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math/0005150
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The standard matrix units basis MATH of MATH also satisfies the hypotheses of REF with MATH.
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math/0005150
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Suppose this were false. Then we could choose MATH and MATH subspaces of MATH so that MATH is a MATH-space for all MATH. Choose then MATH a MATH matrix of elements of MATH, satisfying REF, for all MATH. Let MATH denote the linear space of all infinite matrices of scalars with only finitely many non-zero entries. Let MATH be a free ultrafilter on MATH. Define a semi-norm MATH on MATH by MATH . It is easily checked that MATH is indeed a semi-norm; let MATH be its null space; MATH, and let MATH denote the completion of MATH. It follows easily that MATH is a MATH-space. By standard ultraproduct techniques, it follows that MATH is finitely representable in MATH. But then (since ultraproducts of (commutative) MATH spaces are (commutative) MATH spaces and any separable subspace of a MATH space is isometric to a subspace of MATH), MATH isometrically embeds in MATH. This contradicts REF .
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math/0005150
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Let MATH and MATH be as in the statement. REF is a trivial consequence of the fact that MATH is a norm (that is, the triangle inequality). Also, we easily obtain that MATH and in case MATH, MATH . Indeed, if MATH, then MATH which immediately yields the equality in REF. Since compression reduces the MATH norm, we have MATH which gives the inequality in REF. If MATH and MATH, then setting MATH, MATH yielding the inequality in REF is trivial, since for any MATH, MATH . For the non-trivial assertions of the Lemma, we need the following basic identities (compare CITE, CITE). MATH . (The final inequality is also an equality, but this follows from the conclusion of our Lemma.) Now let MATH be self-adjoint. Let MATH denote the NAME algebra generated by MATH, and let MATH be a MASA contained in MATH with MATH. Then by our initial remarks, MATH is atomless. Let us identify (as we may), MATH and MATH with an atomless probability space MATH. It follows that we may choose a countably generated MATH-subalgebra MATH of MATH so that MATH is MATH-measurable and also MATH is atomless. Denote the corresponding NAME algebra by: MATH. It then follows that MATH is measure-isomorphic to MATH (where MATH denotes the NAME subsets of MATH and MATH denotes NAME measure on MATH), and moreover the measure-isomorphism may be so chosen that the ``random-variable" MATH is carried over to the decreasing function MATH (compare REF). It now follows that MATH . Indeed, it follows that there exists a set MATH with MATH and MATH (where MATH may be interpreted as the projection in MATH obtained via multiplication). Now we define a quantity MATH (depending on MATH) by MATH . We are going to prove that there exists a MATH with MATH and MATH . Note that the first equality in REF is trivial, since MATH. But then all the equalities in REF for the case MATH, follow immediately, for we have also that then MATH and so trivially MATH and MATH; of course also MATH, hence by REF, MATH. Moreover by the argument for REF we have that MATH. Before proving this basic claim, let us see why it also yields REF for MATH (via the NAME). Still keeping MATH fixed, assume MATH, and suppose MATH with MATH. Now setting MATH, MATH is self-adjoint and ``supported" on MATH, whence it easily follows that MATH for MATH. But now we obtain that MATH . Indeed, MATH (since MATH, MATH, and MATH). Now (temporarily) unfixing MATH, we also have that REF holds for MATH, since MATH for all MATH. Thus the NAME yields that MATH . Hence in view of REF, MATH and so at last MATH . Of course REF combined with REF now yields that MATH and now all the equalities in REF follow for MATH as well. We now establish REF. Using the polar decomposition of MATH and duality, we have that MATH . The last equality follows by a conditional expectation argument from classical probability theory. Indeed, given MATH in MATH with MATH, there exists a unique MATH such that MATH . It follows that then MATH and MATH; this yields the desired equality. Now let MATH be defined: MATH . Then MATH is a weak* compact convex set, thus MATH and moreover MATH . Now we claim that if MATH, MATH is a projection. To see this, again identifying MATH with MATH, we regard MATH as a MATH-measurable function on MATH. Were MATH not a projection, we could choose MATH so that setting MATH, then MATH. Since MATH is atomless, choose a measurable MATH with MATH. Now define MATH by MATH . Then MATH, MATH, and MATH. But then MATH, hence MATH and MATH, contradicting the fact that MATH. (For a proof of this claim in a more general setting, see CITE.) We finally observe that the supremum in REF is actually attained, thanks to the MATH-compactness of MATH. But it then follows that this is attained at an extreme point of MATH, that is, there indeed exists a MATH with MATH, satisfying REF. We may now also easily obtain REF. Letting MATH where MATH and MATH, we have (by the proof of REF) MATH . The first equality in REF follows from the fact that for a general MATH affiliated with MATH, there exists a unitary MATH in MATH with MATH (thanks to the finiteness of MATH). But then MATH and MATH are unitarily equivalent, which yields that MATH for all MATH, and hence the desired equality follows by the final equality in REF. It remains to prove the last inequalities in REF, and the final statement of the lemma. Let MATH with MATH and MATH self-adjoint (and so in MATH). Then MATH . But if MATH or MATH, then MATH . Indeed, if MATH, MATH, then MATH. But MATH and MATH are both self adjoint, hence MATH, yielding REF. Of course REF yield the final inequality in REF. Similarly, in case MATH, MATH since we also have for MATH or MATH, that MATH (by an argument similar to that for REF). To obtain the final assertion of the lemma, let MATH, and let MATH. Now if MATH then since MATH we have MATH. Thus MATH . Hence MATH . Now also by the definition of MATH, MATH, and so MATH . Finally, let MATH be the polar decomposition of MATH. In particular, MATH is a partial isometry belonging to MATH. Then MATH satisfies REF. Indeed, MATH and MATH, so also MATH, and MATH .
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math/0005150
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Once REF is established, the other equivalences in this Proposition follow easily from REF. Indeed, we have by the equalities in REF that MATH whence we have the equivalence of REF - REF . Now trivially MATH since MATH for any MATH and MATH (see REF). Suppose first that MATH satisfies REF . Then given MATH, for each MATH we may choose MATH, MATH, with MATH . But then for any MATH, MATH . Hence MATH, proving REF . On the other hand, suppose REF holds. Let MATH, and choose MATH so that MATH . Also, let MATH. Then setting MATH, it follows by the final statement of REF that for each MATH, we may choose MATH with MATH proving MATH . To prove the equivalences of REF , we use the following classical criterion due to CITE: A bounded set MATH in the predual of a NAME algebra MATH is relatively compact if and only if for any sequence MATH of disjoint projections in MATH, MATH . Now suppose first that MATH is not relatively weakly compact; then choosing disjoint MATH's as in the above criteria, we obtain that MATH . But MATH, since the MATH's are disjoint. It follows immediately that MATH which together with REF, proves that MATH . Finally, to show that MATH , assume instead that MATH, where MATH is given in REF. It now suffices to demonstrate the final assertion of REF, for then MATH is not relatively weakly compact by NAME 's criterion. Let MATH with MATH. By REF, choose MATH with MATH . Then choose (by REF ), MATH with MATH and MATH . Since MATH is integrable, MATH is uniformly integrable, so we may choose MATH so that MATH . Next by REF, choose MATH with MATH . (It is easily seen, thanks to the uniform integrability of finite sets in MATH, that in fact MATH; thus we may insure that MATH may be chosen larger than MATH.) Again using REF, choose MATH with MATH and MATH . Then choose MATH so that MATH . Continuing by induction, we obtain MATH, MATH, and projections MATH in MATH so that for all MATH, MATH and MATH . Now set MATH, for MATH. Evidently the MATH's are pairwise orthogonal. For each MATH, let MATH. Now by subadditivity of MATH, MATH . But MATH . Hence we have MATH . Thus by REF , MATH . Hence MATH .
