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math/9911013 | The first two properties are immediate consequences of REF . The third property follows because MATH is spreading. Let us show that REF. holds. This is accomplished by induction on MATH. The case MATH is easy because MATH. Assuming the assertion true for all MATH, we set MATH and MATH. In case MATH, we obtain that MATH. Thus MATH and hence the assertion holds. Next suppose that MATH. It follows that MATH and MATH are maximal MATH sets. On the other hand, because MATH, the induction hypothesis yields that MATH. But also, MATH and so REF . yields that MATH. The proof is complete since MATH, while MATH. We now prove REF. Let MATH and MATH. Denote MATH by MATH. By REF , MATH. Hence MATH from which the result follows. |
math/9911013 | Suppose first that MATH. Let MATH be scalars and choose MATH so that MATH, where we have set MATH. It follows, by our assumption, that we can find MATH successive MATH sets so that MATH. We now set MATH, for all MATH. It is clear that MATH and moreover, MATH belongs to MATH. Finally, MATH . Thus, MATH. Conversely, assume that MATH-dominates MATH and let MATH. Suppose that MATH. It follows that there exist MATH, successive maximal MATH sets so that MATH. Put MATH . We may write MATH, with MATH, for all MATH. If we apply REF , we obtain MATH as MATH and MATH. Hence, MATH which implies that MATH as MATH was an arbitrary MATH set. |
math/9911013 | By NAME 's theorem, it suffices to show that the sets MATH and MATH are first category MATH subsets of MATH. Indeed, we may write MATH . It is easy to see that each set in the union is closed in MATH and thus it remains to show that MATH has empty interior in MATH. If that were not the case, choose MATH and MATH, non-empty basic clopen subsets of MATH so that MATH is contained in MATH. There exist MATH in MATH so that MATH. Fix MATH, MATH. If MATH, MATH, let MATH. Since MATH, it follows that if MATH, then MATH. By REF this implies that MATH is equivalent to the unit vector basis of MATH which is absurd. Arguing similarly, we also obtain that MATH is first category, MATH subset of MATH. |
math/9911013 | By induction on MATH. Suppose first that MATH. Since MATH, there exist MATH, MATH, open non-empty subsets of MATH whose diameters are smaller than MATH, so that MATH. Further, MATH and thus there exist MATH, MATH, open non-empty subsets of MATH, so that MATH. Of course MATH and MATH satisfy the conclusion of the lemma for MATH. Next assume MATH and that the result holds for MATH. We can therefore choose MATH, open non-empty subsets of MATH with diameters smaller than MATH, so that MATH, for all MATH in MATH. Next, set MATH and choose, as in the case MATH, MATH, MATH, non-empty open subsets of MATH with diameters smaller than MATH, so that MATH and MATH, for all MATH. Set MATH and it is easy to check that MATH is the desired sequence. |
math/9911013 | Since MATH contains no isolated points, MATH is nowhere dense in MATH. Hence, MATH is a MATH dense subset of MATH. We shall therefore assume, without loss of generality, that MATH. Now let MATH be a decreasing sequence of open dense subsets of MATH, whose intersection is MATH. We can assume that MATH, for all MATH. We shall construct a collection MATH of open non-empty subsets of MATH so that the following properties are satisfied for every MATH: CASE: MATH, whenever MATH in MATH. CASE: MATH, for all MATH and every MATH, (MATH), initial segment of MATH. CASE: MATH, for every MATH. CASE: MATH, whenever MATH in MATH. Once this is accomplished, we let MATH . Where MATH, if MATH. It is a standard result that MATH is homeomorphic to the NAME set. REF yields that MATH satisfies the conclusion of REF . The construction is done by induction on MATH. For MATH choose MATH and MATH, open non-empty subsets of MATH so that MATH. MATH and MATH are disjoint since MATH. If we apply REF , for MATH, on the dense open subset MATH and the open sets MATH and MATH, we shall find MATH, MATH, non-empty open subsets of MATH, satisfying REF, for MATH. Now suppose that for every MATH we have constructed MATH, a collection of open non-empty subsets of MATH whose members satisfy REF, for MATH. Let MATH, MATH, be an enumeration of MATH. Another application of REF yields MATH and MATH, non-empty open subsets of MATH, MATH, so that MATH, for every MATH and all pairs MATH of distinct elements of MATH. It follows, since MATH, that MATH. According to the induction hypothesis MATH, for all MATH in MATH, and thus MATH, for all MATH in MATH. We next apply REF , for MATH, on the family MATH and the dense open subset MATH. We shall obtain MATH, a collection of non-empty open subsets of MATH, so that MATH whenever MATH, MATH and MATH in MATH. Evidently, MATH satisfies REF. The inductive step as well as the proof of the proposition are now complete. |
math/9911013 | Let MATH be as in the statement of REF , where we have taken MATH. We can apply REF for the space MATH and the subset MATH to obtain MATH, homeomorphic to the NAME set and such that MATH. REF yield that MATH is the desired subset of MATH. |
math/9911013 | We show that all three conditions are equivalent for the members of MATH. MATH . By induction on MATH. If MATH the assertion is trivial. Suppose now MATH and that the assertion holds for MATH. Let MATH. If MATH, the assertion is again trivial. So assume MATH. Choose MATH in MATH so that MATH and MATH. The induction hypothesis yields that for each MATH there exist MATH in MATH so that MATH belongs to MATH and MATH. Let MATH be an enumeration of the set MATH so that MATH. Note that MATH and so it is a member of MATH as MATH. MATH . This implication is trivial. MATH . By induction on MATH. If MATH the assertion is trivial. Suppose now MATH and that the assertion holds for MATH. Let MATH. If MATH, the assertion is again trivial. So assume MATH. We first apply MATH for the set MATH to obtain MATH in MATH so that MATH and MATH. Set MATH, MATH. Then MATH, by the induction hypothesis since MATH which belongs to MATH. Finally, MATH, as MATH. The latter inequality holds because MATH. The proof for the case of maximal NAME sets requires only minor modifications. Namely, all the sets which belong to an appropriate class MATH, MATH and appear in the previous arguments, can be taken to be maximal members of MATH. |
math/9911013 | Let MATH and MATH be scalars. Choose MATH with MATH such that MATH, where we have set MATH. It follows that MATH, by REF . Hence, MATH and thus MATH. We next show that MATH. Let MATH be maximal and put MATH, for all MATH. We apply REF to find MATH and MATH maximal members of MATH so that MATH and MATH is a maximal member of MATH. We claim that each of the MATH's can intersect at most two of the MATH's. Indeed, assume that for some MATH and MATH we had that MATH, for all MATH. Then MATH because MATH. Thus, MATH and hence MATH, by the maximality of MATH. It follows that MATH which is a contradiction as MATH. Therefore our claim holds and evidently, for each MATH, MATH intersects either exactly one of the MATH's, or exactly two (consecutive) MATH's. We can thus partition MATH in the following two subsets: MATH . Let MATH, for all MATH. We now have the following CLAIM. Suppose that MATH, for all MATH. Assume also that for each MATH there exists MATH so that MATH. Then MATH . Once the claim is established we finish the proof as follows: Observe that our claim yields MATH . On the other hand, if MATH there exist MATH so that MATH and each element of MATH is contained in some MATH. Our claim then yields that MATH . Therefore, MATH. It follows now, since MATH was arbitrary, that MATH. The desired estimate follows now from REF . We proceed now to prove our claim. Let MATH, MATH, and choose MATH such that MATH. MATH . The last inequality holds since MATH implies that MATH, for all MATH and thus MATH. The proof of the lemma is now complete. |
math/9911013 | We first note that MATH, for every MATH. Indeed, if that were not the case, we would find MATH and MATH so that MATH, for every MATH. It follows that MATH, for all MATH. But this contradicts REF because MATH is bounded. Fix MATH. Our first task is to show that MATH, for every MATH. Suppose this is not the case and so MATH, for some MATH. We claim that there exist a sequence of positive integers, MATH, and a sequence of successive maximal MATH-sets, MATH, so that letting MATH, for all MATH, the following is satisfied: MATH . Indeed, choose MATH so that MATH. Put MATH. Because MATH contains at least MATH successive maximal MATH-sets, it is clear that there exists MATH, MATH, so that MATH is a maximal MATH-set. Put MATH and MATH. We can find MATH so that MATH. Now, MATH must contain at least MATH successive maximal MATH-sets. If not, then MATH and thus MATH, by REF . But this contradicts the choice of MATH. We set MATH and arguing as we did in the case MATH, we can find MATH, MATH, so that MATH is a maximal MATH-set. We next put MATH and continue in the same fashion to obtain sequences MATH, MATH satisfying the desired properties. Let MATH. Clearly, MATH and MATH, for all MATH. We now set MATH. Then, MATH and MATH, for all MATH. We observe that if MATH, then MATH, for some MATH. Next write MATH, for all MATH. Here, MATH is a convex combination of the vectors MATH and MATH. Evidently, MATH. Observe that MATH is a convex combination of the vectors MATH and thus MATH, for all MATH. It must be the case that MATH for if not, REF yields MATH. On the other hand MATH, for all MATH. Hence MATH is not bounded contrary to our assumption. Therefore, MATH and so MATH, for all MATH. Recall that MATH and MATH is supported by MATH. Using REF , it is easy to check that MATH is a MATH-spreading model in MATH, and consequently, since MATH, MATH is also a MATH-spreading model in MATH. We conclude, as MATH is bounded, that MATH is a MATH-spreading model in MATH. However, if we apply REF we obtain that MATH is a MATH-spreading model in MATH. But this contradicts with the remark after REF . Hence, MATH, for every MATH. It follows that MATH . It is easily seen that every set in the union is closed in MATH. NAME 's theorem now yields MATH and MATH in MATH so that if MATH, MATH, and MATH, MATH, then MATH, for all MATH. It follows now that there exists MATH so that if MATH, MATH, then MATH. Finally, choose MATH so that MATH, for every MATH, MATH. REF now yields that MATH, for every MATH. To complete the proof we need only take MATH. |
math/9911013 | If MATH the assertion is trivial (MATH). Suppose the assertion holds for some MATH. We will show that MATH works for MATH. Let MATH, MATH. Our hypothesis yields that MATH is contained in the union of MATH-sets and so MATH by REF . Choose MATH so that the set MATH is the union of exactly MATH successive maximal MATH-sets. CLAIM. MATH. Once our claim is proven, we apply REF , and REF (REF . and REF.) to conclude that MATH and MATH. To prove the claim we choose MATH so that the set MATH is the union of exactly MATH successive, maximal MATH-sets. Our task now is to show that MATH. The claim will then follow by applying REF . and REF . We first observe that if MATH is chosen so that MATH, for all MATH, then MATH is contained in the union of MATH-sets. Indeed, MATH belongs to MATH, by REF and the fact that MATH. It follows now that MATH belongs to MATH. To see this let MATH, where MATH and MATH, be an enumeration of MATH. Then MATH, for every MATH. Since MATH belongs to MATH which is spreading, we conclude that MATH belongs to MATH. Our hypothesis (for MATH) yields that MATH is contained in the union of MATH-sets. On the other hand, the cardinality of the set MATH is at most MATH. Our hypothesis (for MATH) now yields that the cardinality of the set MATH is at most MATH. We deduce, since MATH, that MATH is contained in the union of MATH-sets. Hence, MATH is contained in the union of MATH-sets. Next set MATH. If MATH does not exist, then MATH, for all MATH. We obtain, by our previous observation for MATH, that MATH is contained in the union of MATH-sets. If MATH does exist, then MATH is contained in the union of MATH-sets. Indeed, this is obvious if MATH. If MATH the assertion follows from our previous observation by taking MATH. Finally, MATH belongs to MATH, since MATH and MATH is contained in the union of MATH successive maximal MATH-sets. Thus, MATH is contained in the union of MATH-sets. Concluding, in any case, the set MATH is contained in the union of MATH-sets. Hence, applying REF , we obtain that MATH, as desired. The proof of the proposition is complete. |
