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math/0104227 | Let MATH for MATH. Note that from convexity of the cone MATH, we have MATH. Using REF , we have MATH since MATH is positive definite from REF . Therefore MATH is non-decreasing, and MATH, so we have MATH . The negative case is similar. |
math/0104227 | Since MATH is compact, at a minimum of the function MATH we have MATH with MATH positive semidefinite. From the lemma we have MATH and certainly we can choose MATH such that MATH . Similarly, if the maximum of MATH is at MATH, we can choose MATH such that MATH . |
math/0104227 | The metric is parallel, so we have MATH . Therefore we have, at MATH, MATH . Using this, we have MATH . |
math/0104227 | We have MATH and MATH . To satisfy REF we need MATH, MATH, and MATH. So choose MATH. Next we have MATH . Now choose MATH and MATH so large that MATH . Then we have MATH . |
math/0104227 | The proof may be found in CITE. |
math/0104227 | From REF , we have MATH therefore the proof of REF applies. |
math/0104227 | We let MATH . Define MATH . Then all of the estimates in the previous sections hold with MATH replaced by MATH, and MATH replaced by MATH, and it is then not difficult to see that we can choose MATH independent of MATH, since the MATH estimate holds uniformly. |
math/0104227 | We consider the following function MATH where MATH is a function of the form MATH . The proof procedes exactly as before, but we end up with the following analogue of REF MATH . Assume that MATH. Then we may choose constants MATH and MATH depending only upon MATH, and MATH. so that MATH satisfies MATH and MATH . This follows easily from REF . With MATH chosen as above, we let MATH and MATH . From REF , we have MATH . The proof then procedes exactly as before. |
math/0104227 | In order to prove this, it is convenient to write REF in slighty different form. Writing MATH, with MATH, we see that MATH solves the equation MATH . As seen in REF, we must have MATH and therefore since MATH, MATH . Next choose MATH such that MATH and MATH such that MATH. Let MATH be a unit speed minimal geodesic such that MATH and MATH. Letting MATH denote the restriction of MATH to MATH, we have MATH therefore MATH . Let MATH, and MATH. Then MATH satisfies MATH . If we let MATH, then it is easy to verify that MATH satisfies the inequality MATH for MATH. Integrating this, and using the boundary condition MATH, we find that MATH for MATH. Since MATH, we conclude that MATH as long as MATH. Evaluating this at the endpoint MATH, we have MATH that is, MATH which implies the stated inequality for MATH. |
math/0104227 | This follows from the proof of REF , but since we have MATH instead of MATH, the inequalities are reversed. |
math/0104227 | We use the continuity method. For MATH we consider the equation MATH where MATH . Letting MATH, we know from REF that a solution necessarily lies in MATH, and we claim that the map MATH is locally invertible at a solution. From REF above we see that the linearized operator is MATH where MATH. The coefficient matrix MATH is positive definite, but there is a slight difficulty due to the fact that the linearized operator is not formally self-adjoint. Nevertheless, it is still invertible. This was proved for NAME equations in CITE, and the proof given there is applicable in this case. Local invertibility of MATH follows from the implicit function theorem (see CITE). Let MATH be a solution of REF . The matrix MATH satifies REF for all MATH, therefore we have that MATH satisfies the NAME REF . Let MATH be a point where MATH attains a global minimum. We have MATH which implies MATH. By also considering a maximum of MATH, we obtain the estimate MATH. Combining this with the NAME inequality, we obtain an a priori MATH estimate on MATH, independent of MATH. From the work in REF, and NAME, we obtain an a priori bound on the MATH norm of MATH, independent of MATH for some MATH. Standard elliptic theory gives a uniform bound on the MATH norm for each MATH. We consider the equation MATH: MATH . Let MATH be any solution to REF . As before, by going to a maximum and minimum of MATH, we find that MATH. Then from the arithmetic-geometric inequality, we have MATH . We conclude that MATH, which implies MATH. The existence of a solution at MATH now follows from the continuity method. It remains to prove the uniqueness at MATH. To see this, if we have REF distinct solutions MATH and MATH at MATH, we may run the continuity method in reverse. From uniqueness at MATH, the paths we obtain must hit at some time MATH. But since the linearization is invertible at MATH, this contradicts local invertibility. |
math/0104227 | We will employ a fixed point argument using the existence and uniqueness of solutions to REF . For MATH, MATH, and MATH, let MATH denote the unique solution in MATH of the equation: MATH . It is easy to show that for each MATH, the mapping MATH is uniformly continuous in MATH, and we also claim that for each MATH, MATH is a compact operator. For a bounded subset of MATH, the right hand side is bounded in MATH. From the proof of REF , solutions are bounded in MATH. Since MATH is a compact embedding, the claim follows. We next show that for all MATH, solutions of the equation MATH satisfy an a priori bound MATH. As in the proof of REF , we need only obtain a MATH estimate. To this end, let MATH be a fixed point MATH, and MATH be a point where MATH attains a global minimum. Then we have at MATH, MATH which implies MATH for some constant MATH, and we obtain the estimate MATH . Similary by considering a maximum of MATH we obtain MATH for some constant MATH. These estimates, coupled with the NAME inequality in REF , imply the desired uniform MATH estimate. As already seen in the proof of REF , we have that MATH for all MATH, where MATH is some constant. We may then apply a fixed point theorem of CITE: Let MATH be a one-parameter family of compact operators defined on a NAME space MATH for MATH, with MATH uniformly continuous in MATH for fixed MATH. Furthermore, suppose that every solution of MATH for some MATH, is contained in the fixed open ball MATH. Then, assuming MATH, the compact operator MATH has a fixed point MATH. Letting MATH, we find a fixed point MATH at MATH. Standard regularity theory then implies that MATH. Adding a constant if necessary, we obtain a solution to REF . |
math/0104228 | CASE: Since MATH is a faithful normal semifinite weight on MATH, so is MATH on MATH. CASE: Let MATH and MATH be the restriction of MATH to MATH, which is a faithful normal semifinite weight on MATH by assumption. Then, we have MATH . Regarding MATH as an isomorphism from MATH to MATH, we get MATH . CASE: This is CITE. CASE: Let MATH and MATH be the restriction of MATH to MATH. Then, we have common continuous decomposition CITE, CITE MATH such that MATH . Therefore, each non-zero projection MATH is equivalent to REF in MATH (more strongly in MATH), and there exist isometries MATH, MATH such that each MATH is a minimal projection in MATH and MATH forms a partition of unity. Setting MATH, we get the first assertion. Since the minimal expectation for MATH is given by MATH we have MATH thanks to the local index formula for the minimal conditional expectation CITE and the fact that MATH is an isomorphism from MATH to MATH. This implies MATH . Thus, MATH . Since MATH is in the centralizer of MATH, this implies that MATH is an invariant pair. CASE: Thanks to REF , the general case is reduced to the case where MATH and MATH are irreducible, and we make this assumption. If MATH, we have nothing to prove, and so we assume MATH for some unitary MATH. Since MATH commutes with MATH and MATH, MATH is proportional to MATH, and there exists a scalar MATH such that MATH. However, the KMS condition and MATH imply MATH, which shows MATH. |
math/0104228 | Let MATH be a dominant weight. First we assume that MATH is an invariant pair, and MATH and MATH are as in the proof of REF . Thanks to REF , we may consider MATH an isomorphism from MATH onto MATH intertwining MATH and MATH. Therefore, MATH extends to an isomorphism MATH from MATH onto MATH sending MATH to MATH. Since MATH satisfies MATH, we may regard MATH as a subalgebra of MATH identifying MATH with MATH. We show that MATH satisfies the required property. Indeed, MATH . Regarding MATH as an isomorphism from MATH onto MATH again, we have MATH and so MATH . Now, we treat the general case. Thanks to REF , it suffices to show the statement for MATH, where MATH is a unitary in MATH, and MATH and MATH are as before. Indeed, MATH . Therefore, MATH has the desired property. |
math/0104228 | In the proof of REF , we have already shown that for MATH and a unitary MATH, we have MATH . Thus, in order to prove the statements we may replace MATH and MATH with unitary equivalent endomorphisms. Thanks to REF , we may and do assume that MATH and MATH are invariant pairs, where MATH is a dominant weight. Therefore, we have MATH . In this situation, REF , and REF are obvious and REF follows from REF . To prove REF , we consider the common continuous decomposition of MATH. We may and do assume that MATH is of the form MATH, where MATH is a dominant weight on MATH and MATH is the minimal expectation from MATH to MATH. Then, we have the common continuous decomposition MATH . Let MATH be the trace whose dual weight is MATH. Then, thanks to CITE, the canonical endomorphism of MATH scales MATH by MATH. Let MATH and MATH be the modular conjugations of MATH and MATH, and MATH be the corresponding canonical endomorphism of MATH. To define the canonical endomorphism of MATH and MATH, we utilize the modular objects of these algebras naturally coming from the modular objects of MATH and MATH through the crossed products as computed in CITE. Let MATH and MATH be such canonical endomorphisms of MATH and MATH respectively. Then, MATH is an invariant pair CITE. On the other hand, applying CITE twice and using CITE for the computation of MATH and MATH, we know that MATH is an extension of MATH leaving MATH invariant. Therefore, MATH coincides with the canonical extension of MATH. |
math/0104228 | CASE: The cocycle relation of MATH follows from MATH . Let MATH be another implementing system for MATH, and set MATH. Since MATH is a matrix valued function that is unitary, we have MATH. Setting MATH, we get MATH which shows that MATH and MATH are equivalent. Let MATH be a unitary of MATH and MATH. Then, MATH is an implementing system for MATH. Since MATH is fixed by MATH, MATH gives the same cohomology class as MATH. CASE: We take a MATH-valued function MATH satisfying MATH . Then, MATH has a desired property. |
