paper stringlengths 9 16 | proof stringlengths 0 131k |
|---|---|
math/0008133 | REF is an immediate consequence of REF follows from REF . To prove REF , observe that the maximal ideal MATH is generated by a sequence of projections MATH, that is, MATH, with MATH. We know from REF , that MATH is isomorphic to MATH, for some locally compact, totally disconnected topological space MATH. Moreover, if MATH is a MATH - module and MATH is the maximal ideal of functions vanishing at MATH, for some fixed point MATH, then MATH, and hence MATH . Since MATH is metrizable, we can choose a basis MATH of compact open neighborhoods of MATH in MATH. Then, if we let MATH to be the characteristic function of MATH, then MATH are projections generating MATH. By choosing MATH to be decreasing, we obtain a decreasing sequence MATH. |
math/0008133 | The vanishing of MATH in the second half of the proposition follows because MATH in that case. Assume now that MATH consists of functions vanishing at MATH, a semisimple element of MATH. The localization functor MATH is exact by standard homological algebra. The sequence of ideals MATH is an increasing sequence satisfying MATH and MATH. Choose MATH such that MATH and MATH. (This happens if, and only if, MATH.) It follows that MATH . Since all the isomorphisms of REF are compatible with this the localization functor, we obtain that MATH . The only quotient MATH that does not vanish is the one containing (a conjugate of) MATH, and then it is isomorphic to MATH. This completes the proof. |
math/0008133 | By the definition of the localization of a module, the map MATH factors through a map MATH . Since MATH is an isomorphism by REF , we may define MATH and all desired properties for MATH will be satisfied. |
math/0008133 | Note first that the map MATH is well defined, that is, that its range is contained in MATH, by REF . To prove that MATH is an isomorphism, filter both MATH and MATH by the subgroups MATH and, respectively, by MATH, using the ideals MATH introduced in REF. Since MATH is MATH - linear, it preserves this filtration and induces maps MATH . These maps are, by construction, exactly the isomorphisms of REF . Standard homological algebra then implies that MATH itself is an isomorphism, as desired. |
math/0008133 | The argument is standard and goes as follows. Recall first that any filtration MATH by MATH - submodules gives rise to a spectral sequence with MATH, convergent to MATH. Now, associated to the open sets MATH of a nice filtration, there exists an increasing filtration MATH by MATH - submodules such that MATH where each MATH is the orbit of a unipotent element, and MATH has the topology given by the disjoint union of the orbits MATH. Fix MATH and MATH, and let MATH be a unipotent element in MATH (so that then MATH is the orbit through MATH), which implies that MATH. Finally, from NAME 's lemma we obtain that MATH and this completes the proof. |
math/0008133 | The product on MATH is given by the formula MATH . Let MATH denote the multiplication (that is, convolution product) on MATH. Thus, we need to prove that MATH for all MATH. Consider the map MATH, and let MATH be the push-forward of the measure MATH. Then the right-hand side of the above formula becomes MATH . We know that MATH, by assumptions (see the discussion before the statement of this lemma), and then MATH by the invariance of the NAME measure. The lemma is proved. |
math/0008133 | Let MATH and MATH be two left MATH - modules. We can regard MATH as a right module, and then the tensor product MATH is the quotient of MATH by the group generated by the elements MATH, as before. Alternatively, we can think of MATH as MATH. This justifies the notation MATH for a morphism MATH induced by a morphism MATH . We shall prove the theorem by an explicit computation. To this end, we shall use the results and notation (MATH and MATH) of REF . By direct computation, we see that the morphism MATH between NAME complexes, is given by the formula MATH . We now want to realize the map MATH, at the level of complexes. In the process, it will be convenient to identify the smooth MATH-module MATH with a subspace of the space of functions on MATH, using the projection MATH. Consider the MATH - morphism MATH induced by the morphism MATH . Explicitly, MATH . Then the resulting morphism MATH is the morphism MATH on homology corresponding to the MATH - morphism MATH. The MATH - morphism MATH given by the formula MATH REF is well defined and induces an isomorphism in homology, because the only nonzero homology groups are in dimension MATH, and they are both isomorphic to MATH. We have an isomorphism MATH of complexes. This shows that the homology of the second complex in REF is isomorphic to MATH, and that the map induced on homology, that is MATH is the NAME isomorphism. Recall now that the isomorphism MATH is induced by the morphism of complexes MATH defined in REF . From the definition of the morphism MATH and the above discussion, we obtain the equality of the morphisms MATH induced by MATH and MATH. Thus, in order to complete the proof, it would be enough to check that MATH at the level of complexes. Let MATH be the projection. Since MATH is surjective, it is also enough to check that MATH. Let MATH where MATH, as before. Then MATH induces a morphism MATH of complexes satisfying MATH. Directly from the definitions we obtain then that MATH. This completes the proof. |
math/0008133 | The result of the lemma follows from the fact that the map MATH is MATH - linear and the isomorphism of NAME 's Lemma, MATH is natural. Alternatively, one can use the explicit formula of REF . |
math/0008133 | This follows from definitions if we observe that in the sequence of maps MATH the second map is induced by MATH and their composition induces on homology the direct summand MATH of the map MATH. |
math/0008133 | Integration over MATH defines a morphism MATH which commutes with the action of MATH. Then MATH coincides with the morphism of complexes induced by MATH. Consider now the maps MATH defined in the proof of REF . Then MATH, and hence MATH, from which the result follows. |
math/0008133 | Fix MATH, not necessarily semisimple and let MATH be the subgroup of elements of MATH commuting with MATH. We choose a complement MATH of MATH in MATH and we use the exponential map to identify MATH with a subset of MATH. Then the Jacobian of the map MATH is MATH, and from this the result follows. |
math/0008133 | First of all, it is clear that the composition MATH is invariant with respect to the NAME group MATH, and hence its range consists of MATH-invariant elements. The localization of MATH at a regular, diagonal conjugacy class MATH is onto by REF . Next, we know that every orbital integral extends to MATH, and this implies directly that the spectral sequence of REF collapses at the MATH term. This proves that the localization of MATH at MATH is also onto, and hence MATH is onto. The rest of the proposition follows also from REF by localization. |
math/0008134 | It suffices to check that MATH. NAME order to do that, recall that the double transitivity implies that MATH (REF on p. REF). Now the desired inequality follows easily from the existence of the MATH-invariant splitting MATH . |
math/0008134 | In light of REF , we may assume that MATH. In light of REF we may assume that MATH does not divide MATH and therefore MATH . The natural representation of MATH in MATH is irreducible (REF on p. REF). Since MATH is normal in MATH, the MATH-module MATH is semisimple, thanks to NAME 's theorem REF . Since MATH, the action of MATH on MATH is doubly transitive. Applying REF , we conclude that the MATH-module MATH is absolutely simple. (See also CITE.) |
math/0008134 | See CITE, p. REF, p. REF. |
math/0008134 | Assume that MATH is a central simple MATH-algebra of dimension MATH. We need to arrive to a contradiction. We start with the following statement. Assume that MATH is a central simple MATH-algebra of dimension MATH. Then there exist a MATH-dimensional abelian variety MATH over MATH, a positive integer MATH, an embedding MATH and an isogeny MATH such that the induced isomorphism MATH maps identically MATH onto MATH . (Here MATH is the diagonal embedding.) In particular, MATH and MATH are abelian varieties of NAME over MATH. Clearly, there exist a positive integer MATH and a central division algebra MATH over MATH such that MATH. This imples that there exist an abelian variety MATH over MATH with MATH and an isogeny MATH such that the induced isomorphism MATH maps identically MATH onto MATH. We still have to check that MATH . In order to do that let us put MATH. Then MATH and therefore MATH. Since MATH is a division algebra and MATH, the number MATH must divide MATH. This means that MATH divides MATH. On the other hand, MATH . This implies that MATH divides MATH and therefore MATH . Now let us return to the proof of REF . Recall that MATH. We write MATH for the space of differentials of first kind for any smooth projective variety MATH over MATH. Clearly, MATH induces an isomorphism MATH which commutes with the natural actions of MATH. Since MATH, we have MATH. Therefore, the induced MATH-linear automorphism MATH has, at most, MATH distinct eigenvalues. Clearly, the same is true for the action of MATH in MATH. Since MATH commutes with MATH, the induced MATH-linear automorphism MATH has, at most, MATH distinct eigenvalues. On the other hand, let MATH be one of the MATH-invariant points (that is, a ramification point for MATH) of MATH. Then MATH is an embedding of MATH-algebraic varieties and it is well-known that the induced map MATH is a MATH-linear isomorphism obviously commuting with the actions of MATH. (Here MATH stands for the linear equivalence class.) This implies that MATH has, at most, MATH distinct eigenvalues in MATH. One may easily check that MATH contains differentials MATH for all positive integers MATH satisfying MATH if MATH does not divide MATH (REF on p. REF; see also REF on p. REF). Since MATH and MATH, we have MATH for all MATH with MATH. Therefore the differentials MATH for all MATH with MATH; clearly, they all are eigenvectors of MATH with eigenvalues MATH respectively. (Recall that MATH is a primitive MATH-th root of unity and MATH is defined in REF by MATH.) Therefore MATH has in MATH, at least, MATH distinct eigenvalues. Contradiction. Now assume that MATH divides MATH. Then MATH. By REF , MATH is birationally isomorphic over MATH to a curve MATH where MATH is a separable polynomial of degree MATH; in addition, one may choose this isomorphism in such a way that it commutes with the actions of MATH on MATH and MATH. This implies that MATH has, at most, MATH distinct eigenvalues in MATH. On the other hand, MATH and MATH is not divisible by MATH. Recall that MATH. We conclude, as above, that for all MATH with MATH the differentials MATH. Now, the same arguments as in the case of MATH not dividing MATH lead to a contradiction. |