math/9911013 | Following CITE, given two infinite matrices MATH and MATH, we shall call MATH a block diagonal of MATH, if there exist MATH, MATH, increasing sequences of positive integers so that MATH . We can represent MATH as an infinite matrix MATH. Then MATH, for every MATH. Because MATH, for every MATH there exists MATH such that MATH. We can thus define a map MATH so that if MATH, then MATH. Observe that MATH is finite, for all MATH, since MATH is weakly null in MATH. In particular, MATH. Let MATH be the increasing enumeration of MATH. Given MATH, we set MATH. It follows, since MATH is bounded and MATH is unconditional, that MATH is a well defined bounded linear operator from MATH into MATH. Moreover, the matrix representation MATH of MATH with respect to the bases MATH and MATH is given by MATH, for all positive integers MATH. We next consider the matrix MATH given by MATH . Note that there exists a unique non-zero entry in every row of the matrix MATH, while each column contains only finitely many non-zero entries. We can thus find MATH, a permutation of MATH, so that the matrix MATH is a block diagonal of MATH. Since MATH represents the bounded linear operator MATH with respect to the bases MATH and MATH, and MATH is a block diagonal of MATH, REF yields that MATH also represents a bounded linear operator from MATH into MATH with respect to the bases MATH and MATH. Consequently, MATH represents a bounded linear operator MATH with respect to the bases MATH and MATH which evidently satisfies MATH (where MATH), for all MATH. Because MATH, if MATH, and MATH is unconditional, we obtain that there exists a bounded linear operator MATH such that MATH, for every MATH. |
math/9911013 | MATH and MATH are immediate. To prove that REF. implies REF. we first apply REF to obtain a map MATH and a bounded linear operator MATH such that MATH, for every MATH. REF will then yield a constant MATH such that MATH, for every MATH and MATH. (Where MATH is the natural bijection.) The result now follows from REF . |
math/9911013 | If MATH, the assertion follows from the fact that MATH. Assume now that MATH and that the assertion holds for MATH. Let MATH and MATH. We will find MATH so that MATH. Once this is accomplished, we can choose MATH so that MATH, where MATH and MATH. Letting MATH, we obtain that MATH, for all MATH and thus MATH. We now pass to the construction of MATH. By the induction hypothesis we can choose a sequence MATH satisfying the following properties: CASE: MATH. CASE: MATH, where MATH is chosen so that MATH, for every MATH. CASE: MATH, for all MATH. Here we have set MATH. Put MATH. We are going to show that MATH, for every MATH. Note that MATH and MATH, for all MATH. Let MATH. Let also MATH be an enumeration of MATH. Choose MATH and MATH in MATH so that MATH. Then MATH. Further, MATH . Therefore, MATH. It follows that MATH. If we take MATH, we obtain that MATH. The proof of the lemma is now complete. |
math/9911013 | Let MATH be an isomorphic embedding. We apply REF to show that MATH dominates MATH. Indeed, we need only check that MATH. If that were not the case, let MATH be a subsequence of MATH such that MATH. By a standard perturbation result we can assume, without loss of generality, that for some block basis MATH of MATH and a null sequence of positive scalars MATH we have that MATH, for all MATH. It follows that also MATH, and thus REF yields MATH so that MATH, where MATH. But then MATH as well. CASE: This is a contradiction because MATH is equivalent to a subsequence of MATH and thus it is a MATH-spreading model. Hence, MATH dominates MATH completing the proof of REF are immediate consequences of REF . |
math/9911013 | We first let MATH. MATH is easily seen to be closed in MATH and thus it is a Polish space. We next set MATH. Arguing as we did in the proof of REF we obtain that MATH is a MATH dense subset of MATH. If MATH then MATH. However, REF implies that MATH is not isomorphic to a subspace of either MATH or MATH. |
math/9911013 | Given MATH, we let MATH. We also let MATH and MATH. Set MATH. Given MATH we let MATH and MATH. We observe that MATH, for every MATH in MATH. Next, we define a map MATH so that MATH, for every MATH. Note that MATH and MATH satisfy REF . and therefore MATH . Suppose now that MATH and put MATH and MATH. It follows that MATH and MATH, MATH, satisfy REF . and thus MATH . Hence, MATH. The assertion follows since MATH . |
math/9911013 | Let MATH and put MATH. We may write MATH, where MATH and MATH belong to MATH. Note that MATH. We also observe that MATH and hence MATH . Finally, MATH and thus MATH, as desired. |
math/9911013 | Choose MATH, MATH so that MATH and MATH, for every MATH. Assume first that MATH. Then we choose inductively MATH so that MATH, for every MATH, where MATH. For every MATH we can find MATH, MATH, so that MATH. Put MATH. We are going to show that MATH is equivalent to MATH. To this end let MATH and MATH be scalars. We first show that MATH. Indeed, if MATH belongs to MATH then set MATH. We have the following estimate MATH as MATH, by REF . Next, let MATH. REF yields MATH in MATH with MATH belonging to MATH and so that MATH. We shall apply REF in order to estimate MATH. Let MATH and MATH satisfy REF . Then MATH, for every MATH. We choose MATH such that MATH, for every MATH. Fix MATH. MATH . Hence MATH. Note also that MATH belongs to MATH. This is so since MATH, whenever MATH and MATH, and thus MATH, for every MATH and MATH in MATH. In particular, MATH, when MATH in MATH. Since MATH is spreading we obtain that MATH belongs to MATH. It follows now that MATH and hence MATH . We shall now assume that MATH and MATH satisfy REF . Then MATH , for all MATH. An argument similar to that in the preceding paragraph, yields that MATH belongs to MATH. It follows that MATH. Finally, MATH . We deduce from REF that MATH and hence MATH. To complete the proof we need to consider the case MATH. We now choose MATH such that MATH, for all MATH. We are going to show that MATH is equivalent to MATH. Arguing as we did in the case MATH we obtain that MATH, for every MATH and all scalar sequences MATH. Next let MATH and put MATH. Then MATH . Hence MATH. The proof of the proposition is now complete. |
math/9911013 | MATH is normalized in MATH since MATH, for every MATH. We let MATH denote the closed linear span of MATH in MATH. Because MATH, we deduce from REF that every semi-normalized block basis of MATH admits a subsequence equivalent to the unit vector basis of MATH. Indeed, let MATH, MATH, be a semi-normalized block basis of MATH. Note that MATH is bounded since MATH is. But also, MATH, since MATH. Hence MATH and therefore REF (for MATH) yields a subsequence of MATH equivalent to the unit vector basis of MATH. It follows that every semi-normalized weakly null sequence in MATH admits a subsequence equivalent to the unit vector basis of MATH. That is, MATH has property MATH CITE. However, MATH fails property MATH when MATH and MATH. Thus, MATH is not isomorphic to a subspace of MATH for every MATH and MATH. To complete the proof we show that MATH is not isomorphic to MATH. This is accomplished by showing that for every MATH there exists MATH so that MATH is isometrically equivalent to the unit vector basis of MATH. In particular MATH contains uniformly complemented MATH's. It is a well known fact that MATH fails this property. We let MATH and choose according to REF. MATH so that MATH. REF . now yields that MATH and thus MATH. Hence MATH for every scalar sequence MATH. Therefore MATH is isometrically equivalent to the unit vector basis of MATH. |
math/9911013 | We inductively choose a sequence of integer intervals MATH such that for every MATH . Put MATH, for every MATH. We then define MATH . REF . is an immediate consequence of the inductive construction. This condition implies that in fact MATH, for every MATH and thus MATH is indeed a normalized convex block basis of MATH. We also obtain from REF that MATH and so MATH. Hence REF . holds in view of REF . It remains to establish that MATH is complemented in MATH. To this end we define a map MATH by MATH . Clearly MATH is well defined and linear. It is also clear that MATH, for every MATH. Our objective is to show that MATH is bounded with respect to the MATH-norm on MATH, for then MATH will extend to a bounded linear projection on MATH with range equal to MATH. To achieve our goal it suffices to show that if MATH is maximal, then MATH, for every MATH and MATH, MATH, with MATH, MATH. According to REF . of our hypothesis, for every MATH there exist MATH successive MATH sets so that MATH , and MATH. Next let MATH and choose MATH maximal members of MATH so that MATH. Of course MATH is maximal in MATH. We shall apply REF . Let MATH and MATH satisfy REF . Recall that MATH, MATH. For each MATH we choose MATH and MATH such that MATH. We have the following estimate MATH . The last inequality holds because MATH and MATH. Indeed, MATH when MATH in MATH and therefore, as MATH, MATH belongs to MATH by REF . Next assume that MATH and MATH satisfy REF . Then MATH, MATH. We set MATH, MATH. Since MATH is an interval, MATH, for all MATH. Moreover, since each MATH is a maximal MATH set and MATH, we have that MATH, for all MATH. To estimate MATH, choose MATH, for every MATH (we have assumed without loss of generality that MATH). Then, the sets MATH and MATH satisfy REF . We deduce from our preceding work that MATH as MATH, for every MATH. We next choose, for every MATH, MATH with MATH and such that MATH . This choice is possible since MATH. (We make use of the following fact: Let MATH be scalars with MATH, MATH, and let MATH. Then MATH.) We now have that MATH . The latter inequality follows since MATH and MATH. Indeed, the cardinality of the set MATH does not exceed that of MATH since MATH, for all MATH. It follows now, since MATH, that MATH belongs to MATH and thus MATH is the union of two members of MATH. Concluding, MATH . REF now implies that MATH, for every MATH and MATH, MATH, with MATH, MATH. It follows that MATH. The proof of the proposition is now complete. |
math/9911013 | Let MATH denote the closed linear span of MATH in MATH and assume that MATH is a bounded linear projection. Note that since MATH is unconditional our assumptions yield that MATH is semi-normalized in MATH. REF now yields that MATH dominates MATH contradicting REF. as MATH. |
math/9911014 | The vector bundle MATH is a vector bundle of weight MATH so we may form the vector bundle over MATH, MATH, and the sheaf of algebras MATH acts on its left whilst MATH acts on the right. Moreover, it is clear that MATH is isomorphic to MATH where MATH is the sheaf of opposite algebras to the sheaf of algebras MATH and thus our two sheaf of algebras are NAME equivalent as stated. Thus if MATH is the dimension of MATH over MATH, and the rank of MATH is MATH, we see that MATH and hence MATH; let MATH. Thus MATH divides the highest common factor of the ranks of MATH vector bundles of weight MATH over MATH equivariant open subvarieties of MATH. The fibre of MATH over the generic point of MATH is isomorphic to MATH. Therefore, there exists an affine open subvariety MATH of MATH on which MATH for a suitable sheaf of algebras MATH on MATH. The matrix units in this sheaf of algebras which are MATH invariant endomorphisms of MATH show that MATH where MATH is a MATH vector bundle of weight MATH on MATH whose rank is MATH. It follows that MATH is actually the minimal rank of a MATH vector bundle of weight MATH over a MATH equivariant open subvariety of MATH and since it also divides all these ranks it is both the minimum and the highest common factor. For the final part, consider MATH. The fibre above the generic point of MATH is MATH where MATH and all idempotents of the same rank are conjugate, thus there exists an affine MATH equivariant open subvariety MATH on which there is a MATH vector bundle MATH of weight MATH and rank MATH such that MATH is MATH equivariantly isomorphic to MATH and MATH is MATH equivariantly isomorphic to MATH which implies the last sentence of the lemma. |
math/9911014 | Let MATH be the new arrow from MATH to MATH and suppose that MATH and MATH. Let MATH be a MATH-standard family of representations over MATH of dimension vector MATH for the quiver MATH. Then the vector bundle MATH over MATH carries a family MATH of representations of dimension vector MATH over the quiver MATH. Here MATH and MATH for MATH and MATH is the universal map from MATH to MATH. Moreover the orbits of MATH correspond to the isomorphism classes of representations of MATH since MATH acts freely on MATH. The vector bundle MATH is a bundle of weight MATH and hence by REF there exists an open subvariety of MATH on which MATH is MATH-isomorphic to MATH where MATH acts by conjugation on MATH and if MATH is an open affine subvariety of MATH for some integer MATH then MATH is an open affine subvariety of MATH. |
math/9911014 | Take a point MATH and let MATH be an open subvariety of MATH containing a point MATH such that MATH and for all points MATH, there exists a point MATH such that MATH. Let MATH be an open subvariety of MATH such that there exists a point MATH where MATH, and for all MATH, MATH has no summand isomorphic to MATH, the simple representation at the vertex MATH and MATH is isomorphic to MATH for some MATH. This is possible since the last two conditions are open and must hold in some neighbourhood of any point MATH such that MATH. Then if MATH, MATH which shows that MATH is general if MATH is. That it is MATH-general when MATH is follows at once and that it is MATH-standard when MATH is follows from the fact that MATH is a vector bundle of weight equal to that of each MATH. |
