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math/9911112 | First we prove the following analogue of REF : MATH where each MATH is a monomial in MATH, MATH. The proof of this formula is exactly the same as the proof of REF. The rest of the proof is completely parallel to the proof of REF . |
math/9911112 | The proof is completely parallel to the proof of REF . |
math/9911112 | The proof is completely parallel to the proof of REF . |
math/9911112 | If MATH is reducible, then MATH should contain a dominant term other than the product of the highest weight terms. But for that to happen, for some MATH and MATH, there have to be cancellations between some MATH appearing in MATH and some MATH appearing in MATH. These cancellations may only occur if MATH, and MATH, by REF . Moreover, MATH according to REF . |
math/9911112 | Consider MATH. We have MATH and MATH has no dominant monomials. Then MATH by REF . |
math/9911112 | Let MATH and MATH is obtained from MATH by replacing MATH by MATH. Then MATH and MATH are elements in MATH with the only dominant monomial MATH by REF . Therefore MATH by REF . |
math/9911115 | Begin by writing MATH . Condition on MATH and then replace the second factor with its NAME chaos expansion and so obtain an expansion of which the typical term is MATH . We now replace each integrand by its chaos expansion- this must simply result in the chaos expansion of MATH . On comparing the two expansions it may be seen that if MATH is distributed according to MATH then MATH contains exactly k points at positions MATH through MATH with probability MATH but since MATH is just the probability distribution of MATH under MATH we are done. |
math/9911115 | Fix an admissible subset MATH. For each MATH consider the correlated random walk MATH. There is the usual weak convergence in the space of continuous MATH-valued paths to a process MATH, each component of which forms a one-dimensional Brownian motion and their co-variation is simply: MATH . Let MATH be the time at which MATH attains its minimum between times MATH and MATH, and similarly define MATH. Now the correlation of MATH and MATH can be split into the sum of two contributions. One arises when the random walks MATH and MATH attain their minimum (between times MATH and MATH ) values simultaneously - in this case MATH and MATH are equal. The remaining contribution tends to zero for large MATH and so the limits MATH exist and are given by MATH . The two random times MATH and MATH can only be equal if their common value lies in one of the components of MATH. For each such component we consider the common time at which MATH and MATH attain their respective minimum (over that component) and compute the probability that this is actually the global minimum of both Brownian motions. We obtain MATH where MATH and MATH is the law of the triple MATH denoting the time at which MATH attains its minimum MATH . By virtue of the scaling properties of BM we have MATH . A well-known exercise REF confirms that MATH . Putting these into the formula obtained in the previous section and recalling the definition of MATH we obtain the desired result: MATH . |
math/9911117 | Pick any NAME derivative MATH and set MATH for some MATH-form MATH. Then MATH and so the traces differ by MATH . Since MATH it follows that there is a unique MATH such that MATH is trace-free. |
math/9911117 | For any vector field MATH, MATH anticommutes with MATH (since MATH) and is skew (since MATH is skew, and MATH is conformal). Hence MATH, which is symmetric in MATH because MATH is integrable and MATH is torsion-free, is also skew in MATH. It must therefore vanish for all MATH. If we now impose MATH we obtain: MATH . Hence MATH. |
math/9911117 | The first two facts are immediate from MATH and MATH respectively. If MATH then wedge product with MATH is injective on MATH-forms, while for MATH, MATH is the multiple MATH of the weightless volume form, since MATH is antiselfdual. The final formula is a consequence of the first NAME identity: MATH and the last term vanishes since MATH for MATH. |
math/9911117 | Clearly REF implies REF . It is immediate from REF that REF implies REF ; to obtain the stronger result that REF implies REF suppose that MATH is not NAME, that is, at some point MATH, MATH. If MATH is a shear-free geodesic congruence near MATH then by REF , MATH is a multiple of the identity on MATH, and one easily sees that this multiple must be the middle eigenvalue MATH of MATH at MATH. Now at MATH, MATH may be written MATH where MATH with MATH. The directions of MATH and MATH are uniquely determined by MATH and MATH must lie in one of these directions. Hence if MATH is not NAME at MATH, there are at most two possible directions at MATH (up to sign) for a shear-free geodesic congruence. (Note that the linear algebra involved here is the same as that used to show that there are at most two principal directions of a nonzero antiselfdual NAME tensor in four dimensions; see, for instance CITE. Our result is just the symmetry reduction of this fact.) Finally, to see that REF implies REF , we simply observe that given any minitwistor line and any point on that line, we can find, in a neighbourhood of that point, a transverse holomorphic curve. This curve will also intersect nearby minitwistor lines exactly once. |
math/9911117 | The equations of the previous proposition are equivalent to the following: MATH . Applying the star operator readily yields the equations of the theorem. The second equation is clearly a monopole equation, since MATH is closed. It remains to check that the right hand side of the first equation is closed: MATH . Here MATH is the divergence on forms, and so the first term vanishes by virtue of the second NAME identity. The remaining multiple of the orientation form MATH is MATH which vanishes by the previous proposition. |
math/9911117 | The curvature of MATH is easily computed to be: MATH . Now MATH, so REF imply that MATH vanishes if MATH is NAME. [Conversely if there is a MATH with MATH for all MATH, then MATH is NAME.] |
math/9911117 | Since MATH is a weightless vector field, MATH. This vanishes iff MATH. Hence MATH for all of the hyperCR congruences MATH iff MATH. This formula shows that MATH is a conformal vector field, and that MATH is a shear-free geodesic congruence with twist MATH. Also MATH preserves the flat connection MATH, since it preserves the parallel sections. Finally, note that the twist of MATH is determined by the conformal structure from the skew part of MATH, so it is also preserved by MATH. Hence MATH preserves MATH and is therefore a geodesic symmetry. |
math/9911117 | The monopole equation on MATH is equivalent, via the definition of MATH, to the fact that MATH lies midway between MATH and MATH. So it remains to show that under this condition, the NAME equation on MATH is equivalent to the selfduality of MATH. The space of antiselfdual NAME tensors is isomorphic to MATH via the map sending MATH to MATH, and so it suffices to show that MATH iff MATH. Since MATH is basic, as a NAME connection on MATH, MATH. Therefore: MATH . If we now take the antiselfdual part of this equation, contract with MATH and MATH, and take MATH then we obtain MATH . Symmetrising in MATH, we see that MATH iff the horizontal part of the symmetric NAME endomorphism of MATH is a multiple of the identity. The first submersion formula CITE relates the NAME curvature of MATH on MATH to the horizontal NAME curvature of MATH on MATH. If we combine this with the fact that MATH and MATH, then we find that MATH for some section MATH of MATH. Since MATH, MATH vanishes on the plane spanned by MATH, and so, by comparing the lengths of MATH and MATH, we verify that the trace-free part of MATH vanishes. Hence MATH on MATH iff MATH is NAME on MATH. |
math/9911117 | Clearly MATH is invariant and horizontal, hence basic. Let MATH be invariant sections of MATH and set MATH. If MATH is integrable then we have seen above that the NAME connection is of this form. Therefore it suffices to prove that MATH iff MATH on MATH. Since MATH this is a straightforward computation. Let MATH be any vector field on MATH. Then MATH and so MATH . Since the right hand side is vertical, it follows that MATH iff MATH . If MATH is parallel to MATH, this holds automatically since MATH, and so by considering MATH we obtain the theorem. |
math/9911117 | By the mini-Kerr theorem MATH admits a shear-free geodesic congruence. The divergence MATH and twist MATH are monopoles on MATH, which may be used to construct the desired hypercomplex and scalar-flat NAME spaces wherever they are nonvanishing. The hyperKähler case was characterised above by the constancy of MATH and MATH. On MATH, this implies that MATH and MATH are linearly dependent, that is, MATH for constants MATH and MATH. Conversely given an NAME space with a shear-free geodesic congruence MATH whose divergence and twist satisfy this condition, any nonvanishing monopole MATH with MATH and MATH gives rise to a hyperKähler metric (and this MATH is unique up to a constant multiple unless MATH). |
math/9911120 | Let MATH be a tubular neighborhood of MATH in MATH (MATH), and consider a REF-handle added along MATH, that is MATH and a homeomorphism MATH. Then MATH. Let MATH be a natural embedding, then MATH is an epimorphism, REF (any link in MATH can be pushed (ambient isotoped) to MATH). Furthermore any skein relation can be performed in MATH. The only difference between KBSM of MATH and MATH lies in the fact that some nonequivalent links in MATH can be equivalent in MATH; the difference lies exactly in the possibility of sliding a link in MATH along the added REF-handle (that is MATH is moving from one side of the co-core of REF-handle to another); compare REF . The proof of REF is completed. |