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math/0005150
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Let MATH be the subset, and suppose first MATH is relatively weakly compact, yet MATH. Then for each MATH, choose MATH with MATH. It follows immediately that also MATH, hence MATH is not relatively weakly compact by REF . If MATH is uniformly integrable, then MATH is bounded, and then MATH is relatively weakly compact by NAME 's criterion, (stated preceding REF).
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math/0005150
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MATH follows immediately from the (obvious) inequality MATH (stated as part of REF ). MATH . Assume that MATH for all MATH. For MATH sufficiently large, define MATH by MATH . Let MATH. Since MATH, by the final assertion of REF , we may choose MATH a spectral projection for MATH so that MATH . It follows immediately that MATH . Thus REF holds by REF , since MATH as MATH. (Note also that the final assertion of REF does not involve the ``atomless" hypothesis, since MATH is defined in terms of the spectral measure for MATH.) MATH . Given MATH and MATH, choose MATH with MATH . Then for any MATH, MATH . Hence MATH, proving that REF holds, since MATH is arbitrary and MATH for any MATH and MATH, by REF .
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math/0005150
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Suppose MATH is MATH-unconditional. Then MATH is MATH-equivalent to MATH in MATH, so it suffices to prove the same conclusion for MATH instead. Let MATH, where MATH is NAME measure on MATH. We may also assume without loss of generality that MATH for all MATH. Now let MATH, and choose MATH so that MATH (using that MATH is uniformly integrable). By the final statement of REF , we may by REF choose for each MATH a MATH so that MATH with MATH . Then fixing MATH, MATH . But MATH since MATH for all MATH. On the other hand, since MATH is type MATH with type MATH constant REF for any NAME algebra MATH, MATH (This fact follows by NAME 's inequalities - see the discussion in the proof of the next lemma.) We thus have that MATH by REF. Since MATH is arbitrary, the conclusion of the lemma follows.
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math/0005150
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We first note that (using interpolation), MATH satisfies NAME 's inequalities: for all MATH, MATH . It follows immediately by induction on MATH that MATH is type MATH with constant one; that is, for any MATH in MATH, MATH . Now let scalars MATH be given, and let MATH. We obtain from REF that since MATH is MATH-unconditional, MATH . Now fix MATH and set MATH. Then MATH . Thus integrating over MATH and again using unconditionality, MATH . But fixing MATH, since MATH is cotype REF with constant at most MATH, MATH . Thus in view of REF, MATH so REF now imply the conclusion of REF .
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math/0005150
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By REF , (see REF), we have, fixing MATH, that MATH . Hence we may choose MATH with MATH . Define projections MATH and MATH by MATH . Then MATH . Now we have by subadditivity of MATH that MATH, and so again by subadditivity, MATH . Thus MATH. Hence we have MATH . By the same argument, MATH . Thus from REF, we obtain MATH . Of course MATH are pairwise orthogonal; hence REF now immediately yields the conclusion of REF.
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math/0005150
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MATH follows immediately from REF . Assume that REF holds and also assume without loss of generality that MATH for all MATH. Then by REF , MATH . Now REF yields that there is a subsequence MATH of MATH so that MATH where MATH is an absolute constant. Indeed, fix MATH. Choose MATH and MATH so that MATH . Suppose MATH and MATH chosen so that MATH . By continuity of the functions MATH for MATH and the fact that MATH, choose MATH so that MATH . Then choose MATH and MATH so that MATH . This completes the inductive choice of MATH. Setting MATH, then MATH satisfies the hypotheses of REF for all MATH, and hence MATH is MATH-equivalent to the MATH basis by REF. By taking MATH small enough, we obtain MATH in REF.
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math/0005150
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MATH : Assume REF is false. Then by REF there exists a subsequence MATH equivalent to the usual MATH-basis. In particular MATH which contradicts REF . MATH . This follows from REF , since REF implies that MATH is uniformly integrable for any subsequence MATH of MATH.
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math/0005150
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Since MATH is a MATH-homomorphism of MATH onto MATH, we have for any continuous function MATH and any MATH, MATH . Applying this to MATH, we get by the trace REF that MATH . In particular, MATH . Thus MATH extends by continuity to a contraction MATH. Now let MATH belong to MATH, and let MATH. Then choose MATH in MATH so that MATH . It follows from REF that MATH and MATH . Since REF holds, replacing MATH by MATH in its statement, we have from REF that MATH . Since MATH is arbitrary, REF holds for all MATH in MATH.
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math/0005150
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Define for each MATH a function MATH by MATH . For fixed MATH, MATH is MATH-equivalent to the usual MATH-basis, so by REF , MATH is uniformly integrable and MATH . Now REF implies that MATH belongs to MATH. Let MATH be as in the statement of REF and define MATH by MATH . Now we claim that MATH . Indeed, using the hypotheses of REF , we have for any MATH and scalars MATH, that MATH . Now define MATH by MATH . Again by REF , MATH is uniformly integrable in MATH, so by REF we have that MATH . Now let MATH. Since MATH is a quotient map of MATH onto MATH, it follows that fixing MATH, there exists for every MATH, MATH so that MATH . Hence by REF , MATH which implies that MATH . Hence by REF MATH . Since MATH was arbitrary, we get REF.
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math/0005150
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Let MATH, and let MATH be as in REF , and let MATH be a free ultrafilter on MATH. Put MATH . Then MATH is a decreasing function and by REF MATH . We claim that REF imply that for a suitable choice of natural numbers MATH one has MATH . To prove REF put for MATH where for MATH, MATH . By REF each MATH, and hence also MATH for all MATH. Since MATH is a free ultrafilter, each MATH is infinite, so we can choose successively MATH such that MATH for all MATH. Put MATH, MATH and MATH, and put as in REF MATH . To prove REF we just have to show that MATH when MATH (compare REF ). Let MATH. By REF we can choose MATH such that MATH . When MATH, MATH. Hence by REF MATH . Since MATH is dense in MATH we have for every MATH, MATH . Hence, we may choose MATH, such that MATH . By REF, MATH for all MATH. This shows that MATH and hence by REF , MATH is uniformly integrable, that is, REF holds. Thus by the assumption that MATH is unconditional, REF yields that for any subsequence MATH of MATH, MATH . Putting now MATH, we have MATH and REF follows.
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math/0005150
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We may assume without loss of generality that MATH for all MATH. Let MATH be such that MATH, and choose MATH a subsequence of MATH satisfying the conclusion of REF . Let MATH denote the NAME functions on MATH (as defined in REF), set MATH, and let MATH for all MATH. Then MATH is REF-unconditional (over complex scalars) and of course MATH is also uniformly integrable in MATH, whence by REF , MATH . Let MATH, and choose MATH so that if MATH, then MATH and MATH . Now fix MATH, and choose MATH with MATH . Of course then MATH . Now if MATH are given scalars of modulus at most one, then MATH . Indeed, MATH is REF-unconditional by the conclusion of REF (since MATH), yielding REF. On the other hand, MATH . Thus we have MATH . (The last inequality holds by REF; the next to the last by REF and the fact that MATH is REF-unconditional over real scalars.) The uniformity of the limit over all subsequences of MATH follows from the fact that the limit in REF is uniform over all subsequences of MATH, thanks to the proof of REF .