math/0008134 | Clearly, MATH. Since MATH is a NAME, MATH is either a totally real field or a NAME. If MATH is a NAME prime then each subfield of MATH (distinct from MATH itself) is totally real. Therefore, REF follows from REF . In order to prove REF , let us assume that MATH is totally real. We are going to arrive to a contradiction which proves REF . Replacing, if necessary, MATH by its subfield finitely generated over the rationals, we may assume that MATH (and therefore MATH) is isomorphic to a subfield of the field MATH of complex numbers. Since the center MATH of MATH is totally real, the NAME group of MATH must be semisimple. This implies that the pair MATH is of NAME type CITE, that is, MATH acts on MATH in such a way that for each embedding MATH the corresponding multiplicity MATH . Now assume that MATH does not divide MATH. We have MATH and therefore MATH . Since the multiplicity MATH is always an integer, MATH is odd. Therefore MATH. Let us consider the embedding MATH which sends MATH to MATH. Elementary calculations (REF on p. REF) show that for all integers MATH with MATH the differentials MATH; clearly, they constitute a set of MATH-linearly independent eigenvectors of MATH with eigenvalue MATH. In light of the MATH-equivariant isomorphism MATH we conclude that MATH . This implies that MATH. It follows easily that MATH and therefore MATH. This gives us the desired contradiction when MATH does not divide MATH. Now assume that MATH divides MATH. Then MATH and MATH. Again, as in the proof of REF , the usage of REF allows us to apply the already proven case (when MATH does not divide MATH) to MATH with MATH. |
math/0008134 | We may assume that MATH. Let MATH be an absolutely simple MATH-module of MATH-dimension MATH. Let MATH be a MATH-subalgebra containing the identity operator MATH and such that MATH . Clearly, MATH is a faithful MATH-module and MATH . CASE: By REF, MATH is a semisimple MATH-module. CASE: The MATH-module MATH is isotypic. Indeed, let us split the semisimple MATH-module MATH into the direct sum MATH of its isotypic components. Dimension arguments imply that MATH. It follows easily from the arguments of the previous step that for each isotypic component MATH its image MATH is an isotypic MATH-submodule for each MATH and therefore is contained in some MATH. Similarly, MATH is an isotypic submodule obviously containing MATH. Since MATH is the isotypic component, MATH and therefore MATH. This means that MATH permutes the MATH; since MATH is MATH-simple, MATH permutes them transitively. This gives rise to the homomorphism MATH which must be trivial, since MATH and therefore MATH is a subgroup of MATH. This means that MATH for all MATH and MATH is isotypic. CASE: Since MATH is isotypic, there exist a simple MATH-module MATH and a positive integer MATH such that MATH. We have MATH . Clearly, MATH is isomorphic to the matrix algebra MATH of size MATH over MATH. Let us put MATH . Since MATH is simple, MATH is a finite-dimensional division algebra over MATH. Since MATH is either finite or algebraically closed, MATH must be a field. In addition, MATH if MATH is algebraically closed and MATH is finite if MATH is finite. We have MATH . Clearly, MATH is stable under the adjoint action of MATH. This induces a homomorphism MATH . Since MATH is the center of MATH, it is stable under the action of MATH, that is, we get a homomorphism MATH, which must be trivial, since MATH is perfect and MATH is abelian. This implies that the center MATH of MATH commutes with MATH. Since MATH, we have MATH. This implies that MATH and MATH is trivial if and only if MATH. Since MATH, MATH is trivial if and only if MATH, that is, MATH is an absolutely simple MATH-module. It follows from the NAME density theorem that MATH with MATH. This implies that MATH is trivial if and only if MATH, that is, MATH. The adjoint action of MATH on MATH gives rise to a homomorphism MATH . Clearly, MATH is trivial if and only if MATH commutes with MATH, that is, MATH. It follows that we are done if either MATH or MATH is trivial. Now one has only to recall that MATH. |
math/0008134 | Let us split MATH into a product MATH of two positive integers MATH and MATH. In REF either MATH or MATH is MATH and the target of the corresponding projective linear group MATH. In REF either one of the factors is MATH and we are done or one of the factors is MATH and it suffices to check that each homomorphism from MATH to MATH is trivial. Since MATH is simple, each non-trivial homomorphism MATH is an injection, whose image lies in MATH. In other words, MATH is a subgroup of MATH isomorphic to MATH. Since MATH is not isomorphic to MATH, the subgroup MATH is proper and simple non-abelian. It is known (REF on p. REF on p. REF) that each proper simple non-abelian subgroup of MATH is isomorphic to MATH and such a subgroup exists if and only if MATH is congruent to MATH modulo MATH. This implies that such MATH does not exist and settles REF . In order to do REF notice that one of the factors say, MATH does not exceed MATH. This implies easily that the order of MATH does not exceed MATH and therefore the order of MATH does not exceed MATH. Hence, the order of MATH is strictly greater than the order of MATH and therefore there are no injective homomorphisms from MATH to MATH. Since MATH is simple, each homomorphism from MATH is either trivial or injective. This settles REF . REF follows readily from REF . |
math/0008134 | By REF , MATH is absolutely simple and MATH is either MATH or MATH. The group MATH is a simple non-abelian group, whose order MATH is greater than the order of MATH and the order of MATH. Therefore each homomorphism from MATH to MATH is trivial. On the other hand, one may easily check that MATH for all MATH. Now one has only to apply REF to MATH and MATH. |
math/0008134 | Recall that MATH is either MATH or MATH. In both cases MATH . Clearly, MATH is perfect and every homomorphism from MATH to MATH is trivial. We are going to deduce the Corollary from REF applied to MATH and MATH. In order to do that let us consider a factorization MATH of MATH into a product of two positive integers MATH and MATH. We may assume that MATH and say, MATH. Then MATH . Let MATH be a group homomorphism. We need to prove that MATH is trivial. Let MATH be an algebraic closure of MATH. Since MATH, it suffices to check that the composition MATH which we continue denote by MATH, is trivial. Let MATH be the universal central extension of the perfect group MATH. It is well-known that MATH is perfect and the kernel (NAME 's multiplier) of MATH is a cyclic group of order MATH, since MATH. One could lift MATH to the homomorphism MATH . Clearly, MATH is trivial if and only if MATH is trivial. In order to prove the triviality of MATH, let us put MATH and notice that MATH contains a subgroup MATH isomorphic to MATH (generated by disjoint MATH-cycles). Let MATH be a NAME MATH-subgroup in MATH. Clearly, MATH maps MATH isomorphically onto MATH. Therefore, MATH is a subgroup of MATH isomorphic to MATH. Now, let us discuss the image and the kernel of MATH. First, since MATH is perfect, its image lies in MATH, that is, one may view MATH as a homomorphism from MATH to MATH. Second, the only proper normal subgroup in MATH is the kernel of MATH. This implies that if MATH is nontrivial then its kernel meets MATH only at the identity element and therefore MATH contains the subgroup MATH isomorphic to MATH. Since MATH, the group MATH is conjugate to an elementary MATH-group of diagonal matrices in MATH. This implies that MATH . Since MATH, we get a contradiction which implies that our assumption of the nontriviality of MATH was wrong. Hence MATH is trivial and therefore MATH is also trivial. |
math/0008134 | The case of MATH was proven in REF . The case of MATH was done in REF . So, we may assume that MATH. In light of REF we may assume that MATH, that is, MATH. Assume that MATH. Then MATH does not divide MATH and MATH is either a prime or twice a prime. Therefore MATH is either a prime or twice a prime. Now the very simplicity of MATH follows from REF . Assume now that MATH. Then either MATH or MATH. In both cases MATH is a prime. Now the very simplicity of MATH follows from REF . |