math/9911120 | The regular neighborhood, MATH, of MATH in MATH can be projected into REF-disk MATH (then MATH), and we use MATH to present link diagrams, compare REF . In MATH one has sliding relations described in REF (with blackboard framing). These relations can be written as MATH, where MATH is a link in MATH in general position with MATH and cutting it MATH times; CASE: After simplifying the formula, using the NAME bracket skein relations, one gets: MATH and finally MATH that is MATH is a linear combination of links with a smaller than MATH intersection number with REF-sphere MATH. For MATH invertible in MATH, one can eliminate MATH from the set of generators. Using induction, one can eliminate all elements of MATH which cut REF-sphere MATH non-trivially. Thus MATH is an epimorphism. |
math/9911120 | CASE: Because MATH is a REF-disk and MATH, obtained from MATH by adding a REF-handle along MATH, is a REF-disk with a hole, therefore adding this REF-handle does not change the KBSM (LemmaREF ). Therefore in MATH and MATH any relation of the form MATH holds. The embedding MATH induces the homomorphism of the KBSM MATH, thus MATH. REF follows, as assumptions of the lemma are chosen in such a way that any allowed sliding in MATH is also a sliding in MATH. CASE: MATH, being a trivial knot, can be isotoped into MATH without changing an ambient isotopy type of the result of the sliding, which will be also a trivial knot by REF. CASE: Any link MATH in MATH can be written in MATH as a linear combination of elements of MATH. A sliding described in REF does not depend on the presentation of MATH (or elements of MATH) so the lemma folllows. Notice that the sliding relation of REF performed on the link MATH disjoint from MATH holds already in MATH. CASE: Let MATH be a realization of a link MATH in MATH. As MATH is arbitrary, one can assume that sliding has support in MATH. Using relations from MATH and the conclusion of REF together with REF we can see that our slidings are performed on links in MATH and have support in MATH, so by REF they do not introduce any new relation. To visualize the assertion that modulo MATH we need to slide only links MATH in MATH with sliding support in MATH, consider disks MATH and MATH with MATH. Let MATH be an arbitrary link in MATH and MATH a sliding with support in MATH. NAME along MATH and MATH of the type described in REF , yield relations in MATH satisfied in MATH (as MATH is parallel to MATH). These slidings allow us to reduce MATH to a linear combination of links (curve systems) in MATH. NAME the support of sliding MATH is (unchanged) in MATH. |
math/9911120 | By the handle sliding lemma REF one has MATH where MATH is the submodule of MATH generated by slidings MATH. By REF MATH for any generating set MATH of MATH. We can assume that MATH and MATH for an arc MATH properly embedded in the disk with MATH holes, MATH. Let MATH be the basis of MATH as described in REF . Let MATH be a subset of MATH composed of links with geometric intersection number with MATH equal to MATH. By REF , MATH is generated by sliding relations of REF , one relation for each element of MATH for MATH. Because MATH is a basis of MATH, therefore MATH is a basis of MATH. On the other hand MATH is also a basis of MATH, thus MATH is an isomorphism. |
math/9911120 | MATH can be obtained from MATH by adding to MATH some REF-handles (disjoint from REF-handle added along MATH) and some REF-handles. By REF MATH is obtained from MATH by sliding links generating MATH along these REF-handles. Denote these slidings by MATH. Consider any link MATH in MATH and any sliding MATH. We can choose a representative MATH of MATH so that MATH and MATH are identical in the neighborhood of MATH. By REF we can present MATH in MATH as a linear combination of links which are disjoint from MATH and differ from MATH only in a small neighborhood of MATH. Thus the sliding relation MATH is a linear combination of sliding relations in MATH. Therefore MATH is an isomorphism. The proof of REF is complete. |
math/9911120 | REF follows immediately from REF . REF corresponds to the case of REF when MATH separates MATH. MATH and MATH differ only by parts of their boundaries so their KBSM are the same. REF follows from REF by REF . REF follows from REF as MATH is a REF-disk with two holes. REF is also a special case of a general theorem in CITE. |
math/9911120 | Let the set MATH be a basis of MATH. By the Universal Coefficient Property it is also a basis of MATH. By REF the set MATH is a basis of MATH. Therefore it is a MATH linearly independent set in MATH, and therefore MATH is a monomorphism. |
math/9911124 | Since MATH is connected, it remains to prove that MATH is trivial. This follows by appealing to the homotopy exact sequence, and the fact (CITE, CITE) that MATH has trivial MATH group. |
math/9911124 | We claim that in this case the homomorphism MATH identifies with MATH. Then it is clear that MATH is surjective. In order to prove our claim, we shall see first that if MATH and MATH denote respectively the center valued traces of MATH and MATH, then MATH. Indeed, let MATH, then MATH is the (norm) limit of a sequence of elements MATH in MATH (where MATH denotes the convex hull of MATH). Since MATH, it follows that MATH. Now let MATH be an element in MATH with MATH, then there exists a projection MATH such that MATH. Under the identification MATH, the class of the loop MATH in MATH corresponds to the element MATH, MATH sends this element to the class of MATH in MATH, that is, to MATH (see CITE). Therefore MATH. The fact that MATH is simply connected follows from the exact sequence, where MATH. |
math/9911124 | As before, one needs to show that if MATH is the center valued trace of MATH, then MATH. First, pick MATH of the form MATH with MATH mutually orthogonal and MATH. Given MATH sufficiently small, we claim that there exists projections MATH such that MATH. Indeed, otherwise there would be a central projection MATH and an interval MATH such that between MATH and MATH there are no values MATH with MATH projection in MATH. In that case, put MATH . Clearly, MATH. Then one can find sequences MATH and MATH, where MATH decreases to MATH and MATH are projections in MATH with MATH, such that MATH. Then MATH is a projection in MATH satisfying MATH. Suppose now that there exists a projection MATH in MATH with MATH. Let MATH be the spectral projection of MATH associated to the interval MATH, lying in MATH. Then clearly MATH is a projection in MATH satisfying that MATH. Since MATH increases to MATH, there exists MATH such that MATH is non zero, this implies a contradiction with the fact that MATH is the supremum of the above set. Therefore no such MATH should exist, which in turn would imply that MATH is a minimal projection in MATH. Since MATH is maximal abelian in MATH of type MATH, it has no minimal projections, and we arrive to a contradiction. Returning to our original central element MATH, it follows that we can find projections MATH in MATH with MATH. Since MATH and these projections are mutually orthogonal, it follows that MATH is a projection in MATH. Moreover, we have that MATH. Then we can construct an increasing sequence MATH of projections in MATH such that MATH converges to MATH. Pick MATH, clearly MATH. Now, if MATH is any element in MATH with MATH, let MATH be an increasing sequence of positive elements in MATH with finite spectrum, converging to MATH in norm. We can find projections MATH in MATH with MATH. Then MATH is an increasing sequence of projections, put MATH. We obtain that MATH, and the proof is complete. |
math/9911124 | Let MATH be the projections of the center of MATH decomposing it in its type MATH parts, MATH. Pick MATH, and put MATH. Suppose that for each MATH we can find MATH in MATH with MATH. Then MATH is a projection in MATH such that MATH. Therefore it remains to prove our statement in the case MATH of type MATH. Indeed, note that MATH is the centralizer of the (faithful and normal) state MATH of MATH, which is simply the restriction of MATH to MATH. Let now MATH be a minimal abelian projection in MATH. Again, pick MATH in MATH. Now MATH is of type MATH, and the state MATH of MATH given by the restriction of MATH to this algebra has centralizer equal to MATH. By the lemma above, there exists a projection MATH such that MATH where MATH is the center valued trace of MATH, that is, MATH. Since MATH is of type MATH, it follows that MATH. Taking trace in the above equality yields MATH and the statement follows. |
math/9911124 | As remarked before, one can restrict to the case when MATH is of type MATH, for a fixed MATH. In this case, MATH equals MATH, where the isomorphism is implemented by the map sending the class of the curve MATH to the continuous map MATH, for MATH a projection in MATH. In other words, MATH identifies with elements MATH which are of the form MATH, with MATH mutually orthogonal in MATH and MATH are integers. The proof follows, recalling that MATH, and therefore MATH, that is, MATH lies in the image of MATH. |
