it follows, that MATH. Since MATH, MATH, and MATH, the assertions follow also for MATH applying the results just proven. Let now MATH be cyclic for MATH. Then for every MATH there exists exactly one MATH such that MATH, hence MATH. Further MATH is dense in MATH and MATH, hence MATH. Since MATH is finite (it possesses a cyclic trace vector) it follows from this MATH that is, MATH (and MATH) is a core for MATH and, similarly, also for MATH and MATH. CASE: This first part follows in the same way as the corresponding assertions in REF. Now MATH for every MATH and, since MATH, MATH such that MATH.
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math/0003087
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Let MATH with MATH be a partial isometry and MATH, the polar decomposition of MATH. Then we have MATH since MATH is a partial isometry from MATH to MATH. Let now MATH be the spectral measure for MATH. Then MATH . From this we see, that MATH and, since MATH is separating for MATH and MATH, MATH. Since MATH is positive MATH,too, and therefore MATH and MATH.
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math/0003087
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Let MATH and MATH be two operators affiliated with MATH, such that MATH . Then MATH is closable, since MATH is finite (compare CITE), and its closure MATH is affiliated with MATH. Then it follows from REF that MATH, that is, MATH and MATH agree on the intersection of their domains. Since MATH is a core for both MATH and MATH (compare REF), they are equal.
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math/0003087
|
Let MATH be the polar decomposition of MATH and MATH the spectral projection of MATH for the interval MATH. Then MATH, such that MATH . Now MATH since MATH is a core for MATH and MATH.
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math/0003087
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CASE: Since MATH also MATH, where MATH is the polar decomposition of MATH. This is equivalent with MATH where MATH is the spectral measure of MATH and we have used the trace property of MATH and that MATH is in the final space of MATH. CASE: If MATH is cyclic there is an operator MATH, such that MATH and MATH . Now we have the following equality MATH . Since MATH is finite MATH is densely defined and closable and the closure MATH (compare CITE). Furthermore also MATH is closable and its closure MATH is affiliated with MATH. From this, REF it follows MATH since MATH is separating, and therefore MATH is injective. Suppose now MATH is injective, that is, MATH and MATH are both affiliated with MATH, where MATH is the polar decomposition of MATH. Now [compare REF] MATH . So we have the following chain MATH since MATH is cyclic for MATH and MATH is a partial isometry with MATH in its initial space. This means that MATH is cyclic for MATH. CASE: If MATH is separating for MATH it is cyclic for the commutant MATH, that is, MATH . The last equality follows from REF. This shows that MATH has dense range. Suppose now that MATH has dense range. Let MATH and MATH. This means MATH. Since MATH and MATH is finite we know that MATH. And using REF we derive MATH and, since MATH has dense range and MATH is bounded, MATH, such that MATH is separating. CASE: This follows from REF.
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math/0003087
|
The existence and the asserted properties follow from REF and the uniqueness from REF .
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math/0003087
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The necessarity of the trace condition follows from REF and the sufficiency from REF.
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math/0003087
|
Let MATH be the spectral projections of MATH corresponding to the interval MATH. Then MATH is in MATH with MATH, MATH as the polar decomposition. According to CITE MATH is a sequence of cyclic and separating vectors converging (MATH are invertible operators!) to MATH with modular objects: MATH where MATH is the conjugation corresponding to the trace vector MATH. Now the modular conjugation of MATH is MATH since all MATH lie in the same natural (closed) cone. Further the modular groups MATH corresponding to MATH converge in the strong operator topology to the modular group MATH corresponding to MATH (compare CITE). Since MATH and operator multiplication is continuous on bounded sets with respect to the strong operator topology, we have MATH . Since MATH and MATH commute, MATH is closable (compare CITE) and the closure MATH is selfadjoint, such that MATH.
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math/0003087
|
CASE: Suppose that MATH are the modular objects corresponding to a cyclic and separating vector MATH. Since MATH is cyclic and separating there exists a non-singular operator MATH corresponding to MATH such that MATH and MATH (compare REF, and REF). Since the decomposition of MATH is unique up to a positive constant REF we have MATH . CASE: Suppose that MATH. Choosing the non-singular operator MATH (according to REF MATH is densely defined and affiliated with MATH) we have MATH and MATH . Hence there is a cyclic and separating vector MATH corresponding to MATH such that MATH are the modular objects with respect to MATH (compare REF).