math/9911014 | Taking the same precautions as in the proof of the preceding lemma proves this result in the same way. |
math/9911014 | By REF and the above discussion we may assume that MATH. If MATH, then a general representation of dimension vector MATH is isomorphic to MATH. So we shall assume that MATH. If MATH then MATH lies in the fundamental region for the action of the NAME group. If neither of these occur then MATH is a smaller dimension vector and if MATH then this lies in the fundamental region for the action of the NAME group. Otherwise, after dualising (see REF ) we have reached the dimension vector MATH for the quiver MATH. In either case the result follows by induction on the dimension vector. |
math/9911014 | We begin by showing that MATH is a preprojective dimension vector. Firstly, MATH and so there is a unique dimension vector MATH such that MATH and MATH; in particular, MATH . Since MATH, it follows that MATH and so MATH where MATH. Since MATH is in the fundamental region, MATH and the maximal value of MATH on this interval is for MATH. Therefore, MATH . Thus for MATH, MATH and for MATH, MATH. Thus when MATH, the dimension vector MATH is the dimension vector of a projective representation and when MATH it is either the dimension vector of a projective representation or else of a representation of type MATH (note that the dimension vector of MATH is MATH). In either case it is a preprojective dimension vector. Next we see that MATH. To calculate this, we need to show that if MATH is a dimension vector of a subrepresentation of a general representation of dimension vector MATH, then MATH. If MATH is a preprojective dimension vector then this is clear since the possibilities for MATH are all themselves preprojective dimension vectors. If MATH is not a preprojective dimension vector then it must be a NAME root by REF and hence there are stable representations and this implies that when MATH is a dimension vector of a subrepresentation of a general representation of dimension vector MATH, then MATH and since MATH, it follows that MATH. Therefore by REF, if MATH is a general representation of dimension vector MATH and MATH is a general representation of dimension vector MATH then MATH, the unique homomorphism is surjective and the kernel is general of dimension vector MATH and hence must be general preprojective as required. |
math/9911014 | Apply MATH to the short exact sequence MATH . Then by construction the homomorphism from MATH to MATH is an isomorphism. Therefore the first part of this lemma follows. Apply MATH to the short exact sequence MATH . By construction, the non-zero homomorphism from MATH to MATH lifts through MATH and hence the second part of this lemma follows. Since MATH it follows that for a general representation of dimension vector MATH, MATH, MATH and so MATH . We restrict to the open subvariety of MATH where MATH . Since the homomorphism from MATH to MATH is surjective there is an open subvariety where the natural homomorphism from MATH to MATH is surjective. Let MATH be the kernel of the natural homomorphism from MATH to MATH which has dimension MATH on this open subvariety. Since the natural homomorphism from MATH to MATH has image in the kernel of the homomorphism to MATH and is an isomorphism with the kernel when MATH it follows that on a suitable open subvariety the complex MATH is a short exact sequence which proves the third part of the lemma. |
math/9911014 | Let the NAME root be MATH where MATH is indivisible and MATH which we may assume by REF . Let MATH and MATH be the representations constructed in the preceding paragraphs and let MATH. Let MATH be the MATH-th NAME quiver. Then we show that for a general representation MATH of dimension vector MATH, MATH and for a general representation MATH of dimension vector MATH for MATH, MATH . Note that if MATH and MATH are the projective representations of dimension vector MATH and MATH for the quiver MATH then MATH whilst MATH. Let MATH be a general representation of dimension vector MATH for the MATH-th NAME quiver. Then we know that MATH and the kernel of the natural homomorphism from MATH to MATH which is surjective is isomorphic to MATH. In fact, let MATH be the kernel of the homomorphism from MATH to MATH. Then for general MATH we showed in the last lemma that MATH is a short exact sequence. However, MATH is the projective resolution of MATH; tensoring the second of these short exact sequences by MATH shows that MATH . Conversely, let MATH be a general vector subspace of dimension MATH in MATH so that MATH is the projective resolution of a general representation MATH of dimension vector MATH for the quiver MATH. Then since there exist such subspaces for which the homomorphism from MATH to MATH is injective, this remains true for a general MATH and the cokernel of this homomorphism is a representation MATH of dimension vector MATH for the quiver MATH. Since there exists a choice of MATH for which MATH, this remains true for a general MATH, and so MATH is the projective resolution of MATH (noting that MATH is naturally isomorphic to MATH) which means that MATH is isomorphic as required to MATH. |
math/9911014 | This follows at once from the preceding theorem, REF . |
math/9911014 | Let MATH be representations of dimension vector MATH respectively such that MATH for MATH. Then by REF, any homomorphism from MATH to MATH must be either injective or surjective. Assume our conclusion is false; then there is a non-zero homomorphism from MATH to MATH for each MATH and a non-zero homomorphism from MATH to MATH. Each homomorphism in this chain must be either surjective or injective but not both and no surjective homomorphism may be followed by an injective homomorphism since their composition would then be neither injective nor surjective; hence these homomorphisms must be either all surjective or all injective which is absurd. |
math/9911014 | We define a relation MATH on the dimension vectors MATH in the canonical decomposition by MATH if MATH. Then the preceding lemma shows that MATH is a partial order and can therefore be extended to a total order which is the conclusion of the lemma. |
math/9911014 | To begin with, note that MATH for MATH and hence MATH. Recall from REF the algebraic variety MATH which parametrises representations of dimension vector MATH with a distinguished subrepresentation of dimension vector MATH. Then from REF it follows that the morphism from MATH to MATH is surjective since the fibre above a point MATH is bijective with the set of subrepresentations of dimension vector MATH in MATH. Moreover, the conditions above mean that the dimension of MATH equals the dimension of MATH so that the fibre above a general point is finite. Both varieties are irreducible. Moreover, there is a rational section, a morphism defined on an open subvariety of MATH to MATH which sends the point MATH to the point in the fibre above MATH corresponding to the subrepresentation that is the direct summand of dimension vector MATH. But a morphism between irreducible algebraic varieties that is generically finite and has a rational section must be generically bijective (consider the effect of the rational section on the function fields of the two varieties). This implies that the fibre above a general point of MATH consists of MATH point which means that MATH is a rigid sub-dimension vector of MATH and it is uniform by assumption. |
math/9911014 | Let MATH be a general representation of dimension vector MATH; in particular, it has a unique subrepresentation MATH of dimension vector MATH which in turn may be taken to have a unique subrepresentation MATH of dimension vector MATH. Since MATH is a uniform rigid summand of MATH, MATH and also MATH since MATH and MATH is a canonical summand of MATH. So, MATH and hence every representation of dimension vector MATH has a subrepresentation of dimension vector MATH. Also since MATH is a canonical summand of MATH and MATH then MATH so a general representation of dimension vector MATH has a subrepresentation of dimension vector MATH. So let MATH be a subrepresentation of MATH of dimension vector MATH; then MATH has a subrepresentation of dimension vector MATH, MATH where MATH has to be MATH since it is a subrepresentation of dimension vector MATH of MATH. So, MATH is a subrepresentation of dimension vector MATH in MATH and must be MATH. Thus MATH is a rigid sub-dimension vector of MATH. |
math/9911014 | Since MATH and MATH is a uniform rigid summand of MATH, MATH. Also MATH and so MATH. Hence a general representation of dimension vector MATH has a subrepresentation of dimension vector MATH. In turn this has a subrepresentation of dimension vector MATH which must be the unique one and the factor must be the unique subrepresentation of dimension vector MATH. So there is a unique subrepresentation of dimension vector MATH. |
math/9911014 | Our assumptions imply that MATH and hence MATH since MATH and MATH are canonical summands of MATH. If MATH then MATH is a summand of MATH because MATH and so MATH . Thus if MATH is not a canonical summand of MATH then MATH and therefore MATH. It follows that MATH since MATH and so one of MATH and MATH must be MATH by REF. By REF , MATH is a rigid sub-dimension vector of MATH. So let us consider its canonical decomposition. Firstly MATH and MATH. On the other hand, MATH. Let MATH be a general representation of dimension vector MATH. Then MATH has a subrepresentation MATH of dimension vector MATH such that both MATH and MATH are general representations. So MATH and MATH where MATH, MATH, each MATH and MATH is a NAME representation, MATH and also MATH for MATH unless MATH is a real NAME root in which case they are isomorphic, and similarly, MATH for MATH unless MATH is a real NAME root in which case they are isomorphic. By NAME 's simplification process CITE one knows that any summand of MATH must have a filtration by subrepresentations such that the factors are isomorphic to either a MATH or a MATH. No MATH can be a summand of MATH since then its dimension vector MATH would be a summand in the canonical decomposition of MATH. Thus any summand in the canonical decomposition of MATH must be of the form MATH where MATH. If MATH is not a summand in the canonical decomposition of MATH, then every canonical summand of MATH is larger than MATH. If MATH is a summand in the canonical decomposition of MATH, we shall soon see that it cannot be the root of a uniform rigid summand. Thus it follows that MATH and so MATH which implies that MATH is a real NAME root and so each MATH is isomorphic to the real NAME representation MATH of dimension vector MATH. However, there is a summand of MATH that has a proper subrepresentation isomorphic to some MATH and so, as claimed, MATH cannot be a uniform rigid summand of MATH. It follows that a uniform rigid summand of MATH cannot be a multiple of MATH in this case and therefore the root of a uniform rigid summand MATH of MATH must be larger than MATH. Therefore, in either case there is a uniform rigid summand MATH of MATH whose root is larger than MATH. However, by REF , MATH is a uniform rigid sub-dimension vector of MATH which completes our proof. |
math/9911014 | We may assume that the support of MATH is MATH. Let MATH be a sink vertex; then as discussed before MATH is a uniform rigid sub-dimension vector of MATH where MATH is the dimension vector of the simple representation at the vertex MATH. Therefore, if we do not already have a dimension vector MATH we may take MATH. If the factor is uniform we simply need to check the numerical statements. But MATH implies that MATH and so equality follows. Otherwise, assume that MATH is a uniform rigid sub-dimension vector of MATH such that MATH is not uniform and proceed by induction on MATH where MATH is the root of MATH. Since MATH is not uniform, MATH where MATH is a uniform rigid summand of MATH. Since the root of MATH cannot be a summand of MATH one concludes by REF that there is a uniform rigid sub-dimension vector MATH of MATH whose root is larger than MATH. Since MATH it follows that MATH by REF hence MATH and hence MATH must divide MATH by REF since it is a uniform rigid summand of MATH. By induction, the result follows. |