math/9911124 | As noted at the beginning of the section, it suffices to prove the statement in the case when MATH is of a definite (finite) type. Type I case was dealt in REF. Suppose that MATH is of type MATH. Then MATH is finite, and there exist two projections MATH in MATH such that MATH, MATH is of type I and MATH is of type MATH. In CITE it was shown that if MATH is a projection in a NAME algebra MATH, then the unitary orbit MATH is simply connected. This unitary orbit is the base space of a fibration of MATH with fibre MATH where MATH. In other words, the quotient MATH is simply connected. In our case we have that MATH is simply connected. The inclusion MATH can be factorized MATH . The inclusion MATH induces an epimorphism of the MATH groups, by REF . The same happens with the inclusion MATH, by REF . The last inclusion MATH also induces an epimorphism of the MATH groups by the remark above. Therefore MATH is simply connected also in this case. |
math/9911124 | Since the index of MATH is infinite CITE, CITE, there exist elements MATH with MATH, MATH and MATH as MATH tends to infinity. It is straightforward to verify that the distance MATH does not tend to zero with MATH. Let MATH be unitaries such that MATH. Then MATH . Therefore the sequence of the classes of the elements MATH tends to the class of MATH in the modular topology. We claim that MATH does not tend to MATH in the usual topology (induced by the norm of MATH). Suppose not. Then there exist unitaries MATH such that MATH. Then MATH . This implies that MATH, and therefore MATH, an absurd. |
math/9911124 | It only remains to prove that if the above mapping has local cross sections, then the index of MATH is finite. The existence of local cross sections implies that the bijective and continuous map induced in the quotient, MATH is open, and therefore a homeomorphism. On the other hand it holds in general CITE that this same bijection is a homeomorphism between MATH and the modular topology in MATH. It follows by the proposition above, that the index of MATH is finite. |
math/9911124 | Since the space considered is homogeneous and the action of the unitary group is continuous, it suffices to consider continuity at the class MATH of MATH. First, note that MATH in the norm topology of MATH if and only if MATH in MATH. Then it is clear that the map MATH is continuous in the norm topology of MATH. Therefore the map induced in the quotient MATH, that is, MATH, is continuous in what we are calling the usual topology of the quotient. In the modular topology, as noted above, MATH is homeomorphic to the orbit MATH in the norm topology. Therefore MATH if and only if MATH. This implies that MATH. Since the mapping MATH, MATH is a *-isomorphism, it follows that MATH, that is MATH in MATH. In order to see that MATH is a homeomorphism with the modular topology, suppose that MATH such that MATH. Therefore there exists MATH such that for MATH is invertible in MATH. Since MATH is finite, it follows that also MATH is invertible, which implies that MATH is invertible. Then the unitary part MATH of MATH REF satisfies that MATH. Indeed, note that MATH, and then MATH. On the other hand, MATH in MATH in the modular topology if and only if there exist unitaries MATH such that MATH. Put MATH as above, then MATH which tend to zero, and therefore MATH is a homeomorphism in the modular topology of MATH. |
math/9911124 | The second inequality is obvious. In order to prove the first note that for any MATH, MATH . Note that if MATH is unitary, by the NAME inequality we have that MATH and MATH. Applying these inequalities we obtain MATH and MATH . Note that MATH, and analogously for the other term. Thus we obtain MATH . |
math/9911124 | One has continuous local cross sections MATH defined on a neighborhood MATH of MATH. If MATH is a NAME net in MATH, choose MATH such that for MATH, MATH. Then since MATH is continuous, MATH is a NAME sequence in MATH in the norm topology, therefore convergent to a unitary MATH in MATH. Then clearly MATH converges to MATH. |
math/9911124 | An argument similar to the one in the previous proposition, shows that a NAME sequence MATH for the modular topology yields another sequence of unitaries MATH in MATH such that MATH and MATH form a NAME sequence in MATH for the norm MATH (this is clear using the continuous cross sections available for this topology as well). Therefore MATH converge to some element MATH in the unit sphere of MATH. Now using the above result, MATH and therefore MATH is a NAME sequence in MATH. Then MATH converges to a normal state MATH. Then MATH converges to MATH. On the other hand, MATH, which converges to MATH by the continuity of the scalar product. Suppose now that MATH is finite, and fix a faithful tracial and normal state MATH. Pick MATH with MATH a limit of unitaries MATH as above. Then if MATH, one has that MATH, which implies that MATH tends to zero in the norm MATH. In other words, MATH. Note that MATH, and therefore MATH, that is, MATH. |
math/9911124 | Represent MATH and MATH in MATH as in the basic construcion. As in the proposition above, one needs to show that if MATH are unitaries in MATH converging to MATH in the norm MATH, and MATH such that MATH, then it must be MATH. We claim that under this hypothesis MATH tends to zero in the strong operator topology. Note that MATH tends to zero in norm, and therefore MATH for all MATH. On the other hand, if MATH then also MATH in the norm of MATH. Indeed, MATH. Since MATH and MATH generate MATH, it follows that MATH tends to zero in norm for any MATH. The claim is proven, using that the sequence is bounded in norm, and the fact that MATH is dense in MATH. By the previous lemma, MATH is finite, and therefore MATH in the strong operator topology (see CITE). Again using that the sequence is bounded, one has that MATH strongly, that is MATH. Then MATH, which implies that MATH, and therefore MATH. |
math/9911128 | Let MATH be a MATH-dimensional metrizable compactification of MATH. By CITE, there exists a MATH-set MATH, containing MATH, so that CASE: MATH is MATH; CASE: MATH; CASE: For every at most MATH-dimensional Polish space MATH the set of all closed embeddings is dense in MATH. Let us show that MATH. Indeed, let MATH be a map defined on the boundary MATH of the MATH-dimensional disk MATH, MATH. According to CITE, there exists an open cover MATH such that the following condition is satisfied: CASE: If a MATH-close to MATH map MATH, MATH, has an extension MATH, then MATH also has an extension MATH. Since MATH is MATH, it follows by CITE that MATH is locally MATH-negligible in MATH. According to CITE we can find a map MATH which is MATH-close to MATH. Since MATH, there exists an extension MATH of MATH. The above stated property MATH of the cover MATH guarantees that MATH also has an extension MATH. This shows that MATH. Therefore MATH is a MATH-dimensional, separable, completely metrizable MATH-space satisfying property MATH. Topological characterization of the NAME space (see REF) implies that MATH is homeomorphic to MATH. The fact that the set the set MATH is dense in the space MATH follows from CITE. Let MATH be a MATH-subset of MATH such that MATH. Since MATH it follows that the set MATH is dense in MATH. Consequently, MATH is a MATH-subset of MATH. This proves REF . Next observe that since MATH is homeomorphic to MATH it can be identified with the pseudo-interior MATH of the universal MATH-dimensional NAME compactum (see CITE). Let MATH be a MATH-subspace of MATH containing MATH. By REF and CITE, the inclusion MATH is a near-homeomorphism. In particular, MATH is homeomorphic to MATH. This proves REF . REF are proved similarly. |
math/9911128 | It is showm in the proof of REF that there exists an embedding of MATH-absorbing set MATH into MATH with locally MATH-negligible complement of the image. Now observe that the complement MATH as a MATH-set CITE in MATH is locally MATH-negligible in MATH. This obviously implies that the complement MATH is also MATH-negligible in MATH as required. |
math/9911128 | Let MATH be a MATH-dimensional separable completely metrizable space containing MATH as a subspace in such a way that MATH is MATH see CITE. As in the proof of REF , we conclude that CASE: the set MATH is dense in MATH. Next we need the following observation. Claim. A compact subset MATH of MATH is a MATH-set in MATH if and only if MATH is a MATH-set in MATH. Proof of Claim. First let MATH be a MATH-set in MATH. Consider a map MATH and open covers MATH such that MATH refines MATH. By MATH, there exists a MATH-close to MATH map MATH such that MATH. Since MATH is a MATH-set in MATH, there exists a MATH-close to MATH map MATH such that MATH. Since MATH is MATH-close to MATH, it follows that MATH is a MATH-set in MATH. Conversely, let MATH be a MATH-set in MATH. Consider a map MATH and open covers MATH so that MATH refines MATH. For each MATH choose an open subset MATH such that MATH. It is easy to see that MATH is a MATH-set in MATH. Consequently there exists a MATH-close to MATH map MATH such that MATH, where MATH. Let MATH be an open subsets of MATH such that MATH and MATH. By MATH, there exists a map MATH which is MATH-close to MATH. Obviously, MATH is MATH-close to MATH and MATH. This shows that MATH is a MATH-set in MATH and completes the proof of claim. We continue the proof of REF . Let MATH be a compact subset of MATH. Clearly MATH is a compact MATH-set in MATH for each MATH. By the above Claim, MATH is a MATH-set in MATH. This means that the set MATH is open and dense in the space MATH. Since MATH is completely metrizable, the space MATH has the NAME property (see, for instance, CITE) and consequently, the set MATH is also dense in the space MATH. This simply means that MATH is a MATH-set in MATH. By the above Claim, we conclude that MATH is a MATH-set in MATH as well. |