math/0008134 | Recall that MATH is a MATH-dimensional abelian variety defined over MATH. Since MATH is defined over MATH, one may associate with every MATH and MATH an endomorphism MATH such that MATH . Let us consider the centralizer MATH of MATH in MATH. Since MATH is defined over MATH, we have MATH for all MATH. Clearly, MATH sits in the center of MATH and the natural homomorphism MATH is an embedding. Here MATH is the MATH-Tate module of MATH which is a free MATH-module of rank MATH. Notice that MATH . Recall also that REF MATH and MATH acts on MATH through MATH . Since the MATH-module MATH is very simple, the MATH-module MATH is also very simple, thanks to REF . On the other hand, if an endomorphism MATH kills MATH then one may easily check that there exists a unique MATH such that MATH. In addition, MATH. This implies that the natural map MATH is an embedding. Let us denote by MATH the image of this embedding. We have MATH . Clearly, MATH contains the identity endomorphism and is stable under the conjugation via NAME automorphisms. Since the MATH-module MATH is very simple, either MATH or MATH. If MATH then MATH coincides with MATH. This means that MATH coincides with its own centralizer in MATH and therefore MATH is a maximal commutative subalgebra in MATH. If MATH then, by NAME 's Lemma, MATH . This implies easily that the MATH-algebra MATH has dimension MATH and its center has dimension MATH. This means that MATH is a central MATH-algebra of dimension MATH. Clearly, MATH coincides with the centralizer of MATH in MATH. Since MATH respects the theta divisor on the jacobian MATH, the algebra MATH is stable under the corresponding NAME involution and therefore is semisimple as a MATH-algebra. Since its center is the field MATH, the MATH-algebra MATH is central simple and has dimension MATH. By REF , this cannot happen. Therefore MATH is a maximal commutative subalgebra in MATH. |
math/0008144 | Clearly MATH. Put MATH. We have MATH. In particular MATH. This implies that MATH is of the form MATH, namely, MATH. Now MATH is a projection in MATH, with MATH. Indeed, MATH . And MATH, that is, MATH. Now it is clear that MATH, which implies that MATH, Therefore MATH, since all these products equal MATH (because MATH). Finally, MATH implies that MATH. |
math/0008144 | Put MATH as above. First note that if MATH, then MATH. Then MATH. It remains to see that MATH. Clearly MATH. Suppose that MATH is a projection in MATH with MATH (MATH here denotes the normal extension of the former MATH to MATH). Note that MATH is in fact a projection (associated to MATH), and verifies MATH. It follows that MATH. Then MATH, and therefore MATH. |
math/0008144 | Let us start with REF: MATH. To prove REF, suppose that MATH. Then they have the same support, that is, MATH, which implies that there exists a unitary element MATH such that MATH (see CITE). Then MATH . Using REF , this implies that MATH, or MATH. To prove REF, use REF, and note that the unitary element MATH satisfies MATH, that is, MATH. |
math/0008144 | If MATH, then MATH. Now by the NAME inequality MATH, and MATH. Then MATH, and the result follows. |
math/0008144 | Since the spaces are homogeneous spaces, it suffices to show that there exist continuous local cross sections at every point MATH of MATH. Suppose that MATH. Then there exists a unitary operator MATH in MATH such that MATH. Then MATH . It follows that MATH is invertible in MATH. Therefore, one can find MATH such that also MATH is invertible in MATH. Then MATH is invertible. Let us put MATH defined on MATH, where MATH denotes the unitary part in the polar decomposition (of invertible elements) in MATH: MATH. First note that MATH is well defined. If MATH is a vector in the fibre of MATH, then MATH for MATH. Then MATH, where the last equality holds because MATH if MATH is unitary. Next, MATH, and MATH is a cross section for MATH, because MATH is a unitary in MATH. Finally, let us see that MATH is continuous. Suppose that MATH, then there exist unitaries MATH in MATH such that MATH. Then by the continuity of the operations, MATH, that is, MATH. It is clear from REF that the fibre is MATH. Namely, MATH. Note that MATH in MATH if and only if MATH in MATH. |
math/0008144 | That the diagram commutes is apparent. Since MATH and MATH are fibre bundles, it follows using REF that MATH is a fibration. Note that if MATH, then there exists MATH such that MATH. Then MATH, therefore MATH. In REF it was shown that MATH (where MATH and MATH are states of MATH with support MATH) implies MATH. So MATH parametrizes the fibres of MATH, and clearly this set is in one to one correspondence with MATH. Now MATH in MATH if and only if MATH, that is, the class of MATH tends to the class of MATH in MATH (with the quotient topology induced by the norm of MATH). |
math/0008144 | If MATH is finite, it was shown in MATH that MATH is connected. |
math/0008144 | The proof follows by applying the tail of the homotopy exact sequence of the bundle MATH, recalling from MATH that the fibre MATH is simply connected. In the selfdual case, it was proven in MATH that the connected components of MATH are simply connected. |
math/0008144 | This time use the homotopy exact sequence of MATH, and the fact that in this case MATH. |
math/0008144 | In both cases, REF, one has that MATH is contractible (see CITE). Therefore the proof follows writing down the homotopy exact sequence of the fibre bundle MATH. |
math/0008144 | Denote by MATH the class of MATH in MATH. Suppose that MATH converge to MATH in MATH. Then there exist unitaries MATH in MATH such that MATH tends to MATH and MATH tends to MATH in the respective norms. By continuity of the inner product, it is clear then that MATH and MATH, and therefore the assignment MATH is continuous. On the other direction, suppose that MATH tends to zero. There exist MATH and MATH such that MATH and MATH. We have that MATH. Now MATH, MATH, and MATH is a fibre bundle with fibre MATH, therefore there exist unitaries MATH in MATH such that MATH. We may replace the MATH by MATH and MATH by MATH, and still have MATH, with MATH. We claim that MATH. Indeed, if MATH, by a typical argument MATH . The first summand is bounded by (using the NAME inequality) MATH which equals MATH. The other summand equals MATH. It follows that MATH. |
math/0008144 | It suffices to exhibit a local cross section around a generic base point MATH. We claim that there is a neighborhood of MATH such that elements MATH in this neighborhood satisfy that MATH is invertible. Indeed, if MATH, then MATH. If we choose MATH small enough so that MATH lies in the ball around MATH in which a local cross section of MATH is defined, then there exists a unitary MATH in MATH such that MATH. Note that MATH . Then MATH is invertible in MATH, and therefore also MATH. In this neighborhood put MATH where MATH denotes the unitary part in the polar decomposition of invertible elements in MATH as before. We claim that MATH is well defined, is a local cross section and is continuous. Suppose that MATH, then there exits a unitary MATH in MATH such that MATH and MATH. Then MATH. Also, MATH. That it is a cross section is apparent. Let us see that MATH is continuous. Suppose that MATH for MATH in the neighborhood of MATH where MATH is defined. This implies that there exist unitaries MATH in MATH such that MATH and MATH in the norm topologies. The continuity of the inner product implies that MATH. Also MATH. |
math/0008144 | Consider the diagram MATH where MATH is given by MATH. Clearly MATH is a fibre bundle, because it is the composition of the projective bundle MATH with the projection MATH. The map MATH was shown to be a fibration. It follows from REF the MATH is a fibration. The fibre MATH consists of all states MATH with MATH. Then there exists MATH such that MATH, so that one may fix MATH (and not just MATH). Now MATH implies MATH. It follows that the fibre over MATH is the set MATH, which identifies with MATH. |
math/0008144 | It was remarked in the preceding section that MATH is contractible. |
math/0008144 | The first fact follows from the contractibility of MATH, which implies that in the homotopy sequence MATH for all MATH. The second fact follows using that MATH is connected. Using that CITE if MATH is properly infinite, then MATH is contractible, it follows that MATH for all MATH. |
math/0008144 | It was noted before that if MATH is finite, then MATH is connected. |
math/0008144 | The proof follows writing the homotopy exact sequence of MATH. If MATH is properly infinite, its unitary group is contractible. If moreover MATH is selfdual, it was pointed out before that MATH is contractible. |
math/0008144 | It was shown in CITE that MATH . |
math/0008144 | Straightforward: MATH. |
math/0008144 | Suppose that MATH, with MATH as above. Then MATH . That is, MATH. Since MATH are unital and MATH, by the NAME inequality this implies that MATH, for MATH, MATH. Again, using that MATH are unital, this implies MATH, that is, MATH. On the other hand, the states induced in MATH by the vectors MATH and MATH via the representation MATH were shown to be MATH and MATH, where MATH are the states induced in MATH by MATH, respectively, as shown in the lemma above. By REF, MATH implies that there exists MATH such that MATH and MATH. So MATH translates into MATH. The other identity MATH can also be interpreted in terms of these vectors in the cone MATH. Namely, the unique vector in the cone associated to the state MATH is MATH, where MATH denotes the modular conjugation of the standard representation MATH. Indeed, clearly MATH, and MATH. Therefore, by the uniqueness condition (on vectors in the cone inducing states), it follows that MATH. Combining this with MATH yields MATH . This implies that MATH acts as the identity operator on MATH, which is dense in MATH, because MATH is cyclic for MATH. Therefore MATH. Then MATH and MATH. |