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math/0005150
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Suppose the conclusion were false. Then we could find for every MATH, a MATH-tuple MATH of elements in MATH so that MATH is MATH-basic, yet no MATH terms are suppression MATH-unconditional. By homogeneity, we may assume that MATH for all MATH and MATH. Now let MATH be a non-trivial ultrafilter on MATH and let MATH denote the ultrapower of MATH with respect to MATH. (That is, we let MATH denote the subspace of MATH consisting of all bounded sequences MATH in MATH with MATH, and then set MATH.) Since MATH is uniformly convex, so is MATH. Now define a sequence MATH in MATH by MATH, for all MATH, where MATH is the quotient map and we set MATH if MATH. It then follows that MATH is also MATH-basic and normalized; since MATH is reflexive, MATH is weakly null. But then by REF , there exist MATH terms MATH of this sequence with MATH-suppression unconditional. Standard ultraproduct techniques yield that MATH given, there exists a MATH so that MATH is MATH-equivalent to MATH and hence the latter is MATH-suppression unconditional. Of course we have a contradiction if MATH.
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math/0005150
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We use the same notations and assumptions as in the proof of REF (for example, we assume that MATH for all MATH and MATH). Assume that REF does not occur. Then again we have that MATH is uniformly integrable for all MATH, and hence REF holds. This is also the case if MATH is hyperfinite and the unconditional assumption in REF is dropped. For suppose to the contrary that for some MATH, MATH has the property that MATH is not uniformly integrable. Then setting MATH in MATH (as defined in the proof of REF ), MATH is unconditional and again MATH is not uniformly integrable, hence there exist MATH with MATH equivalent to the usual MATH-basis, by REF ). But MATH has an unconditional subsequence MATH by CITE, CITE. Of course then MATH is equivalent to MATH, a subsequence of MATH, whence MATH is equivalent to the MATH basis. Now replace the entire matrix MATH by MATH in MATH (where MATH), where MATH is just a ``renumbering" of MATH via MATH (precisely, let MATH be a bijection, and set MATH). Now MATH for all MATH, and MATH; hence assuming REF in the proof of REF occurs, we have that II of REF holds for the matrix MATH. But then since MATH is uniformly convex, II holds for MATH itself, by REF . Indeed, let MATH be as in II of REF, let MATH be given. Choose MATH satisfying the conclusion of REF for MATH (with MATH, say). Choose MATH and MATH with MATH-equivalent to the MATH basis where we set MATH and MATH for all MATH. Then choose MATH with MATH REF-unconditional. But then MATH is REF-equivalent to MATH, and is hence MATH-equivalent to the MATH basis. If REF in the proof of REF does not occur, we have by REF that there exists a generalized diagonal MATH of MATH so that MATH is uniformly integrable. Hence immediately, MATH is uniformly integrable, and so by REF , MATH has a subsequence MATH (which is of course also a generalized diagonal) satisfying III of REF. This completes the proof of REF . To obtain REF , we need two further ``NAME properties of preduals of NAME algebras. The first one holds in complete generality. Let MATH be a NAME algebra, and let MATH be a bounded sequence in MATH such that MATH is not relatively weakly compact. Then MATH has a subsequence equivalent to the MATH-basis. We give a ``quantitative" proof of this result at the end of this section, using the case for commutative MATH established in CITE. In fact, REF is due to CITE. However, the second result we need is a ``localization" of our proof, which does not seem to follow directly from previously known material. This result yields that if MATH elements of MATH (MATH finite) have mass at least MATH on pairwise orthogonal projections, then MATH of these are MATH-equivalent to the MATH-basis. Here, MATH depends only on MATH, MATH on MATH and MATH. To make this more manageable, let us simply say that MATH elements MATH of the predual of a NAME algebra MATH are MATH-disjoint provided there exist pairwise orthogonal projections MATH in MATH such that MATH . (Here, if MATH and MATH, MATH is defined by: MATH for all MATH. Also, MATH denotes the predual norm on MATH.) (We shall also say MATH are disjoint provided there are pairwise orthogonal projections MATH in MATH with MATH for all MATH. Evidently if the MATH's are normalized, they are disjoint iff they are REF Given MATH, then if MATH, then for all MATH, there is a MATH so that for any NAME algebra MATH and MATH-disjoint elements MATH in MATH, there exist MATH with MATH-equivalent to the MATH basis. We delay the proof of this result, and complete the proof of REF , that is, the case MATH. Again we make the same assumptions and use the same notation as in the proof of REF . Now suppose that REF does not occur. We now have, immediately from REF , that MATH is uniformly integrable for all MATH, and hence again REF holds. Now again assume REF of the proof REF holds. Then the proof of REF II yields that for all MATH, there exists a row MATH and MATH so that MATH is MATH-disjoint, where MATH for all MATH. Indeed, we obtain there (following REF ), that for all MATH, there is a sequence MATH satisfying the assumptions of REF (for MATH and MATH) except for the MATH-unconditionality assumption. But the proof of REF yields precisely that MATH is MATH disjoint; the unconditionality assumption was only used, in invoking REF . Of course we may choose MATH, and so MATH is then MATH-disjoint. Then REF immediately yields REF . Finally, assuming REF of the proof of REF does not occur, we obtain again from the proof of REF that there exists a generalized diagonal MATH of MATH with MATH uniformly integrable. Hence there exists a weakly convergent subsequence MATH of MATH, by REF . But since we assume the generalized diagonals of MATH are basic sequences, MATH must be weakly null. Now REF immediately yields REF .
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math/0005150
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Let MATH and choose MATH with MATH. Let MATH be as in REF , MATH as in the hypotheses of REF, and choose MATH satisfying the conclusion of REF. Then MATH is MATH-equivalent to the MATH basis by REF .
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math/0005150
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Let MATH be less than or equal to the number of terms in the sequence, and let MATH be given scalars with MATH . Let MATH. Since the MATH's are pairwise orthogonal, we have that MATH . Now fixing MATH, MATH by REF and the triangle inequality. Hence using REF, MATH . This completes the proof.
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math/0005150
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Let MATH and MATH be as in the conclusion of REF B. Let the MATH's and MATH's be as in the statement of REF. For each MATH, define MATH in MATH by MATH for all MATH. Then MATH for all MATH. Now the conclusion of B yields MATH so that MATH . Then MATH satisfies the conclusion of REF .
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math/0005150
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Let MATH and let MATH. Let then MATH be given. Of course the conclusion of REF may be restated: There exists a MATH with MATH so that REF holds. Let MATH be disjoint subsets of MATH, each of cardinality MATH, and just repeat the argument for REF, Case I. If REF fails, we repeat again the rest of the argument: that is, we find MATH satisfying REF and set MATH. Now we just choose MATH disjoint subsets of MATH, each of cardinality MATH; if REF fails for MATH, we continue as before, with MATH satisfying REF and MATH, MATH. If REF fails for MATH steps, we obtain finally MATH with MATH for all MATH, so MATH; and for MATH the unique number of MATH, REF holds, whence again MATH, a contradiction.