math/9911128 | Let MATH and MATH be MATH-absorbing sets. REF guarantees that MATH. Embed MATH and MATH into a copy MATH of the universal MATH-dimensional NAME space MATH in such a way that REF - REF are satisfied. The rest of the proof follows the argument presented in the proof of CITE (use the MATH-set Unknotting Theorem for MATH instead of the MATH-set unknotting theorem for MATH at the appropriate place). |
math/9911131 | For some computational convenience we consider, instead of the holomorphic differential in REF , the anti-holomorphic differential MATH . Let MATH. By the definition of MATH-operator we have MATH . Thus the first order term in MATH in MATH is MATH . We recall a formula in CITE (see REF ) MATH . Therefore REF is MATH by REF . Summarizing we find MATH which is the desired formula. |
math/9911131 | It follows from the NAME expansion that MATH where MATH is the reproducing kernel of the subspace MATH of MATH with signature MATH with the NAME MATH and MATH are positive constants; see CITE. Performing the inner product in the NAME space of the element MATH with the function MATH and using REF we find that MATH namely, MATH . We take now MATH. Recall REF , that MATH . Substituting this into the previous formula we get MATH . Since MATH is a polynomial in MATH and MATH of degree MATH, we see that MATH is of the declared form. If MATH, we can then calculate MATH further. Expand MATH again using REF . We have MATH because MATH is of degree MATH and it is orthogonal to those terms of lower degree. But MATH where the rest term is of lower order. Therefore by the same reason and by the reproducing property, MATH . Substituting this into REF we then get REF . |
math/9911131 | We estimate the norm of MATH in MATH by using the above Corollary. The polynomial MATH on MATH is bounded, say MATH. We have MATH . By REF we see that MATH, thus the above integral is finite (see CITE), namely the function is in the MATH-space. That MATH is in MATH follows directly from REF . |
math/9911131 | We use the addition formulas in CITE, Appendix, for the quasi-inverses. As special cases we have MATH and MATH . The first order term in MATH in REF is easily seen to be MATH which proves the first formula in REF . Similarly we can calculate the first order term in REF and prove the second formula; using this formula and MATH we get then REF . |
math/9911134 | An examination of the proof of CITE shows that the map MATH of MATH into MATH is the map discussed at the beginning of CITE. Since second countable transformation groups are automatically quasi-regular (CITE; see CITE for a more general result), it follows from CITE that every second countable transformation group MATH with MATH amenable is NAME regular in the sense of CITE. This applies in particular when MATH is abelian, so the present theorem is a restatement of CITE as it applies to second countable abelian transformation groups; that NAME 's quotient MATH of MATH can be viewed as a quotient of MATH is observed before CITE. |
math/9911134 | For each MATH, we have MATH, so MATH implies that MATH in MATH for every prime MATH. Clearly MATH implies that MATH for at least one prime MATH, and, for such a prime, MATH implies MATH, because MATH is a field. The second assertion is obvious. |
math/9911134 | It is clear that MATH holds, so we concentrate on proving MATH. Let MATH belong to the right-hand side, and notice that it is enough to show that MATH for some integer MATH. We may choose MATH such that MATH, and then we can easily write down a typical basic open neighbourhood of MATH: MATH for MATH a finite set of primes and MATH open sets in MATH. We can further assume that MATH without changing its orbit or the set of primes MATH for which MATH vanishes. Let MATH, and notice that for every MATH we can write MATH for some unit MATH. Then MATH is an open set in MATH, and hence contains a positive integer MATH; we then have that MATH. We now choose MATH large enough to ensure that the ball MATH in MATH is contained in MATH. (Notice that because MATH is open in MATH, and MATH is nonzero in MATH, we have that MATH is open in MATH.) By the NAME Remainder Theorem, there is an integer MATH such that MATH . But now for each MATH, MATH . Next define MATH and notice that for all MATH we have MATH, so that MATH. If MATH, in which case MATH, then MATH, so clearly MATH. If MATH, then MATH is coprime to MATH, and MATH embeds as a MATH-adic integer (that is, MATH), and hence MATH. Thus MATH, and MATH. |
math/9911134 | Let MATH be the quasi-orbit map of MATH to MATH. It follows easily from the characterisation of orbit closures that the map MATH factors through MATH, and that MATH is a bijection. By the Lemma on page REF we know that the map MATH is continuous and open. Hence the sets MATH, as MATH runs through a basis for the topology on MATH, form a basis for the quotient topology on MATH. By definition, a typical basic open neighbourhood of MATH in MATH is MATH with MATH a product of the form MATH in which MATH is a finite subset of MATH and MATH is an open neighbourhood of MATH in MATH for every MATH. After relabelling MATH as MATH for those (finitely many) MATH for which MATH, and enlarging the set MATH accordingly, we see that MATH is a product of the same form as MATH. Since the bijection MATH carries MATH to MATH, we want to prove that the collection MATH with MATH as above is precisely the basis MATH for the topology on MATH. Given MATH as above, the set MATH is finite, and every finite subset of MATH arises for some MATH. Thus it suffices to prove that for every MATH as above, MATH . First we prove MATH: MATH . To prove MATH, let MATH and suppose MATH is a subset of MATH such that MATH. We need to find MATH such that MATH. If MATH choose MATH; if MATH but MATH, choose MATH; finally, if MATH, simply take MATH. It is clear that MATH. To see that MATH, notice that MATH for MATH because MATH, and obviously MATH for MATH. |
math/9911134 | This follows from a direct application of REF, using the characterisation of the quasi-orbit space given by REF . |
math/9911134 | The space MATH is a MATH - MATH imprimitivity bimodule. If MATH is a representation of MATH, the corresponding representation MATH acts on MATH via MATH . But the map MATH defined by MATH is unitary, and intertwines the natural left action of MATH with the restriction of MATH. Thus MATH is equivalent to the compression of MATH to its essential subspace MATH. The result therefore follows from standard properties of NAME equivalence CITE. |
math/9911134 | The preceding discussion shows that the map of MATH onto MATH is the composition of the homeomorphism of MATH onto MATH of REF with the homeomorphism of REF; in particular, it is a homeomorphism. |
math/9911134 | Since the operations in MATH are componentwise, REF is trivially equivalent to REF . Suppose REF holds, let MATH be a finite subset of MATH, and let MATH be an open subset of MATH for each MATH, so that MATH is a basic open set in MATH. Let MATH, and define MATH by MATH . Then MATH is a unit, MATH divides MATH, and MATH for MATH. Since multiplication by a unit is a homeomorphism, MATH is an open set, and since the canonical embedding of MATH is dense in MATH, there exists MATH such that MATH. Thus MATH for every MATH, so MATH, proving REF . Conversely, suppose MATH for some MATH. Then MATH is contained in the proper closed set MATH and cannot be dense, so REF implies REF . |
math/9911134 | By CITE, MATH is faithful if and only if MATH is faithful on MATH. Since the kernel of MATH in MATH is MATH, REF follows from REF. Next we prove REF . If MATH for some MATH, then MATH implements the equivalence between MATH and MATH. Now suppose that MATH is unitarily equivalent to MATH; we have to show that MATH and MATH lie in the same MATH-orbit. Denote the canonical conditional expectation from MATH to MATH by MATH; then MATH is the vector state MATH. If MATH is unitarily equivalent to MATH, then there is a unit vector MATH in the NAME space of MATH such that MATH. Write MATH, where the sum is over the set MATH, and choose a finite subset MATH of MATH such that MATH. If MATH there exists MATH with MATH such that MATH and MATH for every MATH. But then MATH which is a contradiction. So MATH has to be in MATH. |
math/9911134 | We may assume that MATH without changing the orbit. Suppose now MATH is in the closure of MATH, and let MATH be a sequence of nonzero rationals such that MATH as MATH. Fix MATH. Since MATH is a neighbourhood of MATH in MATH, there exists MATH such that for every MATH, MATH . Then for every MATH we have MATH . Since MATH, MATH implies MATH. On the other hand, MATH implies MATH . Since MATH is an integer, the sequence MATH must be eventually constant, and hence MATH. |