math/0008144 | If MATH lies in the fibre MATH, then MATH, or MATH, where as in the previous lemma MATH and MATH are the states of MATH associated to the vectors MATH and MATH. Again, this implies that there exists a unitary in MATH such that MATH and MATH. Then MATH. Now suppose that a net MATH converges to MATH in the NAME space topology (of MATH). This implies that MATH with MATH. In other words, MATH . Equivalently, MATH. This implies that MATH in the NAME space norm (of MATH). Now let MATH, then MATH . That is MATH in a dense subset of vectors MATH. Since MATH, being unitaries, are bounded in norm, this implies strong operator convergence of MATH to MATH. The converse implication is straightforward. |
math/0008144 | The second statement is straightforward, because MATH and the well known fact that the topology of the distance between the vectors in MATH yields a topology which is equivalent to the one given by the norm of the induced states in the conjugate space. Let MATH be a net, and MATH an element in MATH. Then MATH. Next we check that the convergence of the net in the sense described is equivalent to convergence to zero of MATH in the strong topology, where as is usual notation, MATH, for MATH. Since MATH is implemented by the vector MATH, MATH, and therefore convergence in the strong topology implies convergence in the former sense. Suppose now that MATH, and take MATH. Then MATH. The set MATH is dense in MATH, and the operators MATH have bounded norms, therefore MATH tends strongly to zero. Let us prove now that the sphere MATH is closed in this topology. First note that this topology, on norm bounded sets, is induced by the seminorms MATH, MATH, MATH CITE. Then it suffices to see that if MATH with MATH, then MATH. Now, MATH. Indeed, if MATH, then MATH that is, MATH. Therefore MATH . Since this is true for all unit vectors MATH, it follows that MATH. |
math/0008144 | By the result above, it is clear that if MATH in MATH and MATH in MATH, then MATH, where MATH are the vectors in the positive cone inducing MATH. On the other direction, suppose that MATH in MATH. This means that MATH. Then, since MATH, follows that MATH and similarly MATH. Then we get MATH . This implies that MATH . Using REF follows that MATH and so MATH . Note that for every MATH . But MATH . It is easy to prove that the first addend tends to zero using REF. The second one is equal to MATH which again tends to zero using REF. Therefore, MATH, and using REF we get that MATH . Finally, using REF follows that MATH in norm. Equivalently, MATH in MATH. Then MATH implies that MATH. Then MATH, that is, MATH in the MATH topology. |
math/0008144 | Recall that MATH is convex. |
math/0008144 | It was noted that the quotient topology is stronger than the norm topology. Let us check the other implication. Let MATH be a sequence in MATH converging to MATH in norm. Testing convergence in operators of the form MATH, MATH, yields MATH . Note that MATH and MATH. This implies that MATH . In particular, testing this difference at MATH, implies MATH. Therefore, MATH . Coming back to MATH and MATH, note that the vectors MATH belong to the cone MATH, but not necesarilly the vectors MATH. However MATH and we shall see that MATH in norm. Indeed, note that MATH and therefore MATH which tends to zero. Combining these results one obtains that MATH. Now, because the vectors MATH lie in MATH, and the fact that norm convergence of vector states with symbols in MATH implies norm convergence of those symbols, one has that MATH in MATH. In other words, MATH . Suppose now that the states MATH do not converge to MATH in the quotient topology of MATH. This means that the fibres of these states do not near in MATH, that is, there exists a subsequence MATH such that MATH for all MATH. Or equivalently, MATH . Clearly this inequality is preserved by taking convex combinations of unitaries MATH (and leaving everything else fixed), as well as by taking norm limits of such combinations. It follows, using the NAME theorem, that for MATH, MATH, MATH . This clearly contradicts the inequality above, taking MATH for appropriate MATH. |
math/0008144 | In this case, since MATH is finite, MATH is complete in the strong (=strong-MATH) operator topology CITE. Moreover, NAME and NAME proved in CITE that it admits a geodesic structure in the sense of CITE. It has been already remarked that the set function MATH is lower semicontinuous in the norm topology. Therefore REF applies, and MATH has a continuous cross section. |
math/0008144 | In CITE it was proven that the unitary group MATH of such a factor is contractible in the ultra strong operator topology, and therefore also in the strong operator topology. The result follows using the above result, recalling that the fibre of the fibration MATH is MATH with this topology. |
math/0008144 | In this case MATH clearly equals MATH above, and the topology is the MATH (that is ) ultraweak topology of MATH. If MATH the statement is trivial. If MATH it follows from the strong operator contractibility of MATH for such MATH proved in CITE. The case of a proper projection follows from REF and the above remark. |
math/0008144 | Consider the polar decomposition MATH, where MATH can be chosen unitaries because MATH is finite. Note that MATH strongly. Indeed, since MATH, MATH. Therefore, for any unit vector MATH, MATH. Therefore MATH which tends to zero. |
math/0008144 | The key of the argument is again REF. In that result it is shown that the homogeneous space MATH admits a global continuous cross section, where MATH are factors satisfying the hypothesis of REF, and their unitary groups are endowed with the strong operator topology. In our situation, the fibre of MATH (over MATH) is the set MATH. The fibre is not the unitary group of a subfactor with the same unit, nevertheless the argument carries on anyway. Therefore in order to prove our result it suffices to show that in MATH the ultraweak topology (equal to the weak operator topology) coincides with the quotient topology induced by the map MATH. In other words, that the bijection MATH is a homeomorphism in the mentioned topologies. It is clearly continuous. It suffices to check continuity of the inverse at the point MATH. Suppose that MATH is a net of unitaries in MATH such that MATH converges weakly to MATH. Then we claim that there are unitaries MATH in MATH such that MATH converges strongly to zero, which would end the proof. This amounts to saying that there exists unitaries MATH verifying that MATH for all MATH. Now since MATH, one has MATH, the former limit is equivalent to the following MATH . Again, MATH strongly (and the fact that MATH is finite), imply that MATH, MATH, MATH and MATH all converge to zero strongly. Using that MATH are unitaries, these facts imply that MATH strongly. Using the lemma above, for the algebra MATH, and MATH, it follows that there exist unitaries MATH in MATH such that MATH converges to zero strongly. Since MATH also tends to zero, it follows that MATH strongly. Clearly this last limit proves our claim. |
math/0008144 | By the above theorem, MATH has trivial homotopy groups, since it is the base space of a fibration with contractible space and contractible fibre. The same consequence holds for the set of normal states with support equivalent to MATH, using the corollary above. |
math/0008144 | It was noted in REF that when MATH is a finite NAME algebra, then MATH is MATH and MATH is the unitary orbit of MATH. MATH is endowed with the ultraweak topology, which coincides in MATH with the strong operator topology. The rest of the corollary follows using that in this case MATH is (the restriction) of a fibration, and again CITE that for such factors MATH the unitary group is contractible in the strong operator topology. |
math/0008145 | This was originally proved by Professor NAME in REF in finding the number of partitions of a MATH-gon into MATH parts CITE. The recent work of NAME and NAME gives a short and elementary proof while providing relations to knot theory CITE. |
math/0008145 | A MATH-gon with MATH diagonals (that is, a cluster containing MATH cells) represents a codim MATH face of MATH. The type of a face depends on the polytope structure; this is soley determined by the cells forming the cluster, not the way in which they are arranged to form the MATH-gon. Associating a MATH-gon to the integer MATH, observe that partitioning the polygon into MATH cells is identical to partitioning the integer into MATH parts. |
math/0008145 | This is proven by CITE. Their argument is similar to the proof given below for the enumeration of MATH-clusters (see REF), but without the twist operation and with the extra variables MATH in the generating function to mark the numbers of MATH-sided regions. From the recursive equation for the generating function, the coefficients (functions of MATH) are obtained by NAME 's theorem and then expanded using the multinomial theorem. |
math/0008145 | This immediately follows from REF . |
math/0008146 | If one of MATH, MATH fails to be a factor, then their tensor product would fail to be a factor, hence both MATH and MATH must be factors. Similarly, if at least one of MATH and MATH fails to be full, their tensor product would fail to be full. If, say, MATH is of type I or type II, then MATH must be type III, since otherwise MATH would be of type II or type I. Hence if at least one of MATH and MATH is not type III, the situation described in REF or REF must occur. If MATH and MATH are both type III, so that MATH is type MATH and MATH is type MATH, we must prove that MATH. Neither MATH nor MATH can be zero, because then at least one of MATH, MATH would then fail to be full, and hence MATH would fail to be full. Denote by MATH the MATH invariant of NAME (see REF). Since MATH CITE and MATH, we obtain REF . |