math/0104228 | CASE: First we show that MATH is injective. Let MATH satisfying MATH. Then, thanks to REF , there exist implementing systems MATH and MATH for MATH and MATH respectively such that they give the same cocycle. We set MATH, which is a unitary in MATH satisfying MATH. This shows that MATH is injective. Next we show that MATH is surjective. Let MATH be a dominant weight on MATH and MATH be the continuous decomposition. We take the trace MATH on MATH whose dual weight is MATH, and take the implementing one-parameter unitary group MATH for MATH. We assume that MATH acts on a NAME space MATH. Then, the NAME duality theorem CITE implies that there exists an isomorphism MATH from MATH to MATH such that MATH where MATH is the left regular representation of MATH and MATH is the multiplication operator of MATH. Moreover, MATH is identified with MATH under MATH, or more precisely we have MATH. Using MATH, we identify the flow MATH with MATH . Let MATH be a given cocycle, where MATH is understood as the point realization of MATH. We take a system of isometries MATH in MATH satisfying MATH and set MATH . Then, MATH is a unitary satisfying MATH which means that MATH is a MATH-cocycle. Since MATH is stable CITE, every MATH-cocycle is a coboundary and there exists a unitary MATH such that MATH. Let MATH. Then, MATH is an isometry in MATH satisfying MATH and MATH . For MATH we set, MATH and MATH. Then, MATH is an endomorphism of MATH commuting with MATH, MATH is a left inverse of MATH satisfying MATH, and MATH is a MATH-preserving conditional expectation from MATH onto the image of MATH. We introduce MATH that is an extension of MATH to MATH leaving MATH invariant. Indeed, such MATH exists because the following hold: MATH . In a similar way, we define a left inverse MATH of MATH and a conditional expectation MATH from MATH onto the image of MATH by the relations MATH and MATH. Note that MATH, MATH, and MATH are extensions of MATH, MATH, and MATH respectively. We claim that MATH, MATH, MATH, and MATH is an invariant pair. Indeed, it suffices to show MATH, and the other claims can be shown easily. We define MATH by MATH . Then, for MATH and MATH, we have MATH which shows MATH is a one-element NAME basis and MATH CITE, CITE. To show that MATH is minimal, it suffices to show MATH for all MATH CITE. Thanks to the NAME relative commutant theorem CITE, for MATH we have MATH and MATH . Thus, MATH which shows the claims. Since MATH is an invariant pair, MATH is an extension of MATH leaving MATH invariant. Therefore, we have MATH which shows that MATH is a modular endomorphism carrying MATH. REF are easy, and REF follows from the above proof. CASE: Assume that MATH and MATH. Let MATH and MATH be implementing systems for MATH and MATH satisfying MATH . We define two isometries MATH and MATH by MATH . MATH and MATH belong to MATH satisfying MATH . Thus, MATH CITE. Since MATH is a bijection, this also shows that MATH implies MATH and MATH. |
math/0104228 | CASE: First, we show that MATH contains at least one isometry if MATH. Let MATH be the set of partial isometries in MATH. Using the polar decomposition, we know that MATH is not empty. We introduce an order into MATH as follows: For two element MATH, we say that MATH dominates MATH if the following holds MATH . Note that when this is the case, MATH, and MATH, MATH hold. We show that MATH is an inductively ordered set with this order. Let MATH be a totally ordered subset of MATH. Since MATH and MATH are increasing nets of projections in MATH and MATH, they converge to projections in MATH and in MATH respectively in strong topology. We claim that the net MATH converges in strong MATH topology. Let MATH with MATH dominating MATH, and MATH be a vector in the NAME space that MATH acts on. Then, MATH . Therefore, the strong MATH limit exists, and it gives a majorant of MATH, which shows that MATH is inductively ordered. Now, we apply NAME 's lemma to MATH and take a maximal element MATH. We show that MATH is an isometry. Suppose MATH. Since MATH and MATH acts on MATH ergodically, there exists some MATH such that MATH. Therefore, MATH dominates MATH, which is contradiction. Thus, MATH is an isometry. Let MATH be a maximal set of isometries in MATH with mutually orthogonal ranges (such a set exists thanks to NAME 's lemma again), and let MATH. If MATH, we are done, and so we assume MATH. Let MATH be the subset of elements in MATH with range projections orthogonal to MATH. We claim that there exists a non-zero element in MATH. Indeed, since MATH, there exists some MATH such that MATH, which shows that there exists non-zero MATH with MATH. Using the polar decomposition, we get the claim. We introduce an order into MATH as before and take a maximal element MATH. Since MATH is a maximal set, MATH is not an isometry, and we set MATH. We claim MATH. Indeed, by maximality of MATH, MATH holds for MATH, and MATH for MATH as well for all MATH because MATH is globally invariant under MATH. Thus, we get MATH . Since MATH, this implies MATH. Let MATH. Then, MATH have mutually orthogonal ranges such that MATH . Thus, we get MATH . Since there exists a conditional expectation from MATH to MATH satisfying the NAME inequality CITE, CITE, MATH is a finite set, and we identify MATH with MATH. We take MATH satisfying MATH, and set MATH to be the NAME subset corresponding MATH. Let MATH, MATH, MATH, and MATH. Then, MATH and MATH satisfy MATH . We set MATH, which is regarded as a MATH by MATH matrix-valued function on MATH. However MATH satisfies MATH which is contradiction. Thus, we conclude MATH, and MATH . CASE: Let MATH be a modular endomorphism with an implementing system MATH for MATH, and MATH be an irreducible component of MATH. We take a non-zero element MATH. Thanks to REF , MATH belongs to MATH, MATH. Since MATH, there exists some MATH such that MATH is non-zero. Thus, MATH is a modular endomorphism thanks to REF . |
math/0104228 | Thanks to REF , it suffices to show that the minimal subgroup of MATH is irreducible if and only if MATH is never equivalent to a direct sum of two cocycles. Assume that the minimal subgroup of MATH is irreducible. If MATH were equivalent to MATH, MATH, MATH, with MATH, MATH could be considered as a MATH-valued cocycle, and the minimal subgroup of MATH would be conjugate to a subgroup of MATH in MATH thanks to REF . This is contradiction and MATH is never equivalent to a direct sum of two cocycles. The converse also follows from REF . |
math/0104228 | Assume that graph change occurs for MATH. Taking sufficiently large MATH, we may assume MATH. Let MATH be the canonical endomorphism for MATH. Then, thanks to REF , this is equivalent to MATH . Therefore, either of the following two holds: CASE: MATH contains two irreducibles MATH such that MATH and MATH. CASE: MATH contains an irreducible MATH such that MATH. Indeed, let MATH be a partition of unity consisting of minimal projections in MATH. If MATH there exist some MATH and MATH such that MATH . If the irreducible components corresponding to MATH and MATH are different, REF occurs. If they are the same, REF occurs. In REF , the NAME reciprocity implies MATH (though MATH is not a factor, the argument in CITE works). Thanks to REF , this shows that MATH, which is contained in MATH, contains a modular endomorphism. Since MATH does not contain identity, it is indeed a non-trivial one. In REF , we take MATH. Let MATH, MATH be isometries satisfying the usual relation CITE MATH . Then, MATH. Let MATH be an irreducible sector contained in MATH that is not equivalent to identity, and MATH be a basis of MATH. If MATH is not zero for some MATH, we are done thanks to REF . Thus, we assume MATH for every MATH not equivalent to identity. This would imply that MATH. However, since MATH commutes with MATH, we would get MATH which is contradiction. Thus, MATH contains a non-trivial modular endomorphism, and so does MATH. The converse can be shown easily. |
math/0104228 | CASE: This is trivial. CASE: We take isometries MATH satisfying MATH . Assume MATH. Thanks to REF , we have MATH which shows that MATH for every MATH and MATH does not depend on MATH. Next we assume that MATH for every MATH and MATH does not depend on MATH, which is denoted by MATH. Then, MATH which shows MATH. CASE: Since MATH is the dual inclusion of MATH, the first part is obvious. The second part follows from REF because MATH contains identity. CASE: This follows from the definition of modular endomorphisms. CASE: This follows from REF . |
math/0104228 | CASE: This follows from REF . This follows from REF . |
math/0104228 | Since the center of MATH is generated by MATH, we have MATH by assumption, which implies MATH . Therefore, there exists a unitary MATH such that MATH thanks to CITE. This means that MATH is an invariant pair. The rest of the proof is the same as those of REF if the dominant weight is replaced with the generalized trace. |
math/0104228 | We set MATH, MATH, and MATH where MATH is the NAME measure of MATH. Let MATH be as above. For MATH, we define MATH by MATH . Then, MATH has the following formal expansion (see CITE): MATH . More precisely, the above summation converges in the GNS NAME space topology with respect to a MATH-invariant normal state, and MATH completely determines MATH. Using this expansion, we can see that MATH if and only if MATH. Therefore, MATH is minimal if and only if MATH does not contain identity for every non-trivial irreducible representation MATH. However, the latter is equivalent to the statement that MATH is irreducible for very irreducible MATH thanks to the NAME reciprocity for the usual compact group representations and that of sectors CITE. |
math/0104228 | CASE: First we claim that the second dual action MATH on MATH is minimal whenever MATH is so. To show the claim, we may assume that MATH is infinite as usual. Thus, we have a NAME space MATH in MATH for each MATH as before. This implies that there exists a NAME space MATH in MATH such that the restriction of MATH to MATH is equivalent to the regular representation. Therefore, MATH is conjugate to MATH, which shows the claim. Let MATH be a cocycle perturbation of MATH. Since MATH is an irreducible inclusion and there exists a projection MATH such that MATH which shows that MATH is minimal. REF . The first part is easy. Stability in these two cases (and REF as well) follows from REF and NAME 's REF by REF matrix trick CITE. CASE: We show that if MATH is of type II, MATH is either of type II or of type MATH, MATH. Let MATH be a faithful normal semifinite trace on MATH. Then, the centralizer MATH is an intermediate subfactor between MATH and MATH, and in particular MATH is a factor. Therefore, the NAME spectrum MATH coincides with the NAME spectrum MATH CITE, CITE. Thus, MATH is either of type II or type MATH, MATH depending on the period of the modular automorphism group MATH. CASE: Note that part of the proof has been already done in the above. Assume that MATH contains MATH, for some faithful normal semifinite weight MATH on MATH. Then, the centralizer MATH is an intermediate subfactor of MATH. Let MATH be the restriction of MATH to MATH, which is a normal conditional expectation from MATH to MATH. Since the restriction of MATH to MATH is nothing but MATH and it is semifinite, MATH is of type II, and so is MATH as well because MATH is trivial. This finishes the proof of the first statement. Now, we show that MATH is not stable in the above situation with type III MATH . First we assume that MATH is of type MATH. Let MATH be a non-zero finite projection, and MATH be an isometry satisfying MATH. We set MATH, which is a MATH-cocycle. It is easy to show that MATH is finite and MATH is not a coboundary. When MATH is of type MATH, a similar construction identifying MATH with a corner of MATH works. |
math/0104228 | CASE: We set MATH. Let MATH be the normal conditional expectation from MATH onto the fixed point subalgebra of MATH under MATH obtained by average of MATH over MATH. Since MATH leaves MATH fixed for all MATH and MATH, the image of MATH coincides with MATH. CASE: Let MATH be a dominant weight on MATH, and MATH. For each MATH, MATH, we may and do assume that MATH is an invariant pair thanks to REF . Then, the modular automorphism group MATH acts on MATH trivially CITE and MATH leaves MATH invariant. Therefore, MATH is given by MATH . Thus, REF and CITE imply that MATH and MATH is the dual action of MATH. |