math/9911134 | The inclusion MATH is clear, and we only need to show that if MATH is zero whenever MATH is zero, then MATH can be approximated from within the orbit of MATH. Since MATH is the restricted product of the MATH relative to the MATH, there is an integer MATH such that MATH for all MATH, and MATH has the same orbit and the same zeros as MATH. So we may suppose without loss of generality that MATH. The same argument shows that we might as well assume that MATH. Suppose then that MATH satisfies MATH if MATH. As a subspace of MATH, the set MATH has the product topology, so a typical basic open neighbourhood of MATH has the form MATH where MATH is a finite subset of primes and MATH is an open interval in MATH. Let MATH; notice that whenever MATH we trivially have MATH for all MATH. We want to find MATH such that CASE: MATH for MATH, CASE: MATH for MATH, CASE: MATH. For MATH, write MATH as MATH for some unit MATH, and choose MATH such that MATH. Then MATH, and since MATH is open in MATH, there exists MATH such that MATH . We will verify REF by finding MATH in this ball. We deal with two cases separately. CASE: Suppose there exists a prime MATH such that MATH. If MATH, the result follows from REF. So we suppose MATH. Let MATH. Choose MATH such that MATH . By the NAME Remainder Theorem, the congruences MATH have a solution MATH, and the set of solutions is then MATH. We claim that if MATH is one of these solutions and we define MATH by MATH then MATH is in the ball of REF. To see this, we compute MATH . This says precisely that every MATH of the form REF lies in the ball of REF, and hence satisfies REF . To see that such MATH also satisfy REF , note that MATH implies MATH. If MATH, then MATH because MATH does not divide the denominator. Hence MATH for all MATH. Now recall that we chose MATH large enough to ensure that every interval of length MATH contains at least one of the elements of MATH . Therefore there exists MATH such that the corresponding MATH satisfies MATH. This MATH satisfies REF . CASE: Suppose MATH for every MATH. Then, because MATH is not invertible, there must exist infinitely many primes dividing MATH. Let MATH. We can choose a set MATH of primes disjoint from MATH such that every MATH divides MATH, and such that MATH . A similar argument to that of Case I, with MATH replaced by MATH, now yields a rational number MATH satisfying REF . |
math/9911134 | The result is a direct application of REF. |
math/9911134 | Every subset of MATH is the vanishing set of a noninvertible adele, and every MATH is equal to MATH, so MATH maps MATH onto MATH. It remains to show that MATH if and only if MATH for every MATH, MATH in MATH. Suppose MATH. This implies that either MATH and MATH are both invertible, or else they are both not invertible. CASE: If they are both invertible, we have MATH hence MATH, from which MATH. CASE: If they are both not invertible, then MATH and clearly MATH. Suppose now MATH. This also implies that either MATH and MATH are both invertible, or else they are both not invertible. CASE: If they are both invertible, we can write MATH for some MATH, and then MATH because MATH. CASE: If they are both not invertible, then MATH and from this it follows that MATH if and only if MATH, so MATH. This completes the proof. |
math/9911134 | The maps MATH for MATH and MATH are all continuous on MATH, so for each finite subset MATH of MATH the product MATH is continuous, and, the net MATH directed by the finite subsets of MATH under inclusion converges pointwise to the absolute value. For each finite MATH and each choice of MATH for MATH, the set MATH is open in MATH, and every adele belongs to some MATH (given MATH, it suffices to choose MATH and MATH). Moreover, for every MATH, if MATH then MATH. It follows that the tail net MATH is nonincreasing when restricted to MATH, because all the factors greater than MATH have been included in MATH already. Thus, the restriction of MATH to MATH is the infimum of the restricted tail net and hence is upper semicontinous on MATH. Since each MATH is open and their union is all of MATH, this completes the proof. |
math/9911134 | By REF, MATH is closed, and the preceding discussion establishes that MATH is a continuous bijection, so it only remains to show that the inverse is continuous. Suppose MATH in MATH; we need to show that MATH converges to MATH in MATH and that MATH is eventually equal to MATH. Dividing everything by MATH, we may assume that MATH, and show that MATH is eventually MATH. For each MATH, let MATH be the MATH-th prime, and let MATH where MATH is an interval in MATH containing MATH. For each MATH, MATH is eventually in MATH, and then MATH . It follows that MATH is eventually an integer, and that either MATH or MATH (because MATH is the smallest possible prime factor of MATH). Since MATH converges to MATH, we know that MATH converges to MATH. But MATH is bounded away from MATH, so MATH has to remain bounded, and this can only happen if MATH is eventually MATH. |
math/9911134 | First we show that the power-cofinite closure of MATH is always contained in MATH. Let MATH be in the power-cofinite closure of MATH; we will show that that every (basic) open set containing MATH intersects MATH. We can assume that the basic open set is MATH for MATH, where MATH is an open subset of MATH for every MATH in the finite subset MATH of MATH, and that MATH for MATH. We need to find an adele MATH such that CASE: MATH for MATH and MATH for MATH (so that MATH); CASE: MATH is not invertible in MATH (so that MATH); CASE: MATH. The set MATH is finite and disjoint from MATH, because MATH. Thus MATH, and because MATH is in the power-cofinite closure, there exists MATH - that is, MATH. Choose MATH as follows: MATH . Then MATH is not invertible (even if MATH) and MATH. This proves that MATH. Next we show that if MATH is not power-cofinite dense in MATH then the power-cofinite closure MATH of MATH contains MATH. Because MATH is not dense, it misses some basic open set MATH, and MATH. We claim that MATH is also MATH-closed. To see this, write MATH and then observe that the set MATH is a finite union of MATH-invariant closed sets. Thus MATH is MATH-closed, as claimed. This implies that the MATH-closure of MATH is contained in MATH, and indeed is contained in MATH . If MATH is power-cofinite dense in MATH, then MATH is in the power-cofinite closure of MATH. The initial paragraph of the proof, with MATH, shows that MATH; we will show that MATH itself is MATH-dense. Let MATH be any open set in MATH. Choose a non-invertible adele MATH such that MATH for every MATH. REF implies that MATH contains MATH for some MATH. Then MATH. Thus MATH is MATH-dense, and we must have MATH. This completes the description of MATH. Next let MATH, and assume first that MATH. We claim that every singleton MATH belongs to MATH. There is a strictly increasing sequence MATH such that each interval MATH contains MATH for some MATH, and for each MATH we choose one such MATH. Since MATH, we have MATH in MATH. Moreover, for all MATH we have MATH for every finite prime MATH and MATH. Since MATH is compact, we can assume by passing to a subsequence that MATH for every MATH; then MATH vanishes precisely when MATH, and hence MATH. By continuity of MATH, we know that MATH in MATH, and, since MATH, it follows that MATH, as claimed Since every singleton MATH is in MATH, MATH meets every basic open neighbourhood MATH of MATH in MATH, and MATH. Since we have already seen that MATH is MATH-dense, it follows that MATH. Assume now that MATH for every MATH, and let MATH. The inclusion map MATH is continuous, so the closure MATH of MATH in MATH is certainly contained in the MATH-closure of MATH in MATH. On the other hand, because the inverse of MATH is continuous by REF, the set MATH is closed in MATH. Since it is also MATH-invariant, its image MATH is closed in the quotient topology; in other words, it is MATH-closed. Thus MATH, and we conclude that MATH. |
math/9911135 | See for example, REF for the first part. We only need to prove the assertion about extending MATH. Since MATH, given MATH there exist MATH such that MATH, and hence the element MATH is in MATH, proving that MATH is directed by the relation defined by MATH if MATH. An easy argument shows that MATH defines a group homomorphism from MATH to MATH that extends MATH. |
math/9911135 | Verbatim from the proof of CITE, except for the following minor modification of the part of the argument where normality is used to obtain an admissible value for the function MATH. The value MATH used there has to be substituted by any (fixed) MATH, and thus the fourth paragraph there should be replaced by the following one. Suppose now that MATH and MATH, and consider the function MATH defined by MATH for MATH. If MATH is admissible for MATH, let MATH. We will show that MATH is admissible for MATH. For every MATH, MATH, and since MATH is admissible for MATH where the second equality holds by the multiplier property applied to the elements MATH, MATH, and MATH in MATH. This proves that MATH is admissible for MATH, so MATH. |
math/9911135 | The proofs of all but the last statement about uniqueness are as in REF, provided one considers the left-quotients MATH instead of the right-quotients used there. In order to prove the uniqueness statement suppose MATH is another unitary MATH-representation such that MATH and MATH is dense in MATH. It is easy to see that the map MATH is isometric, and that it extends to an isomorphism of MATH to MATH because of the density condition. It only remains to show that MATH intertwines MATH and MATH. Since MATH is an NAME semigroup, for every MATH and MATH in MATH there exist MATH and MATH in MATH such that MATH. Then MATH, so MATH . This shows that MATH for every MATH, hence for every MATH. |