math/0008146 | Since the modular group of MATH is MATH, it follows that MATH has period exactly MATH. Hence the centralizer of MATH is a factor. Choose a decreasing sequence of projections MATH and MATH, with MATH, and isometries MATH, so that MATH, MATH, so that MATH normalizes MATH, and MATH. Then MATH is densely spanned by elements of the form MATH with the convention that MATH if MATH. Using the fact that MATH whenever MATH, we can rewrite MATH as MATH . Let now MATH be as above. One can choose a diffuse commutative NAME algebra MATH, containing MATH, MATH, and so that MATH and MATH. Choose a projection MATH, so that MATH and MATH for some integer MATH. Choose projections MATH, so that MATH and MATH. Choose matrix units MATH, so that MATH, MATH. Let MATH be the NAME algebra generated by MATH. By our choice of MATH, MATH is hyperfinite (notice that MATH, and MATH is in fact the crossed product of MATH by a singly-generated equivalence relation). Let MATH. Then MATH is also hypefinite; in fact, it is the crossed product of MATH by the endomorphism MATH. Notice that MATH contains MATH. Furthermore, for all MATH, there is a MATH and partial isometries MATH, so that MATH. Construct in a similar way the algebra MATH, in such a way that MATH and for all MATH, there is a MATH and partial isometries MATH, so that MATH. Notice that MATH is globally fixed by the modular group of MATH. In particular, this means that MATH. Assume now that MATH. Then MATH. Hence MATH. It follows that MATH can be written in one of the following forms, using the fact that MATH: either MATH or MATH where MATH, MATH and MATH, MATH. In the first case, choose MATH for which MATH. Then, writing MATH we obtain MATH . Reversing the roles of MATH and MATH, we get that in general, MATH is dense in MATH. Since each MATH is hyperfinite, the algebra MATH is also hyperfinite; hence MATH is hyperfinite. It follows that the centralizer MATH of MATH can be written as the closure of the span of MATH, where MATH is a tensor product of two type MATH factors, and MATH is a hyperfinite algebra. Since every hyperfinite algebra can be written as a linear span of the product MATH, where MATH are abelian NAME algebras, it follows that the centralizer MATH is the closure of the span of MATH, with MATH a tensor product of two type MATH factors, and MATH, MATH abelian NAME algebras. Hence by NAME 's result CITE, we get that MATH cannot be isomorphic to MATH. |
math/0008146 | Since MATH is a full MATH factor, we have by REF , that the only possible tensor product decompositions with MATH and MATH diffuse are ones where either exactly one of MATH and MATH is type MATH and the other is of type MATH, or each MATH is of type MATH, with MATH. Denote by MATH the free quasi-free state on MATH. It is known (see CITE) that MATH is a factor, isomorphic to MATH. Let MATH be an arbitrary normal faithful state on MATH, such that MATH is a factor. Then (see REF ), MATH. Since MATH has MATH as its fundamental group (see CITE), it follows that whenever MATH is a state on MATH, and MATH is a factor, then MATH. Assume now that one of MATH, MATH is of type MATH; for definiteness, assume that it is MATH. Choose on MATH a normal faithful state MATH for which MATH is a factor, and let MATH be the unique trace on MATH. Let MATH on MATH. Then MATH, and hence cannot be isomorphic to MATH by a results of CITE and Ge CITE, which is a contradiction. Assume now that MATH is type MATH, with MATH. Then by REF , there is a state MATH on MATH, for which MATH is a factor, but is not isomorphic to MATH, which is a contradiction. |
math/0008146 | Let MATH be a NAME subalgebra. Let MATH be a normal faithful conditional expectation. Let MATH be a normal faithful state on MATH, and denote by MATH the state MATH on MATH. Then MATH is a normal faithful state. Furthermore, MATH, because MATH is MATH-preserving and hence MATH. Since MATH is type MATH, it follows that MATH is inner if MATH. Let MATH be a unitary for which MATH, MATH. Then MATH for all MATH, since MATH. It follows that MATH, since MATH is a MASA. Choose MATH positive so that MATH. Note that MATH is in the centralizer of MATH (which contains MATH). Set MATH for all MATH. Then the modular group of MATH at time MATH is given by MATH. It follows that MATH is a normal faithful state on MATH, so that MATH. It furthermore follows from REF that the centralizer of MATH is a factor of type MATH. By the choice of MATH, its modular group fixes MATH pointwise, hence MATH. We claim that MATH is a NAME subalgebra in MATH. First, MATH, hence MATH is a MASA. Since MATH is a NAME subalgebra in MATH, MATH is densely linearly spanned by elements of the form MATH, where MATH is a unitary and MATH. The map MATH is a normal faithful conditional expectation from MATH onto MATH. If MATH is a unitary, so that MATH for all MATH and MATH, then MATH. Hence MATH . It follows that MATH is densely linearly spanned by elements of the form MATH for MATH and MATH. Let MATH be the polar part of MATH, and let MATH be the positive part of MATH, so that MATH is the polar decomposition of MATH. Since MATH it follows that MATH commutes with MATH and hence is in MATH. Moreover, we then have that MATH so that MATH . Thus MATH is densely linearly spanned by elements of the form MATH for MATH and MATH, hence MATH is a NAME subalgebra of MATH. |
math/0008146 | If MATH were to contain a NAME subalgebra, it would follow that for a certain state MATH on MATH, the centralizer of MATH is a factor containing a NAME subalgebra. Let MATH be the free quasi-free state on MATH. Then by REF, one has MATH . Since MATH (see CITE), and because the fundamental group of MATH is all of MATH (see CITE) we conclude that MATH contains a NAME subalgebra. But this is in contradiction to a result of NAME that MATH has no NAME subalgebras (see CITE). |
math/0008152 | We proceed by showing that MATH given in REF satisfies the relation MATH. The full details of this calculation are not especially enlightening, hence we leave them to the reader. We give an outline of this proof by stating that as an intermediate step for the left hand side of this equation, one has MATH . This may be reduced by repeated applications of REF until one arrives at MATH . |
math/0008152 | Take the recurrence for MATH and divide by MATH. By REF , MATH satisfies the desired recurrence which gives us our result. |
math/0008155 | As above, REF is a well-defined, first-order o.d.e. upon MATH in MATH of the form MATH, where MATH is a homogeneous polynomial of degree MATH. The existence for some MATH of a unique, real analytic solution MATH in MATH with initial value MATH follows easily from standard results on ordinary differential equations. The rest of the proof follows that of REF , given in CITE, with small modifications. The compactness of MATH in REF was used only to prove existence of the family MATH, which we have already established, so we don't need to suppose MATH is compact. The evolution REF is exactly the restriction of REF from MATH to MATH. Thus the proof in REF that MATH is special Lagrangian also applies here, wherever MATH is nonsingular. It remains only to show that MATH lies in MATH, rather than just in MATH. Now MATH as MATH is special Lagrangian, and this implies that MATH for MATH. So REF holds for MATH. But REF is an open condition, and it holds for MATH as MATH. Thus, making MATH smaller if necessary, we see that MATH for all MATH. |
math/0008155 | Let MATH be a set of linear or affine evolution data in MATH, with MATH. Let MATH be a nonzero element of MATH, and define MATH by MATH, where MATH is the natural product MATH. Then MATH is a linear or affine REF-form on MATH. The zeros of MATH form a distribution MATH of hyperplanes in MATH wherever MATH is nonzero. The curvature of MATH is MATH. Now clearly MATH, since MATH is nonzero and tangent to MATH at each point of MATH, so MATH is nonzero along MATH and MATH. Therefore MATH is an integral submanifold of MATH, so the curvature of MATH vanishes along MATH. This shows that MATH. Clearly, this is equivalent to MATH being zero along MATH, that is, zero in MATH rather than restricted to MATH. But MATH is linear or affine and MATH is constant, so MATH is linear or affine. As MATH is not contained in any proper linear or affine subspace of MATH (as appropriate), we see that MATH is zero on all of MATH. There are now two possibilities: CASE: MATH, or CASE: MATH for linearly independent MATH, and MATH at each point in MATH. This is because if MATH is nonzero and not of the form MATH, then MATH if and only if MATH, but we know MATH is nonzero on MATH. In REF , we can write MATH for MATH a quadratic polynomial, which is homogeneous if MATH is linear. Then MATH is constant along MATH (assuming MATH connected), so MATH is a subset of MATH. Thus, REF is one of the quadric examples of REF. In REF , we choose coordinates MATH on MATH with MATH and MATH, and it is then easy to show that we are in the situation of REF . |
math/0008155 | Define a linear map MATH by MATH, where we regard MATH as an element of MATH, and MATH is the natural contraction MATH, so that MATH. Let MATH be the NAME subalgebra of MATH generated by MATH, so that MATH maps MATH. Let MATH be the unique connected NAME subgroup of MATH with NAME algebra MATH. Regard elements of MATH as linear vector fields on MATH. Then at each MATH we have MATH. Since MATH by definition, we see that MATH. So the vector fields MATH are tangent to MATH. But the NAME bracket of two vector fields tangent to MATH is also tangent to MATH. Hence, as MATH is generated from MATH by the NAME bracket, every vector field in MATH is tangent to MATH. Since MATH is nonsingular, we have MATH for all MATH, by definition. Thus the map MATH given by MATH is surjective. So the vector fields in MATH span MATH for all MATH. Therefore the action of the NAME algebra MATH on MATH is locally transitive. It follows that MATH is locally isomorphic to an orbit of MATH in MATH, and as MATH is connected, it must be an open set in a MATH-orbit. Next we prove that MATH is MATH-invariant, which is not quite as obvious as it looks. Let MATH, set MATH, and define MATH. We shall show that MATH, where MATH is the NAME derivative. First observe that MATH is a linear combination of terms MATH with MATH for MATH. It follows easily that MATH. But then MATH . Now MATH is a nonvanishing section of MATH and MATH is tangent to MATH, we see that MATH for some smooth function MATH. As MATH by REF , restricting to MATH gives MATH on MATH, that is, MATH on MATH. Therefore MATH or MATH on MATH. But if MATH then clearly MATH. Thus MATH on MATH. Since MATH lies in no proper vector subspace of MATH, and MATH is linear, this implies that MATH. This holds whenever MATH for MATH. Such forms are a basis for MATH. So MATH for all MATH, and therefore for all MATH. As MATH is connected, this shows that MATH is MATH-invariant. It remains to show that MATH is MATH-equivariant and surjective. The MATH-equivariance is now obvious, as MATH is MATH-invariant. So MATH is a MATH-invariant subspace of MATH, that is, an ideal in MATH. But then MATH is closed under the NAME bracket. As MATH generates MATH we have MATH, and MATH is surjective. |