math/0104228 | Before starting the proof, we simplify the notation a little. We may and do assume MATH thanks to REF . For MATH, MATH, we simply denote MATH and MATH. CASE: We take MATH and an orthonormal basis MATH as before, where MATH is transformed by MATH in the same way as the canonical basis of MATH. Let MATH be the unique normal conditional expectation from MATH onto MATH, and MATH be the natural extension of MATH to MATH that is a normal conditional expectation from MATH onto MATH. As in the proof of REF , for MATH we set MATH . Then, MATH is in MATH if and only if MATH for every MATH and every MATH. This is further equivalent to that there exists MATH such that MATH for every MATH and every MATH. Note that thanks to REF , MATH commutes with every element in MATH. Therefore, in the same way as in CITE, we can conclude that MATH is generated by MATH and MATH, where MATH. Since MATH is generated by MATH and MATH, it is also generated by MATH and MATH as well. Thus, to prove MATH, it suffices to show that MATH generate a commutative NAME algebra. Let MATH be the NAME algebra generated by MATH. We show that there exists a MATH isomorphism from MATH onto MATH sending MATH to MATH. For this, it suffices to check that both MATH and MATH have the same algebraic relations because we have MATH, where MATH . First, we show MATH. Let MATH and MATH be isometries defined by MATH . Then, the definition of the crossed product by a NAME action implies that we also have MATH . Thus, we get MATH . For the same reason, we have MATH which implies MATH . Thus, MATH is abelian. For the same reason again, there exist MATH such that MATH which implies MATH . Note that thanks to the NAME theorem, MATH is obtained by MATH . Thus, we have MATH . This shows that MATH holds as well thanks to the NAME theorem again. Therefore, there exists a desired isomorphism from MATH onto MATH. Let MATH be a faithful normal state of MATH whose restriction is given by the measure MATH, and we set MATH. Then, for MATH, MATH, we have MATH . This shows that MATH. MATH acts on MATH as MATH which shows that the flow is given by the skew product. CASE: Since MATH acts on MATH trivially, we have MATH . This means that the NAME module of MATH is given as in the statement. |
math/0104228 | Let MATH be the modular conjugation of MATH on MATH. Then, MATH is given by MATH where MATH is the modular function of MATH and MATH is the canonical implementation of MATH on MATH. Direct computation yields MATH . Thus, we get MATH . |
math/0104228 | First, we show uniqueness of MATH. Let MATH be a NAME algebra including MATH with a unique conditional expectation MATH from MATH onto MATH. We assume that MATH is generated by MATH and MATH. Then, the linear span of MATH is dense in MATH by assumption. For a faithful normal state MATH on MATH and MATH, MATH, we have MATH . This means that the structure of MATH is uniquely determined by MATH and MATH . On the other hand, MATH completely determine the original inclusion via the NAME duality theorem. Therefore, such MATH as in the statement is unique if it exists. We set MATH as in CITE, CITE. More precisely, let MATH be the standard NAME space of MATH. We introduce an inner product into the algebraic tensor product MATH by MATH . Let MATH be the completion of MATH, and MATH be the natural map from MATH to MATH. MATH and MATH naturally act on the first and the second tensor components of MATH, and we denote by MATH the NAME algebra generated by MATH and MATH in MATH. In terms of direct integral, MATH can be expressed as follows (though we do not use it in the proof): Let MATH be disintegration of MATH and MATH over MATH. Then MATH is given by MATH . Thanks to the NAME relative commutant theorem and REF , we have MATH. We show MATH using this and applying CITE (or alternatively, using the above direct integral expression). To do so, we need to separate MATH and MATH by a type NAME algebra. Indeed, let MATH be the commutant of MATH in MATH. Then, MATH naturally acts on MATH and we have MATH and MATH, where the commutant MATH is considered in MATH. Applying CITE twice, we get MATH . We introduce a REF-parameter automorphism group MATH on MATH extending both MATH and MATH. Let MATH be the canonical implementation of MATH on MATH. We set MATH . Then, MATH extends to a REF-parameter unitary group on MATH satisfying MATH . We define MATH to be the restriction of MATH to MATH. Next, we show that there exists a faithful normal conditional expectation MATH from MATH onto MATH satisfying MATH . Let MATH be a faithful normal state on MATH, and MATH be the cyclic and separating vector of MATH giving MATH. We set MATH, and claim that MATH is cyclic and separating vector for MATH. Indeed, MATH is clearly a cyclic vector. To show that it is separating, we introduce a conjugation MATH on MATH by MATH where MATH is the modular conjugation of MATH. It is easy to show that MATH extends to a conjugation such that MATH. Since MATH is cyclic for MATH, MATH is a separating for MATH. We denote by MATH the state of MATH defined by MATH, MATH. Then, it is a routine work to show that MATH is the modular conjugation for MATH, and the modular automorphism group MATH is given by MATH . Thanks to NAME 's theorem CITE, there exists a MATH-preserving normal conditional expectation MATH from MATH onto MATH. Since MATH for MATH, MATH, MATH satisfies MATH. We set MATH and MATH to be the restriction of MATH to MATH. Since MATH satisfies MATH, MATH is a faithful normal conditional expectation from MATH onto MATH. Therefore, MATH is of type III CITE. Let MATH be the restriction of MATH to MATH, where MATH is a faithful normal semifinite weight on MATH. Then, thanks to the NAME theorem CITE, MATH and MATH is the dual action of MATH. We claim that MATH, and in particular MATH is a factor. Indeed, we have MATH . Since MATH acts on MATH ergodically, we get MATH. Instead of showing that MATH comes from the modular automorphism group directly, we show it for the dual action of MATH. We define MATH, which is a faithful normal semifinite trace on MATH satisfying MATH. Indeed, since MATH is a trace, MATH acts on MATH trivially. On the other hand, since MATH is a faithful normal conditional expectation from MATH onto MATH and MATH, the restriction of MATH to MATH is the same as the restriction of MATH to MATH CITE, which is trivial as we saw before. Thus, MATH is a trace with the scaling property MATH . This means that the dual action of MATH is the modular automorphism group for the dual weight MATH. Thus, identifying MATH with MATH using the NAME duality theorem, we get MATH . Let MATH be the generator of MATH. Then, we have MATH, which implies MATH. Thus, we get MATH, which implies MATH and MATH. |
math/0104228 | We use the same notation as in REF implies that the smooth flow of weights of MATH is of the form MATH with the flow and MATH given by MATH where MATH is a minimal cocycle with MATH. According to the above splitting of MATH, we set MATH and MATH that is the fixed point subalgebra of MATH under MATH. Note that the factor flow MATH and the class of MATH are uniquely determined by MATH and MATH. Since MATH is an equivariant copy of MATH, the NAME 's type theorem for coactions CITE implies that MATH is the crossed product MATH by a coaction of MATH, where we use the fact that MATH is the fixed point subalgebra of MATH under MATH. Moreover, since MATH commutes with MATH, the coaction is trivial and we actually have MATH, where MATH and MATH are identified with MATH and MATH in the right-hand side respectively. This implies MATH, MATH and MATH. Since MATH acts on MATH ergodically, MATH is a factor. If MATH were semifinite, MATH would be a coboundary, which would imply that MATH is trivial and MATH. This is contradiction because MATH is of type III. Therefore, MATH is a type III factor. Now, REF show that MATH is the skew product MATH. Since the NAME group for MATH coincides with MATH CITE,CITE and every member MATH in the NAME group is specified by MATH, we conclude that MATH is the dual action of the skew product. |
math/0104228 | This is an easy consequence of REF , and the fact that AFD type III factors are determined by their flows CITE, CITE, CITE, CITE, CITE. |
math/0104228 | CASE: We assume that REF holds and the irreducible decomposition of MATH is given by MATH . Thanks to CITE, the multiplicity MATH of MATH is always finite. We set MATH . Then, the dimension of MATH is MATH and MATH is generated by MATH and MATH CITE. Let MATH be an orthonormal basis of MATH. We choose a dominant weight MATH on MATH and assume that MATH is an invariant pair for every MATH as before. Then, since MATH is unimodular, CITE imply that MATH is in the centralizer of MATH. By assumption, there exists an implementing system MATH for MATH. As MATH is in the centralizer of MATH, it commutes with MATH, and so MATH . Since MATH is generated MATH and MATH, it is also generated by MATH and MATH. CASE: We assume REF , and set MATH, MATH, MATH, and MATH. Then, MATH is a normal conditional expectation with the trace property MATH, MATH. Let MATH be the basic extension of MATH. As before, we regard MATH as a subalgebra of MATH identifying MATH with MATH. Then, MATH is the basic extension of MATH with respect to MATH CITE, CITE. On the other hand, we consider the inclusion MATH acting on the standard NAME space MATH of MATH. Then, the basic extension of this inclusion is MATH. Let MATH be the canonical implementation of MATH, and MATH be the modular conjugation of MATH. We use the same notation as in the proof of REF , and identify MATH with MATH. Note that MATH naturally acts on MATH through the action on the second tensor component. In the same way as before, we get the following: MATH where everything takes place in MATH. Thus, MATH and MATH where MATH is the fixed point subalgebra of MATH under the relative modular automorphism group MATH CITE. Since MATH is discrete, CITE implies that MATH is a direct sum of matrix algebras and the restriction of MATH to MATH is still semifinite. Therefore, if MATH is the center valued trace MATH of MATH, MATH is an extension of the flow MATH with relatively discrete spectrum. Indeed, let MATH be a faithful normal state on MATH. Then using the spatial derivative, we can show that the modular automorphism group of MATH is implemented by the modular operator MATH of MATH on MATH CITE. However, since MATH is a trace and MATH is trivial, MATH is a trace as well. This means that MATH holds with some density MATH affiliated with MATH. Let MATH be the NAME projection for MATH, and MATH. Then, MATH is considered as a field of scalar multiples of rank one projections in the central decomposition of MATH. Since MATH we actually have MATH. CASE: We assume REF . Thanks to REF , there exists a compact group MATH, a minimal cocycle MATH with MATH, and a faithful ergodic action MATH of MATH on a NAME algebra MATH such that MATH (see REF for the notations). Let MATH be the skew product MATH and MATH be the dual action. We set MATH. Then, thanks to ergodicity of MATH, MATH is an irreducible inclusion of factors. Let MATH be the unique MATH-invariant trace CITE. Then, the restriction of MATH to MATH gives a normal conditional expectation from MATH onto MATH. Thanks to REF , to prove the statement it suffices to show that MATH is generated by MATH and MATH, and MATH . REF show that MATH is actually given by MATH where MATH is the left regular representation. Thus, we have MATH which shows MATH. Note that MATH acts on MATH trivially and it acts on MATH as in REF . Now we assume that MATH acts on MATH, where MATH is the standard NAME space of MATH and MATH. Let MATH and MATH be the canonical implementations of MATH on MATH and MATH on MATH respectively. Then, MATH is given by MATH . We introduce a unitary operator MATH on MATH by MATH which satisfies MATH . Let MATH be the restriction of MATH on MATH, which is an automorphism of MATH. Then, the above relation implies MATH . For MATH, MATH, and MATH we have MATH . Thus, we get MATH . CASE: We assume REF and set MATH, MATH. Let MATH be the unique MATH-invariant trace on MATH, MATH be the GNS NAME space of MATH, and MATH be the natural map from MATH into MATH. We denote by MATH the canonical implementation of MATH on MATH. For each MATH, we denote by MATH the spectral subspace corresponding to MATH, which is the image of the map MATH . It is known that the multiplicity of MATH is finite CITE and there exists an orthonormal basis MATH, MATH, MATH with respect to the inner product of MATH such that MATH . Since