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math/0005150
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It follows easily that we may choose MATH a sequence in MATH and MATH so that MATH . Then given MATH, REF A yields a subsequence MATH of MATH so that MATH is MATH-relatively disjoint. Finally, since MATH may be arbitrarily close to MATH and MATH arbitrarily small, we deduce from REF that given MATH, MATH may be chosen MATH-equivalent to the MATH-basis with span MATH-complemented in MATH.
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math/0005150
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Suppose to the contrary that there exists a MATH with MATH completely isomorphic to MATH. But then MATH is completely isomorphic to MATH. Let then MATH. MATH is again a finite NAME algebra, and now MATH is a subspace of MATH; that is, MATH is completely isomorphic to a subspace of MATH. But MATH is (completely isometric to) MATH; this contradicts our main result.
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math/0005150
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We show MATH, MATH, and MATH in case MATH. Of course MATH and MATH are trivial. So is MATH, in virtue of REF . MATH. Fix MATH. Choosing an ``almost isometric" copy of MATH in MATH by NAME 's theorem, we shall show that for MATH large enough, one of the natural basis elements MATH of this copy is such that MATH is almost equal to REF. Define MATH by MATH . Let MATH, and using NAME 's theorem, choose MATH with MATH-equivalent to the MATH basis. In particular, we have that MATH . Again by the final assertion of REF , we may choose for each MATH a MATH so that MATH . Thus letting MATH be as in the proof of REF , again we have MATH by REF and the fact that MATH is type MATH with constant one. Thus MATH . Since MATH and MATH are arbitrary, we obtain that MATH, proving MATH. MATH. We first note that assuming REF, then given MATH, we may choose MATH with MATH and MATH with MATH so that MATH . Indeed, choose MATH in MATH of norm one so that MATH. Then choose MATH a spectral projection for MATH with MATH. But then since MATH commutes with MATH, MATH whence MATH so MATH as desired. Now since MATH and MATH are unitarily equivalent in MATH, we also obtain the existence of a MATH with MATH so that MATH . Then let MATH. We have MATH . Indeed, the first estimate is trivial; but MATH and so REF follows from REF. Now using that for MATH, MATH of norm REF in MATH and MATH may be chosen satisfying REF we choose inductively MATH in MATH of norm one, MATH, and MATH in MATH so that for all MATH, MATH . To see this is possible, just choose MATH, then choose MATH and MATH thanks to REF. Suppose MATH, and MATH chosen. By uniform integrability of MATH, choose MATH so that MATH. Then choose MATH and MATH satisfying REF for MATH. Now define projections MATH and MATH by REF. The MATH's are orthogonal and by the argument for the last part of REF , fixing MATH, we have MATH . Hence MATH (by REF) and also MATH . Hence MATH . Hence finally we have by REF, MATH . Thus MATH is almost disjointly supported, proving that REF holds. MATH is a standard perturbation argument in NAME space theory. Assuming REF holds, we may choose a normalized disjointly supported sequence MATH in MATH and a sequence MATH in MATH so that MATH . But now MATH is REF-equivalent to the MATH-basis, and a simple perturbation argument gives that given MATH, there is a MATH so that MATH is MATH-equivalent to the MATH basis. (Thus MATH is ``almost isometrically equivalent" to the MATH basis.) MATH. We have that if MATH, MATH contains a subspace isomorphic to MATH by REF , so assume MATH. We may choose a sequence MATH of norm-REF elements of MATH, MATH with MATH and MATH so that MATH . By passing to a subsequence, we may assume without loss of generality that MATH is weakly convergent, with weak limit MATH, say. But MATH and hence MATH . That is, we have now obtained a weakly null sequence MATH in MATH so that MATH . By REF , after passing to a subsequence of MATH, we may assume MATH . Now REF yields that for all MATH, there exist MATH so that MATH is REF-unconditional, and hence MATH . This proves that REF holds. Now assume MATH. MATH. Let MATH and choose MATH with MATH and MATH with MATH so that REF holds. Then of course MATH . Now letting MATH, MATH . Thus MATH . Since MATH and MATH is arbitrary, REF holds. MATH. Suppose REF holds, yet REF were false. Choose MATH so that MATH . Let MATH, MATH. By the last statement of REF , choose MATH a spectral projection for MATH so that MATH with MATH . Then MATH . That is, MATH . REF yields that MATH for all MATH; that is, REF does not hold, a contradiction. This completes the proof of the theorem.
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math/0005150
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REF together with REF yields that MATH has the NAME property. It also yields the first assertion in REF. Suppose MATH is not uniformly integrable and assume (without loss of generality) that MATH for all MATH. Applying REF , we may choose a subsequence MATH of MATH so that for some MATH, MATH and MATH . Suppose MATH is a subsequence of MATH. Then it follows that for all MATH, MATH . Let MATH be a ``large" integer and choose MATH with MATH . Then MATH . Thus MATH . Hence MATH . This completes the proof of REF. But we also have that MATH and so MATH thus MATH . This proves that MATH has the MATH-Banach-Saks property, for any weakly null sequence MATH in MATH either has MATH uniformly integrable (and hence a strong MATH-Banach-Saks subsequence), or a subsequence MATH as above. The final assertion of the Proposition follows immediately from REF .
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math/0005150
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Fix MATH a positive integer. Using the generalized parallelogram identity, MATH . It follows that we may choose MATH for all MATH so that MATH . Now simply choose MATH so that MATH and let MATH.
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math/0005150
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Let MATH be the matrix units basis for MATH, and define for each MATH, MATH . Let MATH be the natural basis projection onto MATH; that is, MATH is the projection with MATH if MATH, MATH if MATH (so MATH). Then MATH is isomorphic to MATH, so by the sub-lemma we may choose MATH in MATH with MATH-valued and MATH . We claim that MATH . Of course REF yields the conclusion of the Lemma. Suppose REF were false. It follows that MATH has a subsequence MATH so that MATH and MATH . (REF follows because MATH may be chosen to be a small perturbation of a ``block-off-diagonal sequence", by REF). Of course MATH converges weakly in MATH as well, hence MATH also converges weakly, a contradiction when MATH since then MATH is equivalent to the MATH-basis. When MATH, letting MATH be the weak limit of MATH, we have that MATH since MATH weakly. Moreover MATH, so letting MATH for all MATH, MATH is a uniformly bounded weakly null sequence in MATH with MATH equivalent to the MATH-basis. Finally, since MATH is also semi-normalized in MATH, MATH has a subsequence MATH equivalent to the usual MATH-basis. (Indeed, we may choose MATH equivalent to the MATH-basis in MATH-norm, and unconditional. But then since MATH has cotype REF, MATH is equivalent to the MATH-basis in the MATH-norm). Still, MATH is equivalent to the MATH-basis; this is impossible since MATH.