math/9911014 | Let MATH be the factor of MATH by the diagonal embedding of MATH. The first step of the proof is to construct a MATH-standard family of representations of dimension vector MATH for the quiver MATH which is MATH birational to MATH. The first thing we need to do this are suitable familes of representations of dimension vector MATH and MATH. We deal only with MATH since the same construction is needed for MATH. We need a family MATH over the algebraic variety MATH such that MATH acts on MATH compatibly with an action of MATH on MATH and the stabiliser in MATH of a point in MATH acts on the corresponding representation as the group of automorphisms of that representation. If MATH is a real NAME root then there is a unique real NAME representation MATH of dimension vector MATH. We take the variety MATH to be a point on which MATH acts trivially and the family MATH will be simply the representation MATH on which MATH acts via its action on MATH. If MATH is non-isotropic then the root of MATH is itself and our assumptions imply that that there is a MATH-standard family MATH of representations of dimension vector MATH over an algebraic variety MATH which is MATH birational to MATH. In the case where MATH is isotropic there is more to do. In this case, our assumptions imply that there is an algebraic variety MATH which is an open subvariety of MATH, the affine line, which carries a family MATH of representations of dimension vector MATH and that this family is MATH-standard which in this situation simply means that different points of MATH give non-isomorphic representations. We wish to construct from this a family of representations of dimension vector MATH on which MATH acts, defined over an algebraic variety MATH on which MATH acts so that MATH is MATH birational to an open subvariety of MATH. We take MATH to be the open subvariety of MATH consisting of MATH distinct ordered points of MATH. This carries in the obvious way a family MATH of representations of dimension vector MATH. The natural action of the symmetric group MATH on MATH has the property that its orbits correspond to the isomorphism classes of representations in the family MATH. Of course, this action of MATH on MATH lifts to an action on MATH but this is not quite all we need; the algebraic group MATH acts on MATH via the action of MATH on MATH and the algebraic group MATH generated by MATH and MATH acts on MATH. The subgroup MATH of MATH is normal with factor group MATH and we shall regard MATH as acting on MATH so that MATH acts trivially and MATH acts as described. Thus the actions of MATH on MATH and MATH are compatible. Now MATH is a subgroup of MATH in the obvious way; it is the stabiliser with respect to the action of MATH by conjugation of the vector subspace of MATH consisting of diagonal matrices. Therefore, we can form the family MATH of representations of dimension vector MATH over the algebraic variety MATH. An identification of MATH with some open subvariety of the vector space of diagonal matrices shows that MATH is MATH birational to MATH, each MATH is a MATH vector bundle of weight MATH and for each point MATH of MATH, the stabiliser in MATH of MATH acts on the corresponding representation as the group of automorphisms of that representation since that is true for points in MATH. Note that there is a short exact sequence of groups MATH . Thus a variety on which MATH acts may have MATH vector bundles of weight MATH. Let MATH and MATH be families of representations of dimension vector MATH and MATH over the algebraic varieties MATH and MATH as constructed above. So MATH is MATH-birational to MATH and MATH is MATH-birational to MATH. There exists a non-empty open MATH-equivariant subvariety MATH of MATH consisting of the points MATH where MATH. Then over MATH there is a MATH vector bundle MATH of weight MATH whose fibre above the point MATH is MATH. However, MATH is a MATH vector bundle of weight MATH over MATH and by REF it follows that MATH is MATH birational to MATH. Further MATH carries a family MATH of representations of dimension vector MATH of the quiver MATH and it is clear that MATH is a general family since every representation of dimension vector MATH has a subrepresentation of dimension vector MATH and it is an open condition that one such subrepresentation should be isomorphic to MATH for some point MATH whilst the factor should be isomorphic to MATH for some point MATH. Since a general representation of dimension vector MATH has a unique subrepresentation of dimension vector MATH, it follows that MATH is a non-empty open subvariety of MATH and we shall see that two points of MATH give rise to isomorphic representations if and only if they are in the same orbit for the action of MATH. To show this we represent the points of MATH as triples MATH where MATH, MATH, and MATH. Then if MATH and MATH are points of MATH that determine isomorphic representations, it follows that MATH and MATH lie in the same orbit for MATH and therefore may be taken to be equal and MATH and MATH determine isomorphic extensions of MATH on MATH. However, the stabiliser in MATH of MATH is the group of automorphisms of MATH and similarly the stabiliser in MATH of MATH is the group of automorphisms of MATH and so MATH and MATH are in the same orbit under the action of the stabiliser of MATH and hence MATH and MATH must lie in the same orbit for MATH as required. Finally, it is clear that for each vertex MATH, MATH is a vector bundle of weight MATH since it has a subbundle isomorphic to MATH such that the factor bundle is isomorphic to MATH. Thus we have constructed a MATH-standard family of representations of dimension vector MATH for the quiver MATH which is MATH birational to MATH. Now we assume that the dimension vector MATH for the quiver MATH is reducible to matrix normal form. Thus we have a family MATH of representations of dimension vector MATH for the quiver MATH over the algebraic variety MATH which is a MATH standard family such that MATH is MATH birational to MATH where MATH and MATH where MATH. By REF , we may assume that MATH where MATH acts trivially on MATH and diagonally on the remaining terms and similarly, MATH with a similar action of MATH. By choosing a basis of MATH and of MATH we obtain a morphism from MATH to MATH such that MATH is the pullback of the standard family on MATH. Moreover, if we regard MATH as acting on MATH and on MATH via its action on MATH, and hence on MATH and the standard family, we see that this morphism from MATH to MATH is MATH equivariant and the morphism between families is MATH equivariant. If we regard MATH as a subvariety of MATH with the family MATH of representations of dimension vector MATH for the quiver MATH then the pullback of MATH to an open subvariety of MATH along the MATH equivariant morphism to MATH is quickly checked to be a MATH standard family of representations of dimension vector MATH and by construction MATH is MATH birational to MATH which completes the proof of this theorem. |
math/9911014 | If the dimension vector MATH is a NAME root for the quiver MATH where MATH then the result follows by induction from REF . This means that we have certain minimal cases to look at. The results of REF deal with the case where both MATH and MATH are zero so we may assume that at least one of them is non-zero. Thus the remaining minimal cases are when MATH is one of MATH, MATH, and MATH since if both MATH and MATH are positive then it is clear that MATH is a NAME root. The first of these in fact reduces to one of the other two cases since if MATH then MATH is a NAME root for the quiver MATH because (for example) the dimension vector MATH for the quiver with vertices MATH, MATH and MATH with MATH arrow from MATH to MATH and MATH arrows from MATH to MATH lies in the fundamental region for the action of the NAME group and if MATH then we may apply REF . The second and third case are essentially equivalent by reversing all the arrows of the quiver and applying REF so we shall deal with the case MATH. If MATH is a NAME root for MATH then MATH. We shall show that the dimension vector MATH for the quiver MATH such that MATH is always a NAME root and determine its moduli space. If MATH then the reflection functor at the first vertex together with REF shows that the moduli space of representations of dimension vector MATH is reducible to the moduli space of representations of dimension vector MATH for MATH which is just MATH by MATH matrices up to simultaneous conjugacy. If MATH we may still apply the reflection functor at the first vertex which preserves the value of MATH followed by duality to ensure that MATH. If MATH or if MATH and MATH, then MATH and we may conclude by induction that MATH is a NAME root for MATH and that it is reducible to matrix normal form for the quiver MATH and hence for the quiver MATH. Thus we may reduce either to the case MATH for MATH which is reducible to MATH by MATH matrices up to simultaneous conjugacy and thus a NAME root with the correct moduli space; or else we reduce to MATH where MATH for the quiver MATH and it is enough to deal with the case where MATH since this is a NAME root as we are about to see. Consider the quiver MATH with MATH vertices MATH, MATH and MATH with MATH arrow from MATH to MATH and MATH arrows from MATH to MATH and the dimension vector MATH where MATH; this is also clearly a NAME root since it lies in the fundamental region for the action of the NAME group. Also a general representation of this dimension vector inverts the first arrow from MATH to MATH and hence the moduli space of representations of this dimension vector is birational to the moduli space of representations of dimension vector MATH for MATH. Now MATH is a uniform rigid sub-dimension vector of MATH and its root is MATH which is a real NAME root. Let MATH be the rigid sub-dimension vector of MATH that is constructed from this one such that both it and MATH are uniform in REF . By induction on the pair MATH it follows that the moduli spaces of representations of dimension vector the root of MATH and the root of MATH satisfy the theorem since their construction from smaller NAME roots must involve a smaller pair than MATH and if MATH and MATH then MATH so again the proof is complete by induction. |
math/9911014 | Let MATH be a NAME root for the quiver MATH and let MATH be the corresponding dimension vector for the double of MATH, MATH. We have seen that a NAME root for a quiver without loops is reducible to matrix normal form and therefore we know that MATH is reducible to matrix normal form. If MATH, then MATH, so we assume that we have a MATH standard family of representations of dimension vector MATH of the quiver MATH, MATH, over the affine algebraic variety MATH such that MATH is MATH equivariantly birational to MATH for some integer MATH. Let MATH be the affine open MATH equivariant subvariety where MATH is invertible. Then MATH carries a family of representations of the quiver MATH of dimension vector MATH by defining MATH and MATH. Using the fact that the category of representations of the quiver MATH such that MATH is invertible is equivalent to the category of representations of the quiver MATH, it is a simple matter to check that MATH is a MATH standard family and by construction it is MATH equivariantly birational to MATH as required. |
math/9911014 | This follows from the known results on matrices up to simultaneous conjugacy. A good summary of the known results may be found in CITE. |
math/9911016 | The following lemma gives a dual formulation of the notion of MATH norm. Let MATH in MATH and MATH be a separable NAME space with a MATH-Lipschitz weak-star NAME norm, then MATH whenever MATH is a weakly null sequence in MATH (MATH). Let us call MATH this property. Let MATH in MATH and MATH with MATH. Without loss of generality, we may assume that MATH and MATH exist with MATH. We will first prove the right hand side inequality. For MATH, pick MATH in MATH so that MATH and MATH. Passing to a subsequence, we may assume that MATH and MATH exists. Then, it follows from our assumption that MATH . Notice now that MATH. So we have MATH . For the left hand side inequality, we only need to show that MATH. So we select now MATH in MATH with MATH and MATH and we assume that MATH and MATH exists. Again , we have MATH . Since MATH, we also obtain MATH and therefore MATH. We can write MATH. So, passing to the limit we obtain MATH. Then we conclude by using the fact that MATH. Remark. The best constant MATH is not crucial for the proof of REF that will be achieved with the trivial value MATH. However it will be used in the proof of REF and it helps us to relate this with REF which states, in the particular case MATH, that a space satisfying the property MATH embeds almost isometrically into MATH. Our next Lemma is the analogue of REF . CASE: If MATH is a finite dimensional subspace of MATH and MATH, then there is a finite dimensional subspace MATH of MATH such that MATH . CASE: If MATH is a finite dimensional subspace of MATH and MATH, then there is a finite dimensional subspace MATH of MATH such that MATH . Since the norm of MATH is MATH, MATH is separable. Then the proof is identical with the proof of NAME REF . We will now proceed with the proof of REF , which is only a slight modification of the proof of REF . So let MATH and pick a positive integer MATH such that MATH. Let also MATH be a sequence of positive real numbers satisfying MATH . Finally, let MATH be a dense sequence in MATH. Following the ideas of NAME and NAME, we then construct subspaces MATH, MATH of MATH and MATH of MATH so that: REF dim MATH, dim MATH for all MATH. CASE: MATH is weak-MATH-closed and MATH. CASE: MATH. CASE: If MATH and MATH, then MATH . CASE: If MATH and MATH, then MATH . CASE: MATH and MATH if MATH. CASE: If MATH, then there exists MATH so that MATH and MATH. Now, as in CITE , we define, for MATH and MATH by MATH and MATH. And also MATH by MATH . Then we get MATH . Still following CITE we can also show that if MATH in MATH satisfies MATH, then MATH. Then a NAME argument yields MATH . Since MATH and MATH, we have MATH . Therefore MATH . Thus, for our initial choice of MATH we obtain MATH . Since we have on the other hand that MATH, we get that MATH . As a MATH-sum of finite dimensional spaces, MATH embeds almost isometrically into MATH. Then, by NAME 's theorem CITE, so does MATH. This concludes our proof. |