math/9911135 | By right - reversibility, MATH is directed by MATH so one may follow the argument of CITE. However, extra work is needed here: since MATH need not be abelian, the choice of embeddings in the directed system must be carefully matched to the choice of right-order MATH on MATH. Consider the directed system of C*-algebras determined by the maps MATH from MATH into MATH, for MATH and MATH, that is, for MATH in MATH. By CITE there exists an inductive limit C*-algebra MATH together with embeddings MATH such that MATH whenever MATH, and such that MATH is dense in MATH. The next step is to extend the endomorphism MATH to an automorphism of MATH. For any fixed MATH the subset MATH of MATH is cofinal, so MATH is also the inductive limit of the directed subsystem MATH, and, for this subsystem, we may consider new embeddings MATH defined by MATH for MATH and MATH. By CITE there is an automorphism MATH of MATH such that MATH for every MATH. Since MATH and MATH, the choice MATH gives MATH so that REF holds with MATH and MATH. Since MATH, REF also holds. Uniqueness of the dilated system follows from CITE: MATH is the closure of the union of the subalgebras MATH with MATH, if MATH is another minimal dilation with embedding MATH then there is an isomorphism MATH given by MATH for MATH and hence which intertwines MATH and MATH. |
math/9911135 | We work with the dense subspace MATH of MATH and the dense subalgebra MATH. If MATH there exists MATH such that MATH; assume MATH, since we want MATH to be covariant, the only choice is to define MATH by MATH because MATH restricted to MATH and cut down to MATH has to be equal to MATH. Of course we have to show that this actually defines an operator MATH on MATH for each MATH, that MATH extends to a homomorphism from all of MATH to MATH, and that MATH is covariant. The first step is to define MATH on MATH for a fixed MATH. We begin by fixing MATH, MATH and MATH such that MATH. For MATH with MATH in the cofinal set MATH, we let MATH . If MATH then MATH, and MATH . So the definition of MATH could have been given using any MATH in place of MATH. Next we show that MATH is also independent of MATH and MATH, in the sense that if MATH is also equal to MATH then MATH is equal to MATH for MATH in a cofinal set. To see this let MATH. Then MATH, and since the embedding MATH is injective, it follows that MATH. The map MATH is clearly linear, and since the endomorphisms are injective, MATH. Thus MATH can be uniquely extended to a bounded linear operator (also denoted MATH) on all of MATH such that MATH. For any MATH the map MATH is a *-homomorphism on MATH, and by cofinality of MATH, for any MATH and MATH in MATH there exist MATH and MATH and MATH in MATH such that MATH and MATH. It follows easily from REF that MATH is a *-homomorphism which can be extended to a representation MATH of MATH on MATH. Putting MATH in REF shows that MATH is nondegenerate and there only remains to check that MATH is a covariant pair for MATH. Suppose first MATH and MATH; we can assume that MATH for some MATH and MATH. Let MATH; we can assume MATH, and we observe that MATH. Then MATH and since MATH is dense in MATH and MATH is dense in MATH, the pair MATH satisfies the covariance relation. |
math/9911135 | Let MATH be the projective unitary representation of MATH in the multiplier algebra of MATH, and notice that MATH because MATH is invariant under MATH. Define MATH. Then MATH and MATH, so MATH is a projective isometric representation of MATH with multiplier MATH. Since MATH is generated by the elements MATH, the isomorphism will be established by uniqueness of the crossed product once we show that the pair MATH is universal. Suppose MATH is a covariant representation for the twisted system MATH, and let MATH be the corresponding dilated covariant representation of MATH given by REF. By the universal property of MATH there is a homomorphism MATH such that MATH . Let MATH be the restriction of MATH to MATH, cut down to the invariant subspace MATH. By REF MATH while MATH . Thus MATH and MATH, so MATH is universal for MATH. Finally we prove that the corner is full, that is, that the linear span of the elements of the form MATH with MATH is a dense subset of MATH. It is easy to see that the elements of the form MATH span a dense subset of MATH because MATH, where MATH may be replaced with MATH by minimality of the dilation. Thus the elements MATH with MATH and MATH span a dense subset of MATH, and since MATH, the proof is finished. |
math/9911135 | The embedding clearly intertwines MATH and MATH, in the sense that MATH, and the union of the compact subgroups MATH is dense in MATH, so the union of the subalgebras MATH is dense in MATH, and the result follows from REF. |
math/9911135 | The action of MATH on MATH is by homotheties, which are group automorphisms, so MATH is isomorphic to the crossed product MATH. Moreover, the self-duality of the additive group of MATH satisfies MATH for MATH, thus MATH is covariantly isomorphic to MATH, so MATH is isomorphic to MATH, and the claim follows from REF . |
math/9911137 | MATH. Let MATH, MATH, and let MATH be the homomorphism such that MATH. If MATH and MATH is an inclusion, there is MATH such that MATH. Thus, MATH is a monomorphism. MATH. Let MATH be a homomorphism from the finitely generated module MATH to the finitely presented module MATH. By assumption, there exists a monomophism MATH. Then MATH, and so there is MATH such that MATH. MATH. Suppose MATH is the monomorphism constructed in the proof of the implication MATH. Then MATH. We may take MATH. |
math/9911137 | MATH. According to CITE MATH in MATH. Consider a non-zero homomorphism MATH with MATH and MATH. Suppose MATH; then MATH is a finitely generated subobject of the coherent object MATH. Therefore MATH. Assume that MATH for every MATH. Consider an arbitrary MATH-morphism MATH. Since MATH is a MATH-injective object, there exists MATH such that MATH. But MATH, and so MATH. Whence we obtain that MATH, and thus MATH. We see that MATH, which yields MATH, a contradiction. MATH. By assumption, the module MATH is a MATH-cogenerator. MATH. Since a direct product of MATH-flat modules is a MATH-flat module (see CITE), our statement follows from REF . MATH. Let MATH be an embedding of a module MATH in a MATH-flat module MATH. Denote by MATH. Let MATH, where each MATH. Since MATH is a subobject of the coherent object MATH, the object MATH is coherent itself. Because MATH, each MATH. Consequently, MATH. Let us apply now the left exact MATH-torsion functor MATH to the exact sequence MATH . Since MATH (see CITE), one gets that MATH. Now let MATH; since MATH is a subobject of MATH for some MATH, from the relation MATH we deduce that MATH, whence the ring MATH is right MATH-injective by CITE. MATH. Let MATH; then MATH and since MATH is a MATH-injective object, there is an exact sequence of abelian groups MATH . Because MATH is an epimorphism, we conclude that MATH. Thus MATH, and hence MATH is a monomorphism. MATH. Let MATH; then there is an exact sequence MATH which induces an exact sequence of the form REF. Since MATH, MATH is an epimorphism, and hence a monomorphism. So MATH. By CITE the module MATH is MATH-injective. MATH. This follows from CITE. MATH. Obvious. MATH. This is a consequence of CITE. MATH. Straightforward. MATH. Suppose that MATH is a system of representatives for isomorphism classes of simple MATH-modules. Then every MATH is an indecomposable injective MATH-module and the module MATH is a cogenerator in MATH. Because every module MATH is MATH-flat, the module MATH is MATH-flat by CITE. MATH. Easy. MATH. It suffices to observe that the injective hull MATH of the module MATH is a MATH-flat module. MATH. This is a consequence of CITE. MATH. Since the module MATH is flat, it is MATH-injective, and so is MATH-injective. |
math/9911137 | The proof is similar to that of CITE. |
math/9911137 | By REF we have MATH for every finitely generated right ideal MATH of the ring MATH. From CITE we deduce that MATH for every MATH, that is, MATH is right MATH-injective. |
math/9911137 | MATH is trivial. MATH. Let MATH be an embedding of a module MATH in a flat module MATH. By theorem of CITE the module MATH is a direct limit MATH of the projective modules MATH. By CITE there is MATH such that MATH factors through MATH. Therefore MATH is a submodule of MATH. It remains to observe that MATH is a submodule of some free module. MATH. This follows from CITE. MATH. Obvious. MATH. If MATH is a MATH-injective left MATH-module, then the sequence MATH in which MATH, is pure. Since MATH is a flat module, the module MATH is flat by CITE. Let MATH denote the character module of MATH. By CITE the modules MATH and MATH are injective, and so the exact sequence MATH splits. Consequently, the module MATH is injective, and so MATH is flat by CITE. |
math/9911137 | By assumption, every MATH embeds in a free module (and so in a finitely generated free module as well). One has the following exact sequence MATH in MATH. Since the module MATH is MATH-injective, one gets an exact sequence MATH in MATH, hence MATH. By REF the ring MATH is right coherent. |
math/9911137 | Clearly, an almost regular ring is regular if and only if it is left or right coherent. Therefore our assertion immediately follows from REF . |
math/9911137 | MATH. This follows from CITE. MATH. Apply REF . MATH. Since any left and right MATH-ring is a two-sided MATH-injective ring, our statement follows from REF . MATH. Straightforward. MATH. Since for every cyclic MATH the dual module MATH, the proof of right MATH-injectivity of the ring MATH is similar to that of CITE. Let us show that the ring MATH is right coherent. In view of REF it suffices to prove that MATH for every MATH. We use induction on the number of generators MATH of the module MATH. When MATH, considerating exact sequences REF for MATH, one gets MATH. If MATH is finitely presented on MATH generators, let MATH be the submodule of MATH generated by one of these generators. Since MATH is left coherent, the modules MATH and MATH are finitely presented on less than MATH generators. Because MATH is left MATH-injective, one has an exact sequence MATH where both MATH and MATH are finitely presented by induction. Thus MATH. |