math/0008155 | Let MATH be the minimum and maximum values of MATH. Then MATH and MATH, and using the ideas of REF we find that MATH . As MATH we have MATH and MATH. Also, the factors MATH in REF tend to zero, except near MATH and MATH. Hence, as MATH, the integrands in REF get large near MATH and MATH, and very close to zero in between. So to understand the MATH as MATH, it is enough to study the integrals REF near MATH and MATH. We shall model them at MATH. Then near MATH we have MATH . Since MATH this gives MATH, so that MATH. Therefore, when MATH we have MATH, so when MATH is small and MATH we have MATH changing variables to MATH, where in the second line some surprising cancellations happen, and we have also approximated the upper limit MATH by MATH. When MATH and MATH is small we have MATH and similarly when MATH and MATH is small we have MATH . So on MATH the integrals REF are close to MATH for MATH and REF for MATH for small MATH. In the same way, on MATH the integrals REF are close to MATH for MATH and REF for MATH for small MATH. This proves REF . Next consider the behaviour of MATH as MATH. When MATH is close to MATH, MATH is small and MATH close to REF, so MATH remains close to MATH. Write MATH, for MATH small. Then, setting MATH and MATH taking only the highest order terms, REF become MATH so that MATH and MATH undergo approximately simple harmonic oscillations with period MATH. Then REF shows that MATH . So MATH, as MATH is approximately constant. This proves REF . |
math/0008155 | From REF we know that MATH when MATH and MATH when MATH, and MATH, so that MATH does map MATH to MATH. As MATH and MATH are clearly real analytic functions of the MATH and MATH, we see from REF that MATH is real analytic. When MATH we must have MATH, and going back to REF we see that the equations on MATH are MATH and MATH, which are real linear, and admit simple harmonic solutions with period MATH for any nonzero initial data. Translating this into the notation of REF , we find that MATH and MATH have period MATH and MATH, MATH for any initial data in MATH. Now the limits of MATH in REF satisfy MATH. Thus, from REF we see that the closure MATH contains a nonempty open subset of the MATH-dimensional real hypersurface MATH in MATH. This implies that MATH is at least MATH-dimensional. Since MATH is real analytic and MATH is connected, there are only two possibilities: CASE: MATH is MATH-dimensional, or CASE: MATH lies in the real hypersurface MATH in MATH. However, when MATH we can use REF to eliminate possibility REF . For MATH must contain the limit in REF , which satisfies MATH. This lies in MATH only if MATH, that is, if MATH, since MATH. Now the ranges of MATH and MATH are MATH and MATH. When MATH we are forced to take MATH and we cannot eliminate possibility REF , as it is actually true. But when MATH we are always free to choose MATH or MATH, so MATH contains a point not on the hypersurface. Thus REF is false, so REF is true, and MATH is MATH-dimensional. This shows that MATH must be surjective at some MATH, and as this is an open condition and MATH is real analytic, MATH is surjective for a dense open subset of MATH. |
math/0008155 | We saw at the beginning of REF that if MATH then the time evolution of REF exists for all MATH, and is periodic with period MATH for some MATH. But by REF we have MATH for a dense subset of MATH. Thus, the construction above yields a countable collection of families of SL MATH-folds, locally parametrized by MATH with MATH. Choose one of these families, and define MATH by REF . Then MATH is special Lagrangian by REF . Let MATH be the quadric MATH in MATH. Then MATH is the image of a map MATH taking MATH to the point in MATH defined in REF . As the factors MATH are always nonzero, MATH is an immersion except when MATH, which happens only when MATH. Thus, MATH is a nonsingular immersed submanifold when MATH, and when MATH it has just one singular point REF as an immersed submanifold. Since MATH is periodic in MATH with period MATH, we can instead regard MATH as a map MATH, where MATH. It is also not difficult to see that the image MATH of MATH is closed, provided MATH is periodic. Now MATH is diffeomorphic to MATH when MATH, to MATH when MATH, and to the cone on MATH when MATH. Thus MATH is diffeomorphic under MATH as an immersed submanifold to MATH when MATH, to MATH when MATH, and to the cone on MATH when MATH. It remains only to discuss the parts about free quotients by MATH in REF - REF . We could have left these bits out, as the result is true without them. The point is this: suppose MATH, for integers MATH and MATH with MATH and MATH. Then MATH, so that MATH has period MATH if MATH is even, and MATH if MATH is odd. Thus MATH satisfies MATH . Since MATH is invariant under MATH, the family of quadrics making up MATH has period MATH. But it doesn't simply repeat after time MATH, but also changes the signs of those MATH with MATH odd. Let us regard MATH as mapping MATH, where MATH. Then MATH is generically REF:REF, and filters through a map MATH, where the generator of MATH acts freely on MATH by MATH . This completes the proof. |
math/0008155 | A point MATH in MATH lies in MATH if and only if MATH. Substituting in REF , this is equivalent to MATH . But by REF, and thus MATH. |
math/0008162 | Since MATH is a symplectic leaf of MATH, the bivector field MATH is nondegenerate on subspaces MATH at points MATH sufficiently close to MATH. Hence MATH is a horizontally nondegenerate on a tubular neighborhood MATH of MATH in MATH. By REF , MATH is a coupling tensor on MATH associated with geometric data MATH. Here the intrinsic connection MATH and the bivector fields MATH, MATH are defined by REF respectively. REF follows form MATH-compatibility assumption. |
math/0008162 | We will use a contravariant analog of the homotopy method due to CITE and CITE see also CITE. CASE: Homotopy between coupling tensors. By REF, without loss of generality we can assume that the vertical parts of MATH and MATH coincide, MATH-on MATH and MATH. By the MATH-compatibility assumption we deduce that MATH at MATH. It follows from this property and REF that we can choose MATH in REF so that MATH . Consider the following MATH-parameter families of forms: MATH . Then MATH is a time-dependent connection MATH-form on MATH. By REF and the nondegeneracy of MATH, there is a neighborhood MATH of MATH in MATH such that MATH is nondegenerate on MATH for all MATH. This means that for every MATH and MATH the horizontal lift MATH induces a nondegenerate bilinear form on the qoutient space MATH. Moreover, we observe that the triple MATH defines a geometric data on MATH satisfying REF - REF for every MATH. Thus the time-dependent coupling tensor MATH REF associated to MATH gives a homotopy from MATH to MATH, MATH. CASE: Homological equation. By the nondegeneracy of MATH on MATH, there exists a unique solution MATH of the following equation MATH . Clearly MATH . One can associate to MATH the time-dependent horizontal vector field MATH defined by MATH for all MATH. REF implies MATH . MATH satisfies the equation MATH . Here MATH is the NAME derivative of a bivector field MATH along a vector field MATH, that is, the NAME bracket MATH. The proof of REF is given in REF. CASE: Let MATH be the flow of the time-dependent horizontal vector field MATH: MATH, MATH. By REF and the usual properties of the NAME derivative (see, for example, CITE ) we get MATH. Because of REF for every MATH, we have MATH for all MATH. Hence there exists a neighborhood MATH of MATH in MATH that lies in the domain of the flow MATH for MATH. Finally, the time REF flow MATH of MATH generates a diffeomorphism MATH satisfying REF . |
math/0008162 | Let MATH and MATH be two MATH-compatible NAME tensors on MATH satisfying the above hypotheses. Let MATH and MATH be the geometric data associated with MATH and MATH, respectavely. Thus, NAME connections MATH and MATH are MATH-compatible and hence REF holds. Pick a MATH in REF and define MATH . It follows from REF and the curvature identity REF for MATH and MATH that MATH. Using REF , we deduce: MATH is a REF-cocycle, MATH whose cohomology class does not depend on the choice of MATH in REF . If this class vanishes, then MATH for a MATH and MATH and MATH satisfy REF for MATH and MATH. |