MATH is given by the average of MATH over MATH, the NAME theorem implies MATH . Note that MATH form an orthonormal basis of MATH. Though expansion of MATH with this basis does not converge in weak topology in general, we can still show that MATH generate MATH using the same argument as in CITE. The point of the argument is that when a NAME subalgebra is globally invariant under the modular automorphism group of a state, the corresponding subspace in the GNS subspace determines the subalgebra. We use this principle in the rest of the proof without mentioning it. Let MATH be the unique normal conditional expectation from MATH onto MATH, MATH be a faithful normal state on MATH, and MATH. We denote by MATH, MATH, and MATH the GNS NAME space of MATH, the natural map from MATH into MATH, and the canonical implementation of MATH on MATH respectively. We regard the subspace of MATH-invariant vectors in MATH as the GNS NAME space MATH of the restriction of MATH to MATH. Let MATH be the NAME action of MATH on MATH whose dual action is MATH, and MATH, MATH be an orthonormal basis of MATH as before. Thanks to REF , MATH is a modular endomorphism for all MATH. Note that MATH is globally invariant under the modular automorphism group MATH CITE. Let MATH . Thanks to CITE, if MATH is generated by MATH and MATH, MATH is a discrete inclusion. Now we set MATH . Then, MATH are isometries with mutually orthogonal ranges. The definition of the crossed product by a NAME action implies MATH . Therefore, direct computation using the orthonormal basis MATH yields MATH . Thus, MATH is generated by MATH and MATH, and so MATH is a discrete inclusion. Let MATH be the restriction of MATH to MATH, which is a conditional expectation from MATH onto MATH. Then, we have MATH . This shows that the inclusion is unimodular CITE. When MATH is AFD, MATH is AFD and MATH is AFD thanks to the proof of REF . Therefore, MATH is AFD. Since there exists a conditional expectation from MATH onto MATH, the converse also holds CITE. |
math/0104228 | CASE: Suppose that there exists MATH such that MATH does not have a NAME module. Let MATH, and we consider the NAME decomposition of MATH CITE, CITE, CITE as follows: MATH where MATH and MATH are determined by the conditions MATH . Since MATH does not have a NAME module, either MATH or MATH occurs. We may assume MATH by considering MATH if necessary. Then, MATH contains the the canonical endomorphism MATH for MATH. Note that the dual inclusion MATH satisfies the condition of REF and MATH is a finite to one extension of the flow MATH CITE, CITE. Thus, REF implies that MATH generates a NAME action of the dual MATH of a finite group MATH, which is a subset of MATH. We regard MATH as a subset of MATH. We define a closed normal subgroup MATH by MATH . Since MATH is a finite set of irreducible representations closed under complex conjugate and irreducible decomposition of products of any members in MATH, the quotient group MATH is finite CITE. However, this is contradiction because MATH is connected. Therefore, MATH has a NAME module for every MATH. CASE: For MATH, we consider a family of unitary operators MATH . Then, thanks to REF and the NAME duality theorem CITE, there exists a unique element MATH such that MATH holds. Since MATH is a scalar for every irreducible MATH, MATH is in the center MATH. MATH is clearly a homomorphism from MATH into MATH, which is injective as the set of irreducible representations separates points of MATH. MATH is continuous because the topology of MATH is the same as that induced by the weak topology of the image of MATH by the direct sum representation of all members in MATH. |
math/0104228 | Since the center of a semisimple NAME group is a finite group, REF implies that MATH is of type III. As before, we regard MATH as a subalgebra of MATH using a unique conditional expectation MATH. Let MATH be the NAME action of MATH on MATH whose dual action is MATH. We first assume that MATH is of type MATH, MATH. Since every non-trivial modular endomorphism of MATH is a composition of an inner automorphism and a modular automorphism in this case, no non-trivial modular endomorphisms appear in MATH because a connected semisimple NAME group has no non-trivial REF dimensional representation. Thus, the same computation as in the proof of REF implies MATH . Thanks to REF , MATH has a NAME module for every MATH, which implies MATH . This shows that if MATH is of type MATH, so is MATH. When MATH is of type MATH, MATH, we have MATH. Since MATH is a finite group, MATH is finite and its dual group MATH is a finite subgroup of MATH. Thus, MATH is a finite cyclic group and so is MATH. Now we assume that MATH is of type MATH. Then, in the same way as above we can show MATH . Since MATH is a finite group, MATH is a non-transitive ergodic flow and so is MATH. Therefore, MATH is of type MATH. |
math/0104228 | Considering the NAME spectrum of an appropriate weakly dense separable MATH-subalgebra of MATH, we may and do assume that MATH is a compact metric space, on which MATH acts continuously, and that MATH is invariant under MATH. Since MATH is compact, every MATH-orbit is closed and there exists a NAME subset MATH that meets each MATH-orbit exactly once (see, for example, CITE). Let MATH be the set of all closed subgroups of MATH. Then, MATH has a Polish space structure such that MATH continuously acts on MATH by MATH, MATH, MATH CITE. We denote by MATH the class of MATH in the quotient space of MATH. For each MATH, let MATH be the stabilizer subgroup of MATH. Then, thanks to CITE, the map MATH is a NAME map satisfying MATH for MATH. Thanks to the ergodicity of the MATH-action, there exist MATH and MATH-invariant NAME null set MATH such that for every MATH, we have MATH. We set MATH. Let MATH be the normalizer subgroup of MATH in MATH. Choosing a NAME cross section from MATH to MATH, we have a NAME map MATH such that MATH. We set MATH . Then, MATH is a NAME subset of MATH that meets each MATH-orbit exactly once and MATH for all MATH. Choosing a NAME cross section MATH, we set MATH . Thanks to CITE, MATH is a NAME isomorphism from MATH onto MATH. Since MATH is MATH-invariant, MATH is of the form MATH, where MATH is the (normalized) NAME measure of MATH. We introduce a MATH-action on MATH from the action on MATH through MATH. When we regard MATH as the factor space of MATH by the MATH-action, we denote the action of MATH on MATH by MATH, MATH. Then, for each MATH and MATH, there exists a unique MATH such that MATH. Using a NAME cross section from MATH to MATH again, we can take MATH to be a NAME map from MATH to MATH. Let MATH be the quotient map. Commutativity of the MATH-action and MATH-action implies that for each fixed MATH, MATH takes its values in MATH almost everywhere. Moreover, MATH is a MATH-valued cocycle and MATH . This implies that for every subset MATH, MATH is MATH-invariant. Thus, thanks to the ergodicity of the MATH-action, we conclude that whenever MATH is an open subset of MATH, MATH which implies MATH, that is, MATH is a normal subgroup of MATH. Since the MATH-action is faithful, we get MATH. Uniqueness of MATH up to equivalence is obvious. |
math/0104228 | Let MATH be the canonical implementation of MATH on MATH. We introduce a unitary MATH, MATH on MATH by MATH . Then, MATH is a representation of MATH, which is a NAME map from MATH to the unitary group of MATH with respect to the NAME structure coming from the weak (and strong ) topology. Since any unitary NAME representation of a locally compact group is continuous, (or more strongly, any NAME homomorphism between Polish groups is continuous CITE), so is MATH. MATH is defined by the restriction of MATH to MATH. |
math/0104228 | CASE: By assumption, we have a direct sum decomposition MATH where MATH is a type I factor. We take a system of matrix units MATH of MATH with MATH either a natural number or possibly infinite such that MATH form a partition of unity consisting of minimal projections of MATH. Since MATH, MATH normalizes MATH, it does MATH as well, and we set MATH to be the restriction of MATH to MATH. Then, we have MATH and in particular, MATH . Thus, by ergodicity MATH is a constant, say MATH. In what follows, when we claim that ``some statement involving MATH and MATH holds for almost all MATH", we always mean that for each fixed MATH, there exists a conull set MATH such that the statement holds for all MATH. The annoying fact that MATH varies according to MATH is usually taken care of by NAME 's argument in CITE. We consider the disintegration of MATH over MATH . Then, MATH means that the dimension of MATH is MATH for almost all MATH. We take MATH, MATH such that for almost all MATH is an orthonormal basis of MATH. Since MATH is a unitary satisfying MATH for MATH, for MATH and MATH we have, on one hand, MATH and on the other hand, MATH . This implies, MATH for almost all MATH. Thus, MATH is an orthonormal basis of MATH for almost all MATH. We set MATH . Then, MATH is an element of MATH. We define a compact group MATH and a cocycle MATH by MATH . Changing the system MATH if necessary, we may and do assume that MATH is minimal. We denote by MATH the minimal subgroup of MATH. Note that MATH is minimal for all MATH and that if MATH, MATH and MATH are not equivalent (otherwise, MATH would be bigger). Let MATH and MATH be NAME spaces with basis MATH and MATH respectively. We set MATH . Since MATH is an orthonormal basis of MATH for almost all MATH, we identify MATH with MATH, and MATH with MATH. There exists a natural embedding of MATH in the group MATH of all unitaries MATH as those unitaries acting ``only on MATH, MATH". Through this embedding, we also regard MATH as a subgroup of MATH. Then, we have MATH for almost all MATH. Let MATH be the disintegration of MATH over MATH. Then, the map MATH is measurable with respect to the NAME structure of the set of NAME algebras on MATH CITE, CITE. For MATH, REF implies MATH for almost all MATH. Since MATH has countable generators (we assume that NAME spaces are separable throughout this paper), this implies MATH for almost all MATH. Using the NAME structure in the same way as in the proof of CITE and passing to an equivalent cocycle if necessary (and changing MATH and MATH on a null set), we may and do assume that there exists a NAME algebra MATH such that MATH and MATH normalizes MATH for all MATH. Since MATH is a minimal cocycle with MATH, MATH normalizes MATH, and so we define MATH to be the restriction of MATH, MATH to MATH. Then, REF implies MATH. Since MATH is ergodic, so is MATH. CASE: Thanks to CITE, MATH is injective and so is MATH. Let MATH be a normal conditional expectation from MATH onto MATH. Then, since MATH is the standard NAME space of MATH, there exists MATH whose vector state is MATH, where MATH means the state of MATH corresponding to the measure MATH. Thus, for all MATH and MATH we have MATH . We introduce a measurable field of normal states MATH of MATH by MATH . Then, REF implies MATH for almost all MATH. Moreover, if MATH commutes with MATH, we get MATH for almost all MATH. Let MATH be the set of all normal states of MATH, which is a Polish space in the norm topology. Since MATH is continuous in MATH-topology, MATH acts on MATH as a continuous transformation group. Applying the same argument as above to MATH and passing to an equivalent cocycle (and changing MATH and MATH on a null set), we may assume that there exists MATH such that MATH for all MATH and MATH holds. Thus, MATH is a MATH-invariant state, which is unique because of ergodicity of MATH. Thus, MATH is uniquely determined. MATH has the trace property as MATH is a trace thanks to CITE. CASE: We assume that MATH are also given by MATH, MATH. Let MATH be the unique MATH-invariant trace, MATH be the GNS NAME space of MATH, and MATH be the canonical implementation of MATH, MATH. Then, we have MATH and MATH is given by MATH . Since MATH is a compact group, we have the irreducible decomposition MATH where MATH is the multiplicity space, which may be zero. Since the MATH-action on MATH is ergodic and MATH is a minimal cocycle with MATH we have MATH . Thus, we identify MATH with the subset of MATH consisting of the irreducibles appearing in MATH, and identify MATH with MATH when MATH is identified with MATH. In what follows, we abuse notation and use only MATH, instead of MATH, if there is no possibility of confusion. We set MATH and regard it as a subgroup of the unitary group MATH as before. Since MATH is faithful, we identify MATH with MATH. Let MATH be the measurable field of unitaries describing the identity map on MATH with respect to the two distinct splitting MATH and MATH. MATH is decomposed as MATH where MATH is a measurable field of unitaries. Then, we have MATH for almost all MATH. Thanks to uniqueness of the minimal subgroup up to conjugacy, there exists a unitary MATH from MATH to MATH such that if we set MATH then MATH. We denote by MATH the restriction of MATH to MATH. Let MATH, and MATH . Then, we have MATH for almost all MATH. Since MATH and MATH are mutually equivalent minimal MATH-valued cocycles with MATH, CITE implies that there exist MATH normalizing MATH and a NAME map MATH such that MATH for almost all MATH. Let MATH, then we get MATH for almost all MATH. Let MATH be the restriction of MATH to MATH. Then MATH and MATH have the desired property. |