math/9911137 | MATH. This follows from CITE. MATH. Apply REF . MATH. This follows from CITE. MATH. Obvious. MATH. Let MATH be a cyclic finitely presented left MATH-module. In view of REF the module MATH, and so there is an epimorphism MATH. Hence MATH is a monomorphism. MATH. Let MATH be a finitely generated left ideal of the ring MATH. By assumption, the module MATH embeds in a free module. By CITE there exists a finite subset MATH of MATH such that MATH, i. e., MATH is an annulet ideal. By symmetry, every finitely generated right ideal is annulet. MATH. In view of REF it suffices to show that for arbitrary finitely generated right ideals MATH and MATH of the ring MATH the following equality holds: MATH. Since MATH is coherent by assumption, by the NAME theorem CITE both MATH and MATH are finitely generated ideals. One has MATH . Applying MATH, one gets MATH . Thus MATH is MATH-injective. Likewise, MATH is MATH-injective. |
math/9911137 | CASE: It is easy to see that MATH is a left NAME ring. Thus we must show that the module MATH is noetherian. Suppose MATH is a left ideal of the ring MATH. By assumption, the module MATH is a submodule of a free MATH-module MATH for some MATH. Since the ring MATH is left coherent, the module MATH is coherent, and hence the module MATH is finitely presented, that is, MATH is a finitely generated ideal. CASE: It is necessary to observe that over a noetherian ring every finitely generated module is finitely presented and also make use of the first statement. |
math/9911137 | MATH. This follows from REF . MATH. Apply REF . MATH. This is a consequence of REF . MATH. Since MATH is a left noetherian ring, every finitely generated left MATH-module is finitely presented. By REF every finitely presented left MATH-module is a submodule of the module MATH. Because the ring MATH is right coherent, the module MATH is flat by CITE. By REF MATH is a left MATH-ring and by REF MATH is also a left NAME. MATH. In this case the proof is similar to the proof of the implication MATH if we observe that every finitely generated left MATH-module is a submodule of the flat module MATH. |
math/9911137 | MATH. Any left artinian ring is left perfect, and so is right semiartinian by CITE. MATH. Our assertion follows from CITE. MATH. Easy. MATH. Since MATH is a finitely generated left ideal of MATH, our assertion follows from CITE. MATH. Obvious. MATH. By the preceding lemma MATH is left semiartinian and since MATH is left noetherian, our assertion follows from CITE. |
math/9911137 | The implications MATH, MATH are obvious. MATH. Our assertion follows from CITE. MATH. Over a right NAME ring the module MATH for every non-zero cyclic right MATH-module MATH. Since MATH is a left NAME, by REF MATH is a right MATH-injective ring. Therefore MATH is a left MATH-injective ring by CITE. MATH. Apply REF . MATH. By REF the ring MATH is left artinian. Since any left artinian ring is right perfect, our assertion follows from CITE. |
math/9911137 | It is directly verified that MATH is indeed a monomorphism of MATH-modules. |
math/9911137 | Suppose MATH is left MATH-injective and MATH. Then the module MATH and, in view of REF , there exists a monomorphism MATH with MATH some set of indices. Then the composition MATH of morphisms MATH and MATH is a MATH-monomorphism, where MATH, MATH is a unit of the group MATH. Since MATH is a free MATH-module, the MATH-module MATH is MATH-flat by CITE. REF implies that the ring MATH is left MATH-injective. Let us show now that the group MATH is locally finite. Let MATH be a non-trivial subgroup of the group MATH generated by elements MATH. Then the right ideal MATH of the ring MATH generated by the elements MATH is non-zero. By REF MATH, and so MATH is finite by CITE. Now let MATH be left MATH-injective and the group MATH locally finite. To begin, let us show that the ring MATH is left MATH-injective if MATH is finite. Suppose MATH. Because the group MATH is finite, MATH. Since the module MATH is a MATH-cogenerator by REF , by REF MATH is a submodule of MATH, where MATH is some set. By REF MATH is a submodule of MATH. Consequently, MATH is a MATH-cogenerator, and hence MATH is a left MATH-injective ring by REF . Next, suppose that MATH is an arbitrary locally finite group and MATH. Then there is a short exact sequence of MATH-modules MATH . Let MATH be a finite set of generators for MATH. Because MATH is locally finite, there is a finite subgroup MATH of MATH such that MATH. We result in the short exact sequence of MATH-modules MATH where MATH, MATH, MATH is a finite set of generators of the module MATH. NAME sequence REF on MATH, one gets the following commutative diagram with exact rows: MATH in which MATH is an isomorphism, MATH. Clearly, MATH is an isomorphism, and hence MATH is an isomorphism. If we showed that every MATH is extended to some MATH, we would obtain that MATH, as required. So suppose that MATH and MATH, MATH. Consider MATH. Because MATH is a left MATH-injective ring, there is MATH such that MATH. One gets MATH where MATH is the required homomorphism. |
math/9911137 | If MATH is a left MATH-ring, it is also right MATH-injective by REF . By the preceding theorem the group MATH is locally finite. Similar to the proof of REF , given MATH there is a composition of MATH-monomorphisms MATH with MATH. Since MATH is a free MATH-module, the module MATH is a submodule of the free MATH-module MATH. Conversely, let MATH be a left MATH-ring and MATH a locally finite group. First, let us prove that MATH is a left MATH-ring if MATH is a finite group. For this consider MATH. Since MATH is finite, MATH, and so MATH is a submodule of MATH for some MATH. By REF MATH is a submodule of MATH, i.¥., MATH is indeed a left MATH-ring. If MATH is an arbitrary locally finite group, then for any MATH there is a finite subgroup MATH of the group MATH such that MATH with MATH a finite set of generators for MATH (see the proof of REF ). By assumption, the MATH-module MATH is a submodule of a free module MATH for some MATH. Thus, MATH is a submodule of the free module MATH. |
math/9911137 | REF implies that any NAME is a left and right MATH-ring. Therefore our statement immediately follows from REF . |
math/9911143 | Since MATH is a finite set and MATH, every vertex of MATH is eventually periodic, and there is a positive integer MATH such that MATH for every MATH and every positive integer MATH. If MATH is an edge of MATH beginning at MATH, and MATH is the partition of MATH for MATH, then MATH such that MATH. Since MATH is a finite set and MATH is a fixed point of MATH, there is a positive integer MATH such that, for every positive integer MATH, if MATH has partitions MATH for MATH and MATH for MATH, then MATH. This shows that MATH for every positive integer MATH. By the same argument, we can choose a positive integer MATH for MATH such that MATH. Let MATH be the least common multiple of MATH and MATH for each edge MATH, and MATH the least common multiple of these MATH's. Then we have MATH and MATH for every positive integer MATH. |
math/9911143 | For convenience, we will take the metric MATH on MATH so that MATH is the length of the shortest path between MATH and MATH, as explained in REF . Let MATH be the integer given in REF . So each edge MATH has the partition MATH for MATH as in Notation REF. Without loss of generality, we suppose each MATH. Let MATH be the collection of the intervals MATH. First choose MATH so small that CASE: each interval MATH has length larger than MATH, CASE: if MATH, MATH is the initial point of MATH, and MATH, then MATH, and CASE: if MATH, MATH is the terminal point of MATH, and MATH, then MATH. Then choose MATH such that MATH and for every MATH in the compact set MATH and every MATH, if MATH, then CASE: MATH and CASE: the interval MATH contains at most one vertex. Note that MATH comes from the NAME Axiom. If MATH and MATH, then MATH and MATH lie on the same or adjacent elements of MATH. So we have two cases: CASE: MATH and MATH are end points of an interval of length less than MATH containing a vertex MATH, or CASE: the interval MATH of length less than MATH does not contain any vertex of MATH. For REF , by REF , MATH implies that MATH and MATH are contained in two different intervals among MATH, MATH, MATH, and MATH. For brevity, let's assume MATH and MATH. Then by REF , MATH and MATH. If MATH is the vertex of MATH contained in MATH, then MATH maps MATH into MATH. So MATH where MATH and MATH are the expansion constants. Similarly MATH. Let MATH be the smallest positive integer such that MATH or MATH. Then by induction using REF , we have for MATH that MATH sends MATH injectively into MATH and MATH injectively into MATH. Therefore we have MATH. For REF , let MATH be the smallest positive integer such that MATH contains a vertex. It follows from the NAME Axiom that MATH. If there exists MATH, MATH, such that MATH, then we are done, so suppose not. Then MATH and MATH are endpoints of an interval of length less than MATH containing a vertex. Hence we may conclude the proof by applying the argument of REF . |
math/9911143 | For a pair of points MATH, there is a nonnegative integer MATH such that MATH for all MATH. Let MATH and MATH be given in REF . Then MATH and MATH imply that there exists MATH such that MATH by REF . Therefore we have MATH and this proves that MATH is expansive. |
math/9911143 | For the vertex sets MATH and MATH, let MATH be the set of enlarged vertices in MATH and MATH. Then MATH prove the Lemma. |