math/9911016 | We recall a fundamental lemma due to NAME REF and that can also be found in CITE . For every MATH and MATH, there exists a compact subset MATH of MATH such that, whenever MATH is a continuous map from MATH to MATH satisfying MATH for any MATH in MATH, then MATH. Now, fix MATH such that MATH. Let MATH, where MATH is the compact set obtained in REF . Consider now MATH in MATH and the map MATH from MATH to MATH defined by MATH. It is clear that for any MATH in MATH, MATH. Then, it follows from REF that there exists MATH so that MATH. This concludes our proof. |
math/9911016 | Let MATH be a NAME isomorphism from a subspace MATH of MATH onto the NAME space MATH. REF indicates that we need to build an equivalent MATH norm on MATH. This norm will be defined as follows. For MATH in MATH, set: MATH . Since MATH and MATH are NAME maps, MATH is an equivalent norm on MATH. It is clearly weak-MATH lower semicontinuous and therefore is the dual norm of an equivalent norm on MATH that we will also denote MATH. Consider MATH, MATH and MATH such that MATH and MATH for all MATH. Fix MATH and then MATH and MATH in MATH so that MATH . By using translations in order to modify MATH, we may as well assume that MATH and MATH. Since MATH is a subspace of MATH, it admits a finite codimensional subspace MATH such that MATH . Let MATH be the NAME constant of MATH. By REF , for every MATH there is a compact subset MATH of MATH such that MATH. Since MATH converges uniformly to MATH on any compact subset of MATH, we can construct a sequence MATH such that: MATH . We deduce from REF that MATH and therefore MATH. Using again the fact that MATH, we get that: MATH . Since MATH is arbitrary, by using the definition of MATH and REF , we obtain MATH. This proves that MATH is MATH, and concludes the proof of REF . |
math/9911016 | We only need to prove the ``if" part. So let MATH be a NAME space which is NAME isomorphic to MATH. REF asserts that MATH is linearly isomorphic to a subspace of MATH. Besides, it is known that the class of all MATH spaces is stable under uniform homeomorphisms CITE and that a MATH subspace of MATH is isomorphic to MATH CITE. This establishes REF . |
math/9911016 | Let us first mention that for values of MATH and MATH close to MATH, the result follows directly from a work of NAME REF , who proved that if MATH is a MATH subspace of MATH with MATH, then MATH . Then, it is easily checked that in our setting, if we assume moreover that MATH, we get MATH . For the general case we do not have an explicit function MATH. We will just reproduce an argument by contradiction used in ( CITE , p. REF). Indeed, if there is no such function, then there exist MATH in MATH and a sequence MATH of separable MATH spaces with a MATH norm such that, for all MATH, MATH. But the space MATH is MATH with a MATH norm and thus by REF and CITE it is isomorphic to MATH. So, the MATH's being uniformly complemented in MATH, their NAME distance to MATH should be bounded, a contradiction. |
math/9911016 | Let MATH. We set MATH. It is easily seen that if MATH is a continuous map such that MATH for all MATH, then there exists MATH such that MATH for all MATH. Indeed, if MATH is the natural projection and MATH, then MATH and by NAME 's theorem, there is MATH with MATH. Hence MATH for all MATH, and thus MATH, where MATH . If we now reproduce the proof of NAME 's Principle REF , using the compact set MATH and the space MATH defined above (with an appropriate choice of MATH), we find in the notation of the proof of REF that for any MATH, there is a compact subset MATH of MATH such that MATH and it follows that the norm MATH is MATH. Now REF shows that the distance from MATH to the subspaces of MATH is at most MATH. Since the distance between the original norm MATH of MATH and MATH is less than MATH, it follows that the NAME distance from MATH to the subspaces of MATH is at most MATH. We now observe the following There is a function MATH with MATH, and such that if MATH satisfies the assumptions of the proposition, then MATH is a MATH space. By the ultrapower version of the local reflexivity principle, MATH is isometric to a REF-complemented subspace of some ultrapower MATH. We set MATH. Clearly, there is a bi-Lipschitz map MATH with Lip-MATH . Lip-MATH. It follows that there are maps MATH and MATH with Lip-MATH . Lip-MATH and MATH. By REF , there is MATH extending MATH and such that Lip-MATH=Lip-MATH. The space MATH is isometric to the dual of a MATH-space, hence it is a MATH space (see CITE , p. REF). Since MATH, it follows that if MATH is a metric space, MATH a subspace of MATH and MATH a NAME map, there exists a NAME extension MATH with LipMATH-Lip-MATH. In particular, MATH is isometric to a linear subspace MATH of MATH on which there exists a NAME projection MATH with Lip-MATH. Since MATH is REF-complemented in its own bidual, it follows from (CITE, REF to REF ) that there exists a linear projection MATH with MATH. Therefore, MATH is a MATH space. By ( CITE , see p. REF), MATH is therefore a MATH space, and so is MATH. Moreover REF , when MATH is a finite dimensional MATH space with MATH, then if we let MATH (see REF ) MATH . In the above notation, any finite dimensional subspace of MATH is contained, up to MATH arbitrary, in a space MATH, where MATH is isometric to a finite dimensional MATH. Such a MATH is MATH; therefore REF guarantees the requirement MATH. We now proceed with the proof of REF . We know that MATH is a MATH space whose NAME distance to the subspaces of MATH is at most MATH. Any MATH subspace MATH of MATH is isomorphic to MATH CITE and by using contradiction (see CITE , p. REF) we show the existence of a function MATH such that MATH. Finally according to REF , there is such a function MATH which satisfies MATH (see proof of REF above). For any MATH, MATH is MATH isomorphic to a subspace MATH of MATH which is a MATH space with MATH; the existence of MATH as claimed in the proposition clearly follows. |
math/9911016 | The proof relies heavily on CITE , from which we take the following notation: if MATH denotes the NAME distance between two subsets MATH and MATH of a metric space MATH, the NAME distance MATH between two NAME spaces MATH and MATH is MATH where the infimum is taken over all linear isometric embeddings MATH of MATH into an arbitrary common NAME space MATH. The NAME distance MATH is the infimum of the NAME distances MATH over all isometric embeddings of MATH and MATH into an arbitrary common metric space MATH. By REF , we have MATH where MATH is a bijective map. It follows easily that for any MATH, there is MATH such that if there is MATH a NAME isomorphism with Lip-MATH . Lip-MATH, then MATH. Obviously, one has MATH (and in general these two distances are not equivalent: for instance CITE, MATH while MATH for all MATH). We recall that a NAME space MATH is a MATH-space CITE if there exists MATH such that whenever MATH is a homogeneous function which is bounded on MATH and satisfies MATH then there exists MATH such that MATH . It is shown in CITE that MATH is a MATH-space. By REF , if MATH is a MATH-space and MATH is such that MATH, then MATH. Since MATH is a MATH-space as isomorphic to MATH, for any MATH there is MATH such that MATH implies MATH. Let MATH be a NAME space which contains isometric copies of MATH and MATH with MATH. We may and do assume that MATH is separable. By NAME 's theorem, MATH is linearly complemented in any separable super-space MATH, and the norm of the projection MATH is bounded independently of MATH. It easily follows that given MATH, there is MATH such that if MATH then MATH. Indeed, the restriction to MATH of MATH provides the required linear isomorphism. This concludes the proof. |
math/9911016 | By REF , it suffices to show that MATH is a MATH-ideal in its bidual. But REF , asserts that to be a MATH-ideal in its bidual is a separably determined property. So let MATH be a separable subspace of MATH and MATH be the canonical projection. Pick MATH with MATH, and write MATH with MATH. By definition of a MATH-ideal, what we need to show is that MATH . There is a net MATH in MATH such that MATH in MATH and then MATH in MATH. Since MATH, we have MATH and since MATH is metric-MATH, this implies that MATH. This shows REF since MATH by the triangle inequality. |
math/9911016 | It clearly suffices to show that for any MATH, there is MATH separable such that if MATH . Assume, for instance, that for any separable MATH, there is MATH such that MATH . We construct inductively an increasing sequence MATH of separable subspaces of MATH, and MATH in MATH such that for all MATH, REF if MATH, then MATH. CASE: MATH. CASE: MATH. We let MATH. Since the weak-MATH and norm topologies coincide on MATH, it follows from REF that MATH is dense in MATH. Then REF implies that MATH. But now REF contradicts REF . This proves the lemma, since we can clearly proceed along the same lines with the left hand side of the inequality. |
math/9911016 | The natural norm of MATH is metric-MATH and every subspace of MATH is w.c.g. (CITE, see also CITE , REF). Conversely, if MATH is w.c.g. and has a MATH norm, then MATH is a w.c.g. NAME space and thus ( REF , see also CITE , REF ) MATH has a shrinking P.R.I. MATH. Using REF , we construct by induction on MATH, ordinals MATH such that MATH and such that if MATH and MATH, then MATH . If we let MATH, then MATH is isomorphic to MATH. By REF , the spaces MATH are (uniformly in MATH) isomorphic to subspaces of MATH; this concludes the proof. |
math/9911016 | REF easily follows from the fact that the natural norm of MATH is metric-MATH. REF relies on If MATH has a metric-MATH norm, there exists a P.R.I. MATH on MATH such that for any MATH, if MATH are such that MATH and MATH, then MATH. Indeed by REF we know that MATH is w.c.g. Then REF shows that for all MATH, there is MATH a separable subspace such that MATH for every MATH. We now use the technique of REF : using the same notation, we prove along the same lines that if MATH and MATH are subsets with density MATH, there exist norm closed subspaces MATH and MATH such that REF MATH, MATH. CASE: MATH, MATH. CASE: For all MATH, MATH. CASE: For all MATH, MATH. CASE: For all MATH, for all MATH, MATH. Note that REF shows that MATH, while the choice of MATH and REF shows that MATH for all MATH and MATH. REF now follows by a simple transfinite induction argument, as in REF . Finally, REF follows immediately by transfinite induction from REF , since REF proves it in the separable case and allows us to start the induction. |
math/9911016 | REF was proved in [D-G-Z REF] and the argument for the converse can be found in CITE REF . The equivalence between REF follows easily from the proof of (CITE, REF ). |
math/9911016 | REF implies REF : Since MATH is NAME, MATH is w.c.g. see CITE , REF. By compactness, MATH implies that there is MATH in MATH such that MATH. We proceed by induction on MATH. If MATH, MATH is finite and the implication is obvious. Assume it holds when MATH and pick MATH such that MATH. We let MATH and MATH. The space MATH is clearly isometric to MATH; while MATH is isometric to MATH, and thus isomorphic to a MATH space by our assumption. We observe now that MATH is complemented in MATH, since any MATH space is REF-complemented in any w.c.g. space. For checking this, let us call MATH a subspace isometric to MATH of a w.c.g. space MATH. Using the notation of CITE , REF, we can choose the map MATH from REF in such a way that for any MATH and any MATH: CASE: MATH . CASE: MATH . CASE: MATH, where MATH is a countable subset of MATH. Then ( CITE , REF ) provide a P.R.I. MATH on MATH such that for all MATH: CASE: MATH. CASE: There exists MATH such that MATH for all MATH. By NAME 's theorem, MATH is REF-complemented in any separable super-space. Then we proceed by induction on dens-MATH: if it is true for all w.c.g. MATH with dens-MATH dens-MATH, we consider MATH which satisfies REF above. Since MATH, there is a projection MATH such that MATH. Let MATH and MATH. It is easily checked that MATH is the required projection from MATH onto MATH with MATH. To conclude the proof of REF , we simply observe that since MATH is complemented in MATH, we have that MATH . REF implies REF is clear since the natural norm on MATH is metric-MATH. REF implies REF : By REF , any NAME space which has a metric-MATH norm has a shrinking P.R.I. and thus is w.c.g. The condition MATH follows immediately from REF . |