math/0104228 | As in CITE, it suffices to show the statement for a MATH-action instead of the MATH-action thanks to NAME 's theorem CITE. Let MATH be a given non-singular and non-transitive ergodic transformation on MATH. We show that there exists a minimal cocycle MATH with MATH. We take a non-transitive ergodic transformation MATH on MATH such that MATH is MATH-invariant. Thanks to CITE, there exists a minimal cocycle MATH such that MATH. Let MATH be the MATH-action on MATH given by MATH. We define a cocycle MATH on MATH by MATH which is a minimal cocycle with MATH. Since MATH is orbit equivalent to MATH CITE, we get the result. |
math/0104231 | We prove the proposition by induction of MATH and MATH. For MATH, we can prove the proposition directly. We consider the following commutative diagram whose rows are exact: By the hypothesis of the induction, MATH is injective. By using the NAME spectral sequence and inductive hypothesis, MATH is also injective. By REF MATH, MATH and MATH are generated by MATH, MATH and MATH, where MATH . Here we used notations REF . The horizontal homomorphisms MATH are obtained by NAME residue with respect to the divisor MATH. We prove that MATH is surjective and the kernel of MATH is equal to MATH. Let MATH with MATH. We use the notation MATH in REF . We define MATH and MATH as the subspace of MATH and MATH generated by MATH and MATH with MATH, respectively. Then we have MATH and MATH for MATH. We define MATH by the subspace of MATH generated by MATH . We compute the residue MATH with respect to MATH by the relation MATH for MATH. It is easy to see that MATH for MATH. For MATH, we have MATH and MATH. Therefore it is enough to prove that MATH and MATH REF are surjective and MATH . CASE: The morphism MATH. For MATH, MATH. Therefore MATH is surjective. Suppose that the element MATH is contained in the kernel of MATH. For a subset MATH such that MATH, the element MATH in MATH contains only one term MATH such that MATH, and in this case, MATH. Therefore there exist an element MATH such that MATH contains no term with MATH in REF . The relation MATH and REF implies MATH. CASE: The morphism MATH. Let MATH. Then we have MATH and the surjectivity follows from this equality. Suppose that the element MATH is contained in the kernel of MATH. Let MATH. The boundary MATH contains only one term with MATH. By using the relation MATH, we may assume that there is no term with MATH in the expression of REF . Then by REF and MATH, there is no term with MATH in the expression of REF . Therefore MATH. |
math/0104231 | The subspace of MATH generated by differential forms of word type MATH is denoted by MATH. Then, by REF , we have MATH . We can choose the local coordinate of MATH as MATH if MATH. Then the differential MATH of the spectral sequence MATH of REF preserves the word type and as a consequence, the complex MATH is a direct sum MATH of the complices MATH, where MATH . Here we put MATH. Therefore it is enough to prove the following proposition. |
math/0104231 | Let MATH be subsets of MATH such that MATH and MATH. Set MATH and MATH. We put MATH and MATH. Then the restriction MATH is given as follows. CASE: If MATH, MATH if MATH and MATH. CASE: If MATH, MATH if MATH and MATH. CASE: If MATH, MATH if MATH and MATH. For an integer MATH and MATH, we define the filtration MATH of MATH to be the subspace generated by MATH for MATH and MATH. Then we can check that the restriction MATH preserves the filtration and MATH is a filtration of complex MATH. The differential on MATH induced by MATH is also denoted by MATH. It is enough to prove the following claim for the proof of the proposition. Claim Let MATH. CASE: If MATH then MATH, and REF MATH. We put MATH. We identify MATH with the space MATH, where MATH . The set MATH is identified with MATH. The class of MATH in MATH is denoted by MATH, where MATH and MATH. The set MATH forms a base of MATH. Let MATH be an element of MATH such that MATH. The element MATH is denoted by MATH. Then the MATH component of MATH is CASE: MATH if MATH, CASE: MATH, where MATH if MATH, MATH and MATH, CASE: zero if MATH, MATH, and MATH, and CASE: zero if MATH. We introduce a partial order on the set MATH so that MATH if and only if MATH for MATH. We choose a numbering MATH on MATH such that MATH implies MATH. We define MATH to be the subspace of MATH generated by MATH for MATH. Then MATH defines a decreasing filtration on MATH preserved by the differential MATH. It is easy to see that the complex MATH is exact if MATH. If MATH, then MATH if MATH and MATH is a one dimensional vector space generated by MATH. Therefore we have the claim. |
math/0104231 | We prove the proposition by the induction on MATH. Let MATH be an element of MATH. We may assume that MATH. Let MATH REF and MATH. Then MATH. By the hypothesis of the induction, MATH can be obtained by the subquotient of MATH, where MATH and MATH. That is, there exists an object MATH and surjective and injective morphisms MATH and MATH. Since MATH is surjective, the object MATH is a subquotient of the object MATH defined by the exact sequence MATH . Suppose that MATH is an object corresponding to MATH. Let us write MATH, where MATH and MATH. Let MATH be the element corresponding to MATH. CASE: Since MATH and MATH, the extension MATH corresponding to MATH is an element in MATH. Note that MATH. CASE: By the definition of MATH, the image MATH of MATH is a generator of MATH. Thus there exist rational numbers MATH such that MATH. Therefore an extension MATH corresponding to MATH splits in level MATH. Therefore MATH is a object in MATH by REF . CASE: The extension MATH corresponding to MATH is isomorphic to REF MATH if MATH, and REF MATH if MATH. Since MATH is a subquotient of MATH as in REF , we have the proposition by induction. |
math/0104231 | The space MATH (respectively, MATH) is generated by the set MATH (respectively, MATH). For an element MATH, we have MATH. Therefore MATH. |
math/0104231 | For the first assertion, it is enough to prove that MATH, MATH REF and MATH REF does not intersect with MATH. CASE: Proof of MATH. By the natural map MATH, MATH does not intersect with MATH by the inductive hypothesis. For MATH, MATH is equal to MATH with the coordinate of REF . Therefore MATH does not intersect with MATH. CASE: Proof of MATH. The proof is similar. CASE: Proof of MATH. Since MATH for MATH and MATH, MATH does not intersect with MATH by the inductive hypothesis. The second assertion is a direct consequence of the first assertion. |
math/0104231 | Let MATH and MATH be the complices defined similarly to REF . Then the triangle MATH is isomorphic to MATH . By taking the NAME realization of this triangle, morphism REF is an injective morphism between abelian objects. |
math/0104231 | CASE: This equality comes from the coproduct formula. CASE: By the coproduct formula, we have MATH and the proposition follows from this equality. CASE: We use the notation MATH. By the coproduct formula, we have MATH . Let MATH (respectively, MATH) be the minimal number such that MATH (respectively, maximal number such that MATH). If MATH, that is, there exists no MATH such that MATH, then all the terms in REF vanish. Therefore we have the proposition. We may assume that MATH. We can easily see that MATH or MATH. Suppose that MATH. If MATH, (respectively, MATH or MATH) then MATH is equal to MATH (respectively, MATH). CASE: If MATH, we have MATH . Therefore the sum REF is contained in MATH. CASE: If MATH, we have MATH . Again the sum REF is contained in MATH. For the case MATH and MATH, the proof is similar. |
math/0104231 | We prove the corollary by induction. For MATH, we can prove the proposition directly. By the duality and the first statement of REF , we have MATH. The following diagram is commutative by REF . Here the columns are exact. Homomorphism MATH is an isomorphism by the assumption of induction. Using the duality, homomorphism MATH is an isomorphism by REF . |
math/0104231 | By the inductive structure introduced in REF , we may assume MATH. We compute the extension class of REF in MATH by the recipe in REF. By REF , the differential form MATH with MATH REF forms a MATH base of MATH via the isomorphism MATH . On the other hand, MATH form a basis in MATH under the morphism MATH . Therefore the pairing MATH and the extension class of REF vanishes. |
math/0104231 | The first statement is a direct consequence of REF . We prove the second statement by the induction on MATH. Since MATH splits in level MATH, by REF , there exist an object MATH and MATH in MATH such that the following statement holds. CASE: MATH is a subquotient of MATH. CASE: MATH is an object of MATH. MATH is isomorphic to the direct sum of copies of MATH. CASE: MATH is isomorphic to the direct sum of copies of MATH. By the assumption of induction, MATH (respectively, MATH) is a subquotient of a direct sum of objects in MATH with MATH (respectively, MATH). Therefore MATH and MATH are subquotients of a direct sum of objects in MATH with MATH and MATH, respectively. |
math/0104231 | If every MATH is an odd number greater than REF, MATH. Since MATH, the set MATH is contained in MATH. Since MATH and MATH, MATH is at most one dimensional. Suppose that MATH is even. Let MATH. Since MATH, MATH splits in level MATH. Therefore MATH can be written as a subquotient of a direct sum MATH with MATH and MATH by REF . Therefore we have the second statement. |
math/0104231 | We consider variables MATH, where the degree of MATH is MATH. We put MATH and MATH and MATH, where MATH is a variable of degree REF. Then the poincare series of MATH is equal to MATH . Therefore the coefficient MATH is equal to MATH. |
math/0104233 | Since MATH, the NAME - York tensor of any NAME surface can be written as follows: MATH . (In order to obtain the last line from the preceding one, we use the fact that MATH.) We then have: MATH or, equivalently MATH . The anti-selfdual component of REF gives identity REF; the last statement follows immediately. |
math/0104233 | MATH is a twistor MATH-form if and only if there exists a MATH-form MATH such that, at each point of MATH, MATH, where the triple MATH is any positively oriented, orthonormal frame of (anti-selfdual) almost-complex structures at that point, and MATH is the NAME form of MATH. If MATH is a non-vanishing twistor MATH-form on MATH, then, by choosing MATH, we have MATH . Since the norm of MATH is constant, this equality implies that MATH. Now observe that the equation MATH is equivalent to MATH being parallel with respect to the NAME connection of MATH. Hence if MATH is a twistor MATH-form MATH is NAME. Conversely, if MATH is NAME then REF holds, from which it follows that MATH is the section of MATH corresponding to the MATH-form MATH. |