math/0008164 | In order to see REF recall first that, since MATH is a MATH-homomorphism, according to basic MATH-theory, the unit ball MATH of MATH is related to the unit ball MATH of MATH through MATH. From this by the NAME density theorem MATH is seen. Owing to normality of MATH we then may conclude as follows: MATH . To see REF, apply REF to MATH and MATH in respect of the identity representation of the MATH-algebra MATH. Under these premises REF yields MATH, for each MATH and MATH. In view of REF this is MATH as asserted by REF when applied with respect to MATH, MATH, MATH and MATH. Relating REF, note that for each invertible MATH obviously MATH holds. From this by REF and positivity of all terms the assertion follows. To see REF, note that MATH is equivalent to MATH, by REF , which is the same as MATH, for all MATH. Hence, MATH is equivalent to MATH. Therefore, since MATH and MATH holds, with the support orthoprojections MATH and MATH of the normal positive linear forms MATH and MATH (see REF), MATH is equivalent to orthogonality of MATH with MATH. The latter is the same as MATH. In view of REF equivalence of MATH to MATH follows. |
math/0008164 | For MATH, MATH consider MATH as defined in REF. Then MATH holds, see REF. Hence, in the notations from above MATH . It suffices to show that partial isometries MATH exist which obey MATH, MATH and MATH. In fact, in view of REF, for MATH and MATH one then has MATH and MATH. By the above and in twice applying REF , we may conclude as follows : MATH . Hence MATH, from which MATH follows. In view of REF, the constructed MATH and MATH can be taken to meet our demands. Recall that for a MATH-algebra MATH on a NAME space MATH there is a largest central orthoprojection MATH such that the MATH-algebra MATH over MATH is finite. Thus, in case of MATH, MATH is properly infinite over MATH. Moreover, on MATH one has MATH for each central orthoprojection MATH (the outer commutant refers to MATH). Applying this to MATH with MATH or MATH, and having in mind REF, we see that we can content ourselves with constructing the partial isometries in question only for MATH-algebras with either finite or properly infinite commutants. In line with this, let MATH be the MATH-algebra in question, with MATH either finite or properly infinite, and let MATH and MATH be implemented by vectors MATH and MATH. We then will make use of the notational conventions of REF, and are going to construct MATH and MATH under these premises now. Suppose MATH to be finite. By REF we have MATH. In a finite MATH-algebra, by another standard fact, see CITE, this condition implies the following to hold, in any case: MATH . Hence, there is MATH obeying MATH and MATH. Define MATH. Then, MATH is unitary, MATH, with MATH. Since by polar decomposition MATH is fulfilled, MATH can be chosen. Suppose a properly infinite MATH. Then MATH with MATH exists. By REF, MATH is an orthoprojection. Hence MATH and MATH (MATH be the NAME comparability relation). Hence, there are MATH with MATH, MATH, MATH, MATH. Define MATH, MATH. Since MATH holds, the above and REF imply MATH and MATH. With the help of REF again, we finally see that MATH holds. |
math/0008164 | By REF we may content ourselves with proving the assertions for a MATH-algebra MATH and normal positive linear forms MATH and MATH with nontrivial fibres. For normal forms orthogonality simply means orthogonality of the respective supports. Hence, MATH and MATH imply MATH and MATH. By maximality of MATH then MATH, and thus the first of the relations of REF is seen. Note that MATH, with MATH. From the former owing to MATH then MATH is seen, and thus the two orthogonality relations may be summarized into MATH. Hence, in view of REF then MATH follows. On the other hand, by definition, MATH is subordinate to MATH and must also be orthogonal to MATH. Hence, MATH by REF when applied to MATH. In summarizing the second relation of REF follows. Let MATH and MATH be chosen as in REF . By REF one has MATH, with MATH. Note that MATH. By REF MATH follows. By orthogonality of MATH with MATH we especially must have MATH. Hence also MATH, and thus we obtain MATH . We may apply this to MATH accordingly, with the result MATH. By REF and by symmetry of MATH, substitution of the latter into the former relation yields MATH. Since MATH and REF imply MATH, from this REF follows. In view of REF is seen. Finally, by REF, MATH as well as MATH must be fulfilled. Therefore, MATH implies MATH. By REF then MATH follows. Since MATH holds, in view of REF MATH follows. Hence, MATH. Proceeding analogously for MATH instead of MATH yields MATH. Thus also the final assertion is true. |
math/0008164 | Note that MATH, by REF, and MATH by REF. Hence REF follows. Relating REF, remark that by REF MATH obeying MATH and MATH exists and is unique. In view of this and REF, MATH is a partial isometry of MATH, with MATH. From this and MATH by uniqueness of the polar decomposition the validity of REF follows. Let MATH be as in REF . By REF, MATH, and according to REF, MATH. By REF one has MATH, with MATH, MATH, MATH, MATH and obeying MATH, MATH. The partial isometry MATH of the polar decomposition of MATH has to obey MATH. Hence, MATH follows. From this together with the other properties of MATH one gets MATH . By MATH one has MATH. Thus, from MATH in view of MATH and REF also MATH follows. In view of REF then MATH is obtained. From this REF follows. |
math/0008164 | Suppose MATH satisfies REF for some MATH, and given MATH. By REF the same premises then hold with respect to any other vector MATH in the MATH-fibre of MATH. Also, by REF, one has MATH. From this and REF in view of REF then MATH follows. By REF , MATH. This is the same as MATH, and thus in view of REF and by construction of MATH, in accordance with REF MATH. By REF we see MATH, for MATH of REF. Note that our situation with MATH and MATH can be easily adapted to a context where relation REF can be used. In fact, choosing MATH there, in the notation of the previous proof one has MATH, and in view of MATH (see above and REF) the relation REF yields MATH, for each other MATH. Hence, MATH of REF does not depend on the special MATH, provided MATH, MATH are held fixed. To see that REF is necessary, let MATH obey MATH. By REF, MATH, and thus MATH, by REF. Obviously, for MATH REF can be satisfied with MATH. In view of the above this completes the proof. |
math/0008164 | Let MATH be defined as in REF . Then, by MATH one especially has MATH. Hence, in view of REF the intertwining relations REF can be applied and show that MATH and MATH are fulfilled, and therefore in view of the above MATH has to hold. On the other hand, we have MATH, by REF. The assertion now will follow upon combining together the last two relations. |
math/0008164 | In the following, we let MATH, MATH. To prove REF, note first that, for MATH in the fibre of MATH, with MATH, by REF, also MATH. Also, for MATH with MATH and MATH in accordance with REF we see that MATH and MATH have to be fulfilled. Now, suppose MATH. Then MATH cannot vanish. Hence, for each MATH, also MATH, with MATH. Hence, MATH for all MATH, by REF. For MATH obviously MATH. By REF this especially applies to MATH, and thus for MATH nonuniqueness follows. On the other hand, in case of MATH, in view of the above and REF, one has MATH. The above condition on MATH in this case yields MATH, and therefore MATH, that is, uniqueness holds. To see REF, note first that MATH holds, by REF. Let MATH and MATH. Obviously we then have MATH. By REF, MATH and MATH have to be fulfilled. Owing to MATH, MATH, by REF. Owing to MATH, REF can be applied to the pair MATH, and then yields MATH. Substituting the latter into MATH then gives REF. Note that since MATH is a (convex) cone, the first two of the REF -implications and the third REF -implication within REF, as well as the implication MATH, follow from REF. Owing to this and MATH, to see the remaining other implications, it suffices to prove the implication MATH. In line with this, assume MATH. REF in accordance with REF then tells us that MATH holds. Also, in view of REF and MATH, one has MATH. Now, remind the fact mentioned on in context of REF and saying that in this case MATH has to be fulfilled with vectors MATH, with MATH. Let MATH be the positive linear forms implemented by MATH. By uniqueness of the map of REF one has MATH, and by orthogonality MATH and since MATH is fulfilled, from estimate REF the relation MATH can be inferred. Hence, MATH, that is MATH holds. With the help of REF then also MATH. Hence, MATH. The latter implies MATH. Since by REF holds, in reasoning with REF once more again from this in view of the above MATH can be seen. But then especially MATH, from which MATH has to be followed. Now, by REF, when applied to the pair MATH, positivity of MATH can be seen. In line with the previously derived then especially MATH has to hold, for MATH. Since the scalar product between vectors of the natural positive cone has to be always a nonnegative real, MATH is inferred to hold. As argued above, orthogonality among vectors of the natural positive cone then implies orthogonality of the associated MATH- and MATH-projections, respectively. Thus especially MATH. This in view of MATH implies MATH. Hence MATH, and thus MATH holds, and which is REF. |
math/0008164 | In view of the symmetry of the assertion, we may content ourselves with proving for example, that REF implies REF. Suppose REF. In line with this and REF , MATH. By REF we then have MATH. From this with the help of NAME 's theorem and owing to REF for each MATH, and corresponding MATH with MATH, we infer that MATH. According to REF we then also have MATH. This is the same as MATH. The conclusion of REF when applied to MATH then yields MATH, that is, also MATH commutes with MATH. |
math/0008164 | The implications MATH follows at once from REF . Suppose REF to be fulfilled, and be MATH. Then, by REF, MATH, and REF is seen. In case of MATH the same reasoning by REF for the pair MATH will yield MATH, which in view of REF amounts to REF again. |