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math/0005150
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Suppose to the contrary that MATH is a finite NAME algebra and MATH is an isomorphic embedding. Of course we may assume that MATH; let MATH. Thus we have MATH . Let MATH be the projection of MATH onto MATH with kernel MATH, and set MATH. Also, for each MATH and MATH, let MATH be the natural projection of MATH onto the space MATH . (As before, MATH denotes the MATH matrix unit for MATH. Visualizing MATH as matrices of scalars and MATH as all matrices MATH of functions in MATH with MATH then MATH. MATH is just the space of matrices with all entries zero except in the MATH slot). Now fix MATH and MATH. Of course MATH is isometric to MATH. Thus by REF , we may choose MATH with MATH-valued so that MATH . Now letting MATH, then MATH is a MATH space, in the terminology of the Introduction. That is, every row and column of MATH is REF-equivalent to the MATH basis, while every generalized diagonal is REF-equivalent to the MATH basis. Hence MATH is not isomorphic to a subspace of MATH by our Main Theorem (that is, REF ). However MATH . Indeed, if MATH with only finitely many MATH's non zero, and MATH, then MATH for all MATH and MATH (since the MATH's are contractive and MATH for all MATH and MATH), and so MATH using REF and our assumption that MATH is a contraction. Hence MATH . This proves REF, and completes the proof by contradiction.
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math/0005150
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We have that MATH is (linearly isometric to) MATH. Thus it suffices to prove that MATH where MATH and MATH denotes the NAME distance-coefficient (defined just preceding REF ). Now fix MATH, and let MATH be an isomorphic embedding onto MATH, with MATH . Using the notation and reasoning in the proof of REF , and setting MATH, we may choose for each MATH and MATH with MATH, a MATH-valued MATH satisfying REF. We thus obtain that MATH by REF. Hence for all MATH, MATH using also REF. That is, setting MATH, we have that MATH . Now MATH is a MATH-space; thus MATH (in the notation of REF ), so REF holds by REF .
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math/0005150
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By REF , it suffices to prove that MATH does not embed in MATH if MATH. If MATH did embed, then since it does not embed in MATH, it would have a subspace isomorphic to MATH, by REF . However it is a standard fact that every infinite-dimensional subspace of MATH is either isomorphic to MATH or contains a subspace isomorphic to MATH, a contradiction. We next sketch the proof of REF (which also yields the above mentioned standard fact). Let MATH be a given matrix in a linear space MATH. Call a sequence MATH in MATH a generalized block diagonal of MATH if there exist MATH and MATH so that for all MATH, MATH . Now if MATH is a generalized block diagonal of the matrix MATH consisting of non-zero terms, MATH the matrix units for MATH (as above), then MATH is isometrically equivalent to the MATH-basis. But then it follows immediately that if MATH is a normalized generalized block diagonal of MATH (in MATH) consisting of non-zero terms, MATH is also isometrically equivalent to the MATH-basis. Indeed, for any scalars MATH with only finitely many non-zero terms, and any MATH, MATH . Hence MATH . Now fix MATH, and let MATH be the subspace of MATH defined in the proof of REF (specifically, in REF). Standard results yield that MATH embeds in MATH (actually, MATH is isomorphic to MATH if MATH), and of course MATH is a projection onto MATH with MATH (MATH as defined in the proof of REF). Now let MATH be as in REF , and suppose MATH does not embed in MATH. Then for each MATH, we may choose a MATH with MATH . But it follows that for any MATH, MATH . A standard travelling hump argument now yields a normalized generalized block diagonal MATH of MATH and a subsequence MATH of MATH so that MATH . It follows immediately that MATH is equivalent to the MATH-basis.
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math/0005150
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Let MATH be the normal faithful tracial state in MATH. By REF , MATH is uniformly integrable. Let MATH, to be decided later. Choose MATH so that MATH . Let MATH be elements of MATH. By the final statement of REF , we may choose for each MATH a MATH so that MATH with MATH . Then MATH . Since MATH is REF-unconditional and MATH is type MATH with constant one, MATH . Now MATH . Thus REF - REF yield that MATH . Evidently we now need only take MATH; then choose MATH so that MATH; the identical argument for MATH now yields that REF holds for all MATH.
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math/0005150
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REF follows immediately from REF , and the latter also immediately yields REF . If MATH is a generalized diagonal of the matrix, then there exist projections MATH, MATH in MATH so that the MATH's and the MATH's are pairwise orthogonal, with MATH for all MATH. (That is, MATH is ``right and left disjointly supported".) It then follows that for any MATH and scalars MATH, MATH which immediately yields REF since MATH is semi-normalized.
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math/0005150
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Let MATH and let MATH be a subspace of MATH so that MATH is isomorphic to MATH (using NAME 's result, formulated as REF above). We claim that MATH is not isomorphic to a subspace of MATH (which of course proves REF ). Suppose to the contrary that MATH is an isomorphic embedding. Assume without loss of generality that MATH. Let MATH be chosen so that MATH for all MATH. Let MATH denote the projection of MATH onto MATH, with kernel MATH; and set MATH. Now fix MATH and MATH. Then of course MATH is isometric to MATH. Thus by REF , MATH cannot be an isomorphic embedding (that is, MATH does not embed in MATH). Hence we may choose MATH with MATH . Now let MATH. Since MATH does not embed in MATH, the conclusion of REF holds for the matrix MATH. It follows from REF that MATH . Hence we obtain that MATH . Thus MATH is isomorphic to a subspace MATH of MATH. Let MATH for all MATH and MATH. Now since MATH is an isomorphism, REF yields that there is a MATH so that every row and column of MATH is MATH-conditional, every generalized diagonal of MATH is equivalent to the MATH-basis, yet neither the rows nor the columns of MATH contain MATH-sequences. This is impossible by REF .
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math/0005150
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The hypotheses imply (via known results, compare CITE) that MATH is isomorphic to a NAME subalgebra of MATH, which is the range of a normal conditional expectation, whence MATH is completely isometric to a subspace of MATH. Since MATH is separable, we can assume without loss of generality that MATH acts on a separable NAME space. Then if MATH fails the conclusion, there exists a finite NAME algebra MATH so that MATH is isomorphic to a subalgebra of MATH, and hence MATH is completely isometric to a subspace of MATH. But then MATH does not NAME embed in MATH, since MATH does not embed in MATH by REF .
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math/0005150
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By the assumptions MATH is a properly infinite NAME algebra, that is, MATH as NAME algebras (where MATH is the standard NAME algebra tensor product). In particular MATH is isometrically isomorphic to MATH for some crossnorm MATH on the algebraic tensor product MATH, and therefore MATH imbeds isometrically in MATH. By REF , MATH does not NAME space imbed in MATH.