math/9911016 | It is obvious that REF implies REF . REF implies REF : We use the notation from the proof of REF . Through an easy separable exhaustion argument we can ensure that the spaces MATH are (uniformly in MATH) MATH spaces. By restriction, they have (uniformly in MATH) MATH norms. Hence by REF they are uniformly isomorphic to MATH. This clearly implies REF . REF implies REF follows immediately from REF . |
math/9911016 | The proof of REF shows that if MATH is NAME isomorphic to a subspace of MATH, then MATH has an equivalent MATH norm. Indeed the MATH property is separably determined by definition and an easy exhaustion argument shows that if MATH is any separable subspace of MATH, there is a separable space MATH with MATH and MATH is NAME isomorphic to a subspace of MATH. Now REF follows from REF from REF and the fact that being a MATH space is stable under NAME isomorphisms CITE. |
math/9911016 | Since MATH is a MATH-ideal in MATH, it is w.c.g. and it admits a shrinking P.R.I. MATH by REF . Let MATH be such that MATH is MATH. For any sequence MATH in MATH with MATH and MATH, there exists MATH such that: CASE: MATH for every MATH. CASE: MATH is a MATH space. Since MATH is a separable MATH space which is MATH-ideal in its bidual, we have by ( CITE , REF , p. REF) that MATH, where MATH depends only upon MATH. It follows that there exists a cluster point to the sequence MATH in MATH, say MATH, such that MATH, where MATH depends only on MATH (that is, on MATH). If now MATH and MATH, with MATH, there is, by the above, MATH in MATH with MATH and MATH in MATH. Since MATH, one has MATH and it follows that MATH. Recapitulating, we have shown that any separable subspace of MATH is MATH. Finally, REF yields the conclusion. |
math/9911017 | Note that if MATH then for any MATH, MATH . Hence MATH where MATH . Iterating we obtain MATH and so MATH . This implies the result for the NAME index and the convex version is similar. |
math/9911017 | For the first part, let us write MATH where MATH and MATH with MATH a nonnegative integer, or MATH and MATH . Then MATH . Now observe that MATH and so MATH . This immediately gives the conclusion of REF . For the second part, first suppose MATH is a weak-MATH-compact and convex subset of MATH. We will argue that if MATH and MATH then MATH . Indeed suppose MATH . Then there is a sequence MATH with MATH and MATH weak-MATH. Now for any MATH we have MATH . Letting MATH we obtain that MATH . Letting MATH and repeating we obtain that MATH . In particular we observe that MATH . Now suppose MATH and MATH . Suppose MATH and let MATH . Then MATH . Now MATH . But MATH yields MATH . Hence MATH . |
math/9911017 | We observe that by REF we have MATH if MATH is any finite-dimensional subspace of MATH. Hence, since MATH is separable we find an increasing sequence (not necessarily strictly increasing) of finite-dimensional subspaces MATH so that MATH is dense in MATH and MATH . There exist MATH with MATH and MATH and this gives the conclusion. |
math/9911017 | Note that REF implies that MATH and this immediately yields MATH . Then the last statement follows from REF . First assume REF holds and MATH are chosen as in REF . It is enough to show it is impossible that MATH . Suppose this holds. Then we can pick MATH with MATH . Passing to a subsequence we can suppose MATH converges weak-MATH to some MATH and then put MATH . We can assume that MATH exists. If MATH then MATH . On the other hand if MATH then we have, using the convexity of the norm, MATH . Hence MATH and so MATH . This gives us the required contradiction. Assume REF holds. Suppose that MATH are chosen as in REF . It will be enough to show the conclusion for some subsequence. Then, given MATH we can choose MATH with MATH and MATH . Now by REF , we can, by passing to a subsequence, assume there exist a weakly null sequence MATH with MATH and MATH . Now MATH and so MATH . Hence letting MATH we have MATH . |
math/9911017 | Suppose MATH is of norm one and MATH is a norming subspace of MATH . Then by the NAME theorem CITE there is a weakly NAME sequence MATH in MATH with MATH and MATH weak-MATH. If MATH then, for any MATH and for each MATH we can choose MATH with MATH, MATH and MATH . Then we can find MATH so that MATH . We conclude that MATH . Letting MATH, since MATH weakly we have MATH . Thus MATH . Then MATH we have MATH which is a contradiction. |
math/9911017 | REF follows trivially from convexity considerations. CASE: It follows from REF that if MATH is not separable, then MATH for all MATH and then it is obvious that MATH REF-dominates MATH. If MATH is separable, we pick MATH in MATH such that MATH and MATH . By REF there is a weakly null sequence MATH with MATH and MATH. It follows easily that MATH so that MATH . Now since MATH we obtain MATH . Hence MATH REF-dominates MATH . The same considerations show that MATH REF-dominates MATH, MATH REF-dominates MATH and MATH REF-dominates MATH . Now if MATH pick MATH . Then by REF we have MATH . Hence MATH . Thus MATH REF-dominates MATH . The proof for MATH in place of MATH and MATH in place of MATH is similar. CASE: We deduce from REF that MATH is REF-equivalent to MATH. Next let MATH . Then MATH . Hence since MATH is convex we have MATH . Hence MATH is REF-equivalent to MATH . The argument for MATH and MATH is similar. CASE: This is an immediate deduction from REF . |
math/9911017 | For each countable ordinal MATH we define a subset MATH of MATH as follows. If MATH let MATH if MATH . Then inductively if MATH we say MATH if MATH has infinitely many successors MATH with MATH . Let MATH . If MATH then an easy induction argument produces a full tree MATH with MATH for every MATH . Otherwise the set MATH is a full tree with the property that MATH for every MATH . |
math/9911017 | One easy way to prove this is to consider MATH with canonical basis MATH and then the tree-map MATH if MATH and MATH if MATH . The lemma follows from the NAME theorem and repeated applications of REF . |
math/9911017 | Let MATH be the set of MATH so that if MATH then MATH . It is not difficult to see that this is a full tree and that if MATH is a branch in this tree then MATH . |
math/9911017 | First assume MATH . Then MATH . Now there exists a sequence MATH converging to MATH weak-MATH so that MATH and MATH since otherwise there is a weak-MATH-open neighborhood of MATH relative to MATH of diameter less than MATH . Then for each MATH we find a sequence MATH so that MATH weak-MATH, MATH and MATH . This procedure can then be iterated to define MATH if MATH. Setting MATH if MATH we obtain the desired tree-map. The converse is equally easy. Obviously if MATH is the given tree-map then we have MATH for all MATH . It then follows inductively that MATH whenever MATH . Setting MATH gives the result. |
math/9911017 | Define MATH and then for MATH define MATH to be the infimum of all MATH so that whenever MATH is a weakly null tree-map of height MATH with MATH for all MATH then there is a full subtree MATH so that MATH for every branch MATH . We observe first that MATH is an increasing sequence, and that MATH . Next notice that MATH by an elementary calculation, which we omit. We also observe that MATH where MATH . We next claim that MATH is convex. Indeed let MATH where MATH . Suppose MATH and MATH . Let MATH be any weakly null tree-map of height MATH with MATH for all MATH . Then we can find a full subtree MATH so that for every branch MATH we have MATH and then a full subtree MATH so that for every branch MATH . Obviously for every branch MATH so that MATH . Next we note that if MATH and MATH weakly then MATH for all MATH . Indeed for MATH this is obvious. If MATH assume that MATH . By passing to a subsequence we can suppose MATH for every MATH . Then for each MATH there is a weakly null tree-map MATH of height MATH so that MATH for all MATH and MATH for every branch MATH . Now let MATH be the tree consisting of all sets MATH where MATH such that if MATH then MATH . We define a weakly null tree-map of height MATH by MATH . Then for every branch MATH we have MATH so that MATH. This implies our claim. Now let us set MATH . Then MATH is convex, and MATH . Further if MATH is weakly null with MATH for all MATH then MATH . Let MATH be the NAME functional of the set MATH . Then it is clear that MATH . Suppose MATH and MATH is a weakly null sequence with MATH . Then MATH . Thus MATH . Now MATH and so from the convexity of MATH it follows that MATH whence MATH . |
math/9911017 | The proof is almost identical to the proof of the preceding REF , except that one considers MATH for arbitrarily large choices of MATH . We omit the details. |
math/9911017 | Suppose MATH is a weakly null tree-map of height MATH with MATH and so that MATH for every branch MATH . Fix MATH so that MATH . For any MATH with MATH we choose MATH with MATH and MATH . If MATH set MATH where MATH and MATH. We now define MATH by backwards induction so that for each MATH, MATH is a weak-MATH-cluster point of MATH . It is then easy to apply induction to produce a full tree MATH so that MATH . Now let MATH when MATH, and MATH so that MATH is a weak-MATH-null tree-map in MATH of height MATH . Let MATH so that MATH . We thus have on every branch MATH of MATH that MATH and so MATH . Now using REF we can pass to a further full tree MATH so that for suitable MATH we have MATH if MATH for MATH . Then by REF we have that MATH . Next it is clear that we can pass to a further full subtree MATH so that for every MATH we have MATH (since MATH is weakly null). Finally we use REF to produce a full tree MATH so that for any branch MATH . Now for any branch MATH in MATH we have MATH . Hence MATH . Thus we have MATH . |
math/9911017 | Suppose MATH. Then by REF and the preceding REF we have that if MATH then MATH . Thus MATH . Then for any MATH if MATH we have MATH . Hence MATH . Since MATH is convex this implies that MATH REF-dominates MATH . Now by REF we have that MATH REF-dominates MATH and hence MATH REF-dominates MATH. Thus MATH NAME MATH with MATH. By the remarks after REF this means that MATH and MATH are NAME. Now recall that MATH for MATH by REF . Thus for MATH we have MATH . It follows easily that MATH is REF-equivalent to a function MATH with the property that MATH is increasing, which is then REF-equivalent to a convex function. Hence MATH is REF-equivalent to a convex function and hence also to MATH . Now MATH REF-dominates MATH and hence REF-dominates MATH . Thus MATH is REF-dominated by MATH and thus MATH-dominated by MATH . Hence MATH is MATH -dominated by MATH and so MATH is MATH -equivalent to MATH . |
math/9911017 | Let MATH . Then arguing as with MATH we have that MATH is REF-equivalent to MATH . We next note that by REF if MATH we have that if MATH then MATH . Reasoning as above gives MATH . Hence MATH REF-dominates MATH. By REF this implies that MATH is MATH-dominated by MATH for a suitable absolute constant MATH . This clearly yields the result. |
math/9911017 | Fix MATH . In this case we have that MATH for some constant MATH . By REF we have a similar estimate MATH . Hence, for a suitable constant MATH for each MATH there is a norm MATH on MATH which is REF-equivalent to the original norm and such that if MATH, MATH with MATH weak-MATH-null then MATH . Now define the (dual) norm MATH on MATH by MATH . This clearly defines an equivalent dual norm on MATH. Thus there is a uniform constant MATH so that for every MATH we have MATH . Suppose MATH and MATH with MATH weak-MATH-null. Pick MATH so that MATH . Then MATH and MATH . Hence MATH . This implies that MATH for suitable MATH . |
math/9911017 | We begin by noting that MATH is separable. From REF and the fact that MATH for all MATH it follows that we can replace the original norm with an equivalent norm on MATH so that there exists MATH with the property that if MATH and MATH is weakly null then MATH . Using this property we make an estimate of MATH . If MATH there is weakly null tree-map MATH of height MATH so that MATH for every MATH but MATH on every branch. We define a second tree-map MATH by MATH if MATH and MATH if MATH and MATH if MATH . If MATH then MATH . It follows easily from REF that there is a full subtree MATH so that on every branch we have MATH . In particular MATH . Iterating we have MATH for all MATH. The estimate on MATH clearly implies an estimate of the type MATH where MATH . By REF we obtain a dual estimate MATH where MATH . Note that MATH . We now use REF . First note that for a suitable constant MATH we have MATH . Recall we also have an estimate from REF MATH . Now pick MATH so that for every MATH we have MATH . This is possible by the growth condition on MATH as MATH . Then MATH . This implies the Theorem. |
math/9911017 | We first note that MATH follows directly from the second part of REF . Next we prove that REF . Assume that REF does not hold. Let MATH . Then we have MATH . Thus assuming REF we conclude an estimate MATH which gives a contradiction. Next we note that REF is immediate from REF . It remains therefore to show that REF . Note first that we can assume that MATH is a monotone increasing function. It will be sufficient to prove the result with MATH replaced by MATH . We note that by REF we have that MATH for some MATH . Hence if MATH there is, for each MATH a REF-renorming MATH of MATH so that MATH for MATH and MATH whenever MATH and MATH is a weak-MATH-null sequence in MATH with MATH . Now using the same idea as in the preceding proof we define a dual norm on MATH by MATH . It follows from our definition that MATH is well-defined and REF-equivalent to the original norm. Now suppose MATH are such that MATH, MATH and MATH is weak-MATH-null. Pick MATH so that MATH . Then if MATH we have MATH and MATH . Hence we obtain by convexity MATH . Summing over MATH we have MATH . Hence for a suitable constant MATH we have MATH . This completes the proof. |