math/0104233 | It is easily checked that the right hand side of REF is (the contraction with MATH of) the general form of a section of MATH. |
math/0104233 | This is immediate from REF : in REF, MATH if and only if MATH. |
math/0104233 | CASE: Identity REF can be written in terms of MATH as MATH . Differentiating again and skew-symmetrizing, we get MATH . Since MATH is MATH-invariant in MATH, it follows that MATH hence MATH where MATH . As an algebraic object, MATH is a symmetric, MATH-anti-commuting, endomorphism of MATH; hence by contracting REF with a vector field MATH and taking the trace over MATH and MATH, we see that MATH, and therefore, MATH is a holomorphic potential. CASE: From REF we derive MATH hence MATH . From this, we infer (again using REF): MATH . The second term of the right hand side of REF is clearly MATH-invariant; the first term is MATH-invariant as well since MATH is a holomorphic potential and MATH is MATH-invariant. Hence MATH is also a holomorphic potential. CASE: By contracting REF with MATH, we see that MATH and MATH . NAME. |
math/0104233 | On each component of MATH where MATH is non-zero, the set MATH where MATH is non-vanishing is dense, hence MATH is non-vanishing on MATH and so the zero set of MATH is dense in this connected component. On the other hand if MATH is identically zero on a component, then MATH is parallel on that component, hence identically zero or everywhere non-zero. |
math/0104233 | The contraction of REF with MATH yields MATH and hence MATH . Since MATH, we deduce that MATH and therefore MATH . This means that REF-forms MATH and MATH, wherever they are non-zero, are eigenforms for the symmetric endomorphism MATH, corresponding to the eigenvalues MATH and MATH, respectively; in particular, they are orthogonal on the open set MATH where MATH (that is, MATH) is non-zero. However, by REF this open set is empty or dense in each connected component of MATH, and the result follows. |
math/0104233 | We apply the preceding argument to the NAME pair MATH on MATH and to MATH; by hypothesis, MATH satisfies MATH, where MATH is actually the selfdual NAME tensor of MATH for the orientation induced by MATH; from the above mentioned uniqueness property, it follows that, wherever MATH is non-zero, MATH coincides with MATH up to rescaling, that is, that MATH is a locally constant multiple of MATH. However, the same holds on the interior of the zero set of MATH. Hence by the continuity of MATH on MATH, MATH for some constant MATH on each connected component of MATH. |
math/0104233 | If MATH is constant, then by REF the NAME form is parallel. Hence either MATH is locally irreducible, and is NAME, or MATH is locally the NAME product of two NAME surfaces of constant curvatures. If MATH is not constant, then by REF the open set MATH where MATH does not vanish is an open dense subset of MATH. However, by REF , MATH is constant. If this constant is zero, then MATH must vanish identically and MATH is selfdual; otherwise MATH and MATH have no zero on MATH, so MATH, the NAME pair MATH is defined on MATH and MATH is degenerate, but nonzero everywhere. As observed in REF, for a weakly selfdual NAME surface the NAME form MATH is a multiple of MATH. Since MATH is a conformally covariant tensor, it follows that NAME tensor of MATH is MATH-invariant, showing that MATH is an extremal NAME metric. |
math/0104233 | CASE: The NAME form may be written MATH which is certainly closed. If is also immediate that MATH and MATH are hamiltonian Killing vector fields with NAME momentum maps MATH and MATH. Since MATH and MATH are closed, MATH is integrable, and the NAME surface is clearly ortho-toric. CASE: Conversely, suppose that MATH is an ortho-toric NAME surface with Killing vector fields MATH, MATH. Since the dual frame to MATH consists of closed MATH-forms, we may write it as MATH, where MATH and MATH are locally defined up to an additive constant. Note also that MATH are linearly independent MATH-forms - where MATH and MATH are the momentum maps of MATH and MATH - so we may use MATH as a coordinate system. Since MATH and MATH we may write MATH for some functions MATH and MATH (of MATH and MATH). The equations MATH give MATH and MATH, using the fact that MATH and MATH are orthogonal. A similar argument tells us that MATH for the same functions MATH and MATH. Now since MATH and MATH are closed, we obtain MATH and MATH, so that MATH and MATH. The NAME form MATH is evidently given by REF, and since we know MATH and MATH, we readily obtain REF, and hence the metric REF. |
math/0104233 | Clearly MATH and MATH are closed, so MATH is integrable. From REF, we easily infer that the NAME form MATH of the Hermitian pair MATH, defined by MATH, is MATH . It follows that MATH is closed, that is, the pair MATH is NAME. |
math/0104233 | From REF, we infer that the volume-form MATH of MATH is given by MATH since MATH and MATH are the real parts of MATH-holomorphic coordinates. By putting MATH, we have MATH . Now according to REF, MATH and hence MATH this implies MATH from which REF follow easily. From REF, we get MATH on the other hand, we compute that MATH and we obtain REF. |
math/0104233 | Since the scalar curvature MATH is a function of MATH and MATH, MATH belongs to the span of the Killing vector fields MATH and MATH and commutes with them; if it is itself a Killing vector field, it has to be a linear combination of MATH and MATH with constant coefficients, that is, MATH, where MATH are constants. By REF, this implies REF. Finally, using REF, we easily compute that the anti-selfdual MATH-form associated to the NAME tensor of an ortho-toric extremal NAME surface is MATH . Since MATH is also MATH-invariant, the NAME metric MATH is extremal as well (see REF). |
math/0104233 | Since a bi-extremal surface is extremal, we may apply REF . We then compute MATH . As in the proof of REF , MATH cannot be a MATH-holomorphic potential unless it is a linear combination of MATH and MATH with constant coefficients; this in turn is equivalent to the condition MATH. Since MATH and MATH it follows that MATH. Hence MATH is a hamiltonian and so MATH is weakly selfdual by REF. By substituting in the expression of MATH given by REF we get MATH; since the condition MATH is equivalent to MATH, the characterization of the selfdual case follows. Also MATH is NAME if and only if MATH, and so the last two assertions are immediate. |
math/0104233 | Suppose there exist a third infinitesimal symmetry of MATH, say MATH, which does not lie in the plane spanned by MATH and MATH; then, we must have MATH and MATH, for MATH; this implies that MATH and MATH are colinear; we then infer from REF that we have MATH and MATH in REF, so that MATH is NAME - NAME by REF . If, in addition, MATH, we obtain REF and MATH is then a selfdual NAME - NAME surface, that is, a complex space form. If MATH does not vanish identically, on the open set where MATH, the Hermitian pair MATH is invariant under the action of the Killing vector fields MATH's, as MATH is determined by the eigenform of MATH corresponding to its simple eigenvalue, compare CITE; then, MATH (a constant multiple of the simple eigenvalue of MATH), the square-norm MATH of the NAME form MATH of the pair MATH as well as MATH are also invariant under the action of MATH's, MATH; we then have MATH; by using REF , we can check that this implies that MATH and MATH satisfy REF; by REF , the corresponding ortho-toric NAME surface is NAME and NAME; it then follows from REF that MATH; in particular, MATH; by CITE, this implies that the metric is of cohomogeneity one. Since MATH is non-zero, it is not constant, and so the metric is not homogeneous. Evidently the two cases overlap when MATH and MATH, in which case MATH is flat. |
math/0104233 | The proof follows NAME 's description CITE of NAME metrics with a hamiltonian Killing vector field MATH. Supposing first that MATH is only almost Hermitian, note that MATH is a holomorphic vector field, so that the complex quotient is locally a NAME surface MATH. Introducing a local holomorphic coordinate MATH on MATH, we may write MATH where MATH, MATH is an invariant MATH-form with MATH, and MATH are functions of MATH. The almost Hermitian structure MATH is given by MATH with NAME form MATH . We now impose the condition that MATH and MATH are NAME for some non-vanishing function MATH. Now MATH if and only if MATH while MATH is closed if and only if MATH . In the presence of REF is equivalent to MATH which holds if and only if MATH and MATH for some function MATH. Let MATH be the complex MATH-form MATH. Then, since MATH is closed, the complex structures MATH and MATH are integrable if and only if MATH belongs to the ideals generated by MATH and by MATH respectively. Since MATH, these conditions force MATH and MATH to vanish. Hence MATH is integrable if and only if MATH while MATH is integrable if and only if MATH here MATH is an arbitrary function. Hence MATH and MATH are both integrable if and only if MATH and MATH is necessarily a function of MATH only. Putting together REF, and REF, we see that MATH is of NAME type, with Killing vector field MATH, if and only if MATH, with MATH, MATH and MATH, so that MATH for constants MATH and MATH. Using the freedom in the choice of MATH, we may then assume MATH is a MATH-form on MATH, while MATH is a metric on MATH, and the result follows. |
math/0104233 | The NAME form MATH is given by MATH. The first term is MATH, and we compute the second term as follows: MATH . Evidently we may write this as a linear combination of MATH and MATH, and we readily obtain REF. The conformal scalar curvature is most easily computed by noticing that the conformal NAME metric MATH is also of NAME type, with MATH and MATH. Hence its scalar curvature is MATH from which REF follows, since MATH. |
math/0104233 | From REF, we have MATH . If MATH is an affine function of MATH, then both sides of this equation must be constant. Hence MATH and MATH must be a quartic in MATH with quadratic term MATH. The formulae for MATH and MATH are immediate, as are REF - REF . (For REF we use the fact that MATH is integrable and MATH is closed, so that weak selfduality is equivalent to the equation MATH.) REF The pfaffian MATH of the normalized NAME form is given by MATH in particular, MATH is a rational function of MATH; since MATH is a holomorphic potential, MATH is a holomorphic potential if and only if it is an affine function of MATH, if and only if MATH; MATH is then equal to MATH. CASE: For any extremal NAME surface the NAME tensor is MATH-invariant and the corresponding anti-selfdual MATH-form, is expressed by REF which yields MATH . |