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math/0003087
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CASE: Let MATH be a type MATH factor (MATH), that is, it is isomorphic to MATH with a finite dimensional MATH. Since MATH is finite dimensional all linear operators are bounded and have finite trace, that is, REF is always satisfied. CASE: compare REF .
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math/0003087
|
CASE: Let MATH. Then there exists a unitary MATH which commutes with MATH and MATH, such that MATH and MATH. Setting MATH we have MATH and MATH commutes with MATH (since MATH and MATH do). Then we can calculate MATH . Since MATH commutes with MATH we have MATH . With this, REF, and REF follows MATH . CASE: We can assume without loss of generality that MATH where MATH unitary and MATH and MATH commutes with MATH. Then MATH . Since MATH, we define MATH, such that MATH commutes with MATH and MATH. Further MATH is cyclic and separating for MATH and MATH are the modular objects for MATH. This means that MATH are the modular objects for MATH. But MATH are the modular objects for MATH, too. From this follows, that MATH, and MATH. CASE: Analogous to REF , compare CITE.
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math/0003087
|
Let MATH be a unitary, such that MATH and MATH where MATH. Then from REF follows MATH . This shows MATH .
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math/0003087
|
The relation defined in REF is an equivalence relation, since it is reflexive (choose MATH), symmetric REF and transitive: Let MATH and MATH; this means hat MATH and MATH, where MATH and MATH are unitaries with the properties described above. Then with MATH we have MATH and also MATH have the right properties, such that MATH.
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math/0003087
|
Let MATH be two members of this class. Then there exist unitaries MATH, such that MATH, MATH commutes with MATH and MATH, and MATH (compare REF). Now define MATH . Then MATH and MATH such that the conditions of REF are fulfilled and MATH.
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math/0003087
|
Since MATH and MATH are solutions of the inverse problem, there are two unitary operators MATH and MATH both commuting with MATH, such that MATH, MATH are the modular objects for MATH (MATH). CASE: Let MATH. Then there is a unitary MATH commuting with MATH and MATH, such that MATH. Setting MATH an easy calculation gives MATH and MATH commutes with MATH. Also we can calculate MATH . REF now shows that MATH (note our normalization REF ). CASE: Suppose now MATH. Since MATH possesses a cyclic and separating vector there is a unitary MATH, such that MATH. Further MATH has modular objects MATH. Now there is a cyclic and separating vector MATH in the natural cone of MATH and a unitary MATH, such that MATH, and MATH has modular objects MATH, that is, MATH commutes with MATH and, since MATH, MATH. Now define MATH. Then MATH commutes with MATH and MATH, for: MATH . Also MATH, and, since MATH has modular objects MATH, MATH has modular objects MATH. Now, since the cyclic and separating vector is (up to the sign) uniquely determined by the modular objects (s. CITE), MATH, that is, MATH, and MATH is the unitary required by REF.
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math/0003087
|
Since MATH and MATH has no non-zero Abelian projection there exist for every MATH orthogonal families MATH and MATH of non-zero projections, such that MATH (compare CITE). Now MATH where the projections MATH are pairwise orthogonal projections, since MATH commutes with MATH, they are not MATH, since MATH is a factor (compare CITE), the latter means that they have at least NAME space dimension MATH. This means that the dimension of MATH is at least MATH and, since MATH was arbitrary, infinite.
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math/0003087
|
The first assertion follows directly from REF and the fact, that MATH since MATH is a factor (compare CITE). The second assertion follows from MATH and MATH in the type MATH case and from REF in the type MATH case.
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math/0003087
|
Since MATH are two selfadjoint operators having the same eigenvalues MATH REF we can write MATH where MATH are the corresponding (orthogonal) eigenprojections (and MATH) and the convergence is understood in the so-topology. Since MATH and MATH have the same multiplicities we have MATH (MATH), where MATH is the unique dimension function on MATH. This means that there are partial isometries MATH, such that MATH and MATH (compare CITE). Setting MATH we get a unitary in MATH, such that MATH REF and MATH.
|
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