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math/0005150
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CASE: Let MATH and put MATH, MATH. Since MATH is a properly infinite NAME algebra, there exists two isometries MATH, such that MATH and MATH are two orthogonal projections with sum REF. Define now MATH and MATH . Then MATH and MATH. Since MATH and MATH are normal (that is, continuous) in the MATH-topologies on MATH and MATH and also are completely bounded maps, it follows that MATH. Similary we have MATH. Thus the pair MATH satisfies REF . We next check REF . Since MATH as NAME algebras (compare CITE), we can without loss of generality assume that MATH where MATH. Let MATH be a normal faithful state on MATH and define MATH and let MATH be the left slice map given by MATH, that is, the unique normal linear map MATH for which MATH . Thus MATH and MATH. Hence MATH has a completely bounded factorization through MATH, that is, MATH is MATH-isomorphic to a MATH-complemented subspace of MATH. To prove the converse, we use that if MATH is a normal faithful state on the MATH-factor MATH and MATH is the moduluar automorphism associated with MATH at MATH, then the crossed product MATH is a factor of type MATH (compare CITE). Moreover injectivity of MATH implies that the crossed product is injective (compare CITE). Hence MATH as NAME algebras, so in this part of the proof we may assume that MATH. Further, after identifying MATH with its natural imbedding in the crossed product, we have that MATH is generated as a NAME algebra by MATH and a certain unitary group MATH coming from the crossed product construction (compare CITE). Let MATH be the imbedding and let MATH be the unique normal faithful conditional expectation of MATH onto MATH for which MATH, for MATH (see again CITE). Then MATH and MATH are normal maps and MATH, so as above, we obtain that MATH is MATH-isomorphic to a MATH-complemented subspace of MATH. Hence REF follows from REF . CASE: Put again MATH and let MATH be a dense countable subgroup. Let MATH be a normal faithful state on MATH and put MATH where MATH is the restriction of the modular automorphism group MATH to MATH. It follows from CITE (see the proof of CITE) that MATH is a factor of type MATH, which is also injective (by CITE). Moreover MATH, where MATH is NAME MATH-invariant. Hence MATH implies, that MATH and MATH are not NAME isomorphic. It is easy to check, that there are uncountably many dense countable subgroups of MATH. Put MATH. Since MATH for MATH, we have MATH, MATH, which by CITE implies that MATH is a factor of type MATH. Since MATH is also injective we have MATH as NAME algebras. As in the proof of REF , it now follows, that MATH is MATH-isomorphic to a MATH-complemented subspace of MATH. Moreover, since MATH is a crossed product with respect to a discrete group, there is again an embedding MATH and a normal faithful conditional expectation MATH, and the rest of the proof of REF follows now exactly as in the proof of REF .
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math/0005150
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Let MATH, MATH and put MATH and MATH. Since MATH is isomorphic to a subgroup of MATH and vice versa, MATH is NAME isomorphic to a subfactor MATH of MATH and MATH is NAME isomorphic to a subfactor MATH of MATH (see CITE for details). Moreover, let MATH and MATH be the unique normal faithful tracial states on MATH and MATH respectively. Then there is a unique normal faithful conditional expectation MATH preserving the trace MATH (respectively, a unique normal faithful conditional expectation MATH, preserving the trace MATH). As in the proof of REF , this implies that MATH and MATH satisfy REF . We next prove that REF is satisfied with MATH. Since MATH is a MATH-factor, we can choose a sequence of orthogonal projections MATH in MATH, such that MATH and MATH (convergence in the strong operator topology). By NAME 's result quoted above, MATH for MATH as NAME, which implies that MATH as NAME. Indeed, NAME 's result yields that there are orthogonal equivalent projections MATH in MATH with MATH so that MATH. It follows (by uniqueness of MATH) that MATH, for all MATH and MATH, and so MATH. Since also MATH and MATH is a finite factor, MATH and MATH are equivalent, and hence MATH as desired. Put MATH . Then MATH is a NAME algebra isomorphic to MATH, which is a NAME subalgebra of MATH. Moreover, there is a MATH-preserving normal faithful conditional expectation MATH. Hence MATH is MATH-isomorphic to a MATH-complemented subspace of MATH. Put as above MATH. Then MATH as operator spaces. Hence we have shown that MATH completely factors through MATH, so MATH and MATH are completely isomorphic by REF . This proves REF iin REF with MATH. Hence MATH and MATH are completely isomorphic.
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math/0005151
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Suppose that MATH and MATH such that MATH. Then there exists a map MATH given by MATH with MATH such that MATH is non-orientation reversing. Define MATH by MATH. Then we have MATH and MATH. For every orientation preserving parameterized curve MATH, MATH does not move clockwise on MATH as MATH increases. So MATH is non-orientation-reversing as MATH is a positive integer. Therefore MATH, and MATH is unperforated.
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math/0005151
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Suppose that MATH is an orientation of MATH where MATH is a branched matchbox with the projections MATH. Let MATH be a light uniformly simplicial presentation of MATH given by REF , and MATH the canonical projection to the MATH-th coordinate space. If MATH is an edge of MATH with MATH, then give the direction to the set MATH so that, for every curve MATH, MATH is increasing if and only if MATH is increasing. Since MATH is an orientation of MATH, we can extend this direction on MATH to MATH, and each edge MATH has a direction induced by the orientation of MATH. Suppose that MATH is a point in MATH such that MATH is a vertex and that MATH is a branched matchbox such that the domain of MATH contains MATH. Then there is a match MATH containing MATH such that MATH for some MATH. Since MATH and MATH are nonempty sets in MATH, there exist an edge MATH such that MATH, which is incoming to MATH, and an edge MATH such that MATH, which is outgoing from MATH. Therefore MATH is nondegenerate. Suppose that MATH and MATH are two edges such that MATH, and that MATH is a chart such that MATH. Then we have MATH, and for every every curve MATH, MATH. Let MATH be given by MATH. Then we have MATH, and MATH is increasing as MATH is increasing. Therefore MATH is orientation preserving.
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math/0005151
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Define MATH by MATH where MATH is given by MATH. Then MATH is well-defined and MATH is a zero dimensional set. Since MATH, an arc component of MATH, is given by MATH where MATH, MATH given by MATH is an orientation preserving homeomorphism.
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math/0005151
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Suppose that MATH is an element of MATH. For each vertex function MATH defined at the vertex MATH of MATH, define a map MATH for MATH by MATH . Then MATH, MATH, is a homotopy between MATH and MATH. Now suppose that MATH and MATH are homotopic on MATH. Since the winding number of the restriction of MATH on every cycle in MATH is a homotopy invariant and MATH is the winding number for every cycle MATH in MATH, we have that MATH is zero on every cycle, and MATH is an element of MATH by REF .
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math/0005151
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Since MATH, MATH is a group homomorphism. By REF , MATH is homotopic to a constant function MATH if and only if MATH is a vertex coboundary. So we have MATH is injective. To obtain an inverse of MATH, suppose that MATH belongs to MATH. Then we can choose a map MATH where MATH is the vertex set of MATH such that MATH for every vertex MATH of MATH. Define MATH by MATH . Then MATH is homotopic to the constant map MATH by MATH for MATH, MATH is homotopic to MATH, and for every vertex MATH of MATH, MATH. For each edge MATH, let MATH be the number of times the loop MATH winds around MATH as MATH moves on MATH. Since MATH for every vertex MATH of MATH, MATH is well-defined for each edge MATH. Then MATH is an element of MATH, and MATH wraps around MATH the same number of times as MATH. Therefore MATH is homotopic to MATH, and MATH gives the desired inverse to MATH. Clearly if MATH, then MATH is a positive element in the winding order. Conversely if MATH is a positive in the winding order, then there exists a map MATH such that MATH is non-orientation-reversing. It follows that MATH has to be nonnegative on cycles, and we have MATH by REF . Therefore MATH is an isomorphism of preordered groups.
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math/0005151
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For every MATH and every cycle MATH in MATH, MATH is a cycle in MATH and MATH by REF . Therefore MATH is an element of MATH, and the map MATH given by MATH is a well-defined homomorphism. That MATH induces a homomorphism follows from the definition of the NAME group.