math/0104233 | By REF , we know that MATH is either selfdual, hence, by the above mentioned result of NAME, described by REF or REF , or of constant scalar curvature, hence, again, described by REF or REF , or of non constant scalar curvature. In the latter case, the (negative) NAME structure MATH is globally defined; it then follows from a result of CITE that the signature of MATH is zero CITE; moreover, since the (real) holomorphic field MATH has non-empty zero set, we know by CITE that the NAME dimension of the NAME surface MATH is MATH, hence MATH is a ruled surface which is the projectivization MATH of a rank REF holomorphic vector bundle MATH over a compact complex curve MATH CITE. If MATH, MATH is a NAME surface MATH, where MATH is a positive integer, or the product MATH; the only extremal NAME metrics of MATH are the (symmetric) product metrics, which are of constant scalar curvature; on the other hand, any maximal compact subgroup of MATH is conjugate to MATH, and therefore any extremal NAME metric must be a cohomogeneity-one MATH metric CITE, hence, locally of the form REF with MATH (compare the end of the preceding section for the case when MATH); as shown by NAME in CITE, for each MATH, any NAME class of MATH carries a unique extremal NAME metric (up to a re-parameterization); each one can be put in the form REF, where the polynomial MATH, in the notation of REF, is determined by MATH where MATH are the parameters of the NAME class; according to REF , MATH is weakly selfdual precisely when MATH; in the present situation, this is equivalent to MATH and happens if and only if MATH and MATH, that is, if MATH is the first NAME surface equipped with a NAME weakly selfdual NAME metric. We now show that a compact ruled surface MATH whose base MATH is a compact complex curve of genus MATH at least REF does not carry weakly selfdual NAME metrics of non-constant scalar curvature. We thus assume that MATH carries a weakly selfdual NAME metric of non-constant scalar curvature to get a contradiction. Using an argument from CITE, we first observe that the rank two vector bundle MATH splits as MATH, where MATH stands for the trivial holomorphic line bundle and MATH is a holomorphic line bundle MATH of degree MATH. Indeed, recall the already mentioned result of CITE that the connected component of the isometry group MATH is a maximal compact subgroup in MATH; according to CITE, the group of automorphisms of ruled surfaces can be described as follows: If MATH, there exists an exact sequence MATH where MATH denotes the group of relative automorphisms of the bundle MATH, and MATH is the group of automorphisms of MATH (of course, MATH is finite if MATH); on the other hand, the (non-trivial) homomorphic vector field MATH whose real part is the Killing vector field MATH has a non-empty zero set; since MATH preserves the (unique) ruling MATH, it projects onto a holomorphic vector field on the base MATH; since MATH has at least one zero, the induced vector field on MATH vanishes; it follows that MATH is tangent to the MATH fibers (equivalently, MATH belongs to the NAME algebra of MATH); this shows that the kernel of the group homomorphism MATH induced by the exact sequence REF is a non-trivial compact subgroup of MATH; one can therefore find a MATH in the connected component of the identity of MATH; denote by MATH the induced holomorphic vector field, such that the imaginary part MATH generates the MATH-action, whereas MATH itself generates a MATH-action; as a matter of fact, MATH can be identified to a traceless holomorphic section of MATH, say MATH; note that MATH is of constant determinant; since MATH generates a periodic MATH-action, MATH must be diagonalizable; this shows that we have a holomorphic splitting of MATH into eigensubbundles of MATH; by twisting by a line bundle, we obtain the splitting MATH, where MATH; then, MATH is nothing but the NAME vector field of MATH. If the degree of MATH is zero, then any NAME class contains a locally symmetric NAME metric CITE, so that any extremal NAME metric on MATH is of constant scalar curvature CITE, a contradiction; we thus obtain a splitting MATH where MATH is a holomorphic line bundle of MATH. As MATH is a maximal compact subgroup in MATH, we may assume CITE that (up to a biholomorphism) the metric MATH is invariant under the fixed MATH action generated by MATH. For any non-trivial Killing vector field, MATH, which arises from a real holomorphic potential, the argument already used above shows that MATH must be tangent to the fibers, and therefore MATH. In particular, we get that MATH where MATH are smooth functions defined on an open dense subset of MATH where MATH. But MATH, that is, MATH, and therefore MATH is a constant. By rescaling the metric if necessary we may assume therefore MATH. Similarly, MATH must be a constant multiple of MATH and by REF , MATH must be locally of cohomogeneity one, that is, MATH can be written of the form REF on an open dense subset of MATH. Note that MATH contains exactly two curves fixed by the MATH-action generated by MATH, corresponding to the zero and infinity sections, MATH and MATH, of MATH; moreover, the function MATH appearing in REF makes sense on the whole of MATH as being a momentum map of the corresponding MATH-action (up to multiplication by a non-zero constant); it then follows that MATH maps MATH onto an interval MATH, such that MATH is regular on MATH; therefore, for any MATH, MATH, whereas MATH and MATH (see CITE). By using an argument from CITE, it is shown that MATH is a smooth function on MATH, which satisfies the boundary conditions MATH where MATH is a real constant; for NAME 's metrics REF one has MATH, so that REF read MATH we thus obtain the following values for the coefficients MATH of MATH (notations of REF): MATH according to REF we have MATH, and from REF we also get MATH; by REF it follows that the NAME tensor of MATH has two distinct eigenvalues, equal to MATH and MATH respectively; by REF , MATH nowhere vanishes on MATH, meaning that the scalar curvature MATH nowhere vanishes as well; since MATH is a non-trivial Killing vector field, MATH must be everywhere positive; this shows that the first NAME class MATH of MATH is positive, and therefore MATH, a contradiction CITE. |
math/0104233 | The proof follows the one given in the NAME case, with slight modifications in places where the non-integrability of MATH must be taken into account: the NAME tensor is now written as MATH . Since MATH closed CITE, we have MATH . As an algebraic object, MATH is a skew-symmetric endomorphism of MATH, associated (by MATH-duality) to the section MATH of the bundle of MATH-anti-invariant REF-forms; it then anti-commutes with MATH, and commutes with any skew-symmetric endomorphism associated to a section of MATH; in particular, MATH commutes with the endomorphism corresponding to MATH via the metric (which will be still denoted by MATH). We thus obtain MATH . By using the closedness of MATH we derive MATH . Substituting REF , we finally get MATH . The MATH-component of REF gives the identity REF ; the last part of the lemma is immediate. |
math/0104233 | On MATH, the NAME tensors of both MATH and MATH are MATH-invariant, and therefore MATH is Killing vector field (with respect to both metrics) CITE. Since MATH, MATH and MATH is a Killing vector field on MATH. On the other hand if MATH vanishes identically on an open set MATH then MATH is NAME on MATH, so that MATH is constant, and MATH is a trivial Killing vector field. Hence by continuity MATH is a Killing vector field everywhere. |
math/0104233 | Since MATH we have MATH. Straightforward calculation gives MATH . Since MATH is Killing by assumption, it follows that MATH is Killing if and only if MATH is skew. This is automatic on the open set where MATH vanishes, where MATH and hence MATH. Therefore we can assume MATH is nonvanishing and write MATH, where MATH is a complex structure of the opposite orientation to MATH. Now MATH is skew if and only if MATH is MATH-invariant; since MATH and MATH commute, this means that the MATH-anti-invariant part of MATH must be MATH-invariant. But MATH is Killing, so the MATH-anti-invariant part of MATH is equal to MATH . The latter is MATH-invariant if and only if for any vector field MATH we have MATH . Suppose now that MATH for a vector MATH at some point. Since MATH, like MATH, is a MATH-anti-invariant endomorphism, we can write MATH for some MATH. REF now reads MATH . Since MATH commutes with MATH and anticommutes with MATH, by applying MATH to the both sides we obtain: MATH. However MATH is orthogonal to both MATH and MATH, and so MATH and MATH. Thus, on the open set where MATH is non-integrable and MATH, we have MATH, so MATH and MATH are linearly dependent. |
math/0104233 | Without loss of generality, we take MATH to be a momentum map for MATH (with respect to MATH). As in REF , compare CITE, we may introduce coordinates such that MATH where MATH is the NAME form of MATH with respect to MATH. The integrability of MATH together with the closedness of MATH and MATH yields MATH from which we obtain the integrability condition MATH . The NAME tensor of MATH is MATH-invariant if and only if MATH so that we can write MATH . Since MATH, we obtain MATH . The latter is explicitly integrated, and we get MATH for some functions MATH and MATH. By substituting into REF we discover that either MATH or MATH must be constant. If MATH is constant, then MATH, that is, MATH is of NAME type and MATH is integrable. Consider now the case when MATH is a constant; then, MATH where MATH is a positive harmonic function and MATH is an arbitrary function of MATH; the almost NAME structure MATH takes the form REF - REF where: CASE: MATH; CASE: MATH; CASE: MATH is a positive harmonic function on MATH; CASE: MATH satisfies MATH and we can locally choose MATH so that MATH, where MATH is a harmonic conjugate of MATH and MATH. This almost Hermitian structure MATH is almost NAME with MATH-invariant NAME tensor, since MATH, MATH and MATH solve the required equations. |
math/0104233 | We inspect the possible compact non-integrable almost NAME MATH-manifolds given by REF - REF . The case of constant scalar curvature is described by REF . Our claim then follows by CITE and CITE. Suppose that MATH is not constant, that is, that MATH is a non-trivial Killing vector field by REF . Let MATH be a zero of MATH; then, the isotropy subgroup MATH of the connected group of isometries of MATH is a compact group of dimension at least one; one can therefore take a MATH in MATH. NAME theory implies that on a compact almost NAME manifold any isometry which is homotopic to the identity inside the group of diffeomorphisms is a symplectomorphism (see for example, CITE); hence the chosen isometric MATH-action is symplectic with respect to MATH. Since MATH is a fixed point of the MATH-action, we obtain a hamiltonian MATH-action on MATH CITE. The manifold is then equivariantly (and orientedly) diffeomorphic to a rational or a ruled complex surface endowed with a holomorphic circle action CITE. Moreover, in this case MATH is given by REF or REF . Consider first REF . Since MATH is not constant, the selfdual NAME tensor MATH does not vanish CITE. By the NAME formulae, the signature of MATH is strictly positive, and therefore MATH is diffeomorphic to MATH CITE. Combining the results of CITE and CITE, one sees that on MATH the only selfdual conformal structure with non-trivial (conformal) Killing vector field is the standard one. Thus, modulo diffeomorphisms, we may assume that MATH is conformal to the standard NAME metric MATH. Since MATH and MATH are both harmonic selfdual REF-forms on MATH, and since MATH=REF, we conclude that MATH, showing that MATH is integrable, a contradiction. Suppose MATH is as in REF . Now MATH determines an integrable almost complex structure MATH compatible with MATH and with the opposite orientation of MATH, such that MATH is an extremal NAME metric with MATH. Denote by MATH the smooth manifold MATH endowed with the orientation induced by MATH. Thus, the oriented smooth MATH-manifolds MATH and MATH both admit complex structures. Since MATH is the underlying smooth manifold of a rational or a ruled complex surface, we conclude as in the proof of REF that the complex surface MATH is a ruled surface of the form MATH, where MATH is a holomorphic rank REF bundle over a compact NAME surface MATH of genus MATH, and which splits as MATH for a holomorphic line bundle MATH of degree MATH. We have to prove that MATH. Indeed, if MATH, we obtain the NAME surface MATH, where MATH is a positive integer. As we have already observed in the proof of REF , the extremal NAME metrics on these surfaces are the NAME cohomogeneity-one U REF -metrics, that is, MATH is given by the NAME construction REF - REF; since MATH, it follows that MATH is a cohomogeneity-one metric as well and therefore MATH, showing that MATH. Since MATH is not constant, by REF , MATH is integrable on the open dense subset where MATH, hence everywhere. We thus conclude that MATH. Note that the above local argument applies to any extremal NAME MATH which is locally of cohomogeneity one, so that the last part of the theorem also follows. |