math/0008164 | By REF one has MATH. Thus MATH. By the last part of REF we then see MATH. Hence MATH. Also, since MATH holds, REF may be applied with respect to MATH and then yields MATH. Note that in accordance with REF especially also MATH must be fulfilled. This condition is equivalent to MATH, for all MATH. Now, let MATH be chosen in accordance with REF. By the previous MATH and MATH have to be fulfilled. Also, owing to MATH and MATH and by REF the following orthogonality relations can be followed: MATH and MATH. Hence, both MATH and MATH are orthogonal systems of vectors, and therefore we may conclude as follows: MATH . Hence, in view of REF MATH must be fulfilled. Since MATH and MATH hold, from this by REF equality is inferred, MATH. That is, REF has to be true. Suppose REF is fulfilled. By definition of MATH and in view of REF there has to exist MATH with MATH such that MATH holds. In view of REF then MATH. Hence, MATH, on MATH. From MATH with MATH we then infer that MATH. Let MATH be the orthoprojection MATH. Then MATH, and thus the previous formula reads as MATH, with MATH. Note that MATH, and MATH. Thus MATH is a partial isometry in MATH with initial projection MATH and obeying MATH. By uniqueness of the polar decomposition then MATH follows. By definition of MATH from this MATH follows, which is the condition in REF. Thus, in view of the above REF is equivalent to REF. Note that REF is equivalent to MATH and MATH, for some MATH. This is the same as requiring MATH with MATH. In defining MATH we then have MATH, with MATH. On the other hand, for each MATH with MATH and MATH one has that MATH. Hence, MATH is a partial isometry, with MATH and MATH. Thus REF is equivalent to REF. |
math/0008164 | Suppose MATH first. By REF then MATH. Hence MATH, and thus the condition of REF can be fulfilled in a trivial way, for each MATH, MATH. In view of REF the fact then follows. If the support orthoprojection MATH is finite, then an application of REF yields that also MATH is finite (and vice versa), and so has to be also each other MATH, for MATH. Let MATH, MATH, with MATH. Since also MATH is finite, by CITE MATH extends to a unitary MATH. The partial isometry MATH then obeys MATH and MATH, and thus REF can be applied. |
math/0008164 | According to REF and by positivity of MATH the assumptions on MATH and MATH equivalently read as MATH and MATH. Thus MATH and MATH. Suppose a cyclic MATH in the fibre of MATH to exist. Then, in view of REF there is MATH with MATH and MATH. Thus especially MATH and MATH. It is obvious that REF cannot be satisfied by MATH. Thus MATH, which is REF. To see REF, suppose now MATH to be a factor, and MATH. By REF we then have MATH and infinite MATH. Remind that, MATH being a standard form MATH-algebra, MATH admits faithful normal positive linear forms. Hence MATH is a countably decomposable factor. Thus there can exist only one equivalence class of infinite orthoprojections in MATH, and therefore MATH must hold. From the latter in view of REF the relation MATH follows. Thus, a cyclic vector has to exist in MATH, and MATH by the above. |
math/0008164 | As mentioned, under the given premises MATH. Thus REF equivalently says that MATH holds iff MATH or MATH. The latter is equivalent to MATH, by REF. Since MATH is a countably decomposable factor, this is the same as asserting MATH to be finite. |
math/0008164 | Owing to MATH both MATH and MATH have to be normal positive linear forms which sum up to MATH. Especially, MATH then is subordinate to MATH and is orthogonal to MATH by construction. On the other hand, MATH also implies MATH. Hence, MATH. In view of REF then MATH follows. |
math/0008164 | In view of REF and MATH, MATH follows. On the other hand, MATH implies MATH. Moreover, in view of REF from MATH the relation MATH is obtained. Thus a cyclic vector has to exist in MATH. Application of REF then yields the result. |
math/0008164 | As in the previous proof, MATH and MATH for faithful MATH imply MATH. Note that the latter condition implies MATH, by triviality. Hence, for faithful MATH and MATH, MATH is equivalent to MATH. But then the assertion is a consequence of REF . |
math/0008164 | Let MATH be a cyclic and separating vector for MATH. Note that by assumption on MATH, MATH. Also, by positivity of MATH, MATH. From this then MATH follows. On the other hand, in any case, MATH has to be fulfilled. Thus we conclude MATH. In view of REF then MATH and MATH follow. In the case at hand REF can be applied and yields MATH. Suppose MATH is faithful. Then we can choose a cyclic and separating vector MATH, and will work in the corresponding standard form action of MATH. Suppose that MATH commutes with MATH. By REF , since MATH holds, and since commutation means MATH, we have MATH. But in view of REF also MATH. Hence, MATH by REF. As has been explained in context of REF, this is the same as MATH. Also, since MATH implies MATH, MATH must be separating for MATH. Hence, by REF and with MATH, MATH follows, with densely defined, positive, selfadjoint linear operator MATH which is affiliated with MATH. In view of the above then MATH. Since MATH is separating, from this MATH follows, that is, MATH. On the other hand, for MATH, MATH, in view of the above formula MATH and REF, MATH follows. Applying REF accordingly yields MATH. Hence, MATH commutes with MATH. |
math/0008164 | Let us consider the standard form action of MATH with respect to cyclic and separating MATH. By assumption such a vector has to exist. As mentioned in the previous proof, MATH implies (in fact is equivalent to) MATH, with densely defined, positive, selfadjoint linear operator MATH which is affiliated with MATH. Thus MATH, where MATH is the orthoprojection onto the closure of the range of MATH. By spectral calculus, if MATH is the (left-continuous) spectral resolution of MATH, then MATH. Also, owing to MATH, for MATH with MATH, we have MATH and MATH. Hence MATH. Thus MATH follows. In view of the above then MATH. |
math/0008164 | Let MATH, with orthoprojection MATH, be the canonical decomposition of MATH into a finite component MATH (which might be vanishing) and properly infinite component MATH. By central decomposition techniques, the properly infinite case can be reduced to an appropriately defined field MATH of properly infinite `subfactors' MATH acting over direct integral NAME subspaces MATH, in the sense that each MATH and MATH can be written as appropriately defined central integrals MATH and MATH, with some MATH and MATH, such that MATH can be written as MATH over terms MATH, with MATH, and with respect to some central measure MATH. Thereby, the index MATH refers to the central decomposition which is isometrically defined relative to some direct integral decomposition MATH of the NAME space MATH, see for example, CITE for the details. Since all decompositions refer to the center and MATH is faithful over MATH, for MATH we will have MATH, with faithful MATH over properly infinite subfactor MATH. By REF, MATH cannot be trivial. Therefore, there is an orthoprojection MATH, with MATH. As orthoprojection of the infinite factor MATH, at least one of MATH and MATH must be infinite. We make a choice, and call this infinite projection MATH. Since MATH is countably decomposable by assumption, each of MATH is a countably decomposable infinite factor. Hence, we then must have MATH and MATH, with respect to MATH. Define an orthoprojection MATH in MATH as follows: MATH . One then has MATH, MATH, and MATH with respect to MATH. But then, by assumption on MATH, there is MATH with MATH and MATH. It is easily seen that for each such MATH then MATH must be fulfilled. Hence, MATH with MATH but MATH, for each such MATH. By REF with MATH then MATH follows, for each MATH with MATH and MATH. Now, let MATH be a partial isometry with MATH and MATH. Then, the map MATH is a MATH-isomorphism between MATH and the hereditary MATH-subalgebra MATH. Thus, for MATH running through all MATH with MATH, MATH is running through all faithful MATH. In view of MATH, by the above conclusion for each such MATH and with MATH the result is MATH. But note that MATH. That is, MATH must be fulfilled, for each MATH with MATH. |
math/0008164 | By REF means commutation of MATH with MATH, in the sense of REF . Thus REF follows along with REF . Relating REF, by REF we yet know that MATH is always fulfilled on a finite MATH. On the other hand, for infinite MATH, according to REF there exists MATH commuting with MATH and obeying MATH, which in view of REF yields REF. Since a MATH-algebra is either finite or infinite, this then also completes the proof of REF. |
math/0008164 | By REF implies MATH. REF then may be applied, and yields MATH iff either MATH is finite or MATH, compare also REF . |
math/0008167 | Consider MATH for any MATH and let MATH be a nonzero map. We want to show that MATH. Since MATH is a nonzero map and MATH is simple, MATH is in fact injective and so MATH is a simple submodule of MATH. Hence, MATH is contained in the socle of MATH. By construction, MATH. Hence, MATH. In other words, MATH as claimed. |
math/0008167 | If MATH is projective, then clearly MATH for all MATH. Conversely, suppose that MATH and let MATH be a minimal injective resolution of MATH as above. REF shows in particular that MATH and hence MATH. Since MATH is unipotent, the MATH-fixed points of every non-zero module are non-zero. Hence, we must have MATH. But, that means MATH was injective (equivalently projective) to begin with. |
math/0008168 | Let MATH and consider the operation MATH defined by MATH (or the sum of the MATH in the case MATH). It follows from the NAME formula, MATH, that MATH is an algebra homomorphism. Moreover, from REF , it is the algebra homomorphism defined by MATH for each MATH. Finally, since MATH is stable under the NAME operations, it is stable under MATH, and so the claim follows from the NAME to REF. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.