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math/0005151
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It is not difficult to check, for every MATH, MATH and we have MATH. To show that MATH is order preserving, suppose MATH. Then there exists a MATH such that MATH is non-orientation-reversing. Since MATH is an element of MATH by REF and MATH is orientation preserving, for every orientation preserving parameterized curve MATH, MATH is an orientation preserving parameterized curve in MATH, and MATH does not move in the clockwise direction as MATH increases. Therefore MATH is an element of MATH, and MATH is an order preserving homomorphism. Since MATH is an order preserving isomorphism by REF , MATH is also order preserving.
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math/0005151
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Define a metric MATH on MATH by MATH where MATH, MATH and MATH is a metric on MATH compatible with its topology. Since MATH is a compact NAME space, every element in MATH is uniformly continuous. So, for given MATH and MATH, there exists a nonnegative integer MATH such that for MATH, MATH implies MATH. For MATH, let's denote MATH. Then for all MATH, MATH and we can choose a point MATH such that MATH is the center of the smallest interval containing MATH in MATH. Define MATH by MATH. Then it is clear that MATH and MATH if MATH. Since MATH for all MATH, MATH is homotopic to MATH.
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math/0005151
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To show that MATH is surjective, suppose MATH and that MATH and MATH are given in REF . Define MATH by MATH for MATH. Then MATH is well-defined, and it is trivial that MATH. Therefore MATH is homotopic to MATH, and MATH is surjective. Suppose MATH and that MATH is homotopic to MATH. Then by the surjectivity of MATH, there exist nonnegative integers MATH and MATH, MATH such that MATH is homotopic to MATH and MATH is homotopic to MATH. Since MATH, we have MATH . Hence MATH is injective.
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math/0005151
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(Trivial case). Suppose that all but finitely many MATH has a unique edge, that is, MATH is homeomorphic to a circle MATH with a unique vertex by the standing REF , and that the connection map MATH is the identity map if MATH. Then it is obvious that MATH and MATH is an isomorphism. (Nontrivial case). We have that MATH is a group isomorphism, and clearly MATH. It remains to show that MATH maps MATH onto MATH. So we assume that MATH is an element of MATH. Then there is a MATH in MATH for some MATH such that MATH, and we need to show MATH. That MATH is an element of MATH implies that there is a map MATH such that MATH is non-orientation-reversing. Since MATH is an element of MATH, there is a continuous map MATH such that MATH. For MATH, if MATH for MATH and MATH, then MATH is defined by MATH, Suppose that MATH is a constant map to MATH. Then we have MATH in MATH as MATH is homotopic to the identity element in MATH. Hence the equivalence class of MATH is the identity element in MATH, for MATH is an isomorphism. Next suppose that MATH is not constant on MATH. Then there are nonnegative integer MATH, a small interval MATH contained in some edge MATH of MATH, and MATH such that if MATH is any orientation preserving curve in MATH and MATH, then MATH. Given an arbitrary constant MATH, by the simplicity condition we can choose a sufficiently large integer MATH such that MATH is covered under MATH at least MATH times by every edge in MATH. Define MATH. Then by REF , MATH is homotopic to MATH. For MATH, as MATH moves forward through a directed edge MATH of MATH, its image under MATH moves MATH times around MATH where MATH is the number of times MATH covers MATH under the map MATH. For every edge MATH, MATH. Regard MATH as a curve MATH, MATH, and pick a curve MATH such that MATH. As MATH increases from MATH to MATH, the point MATH moves counterclockwise on MATH from MATH to MATH, covering an arclength MATH in the plane such that MATH . Because MATH, as MATH runs from MATH to MATH the curve MATH wraps around MATH at least MATH times, and therefore MATH. Consequently MATH as required. Since we can choose MATH to make MATH as large as we wish, we can make the choice to guarantee MATH for every edge. Therefore MATH is an element of MATH.
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math/0005151
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CASE: For each MATH and MATH given by MATH, if we represent MATH as MATH, then MATH is isomorphic to MATH and MATH is given by MATH. Hence we have MATH. Since MATH is the set of elements in MATH with range in MATH, MATH is simplicially ordered, and so is MATH. Therefore MATH is order isomorphic to MATH. CASE: Suppose that MATH is a proper order ideal of MATH and that MATH. Then there exist a nonnegative integer MATH and MATH such that MATH. By the Simplicity Condition, there is a nonnegative integer MATH such that, for every MATH and every edge MATH, MATH. If MATH, then we can choose a positive integer MATH and MATH such that MATH. Let MATH. Then MATH and MATH. So we have MATH and MATH. Therefore MATH, and MATH is a simple dimension group. The group MATH is an unperforated ordered group by REF , and its positive set is the image of the positive set of MATH under the quotient map MATH. We claim that with this quotient order, MATH satisfies the NAME Interpolation Property (and therefore by REF is a dimension group.) (We learned this argument from unpublished remarks of NAME. The general line of argument is also implicit in remarks on pp. REF.) Let MATH. Note that if MATH contains a nonzero positive element MATH, then for every MATH we have for some integer MATH that MATH, and therefore MATH, which contradicts the image of MATH being a nontrivial ordered group. Therefore all elements of MATH are infinitesimals. To show the NAME Interpolation Property, suppose that MATH satisfy MATH (MATH). Let MATH and MATH be preimages of MATH and MATH, respectively. Since MATH is a nonzero positive element of MATH, there exists a MATH such that MATH is a nonzero positive element of MATH. Because MATH is an infinitesimal element, it follows that MATH is a nonzero positive element of MATH, and MATH for MATH. Hence by the NAME Interpolation Property for MATH there exists an element MATH such that MATH. Then by definition of the quotient order we have MATH for all MATH as required. Therefore MATH is a dimension group by REF . Suppose that MATH is a proper order ideal of MATH. Then it is not difficult to see that MATH is a proper order ideal of MATH which is a simple dimension group. Therefore MATH is a simple dimension group.
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math/0005153
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We have MATH .
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math/0005153
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First, note that by REF , MATH and MATH are bases for filters of relations. Thus, it is clear that MATH and MATH . Clearly also, MATH ; to show MATH , it suffices to show that the set MATH of operations MATH (of all arities) such that for all MATH, MATH with MATH, is a clone. Clearly, the MATH of MATH projection MATH for all MATH and MATH; given MATH-ary MATH and a MATH-tuple MATH of MATH-ary elements of MATH and given MATH, MATH with MATH, and MATH such that MATH for all MATH, using the fact that MATH is a filter. Then MATH. Thus, MATH is a clone.
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math/0005153
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Given a tuple MATH of uniformities, the meet of the tuple in the lattice of filters of reflexive relations is the filter MATH having as base the set of finite intersections of elements of MATH. MATH clearly satisfies REF through REF . To prove REF , it suffices to show that if MATH is a finite meet of elements of MATH, then there is a MATH such that MATH. Thus, let MATH, where MATH. By REF for the MATH, there are MATH such that MATH. Then MATH and MATH.
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math/0005153
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First, we must verify REF through REF . CASE: Let MATH, MATH. Define MATH. Then MATH. REF is trivial. CASE: For all MATH, MATH. CASE: If MATH, let MATH be such that MATH. Then MATH . Thus, MATH is a base for a uniformity, and it is clear that that uniformity is MATH. If MATH is onto, then MATH for all MATH and if MATH, MATH. Thus MATH and we have MATH.
