paper stringlengths 9 16 | proof stringlengths 0 131k |
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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 M... |
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 satisfy... |
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 in... |
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/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 t... |
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 MAT... |
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... |
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... |
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/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... |
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... |
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 M... |
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 ha... |
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 MAT... |
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 simpl... |
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... |
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).... |
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... |
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/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 a... |
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... |
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 continu... |
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... |
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 t... |
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 represe... |
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 top... |
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 ... |
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 . U... |
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 ... |
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 a... |
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 t... |
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... |
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 inte... |
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 ... |
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 oth... |
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 element... |
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 ... |
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/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 (... |
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 ... |
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... |
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. T... |
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 ... |
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 betwe... |
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/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 M... |
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 t... |
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 MA... |
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 t... |
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, a... |
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 ma... |
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 ... |
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... |
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 constructi... |
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... |
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... |
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. Acc... |
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... |
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 ... |
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 MAT... |
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 an... |
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 MA... |
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/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 ... |
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, tha... |
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... |
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