math/9911143 | By REF , we can choose MATH and MATH such that, for all MATH, if MATH, then there exists a nonnegative integer MATH such that MATH . Choose a nonnegative integer MATH and MATH such that, for MATH and MATH, MATH implies MATH, and MATH implies MATH. Now suppose MATH. From MATH for every nonnegative integer MATH, we have MATH and if MATH and MATH, then for every nonnegative integer MATH . Therefore we have that MATH implies MATH, and equivalently MATH implies MATH. |
math/9911143 | Let MATH be as in REF . Define MATH by MATH. Then MATH is well-defined by our choice of MATH. Now show that MATH for every positive integer MATH. For MATH, MATH . To show the continuity of MATH, let MATH and MATH be as in REF , and choose MATH such that if MATH are elements in MATH with MATH, then there exist MATH with MATH and MATH such that MATH. Then we have MATH and MATH is continuous. That MATH is trivial by the construction of MATH. |
math/9911143 | We will define a sliding block code MATH induced by MATH, and show that MATH is the required conjugacy MATH. Let MATH and MATH denote the edge sets of MATH and MATH, respectively, MATH and MATH the alphabets of MATH and MATH, respectively, and MATH and MATH the semiconjugacies. Then MATH is a subset of MATH such that MATH (ignoring the direction). Note that if MATH, MATH, and MATH, then MATH and MATH for every nonnegative integer MATH. Let MATH and MATH. Then by REF , MATH and MATH are dense in MATH and MATH, respectively. CASE: Show that MATH. By REF , MATH if and only if MATH where MATH is the projection map from the branched solenoids to their MATH-th coordinate spaces. So we have MATH. If MATH for some MATH, then MATH and MATH for some MATH. Since the shift equivalence maps MATH and MATH are graph maps by REF and MATH, we have MATH a contradiction. Hence we have MATH. By the same argument, we can show that MATH. Then MATH and MATH imply that MATH . Therefore we have MATH. Now we have a well-defined bijective map MATH. This map will define MATH on MATH. CASE: Find a block map MATH where MATH is the set of all MATH-blocks in MATH such that for every MATH. For MATH, let MATH and MATH. Then MATH, MATH, and there exists a unique MATH such that MATH . Let MATH be the partition of MATH for MATH, MATH. Then each MATH is contained in a unique MATH, and we have a unique descending sequence MATH such that MATH, MATH, and MATH. Since the shift equivalence maps MATH and MATH are graph maps and MATH, MATH and MATH imply that MATH is contained in a unique edge MATH in MATH and that MATH is contained in a unique path MATH such that MATH and MATH. Define a block map MATH by MATH . Then the sliding block code MATH induced by MATH maps MATH to MATH with MATH. To prove that MATH, we need only show that MATH, that is, MATH. From the descending sequence for MATH is the unique path in MATH such that MATH and MATH. So MATH is contained in MATH, and MATH. Therefore MATH by the definition of MATH, and this proves MATH. CASE: Define a block map MATH with MATH defined on MATH by MATH such that MATH on MATH. We define MATH from MATH just as we defined MATH from MATH in REF . For MATH with MATH, let MATH be the partition of MATH for MATH, MATH, so that MATH and MATH. Then we have a unique descending sequence MATH such that, for MATH, we have MATH and MATH is contained in a unique path MATH in MATH such that MATH. We define MATH by MATH . Let MATH, MATH, and MATH. If MATH and MATH as in REF , then we have MATH and MATH . Therefore MATH is the sliding block code with memory MATH and anticipation MATH induced by a block map defined by MATH and MATH on MATH. CASE: Deduce that MATH gives the required conjugacy MATH. Because MATH maps MATH onto MATH, and these sets are dense in MATH and MATH, it follows that MATH maps MATH onto MATH. Similarly MATH maps MATH onto MATH. Since the continuous maps MATH and MATH agree on the dense set MATH, we have MATH on MATH, and so MATH is a conjugacy. That MATH is a lift of MATH follows because MATH on the dense set MATH. The lifting is unique for it is uniquely determined on the dense set MATH. |
math/9911143 | Clearly every factor of MATH is an element of MATH. We must check that every factor is locally one to one. First consider the case that the image of MATH is a subset of an edge of MATH. Assume that MATH is represented as a product MATH such that each MATH, and MATH for some MATH. Then MATH implies that MATH is not locally one-to-one on the image of MATH, and so MATH is not locally one-to-one. This contradicts the NAME Axiom as we chose the image of MATH to be a subset of an edge. So we have MATH for MATH. Now suppose that MATH contains a vertex MATH of MATH as an interior point. Let MATH be a subset of an edge. Then there is a positive integer MATH and a factor MATH of MATH such that MATH and MATH contains MATH as an interior point. By REF , we have MATH, and MATH is locally one-to-one as MATH is a subset of an edge. Therefore factors of MATH are elements of MATH for all MATH. |
math/9911143 | By REF , MATH or MATH is contained in MATH for every MATH. It remains to check minimality. Suppose that MATH for some MATH. Then by REF , there exists a positive integer MATH such that MATH. If MATH such that MATH and the interior of MATH is contained in an edge MATH, then there exists a subpath MATH such that MATH, and we have MATH is a factor of MATH. Next suppose that MATH where MATH is a path in MATH such that MATH contains a vertex MATH as an interior point. Let MATH be contained in an edge MATH, and MATH. Then for some MATH, MATH is the image under MATH of an interior point of MATH, and by REF , MATH is a factor of MATH. This proves the minimality of MATH, and the uniqueness claim is also clear. It is clear that MATH, and for all MATH there exists a positive integer MATH such that MATH is a factor of MATH. Then the number MATH can be chosen uniformly because for every MATH such that the image of MATH is contained in an edge, if MATH, then MATH is a factor of MATH for every MATH and large MATH. |
math/9911143 | We will show that MATH is part of a shift equivalence, that is, we will define a continuous map MATH and a positive integer MATH such that MATH . Recall MATH for some fixed choice of a subset MATH of MATH as in REF . By the choice of MATH (from REF ), every MATH contains at most one vertex of MATH. It is trivial that if MATH is contained in some MATH and the interior of MATH does not contain any vertex of MATH, then MATH is the unique element in MATH which contains MATH. If MATH is a vertex of MATH, then by REF , MATH or there is a neighborhood MATH of MATH such that MATH is the union of all paths in MATH containing MATH as an interior point and MATH is the image of MATH. Hence, for every MATH, there is a unique path MATH such that if MATH, then MATH. For MATH, let MATH be the edge of MATH corresponding to MATH. We will define MATH to be the appropriate point MATH in MATH satisfying MATH. Let MATH be the continuous function (in the equivalence class MATH) associated to MATH in the definition of MATH and MATH. Fix MATH in MATH such that MATH. For the moment let MATH also denote a specific map MATH. Then there is a homeomorphism MATH such that MATH. Let MATH be the unique number in MATH such that MATH, and define MATH. Then MATH as required. For MATH, we define MATH as the unique point in MATH which MATH maps to MATH. Then MATH is continuous, MATH, and clearly MATH by REF . Consequently MATH. Therefore the two maps MATH and MATH send any given edge (considered as a path) in MATH to paths which MATH sends to the same concatenation of elements of MATH. Such a concatenation has a unique lifting under MATH, therefore MATH. It remains to show that MATH. Because MATH is surjective, it suffices to show MATH, and this is true because MATH . Therefore MATH is shift equivalent to MATH, and MATH is a topologically conjugacy by REF . |
math/9911143 | That MATH is a conjugacy implies that MATH is a finite union of periodic orbits of MATH. Let MATH and MATH be the new graphs with graph maps defined by MATH and MATH, respectively. Then by REF , there exist shift equivalence maps MATH for MATH and MATH, and MATH for MATH and MATH. Let MATH be the conjugacy MATH, which lifts MATH. By the Ladder Lemma, there is a shift equivalence MATH and MATH such that MATH and MATH. Since MATH sends MATH to MATH, MATH and MATH are graph maps. Then it follows from REF that there is a unique conjugacy MATH lifting MATH. Therefore MATH is the unique conjugacy lifting MATH. |
math/9911144 | The equivalence of REF is essentially contained in the usual proof of NAME 's MATH-theorem (compare CITE pp. REF). REF trivially implies REF . Let us now prove that REF implies REF . By the Uniform Boundedness Principle we may suppose MATH bounded and by passing to subsequences and renormalizing we may suppose that MATH for all MATH . Let MATH and suppose MATH is any weak-MATH-cluster point of MATH . If MATH then MATH for all MATH and so MATH . In this case there exists MATH with MATH . If MATH, then we apply REF . Thus REF holds with MATH . Finally we show REF implies REF . Indeed there exists MATH with MATH and so MATH . |
math/9911144 | We may suppose MATH is a MATH-M-ideal for some MATH . Suppose MATH and that MATH is a weak-MATH-null sequence. Then there exists MATH so that MATH . Hence MATH has a weak-MATH-cluster point MATH with MATH . Clearly MATH and so MATH . We can now apply the result of CITE to deduce that MATH embeds into MATH . |
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