math/9911017 | Let MATH be a uniform homeomorphism with MATH . For any MATH define MATH so that MATH . Thus MATH and MATH . Hence for any MATH we have MATH and MATH . It is clear that if we take MATH large enough and MATH small enough, we obtain the lemma. |
math/9911017 | Let MATH . We assume that MATH is a uniform homeomorphism of MATH onto MATH which satisfies: MATH . This is possible by REF . Recall that MATH . We first define a decreasing sequence of dual norms MATH on MATH by MATH . It is clear that we have MATH for every MATH . We will prove the following Claim concerning the norms MATH: Claim: suppose MATH and MATH are such that MATH, MATH is weak-MATH-null and MATH for all MATH . Then if MATH . To prove the claim we first define MATH so that MATH . Let us also for convenience of exposition write MATH . Now for any MATH we can choose MATH so that MATH and MATH . As usual in such arguments we can suppose by using translations that MATH and MATH . Next we apply REF and the separability of MATH. We deduce the existence of a finite codimensional subspace MATH of MATH so that MATH and MATH . Now, by assumption we have MATH if MATH . Hence we apply the NAME principle, REF for MATH by choice of MATH and MATH . We deduce that there is a compact subset MATH of MATH so that: MATH . It follows from REF that there is a sequence MATH so that MATH . Now MATH so that by REF we have MATH . But MATH . Hence we deduce MATH . Combining these estimates gives: MATH . The left-hand side is estimated by MATH . Hence since MATH is arbitrary and MATH . Our choice of MATH gives MATH and so REF is proved. The completion of the argument from the claim is easy. If MATH the original norm will suffice. Otherwise choose MATH so that MATH . Pick MATH so that MATH . Pick an integer MATH so that MATH . Now let MATH which clearly defines a dual norm on MATH with MATH . Assume MATH and that MATH is weak-MATH null with MATH . Then MATH and MATH . Hence for each MATH we have REF . Summing gives MATH . Now MATH so that the theorem is proved. |
math/9911017 | We only provide a sketch. In this case one can use the norm on MATH defined by MATH . The calculations are similar but rather simpler. We leave the details to the reader. |
math/9911017 | We note first that the relevant functionals are separable determined CITE. Hence we may assume that MATH and MATH are separable. The result is then an almost immediate deduction from REF . In fact it is immediately clear that the statement of the Theorem yields that for the original norm on MATH . If we then consider all MATH-equivalent norms on MATH we deduce an estimate of the form MATH for a different constant MATH . However by REF this implies an estimate MATH . This estimate and its converse establish both REF . |
math/9911017 | Since any subspace of MATH has summable NAME index then MATH must also have summable NAME index by REF . For the second part note that MATH must be a MATH-space CITE and so REF yields that MATH is isomorphic to MATH CITE. |
math/9911017 | We use the ideas of CITE: in particular we use the the NAME distance for two NAME spaces MATH for which we refer to CITE. From REF we have the fact that if MATH is a sequence of NAME spaces converging to MATH in MATH then the NAME distance MATH . It therefore suffices to show that MATH implies MATH . To achieve this suppose that MATH . Then for any MATH we use REF to produce a uniform homeomorphism MATH such that MATH and MATH . Now define MATH by MATH and MATH by MATH . Then if MATH and MATH . Now MATH and a similar reverse inequality gives MATH . Since MATH we obtain MATH . This implies CITE that MATH . The result then follows. |
math/9911017 | In both cases we have MATH where MATH . We therefore deduce by REF that MATH . Now we use the ``standard ultraproduct technique" (compare CITE p. REF or CITE). The spaces MATH and MATH are super-reflexive and hence we can find NAME separable spaces MATH so that MATH is one-complemented in an ultraproduct MATH and MATH is one-complemented in MATH . Then MATH embeds complementably into MATH and MATH embeds complementably in MATH . Since MATH and MATH are complemented in their ultraproducts this leads to the fact that MATH is a quotient (respectively a subspace) of MATH. In the case when MATH is a subspace of MATH we can complete the argument very simply. Since MATH if MATH does not embed into MATH then MATH embeds complementably in MATH CITE and so MATH for some MATH which yields a contradiction. Now consider the case when MATH is a quotient of MATH so that MATH is a quotient of MATH . We have MATH . It follows from REF that MATH for a suitable constant MATH . We will argue that this implies MATH is of type MATH-Banach-Saks in the sense of CITE, that is, there is a constant MATH so that every normalized weakly null basic sequence MATH has a subsequence MATH satisfying MATH . To do this we note that by the definition of MATH there is, for each MATH an equivalent norm MATH on MATH so that MATH for MATH and MATH if MATH is weakly null and MATH . Now if MATH is a normalized weakly null basic sequence in MATH, we pass to a subsequence MATH such that if MATH is such that MATH then MATH . For fixed MATH, let MATH . Let MATH be the greatest integer with MATH so that MATH . Provided MATH we have MATH . Hence for MATH we have MATH and so MATH . We can now combine REF (or REF ) of CITE to deduce that MATH is isomorphic to a quotient of MATH . |
math/9911019 | Let MATH . It follows by the second remark after REF , that MATH is closed in MATH and therefore NAME. Hence, there exists MATH such that either MATH, or, MATH. If the former, then MATH, for all MATH. If the latter, then MATH, for all MATH. |
math/9911019 | We first consider the case of MATH being a successor ordinal, say MATH. Let MATH and MATH. We claim that there exists MATH such that MATH . Indeed, if this is not the case, we obtain through REF , MATH such that MATH . Since MATH is MATH large, there exists MATH such that MATH where MATH. We also have that MATH - norms MATH and therefore, there exists MATH such that MATH. It follows that MATH, and thus, since MATH is hereditary, we can assume that MATH and MATH. Set MATH which belongs to MATH, and choose MATH minimal with respect to MATH. Now, MATH, as MATH and so it belongs to MATH. Therefore MATH and thus MATH which contradicts the choice of MATH. Hence our claim holds and we can inductively choose MATH . Next choose MATH with MATH, for all MATH. We set MATH and claim that MATH. Indeed, let MATH and assume that MATH. Then MATH, where MATH belong to MATH and MATH. Applying REF , we obtain a finite sequence MATH, MATH, of consecutive maximal MATH subsets of MATH with MATH and such that MATH . Note that MATH, for MATH and so there exists MATH such that MATH . We now obtain that MATH, as MATH. This completes the proof for the case of a successor ordinal MATH. Let now MATH be a limit ordinal and assume that the assertion of the theorem holds for all ordinals smaller than MATH. Let MATH be the sequence of ordinals associated to MATH. We can now choose by the induction hypothesis, MATH such that MATH, for all MATH. Suppose that MATH, for all MATH, and choose MATH such that MATH and MATH, for all MATH. Set MATH and it is easy to see that MATH. |
math/9911019 | Define MATH . Evidently, MATH is closed in MATH and therefore NAME, for all MATH. Of course, by our assumption, we have that MATH . We may now choose MATH and MATH such that MATH. Let MATH, and clearly MATH. |
math/9911019 | By induction on MATH. The case MATH has been settled in REF . So assume MATH. Choose MATH a countable dense subset of MATH. Clearly, MATH satisfies the same assumptions as MATH does in the hypothesis of REF . Therefore, without loss of generality, we shall assume that MATH itself is countable. Let MATH be an enumeration of the elements of MATH. We claim that there exist: CASE: A sequence MATH of elements of MATH such that MATH and MATH, for all MATH. CASE: A decreasing sequence MATH of infinite subsets of MATH such that MATH and MATH, for all MATH, and all MATH. Indeed, suppose that MATH and MATH satisfying REF have been constructed. Let MATH such that MATH and MATH. Define MATH . Let MATH, where the union is taken over all possible subsets MATH of MATH which belong to MATH and satisfy the relation MATH. Of course, MATH is hereditary and the hypothesis of REF is satisfied by the family MATH, the integer MATH and the set of measures MATH. By the induction hypothesis there exists, for all MATH, MATH such that MATH and MATH. Next choose according to REF , MATH and MATH such that MATH, MATH and MATH. It follows now that MATH, for all MATH. This completes the inductive construction and our claim holds. Let now MATH be a finite subset of MATH and set MATH . We let MATH, if MATH is not contained in MATH, for all MATH. Inductively we construct a sequence of positive integers, MATH, in the following manner: Suppose that MATH and MATH have been constructed. ( MATH is chosen arbitrarily in MATH and MATH.) We set MATH . Next choose MATH such that MATH and MATH, where MATH, for all MATH. Let MATH. We now claim that if MATH is contained in MATH, for some MATH, then MATH, for all MATH such that MATH. Indeed, let MATH. Then MATH and MATH. It suffices to show that MATH for all MATH such that MATH. Our claim will then follow by REF above. To this end, let MATH, such that MATH. Suppose that MATH, where MATH belongs to MATH. Now, for all MATH, MATH . So, MATH . Hence, MATH, as claimed. Finally, consider the hereditary family MATH . Clearly, the hypothesis of REF is satisfied by MATH and the set of measures MATH. We can thus find MATH such that MATH. It is now easily verified that MATH. |
math/9911019 | It suffices to show the following: Claim: Let MATH. For all MATH, MATH and MATH, there exists MATH such that MATH. Indeed, assuming our claim holds, we observe that the set MATH, is closed in MATH and therefore NAME. Our claim now yields the existence of MATH such that MATH, for every MATH. By stability, we obtain the assertion of the proposition. We shall prove our claim by transfinite induction on MATH. If MATH, then MATH and the claim is easily verified. Assuming our claim holds for all ordinals smaller than MATH, let first MATH be a limit ordinal. Let also MATH be the sequence of ordinals associated to MATH. Suppose now that MATH and choose MATH so that MATH . We apply the induction hypothesis on the ordinal MATH and the set MATH to obtain MATH, infinite subsets of MATH such that MATH . By stability property MATH, there exists MATH with MATH and such that MATH . Now, MATH . Hence, MATH . So our claim holds if MATH is a limit ordinal. Suppose now that MATH. If MATH, choose according to the induction hypothesis MATH such that MATH. Let MATH and choose again by the induction hypothesis, MATH, infinite subsets of MATH such that MATH . But once again, by stability, there exists MATH with MATH and such that MATH . Now, MATH, and thus, MATH. The final case to consider is when MATH. Let MATH be the sequence of ordinals associated to MATH. Choose MATH such that MATH. Set MATH. It follows that MATH . Let MATH. Choose according to the induction hypothesis MATH with MATH and such that MATH . Let MATH. Successive repetitions of the previous argument yield MATH such that if MATH for MATH, then MATH . Stability now guarantees the existence of MATH with MATH and such that MATH . Now, MATH and it remains to show that MATH. Indeed, let MATH and choose MATH minimal with respect to MATH. Let MATH and observe that MATH. Choose MATH such that MATH. There exist MATH and MATH consecutive members of MATH such that MATH. Note that MATH . Thus, MATH . Therefore, MATH . Hence, MATH and so MATH. The proof of the claim is now complete. |
math/9911019 | REF and the fact that MATH is MATH large, immediately yield that MATH and MATH satisfy the hypothesis of REF . The assertion of the corollary now follows. |