math/0104233 | Suppose MATH is a compact almost NAME MATH-manifold of constant Lagrangian sectional curvature, but for which MATH is not integrable. According to CITE the scalar curvature MATH of MATH is not constant; then, by REF , the smooth manifold MATH is diffeomorphic to a minimal ruled surface. Since any such surface admits an orientation reversing involution, we conclude that MATH carries a complex structure which is compatible with the orientation induced by MATH. Then, by CITE, the almost complex structure MATH must be integrable, contradicting our assumption. |
math/0104234 | For MATH fixed, the powers MATH, MATH, of the fundamental unit give all solutions MATH of the Pellian equation MATH with MATH by way of the rule MATH . For every such solution we have MATH. Thus REF gives MATH . Since MATH it follows from REF that MATH . Since MATH (this can be seen, for example, from REF and the estimates MATH and MATH; the latter follows easily by partial summation from the orthogonality relation for characters), it follows that for MATH with MATH, MATH . Thus MATH for all MATH; here MATH denotes the divisor function and MATH was used. This implies MATH which proves the lemma. |
math/0104234 | NAME 's inequality gives MATH . For MATH, this gives MATH . The second sum on the right hand side is MATH as MATH (see CITE). This means that the values MATH are constant in the mean when ordered according to the sizes of their fundamental units. Thus the lemma follows. |
math/0104234 | Since MATH for MATH, we see that for MATH the inner sum in MATH is MATH . NAME 's inequality now gives MATH where MATH was used. The last sum is MATH (see CITE or CITE), and the lemma follows. |
math/0104234 | See CITE, estimate REF . |
math/0104234 | Split the first sum in MATH into two sums depending on whether MATH or MATH. Thus MATH. A trivial estimate gives MATH and thus MATH since MATH. NAME 's inequality gives MATH . Thus for MATH, MATH . Applying REF to the innermost sum gives the estimate MATH which together with REF proves the lemma. |
math/0104234 | Choose MATH with MATH. Choose MATH. Define the rectangle MATH . If MATH and MATH has no zeros in MATH then a standard argument (see, for example, REF ) shows that for MATH, MATH, we have MATH with some constant MATH depending on MATH. Together with REF this gives MATH . Next we must show that MATH cannot have a zero in MATH too often. From zero density estimates it follows that MATH (see REF or CITE, estimate REF ). A trivial estimation of REF gives MATH . REF and NAME 's inequality give MATH which proves the lemma. |
math/0104234 | For MATH, choose MATH. Then REF show that MATH . Since the series MATH converges, we have for MATH fixed MATH . Together with REF this proves the proposition. |
math/0104234 | In the definition of MATH, write MATH with MATH. There is a discriminant MATH with MATH iff MATH for all MATH and MATH. Since MATH for MATH, the last condition is equivalent to MATH. If these conditions are fulfilled, we have MATH with MATH, MATH, for MATH. Thus MATH for MATH. This proves the lemma. |
math/0104234 | For MATH with MATH fixed and MATH with MATH it follows from REF that MATH here MATH as MATH since MATH. Furthermore, MATH as MATH, since MATH . Thus MATH. For MATH arbitrary and MATH, we have MATH by NAME 's inequality. Thus MATH for all MATH. Since the MATH-th summand of MATH is MATH-periodic (MATH-periodic in case MATH) and the series representing MATH is uniformly convergent, the function MATH is uniformly limit periodic, that is, MATH; here MATH is the set of all functions which can be approximated to an arbitrary accuracy by periodic functions with respect to the supremum norm. Since MATH is closed under multiplication it follows from REF that MATH for all MATH. This gives MATH for all MATH. |
math/0104234 | From REF it follows that for arbitrary MATH there is some MATH with MATH and MATH for all MATH. For all MATH it follows from REF that there is some MATH and coefficients MATH, MATH, such that MATH where MATH denotes the supremum norm and MATH. From REF it follows that MATH . Thus MATH . For MATH we have MATH. Furthermore, MATH iff MATH for MATH. Therefore the orthogonality relation for the exponential function gives MATH . Similarly, MATH for MATH and thus MATH . This gives MATH . In the next section we will compute MATH and thereby show that MATH. This gives REF and MATH . Since MATH is arbitrary, REF follows. |
math/0104240 | This is a direct consequence of REF . |
math/0104240 | By the work of CITE it suffices to show that the Gamma space MATH is very special, that is, the map MATH induced by the projections MATH and MATH, is a weak equivalence, and that the monoid MATH is a group. By REF below it suffices to show that the Gamma spaces MATH are very special. To see that the map MATH is a weak equivalence, it suffices by the approximation REF to note that by the NAME theorem the map MATH is MATH-connected. To see that MATH is a group, it suffices to note that MATH is a group. |
math/0104240 | It follows from the realization lemma and the fact that realization commutes with products that the resulting Gamma space is special, that is, the map MATH induced by the projections MATH and MATH is a homotopy equivalence. A special Gamma space MATH is very special when the monoid MATH with multiplication induced by the composite MATH is a group. Here MATH is the fold map with MATH for MATH and MATH is a homotopy inverse to the homotopy equivalence MATH. This is equivalent to the map MATH being a homotopy equivalence. (Clearly if this map is a homotopy equivalence, then MATH is a group. Conversely, if MATH is a group, then this map induces an isomorphism on MATH for all MATH, and by the NAME theorem we can conclude that it is a homotopy equivalence.) It follows from the realization lemma that MATH is very special. |
math/0104240 | We consider the cofibre MATH as a bisimplicial set MATH. Since the cofibre of a cofibration of MATH-connected spaces is MATH-connected MATH is MATH-connected for every MATH, and by REF MATH is MATH-connected for every MATH. Using the spectral sequence of CITE we obtain the assertion of the lemma. |
math/0104240 | Let us note that REF gives that MATH is MATH-connected. Replacing REF by REF the proof of REF also proves this lemma. |
math/0104240 | Let us start by showing that the map from the mapping cone of the map MATH to MATH is MATH-connected. If MATH is a cofibration of MATH-connected pointed simplicial sets then by applying the NAME - NAME theorem CITE several times we see that the map from the mapping cone of the map MATH to MATH is MATH-connected. Since the mapping cone construction commutes with geometric realization it follows that the map from the mapping cone of the map MATH to MATH is MATH-connected. Now let MATH denote the map MATH, and let MATH denote its homotopy fibre. It then follows from the NAME - NAME theorem that the map MATH is MATH-connected. From the weak equivalence MATH it follows that the map MATH is MATH-connected. Since MATH is arbitrary, it follows that this map is a weak equivalence. |
math/0104240 | The diagonal induces a map MATH . We will show that this is an isomorphism of simplicial sets. We may assume that MATH is a discrete set filtered by injections. We note that by the pushout diagram in the proof of REF the map MATH is an injection for all MATH, and therefore we have an injection MATH with the convention that MATH and MATH. There is a commutative diagram MATH where the vertical arrows are injections. It follows from the diagram that the map MATH is injective. To see that it is onto, let us pick a representative MATH for a point MATH in MATH fixed under the MATH-action. From the condition MATH, it follows that there exists a MATH such that MATH. Since the image of MATH in MATH is a fixed point, we must have that MATH represent the same element in MATH. Since the map MATH is injective, it follows that MATH represents MATH, and we can conclude that the map MATH is onto. |
math/0104240 | Suppose that the map MATH is MATH-connected for MATH. Using the cofibration sequence MATH we can by induction show that the map MATH is MATH-connected when MATH, for all MATH. It follows that the map MATH is MATH-connected when MATH. Therefore the map MATH is at least MATH-connected, and the map MATH is at least MATH-connected when MATH. |
math/0104240 | Let MATH denote the homotopy equalizer of the diagram MATH where MATH forgets the MATH'th coordinate, and otherwise MATH and MATH are truncations of the maps defining MATH. There is an obvious map MATH induced by projection away from the last factors of the products. Using the norm cofibration sequence the fibre of this map may be identified with MATH. Since homotopy limits commute we have that MATH is the homotopy limit of the sequence MATH . Since homotopy orbits preserve connectivity and MATH, MATH is a sequential homotopy limit of spaces as least as connected as MATH and with homotopy fibres as least as connected as MATH. It follows that MATH is at least as highly connected as MATH. Since homotopy equalizers at most lower connectivity by one we have that MATH is at least as highly connected as MATH. |
math/0104240 | Since MATH, the inclusion MATH induces the trivial map from MATH to MATH. Therefore the homotopy equalizer of the maps induced by MATH and MATH on the quotients of the products in the definition of MATH agrees with the homotopy fibre of the map induced by MATH. Using the norm cofibration sequence we can identify this fibre with MATH. Since MATH, where MATH denotes the norm map, and where MATH denotes the transfer map, we have that MATH. The lemma now follows from REF . |
math/0104240 | In CITE NAME and NAME have established a spectral sequence with MATH-term MATH converging towards MATH. The lemma follows from the fact that MATH and that MATH when MATH. |
math/0104240 | We refer to CITE for the notation used in this proof. It is well known (see for example, CITE, or use the argument below) that there is an isomorphism MATH. To see that MATH also is isomorphic to this MATH group, it suffices to check the usual properties determining MATH groups up to isomorphism (see for example, CITE). Firstly we note that there are isomorphisms MATH. Secondly the representable functors MATH with MATH form a set of projective generators for the category of cyclic objects in the category of abelian groups, that is, every projective object in this category is a quotient of a sum of objects of the form MATH. Since MATH we can use REF to see that MATH, and hence MATH for MATH. Thirdly, given a short exact sequence MATH of cyclic objects in the category of abelian groups, we obtain a cofibration sequence MATH of spectra with an action of MATH. Since homotopy orbits take cofibration sequences to cofibration sequences we obtain a long exact sequence of homotopy groups of homotopy orbits. |
math/0104240 | Recall that MATH is the colimit running over MATH of MATH . If MATH then there exists a MATH such that MATH. Therefore the smash product is zero if MATH, or equivalently if MATH, and hence MATH if MATH. |
math/0104240 | Let us write MATH instead of MATH. We have an isomorphism MATH when MATH and a surjection MATH when MATH. By REF there is for MATH an isomorphism MATH. Splicing these maps with the isomorphism of REF we obtain the asserted isomorphism MATH when MATH and the surjection MATH when MATH. |
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