math/9911019 | Choose first MATH, a sequence of positive scalars such that MATH. Let MATH and MATH, if MATH. By a MATH-tuple of positive integers MATH, we shall either mean the empty tuple, if MATH, or, that MATH, if MATH. Let now MATH and MATH. The MATH-tuple MATH and the infinite subset MATH of MATH, (MATH), are said to satisfy property MATH, provided that MATH, if MATH, and the following statement holds: If MATH, and there exists MATH which is MATH-good for MATH, then there exists MATH which is MATH-good for MATH and such that MATH . Let us also say that MATH and MATH satisfy property MATH, if they satisfy property MATH, for every MATH. We shall inductively construct an increasing sequence MATH of elements of MATH, and a decreasing sequence MATH of infinite subsets of MATH with MATH, if MATH, and so that for every MATH, if MATH, then MATH and MATH satisfy property MATH. The first inductive step is similar to the general one and therefore we shall not explicit it. Assume that MATH and MATH, infinite subsets of MATH, with MATH for MATH have been constructed so that if MATH and MATH, then MATH and MATH satisfy property MATH. Let MATH. Fix MATH and define MATH . Clearly, MATH is closed in MATH and therefore NAME. Suppose that for some MATH, MATH, we had that MATH. Let MATH and set MATH . Since MATH, for all MATH, there exist integers MATH and functionals MATH in MATH so that letting MATH, we have that for all MATH and moreover, if MATH is MATH-good for MATH then, MATH . Next choose MATH such that MATH. We observe that if MATH is MATH-good for MATH and MATH, for all MATH, then MATH is also MATH-good for MATH, for every MATH. Now let MATH and note that MATH . By the induction hypothesis, since MATH and MATH satisfy property MATH, there exists MATH-good for MATH and such that MATH . Thus, MATH . Without loss of generality, since MATH is bimonotone, we can assume that MATH, for all MATH. Our previous observation yields that MATH is MATH-good for MATH, for all MATH, and thus, MATH . Hence, MATH, for all MATH. We have reached a contradiction since MATH is weakly null and MATH was arbitrary. Concluding, there exists MATH such that MATH. By repeating the previous argument successively over all possible subsets of MATH, we obtain MATH such that MATH, for all MATH. This completes the inductive construction. Set MATH. Let MATH and MATH and suppose that there exists MATH which is MATH-good for MATH. Now let MATH and MATH. Our construction yields that MATH and MATH, satisfy MATH. We also have, by stability, that MATH where MATH and therefore there exists MATH which is MATH-good for MATH and such that MATH . But MATH is bimonotone and thus we can assume that MATH for all MATH. Hence, MATH, as desired. |
math/9911019 | Let MATH. Our goal is to find MATH and a constant MATH such that if MATH, MATH, and MATH are scalars in MATH, then MATH for all MATH. If this is accomplished, then a simple diagonalization argument yields MATH which works for all MATH. We let MATH, for all MATH. Inductively we construct a decreasing sequence MATH of infinite subsets of MATH such that for all MATH, MATH satisfies the conclusion of REF for MATH and MATH. Next choose MATH so that MATH, for all MATH. Let MATH. We shall show that MATH is the desired. To this end, let MATH and scalars MATH in MATH. Let also MATH such that MATH, for all MATH. Choose MATH so that MATH and set MATH. We claim that MATH . Assume this is not the case and choose MATH such that MATH . We can further choose MATH such that MATH . Observe that MATH, when MATH, and thus by stability, there exists MATH such that MATH . But now, MATH . Thus, MATH and since MATH is bimonotone, we also have that MATH which is a contradiction. Therefore, our claim holds and hence MATH . Concluding, MATH where, MATH. The proof of REF is now complete. |
math/9911019 | The proof is similar to that of REF . Let us now say that the functional MATH is good for MATH, if MATH, for all MATH. Next choose MATH, a sequence of positive scalars such that MATH. Using the same notation and terminology as in REF , let MATH and MATH. The MATH-tuple MATH and the infinite subset MATH of MATH, (MATH), are said to satisfy property MATH, provided that MATH, if MATH, and the following statement holds: If MATH, and there exists MATH which is good for MATH, then there exists MATH which is good for MATH and such that MATH . Let us also say that MATH and MATH satisfy property MATH, if they satisfy property MATH, for every MATH. We shall inductively construct an increasing sequence MATH of elements of MATH, and a decreasing sequence MATH of infinite subsets of MATH with MATH, if MATH, and so that for every MATH, if MATH, then MATH and MATH satisfy property MATH. The first inductive step is similar to the general one and therefore we shall not explicit it. Assume that MATH and MATH, infinite subsets of MATH, with MATH for MATH have been constructed so that if MATH and MATH, then MATH and MATH satisfy property MATH. Let MATH. Fix MATH and define MATH . Clearly, MATH is closed in MATH and therefore NAME. Arguing as in the proof of REF , we obtain MATH such that MATH, for every MATH. Indeed, we need only modify the definition of MATH in the argument of REF . We alternatively set MATH and observe that if MATH is good for MATH, then MATH is also good for MATH, for every MATH. The argument of REF is now carried over unaltered yielding the proof of REF . |
math/9911019 | Again, the proof is much similar to that of REF . Choose first MATH, a sequence of positive scalars such that MATH. Let MATH be an element in MATH. We shall say that the functional MATH is MATH-good for MATH, where MATH is an infinite subset of MATH, if MATH. Using the same notation and terminology as in REF , let MATH and MATH. The MATH-tuple MATH and the infinite subset MATH of MATH, (MATH), are said to satisfy property MATH, provided that MATH, if MATH, and the following statement holds: If there exists MATH which is MATH- good for MATH, then there exists MATH which is MATH-good for MATH and such that MATH . Let us also say that MATH and MATH satisfy property MATH, if they satisfy property MATH, for every MATH. We shall inductively construct an increasing sequence MATH of elements of MATH, and a decreasing sequence MATH of infinite subsets of MATH with MATH, if MATH, so that for every MATH, if MATH, then MATH and MATH satisfy property MATH. The first inductive step is similar to the general one and therefore we shall not explicit it. Assume that MATH and MATH, infinite subsets of MATH, with MATH for MATH have been constructed so that if MATH and MATH, then MATH and MATH satisfy property MATH. Let MATH. Fix MATH and define MATH . Clearly, MATH is closed in MATH and therefore NAME. Arguing as in the proofs of REF we obtain MATH such that MATH, for every MATH. The inductive construction is now complete and we set MATH. It follows, by our construction, that if MATH and MATH is MATH-good for MATH, then there exists MATH-good for MATH and such that MATH. Let us then say that MATH works for MATH. Finally, let MATH be a finite MATH-net in MATH, and choose MATH which works for every MATH. It is easily verified that MATH is the desired. |
math/9911019 | Let MATH. It is enough to find MATH , MATH, so that if MATH and MATH are scalars satisfying MATH and MATH, then, MATH, for all choices of scalars MATH with MATH and such that MATH, for all MATH. Once this is accomplished, then a simple diagonalization argument yields MATH which works for all MATH. To this end, let MATH be the infinite subset of MATH resulting from REF applied on the sequence MATH for MATH, MATH and MATH. Let now MATH and MATH be scalars such that MATH and MATH. Choose MATH such that MATH and set MATH . Clearly, MATH. Next, let MATH with MATH and MATH, for all MATH. By splitting MATH into four sets in the obvious manner, we find MATH such that MATH . Now choose MATH with MATH, for all MATH, and such that MATH. It follows that MATH . |
math/9911019 | Suppose REF. holds and let MATH such that MATH . Let MATH denote the sequence MATH, where MATH, for all MATH. Then, MATH, for all MATH. It follows that for MATH, the hereditary family MATH is MATH large. REF now yields MATH so that MATH and thus REF. holds. Assume now that REF. holds and choose MATH as in the proof of REF , applied on the sequence MATH for MATH. Let MATH and scalars MATH such that MATH. We claim that MATH, which evidently yields REF. Indeed, by our assumption, there exists MATH such that MATH, for all MATH. Next choose MATH such that MATH . Therefore, MATH, and hence applying REF we obtain that MATH, as claimed. |
math/9911019 | Assume that REF. does not hold. Let MATH and MATH. It is easily seen that the set MATH where MATH, is closed in MATH and therefore NAME. If it were the case that MATH, for some MATH, then the family MATH would be MATH large, and hence, by REF , MATH would in turn be a MATH spreading model, for some MATH contradicting our assumption. It follows now that we can construct MATH, a decreasing sequence of infinite subsets of MATH, such that for all MATH, MATH . Let now MATH be any infinite subset of MATH almost contained in each MATH, and it is easy to verify that MATH is MATH convergent, for all MATH. In order to show that REF. and REF. are mutually exclusive, assume that MATH is a MATH spreading model with constant MATH. We can choose MATH such that MATH where MATH is chosen so that MATH. It follows now that for every MATH, MATH and thus MATH is not MATH convergent. Hence REF. does not hold. |
math/9911019 | Suppose that MATH is a MATH spreading model with constant MATH. It follows that MATH, for all MATH and MATH. Next, choose according to REF , MATH so that MATH is MATH convergent, for every MATH. Evidently, MATH is NAME summable for all MATH. |
math/9911019 | Let MATH be the constant of the MATH spreading model MATH. Let now MATH, MATH satisfying the conclusion of REF for the sequence MATH and MATH. We claim that MATH is MATH unconditional. Indeed, let MATH and scalars MATH be given. Let also MATH, MATH, such that MATH. It follows that MATH and MATH. If there exists MATH so that MATH, then, since MATH is bimonotone, we obtain that MATH. So assuming that MATH, for all MATH, we obtain through REF that MATH. Hence, MATH for every MATH and all choices of scalars MATH. |
math/9911019 | Choose first MATH such that MATH. By passing to a subsequence, if necessary, we can assume that MATH is NAME basic with basis constant MATH. We first show that for every MATH and MATH, there exists MATH, MATH so that MATH for every MATH and all choices of scalars MATH in MATH. Indeed, apply REF to the sequence MATH to obtain MATH, MATH, satisfying the conclusion of that lemma for MATH, MATH, and MATH. Let now MATH such that MATH. Let also MATH and choose MATH such that MATH . Then choose MATH such that MATH . We now have the following estimate MATH and thus, MATH, as desired. We can now choose a decreasing sequence MATH of infinite subsets of MATH such that for every MATH, MATH for all MATH and MATH, and all choices of scalars MATH. Finally choose MATH with MATH, for all MATH, and set MATH. It is easily verified that the subsequence MATH is MATH unconditional. |
math/9911019 | Once again, we assume our sequence MATH is bimonotone. We can also assume, without loss of generality, that MATH . Furthermore, we shall assume that MATH satisfies the conclusion of REF for MATH and MATH. That is, for all MATH. there exists a constant MATH such that for every MATH and all scalars MATH in MATH, if MATH and MATH for all MATH, then MATH . Assume first that REF. holds and choose MATH, MATH, and MATH so that MATH for every MATH and all choices of scalars MATH. We claim that MATH is a MATH spreading model. Indeed, let MATH and scalars MATH. For each MATH, let MATH be the sign of MATH. Of course, MATH. Therefore, MATH and thus REF. holds. Suppose now that REF. holds. Choose, according to REF , MATH, MATH and MATH such that MATH . It follows that MATH . We now claim that for every MATH, the sequence MATH, where MATH, is strongly bounded. Indeed, let MATH and note that by monotonicity we have that MATH, for all MATH. Then, for all MATH, MATH and our claim holds. Let now MATH and choose MATH such that MATH, for all MATH. Since MATH, for all MATH, our initial assumptions on the sequence MATH yield that MATH where MATH. Thus, letting MATH, for MATH, we have that MATH and MATH, for all MATH and MATH. A standard argument now shows that MATH for all MATH and scalars MATH. Hence, REF. implies REF. |
math/9911019 | Assume first that REF. holds. Let MATH be a subspace of MATH and consider the normalized weakly null sequences MATH and MATH in MATH and MATH respectively, with MATH-convergent. Suppose that for some MATH and MATH, it was the case that MATH, for all MATH. It follows that REF is satisfied and therefore MATH admits a subsequence which is a MATH spreading model in MATH. This contradicts with REF , as MATH is MATH-convergent. Hence MATH satisfies the MATH-DP and REF. holds. Now suppose that REF. holds. Let MATH be a normalized weakly null sequence in MATH admitting no subsequence which is a MATH spreading model. In particular, no subsequence of MATH is equivalent to the unit vector basis of MATH, and thus by REF there exists MATH such that the sequence MATH is weakly null in MATH. Where MATH denotes the sequence of the biorthogonal functionals of MATH. Our assumption further yields that REF fails for the space MATH and the weakly null sequences MATH and MATH in MATH and MATH respectively. Thus REF . fails as well and so there exists MATH so that MATH is MATH-convergent, according to REF . But MATH is MATH-DP and thus, MATH which is absurd. |
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