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math/0005153
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Given MATH, let MATH be symmetric and such that MATH. If MATH, then because MATH is dense in MATH and MATH, there exist MATH, MATH such that MATH and MATH. We have MATH. Thus, REF follows. To prove REF , we must show that every set of the form MATH is an element of MATH. Given such an element, let MATH be symmetric and such that MATH and let MATH be a symmetric element of MATH such that MATH. Then as before, we have REF follows.
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math/0005153
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Since MATH, MATH. By REF, this implies MATH.
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math/0005153
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Clearly, if an operation is uniformly continuous, it is uniformly continuous at each argument. To prove the converse, given MATH, we use REF repeatedly to find that there is a MATH such that MATH. Then, for each MATH, there is a MATH such that if MATH and MATH for MATH, then MATH. It follows that if MATH for all MATH, we have MATH, whence MATH.
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math/0005153
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Clearly MATH , MATH , MATH REF , and MATH . To show that MATH , let MATH be a MATH-ary term operation. MATH is trivially uniformly continuous if MATH; suppose MATH, and let MATH. Let MATH . Then MATH is uniformly continuous in the first argument by REF , which implies that MATH is uniformly continuous in the MATH argument. Thus, MATH is uniformly continuous, by REF .
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math/0005153
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Follows from REF .
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math/0005153
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MATH is a base for MATH. Thus, MATH iff MATH for some MATH, iff MATH.
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math/0005153
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Let MATH be a tuple of elements of MATH. Then MATH is the principal filter generated by the union MATH. The divisible elements of this filter all contain MATH, which is itself divisible. Thus, MATH.
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math/0005153
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Denote MATH by MATH. We must show that MATH is compatible, that is, that if MATH is a MATH-ary term for MATH, and MATH, then MATH. Let MATH, MATH, MATH be a dividing sequence of elements of MATH with MATH. For each MATH, we have MATH for some MATH. We have MATH by REF . The MATH form a dividing sequence of elements of MATH, because MATH . Thus, MATH.
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math/0005153
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It is easy to see that MATH is a semiuniformity for each MATH, the MATH are increasing, and that MATH. For some MATH, we must have MATH, which implies that the sequence becomes stationary and that MATH. Thus, MATH is a uniformity and consequently, MATH.
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math/0005153
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We always have MATH and MATH. Thus, if MATH and MATH to not permute, we cannot have MATH. On the other hand, let us assume that MATH and MATH permute, and we will show that REF through REF hold for MATH. CASE: If MATH, MATH, then let MATH and MATH. We have MATH and MATH, whence MATH . REF is clearly satisfied. CASE: If MATH and MATH, then MATH and MATH, and by REF , there are MATH, MATH such that MATH . CASE: If MATH and MATH, then by REF for MATH and MATH, and using REF , there exist MATH, MATH, and MATH, MATH such that CASE: MATH; CASE: MATH; CASE: MATH; CASE: MATH; CASE: MATH. Then we have MATH .
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math/0005153
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Let MATH, MATH, and MATH such that MATH. It suffices to show that MATH as the reverse inequality holds in every lattice. Let MATH, MATH, and MATH; it suffices to show that MATH is contained in the left hand side, because such elements form a base for the right-hand side. Elements of the form MATH, where MATH, MATH, and MATH, form a base for the left hand side. We choose MATH, MATH such that CASE: MATH, CASE: MATH, and CASE: MATH (using the assumption that MATH). If MATH, then we have MATH for some MATH, and MATH. Then because MATH, we have MATH, which together with MATH implies that MATH. It follows that MATH .
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math/0005153
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Clearly, MATH to prove the reverse inequality, it suffices to show that if MATH for all MATH, then there is a MATH such that MATH, because by REF , relations of the form MATH form a base for MATH. We simply choose MATH: it is a congruence because MATH has permuting congruences, and it is an element of MATH.
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math/0005153
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Let MATH be a NAME term for the variety, and let MATH, MATH. There are MATH, MATH such that CASE: if MATH, then MATH for all MATH and MATH, and CASE: if MATH, then MATH for all MATH and MATH. If MATH, then we have MATH. Thus, MATH; this proves REF . The proof of REF is similar.
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math/0005153
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Let MATH, MATH, MATH. It suffices to prove MATH as the reverse inequality holds in any lattice. Using REF , every element of the right-hand side contains a relation of the form MATH, where MATH, MATH, and MATH. Let MATH (MATH, MATH) be such that MATH and MATH (respectively, MATH, MATH), where MATH is a ternary term satisfying the above identities. If MATH, this means that MATH and MATH for some MATH. Then MATH and MATH showing that MATH the desired inequality follows.
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math/0005153
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For the given MATH, let MATH be such that MATH. Let MATH where each MATH is symmetric and such that MATH. Let MATH, MATH, MATH, MATH be given such that MATH and MATH for all MATH. Define MATH and MATH for all MATH. Since MATH and MATH for all MATH, we have for even MATH, MATH, while for odd MATH, MATH. If follows that for all MATH, we have MATH. Thus, MATH.
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math/0005153
|
Follows from REF .
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math/0005153
|
Let MATH satisfy the conclusion of REF . Let MATH, MATH be symmetric and such that MATH. Let MATH be symmetric and such that MATH. Using REF , let MATH be symmetric and such that MATH. Then MATH for all MATH, and MATH for all MATH, whence MATH for all MATH. Then if MATH, it follows from REF that MATH.
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math/0005153
|
Let MATH. Write MATH for MATH. By REF , there exist MATH, MATH, and MATH such that MATH is symmetric and MATH and MATH imply MATH. Let MATH. Then MATH implies that MATH and that there exist MATH, MATH such that MATH, which implies that MATH and that MATH. We then have MATH. Thus, MATH, implying that MATH.
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math/0005153
|
It is trivial that MATH, as this inequality holds in every lattice. Define MATH for every cardinal MATH as MATH, MATH, and for limit cardinals MATH, MATH. By REF and because MATH is generated by a congruence MATH, we have MATH for all MATH. By REF , we have MATH for some cardinal MATH. Thus, MATH.
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math/0005153
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All elements of MATH are reflexive, and MATH has a base of symmetric relations, so MATH is reflexive and symmetric. That MATH is transitive follows from the inclusion MATH which follows from REF of MATH. That MATH is NAME then follows from REF on the form of the direct image uniformity. If MATH is uniformly continuous, and MATH, then MATH. Thus, MATH must factor through MATH, implying REF .
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math/0005153
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Let MATH be a MATH-ary term operation of the variety, let MATH, and let MATH, MATH be such that MATH. Since MATH, we have MATH for every MATH. This implies that MATH. As this holds for all MATH, MATH. Thus, MATH is a congruence. To prove MATH is a uniform algebra, let MATH be a MATH-ary basic operation for MATH, and let MATH. Let MATH, MATH, MATH be such that MATH if MATH for all MATH. Then MATH if MATH for all MATH, where MATH. Thus, MATH is uniformly continuous, and MATH is a uniform algebra.
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math/0005153
|
Follows from REF.
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math/0005153
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Every NAME net in MATH can be shown equivalent to a net MATH where the directed set MATH is MATH, ordered by reverse inclusion.
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math/0005153
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MATH and MATH are bases for filters of relations on MATH, and the filters they generate are independent of the chosen base of MATH, by REF. We must verify REF , and REF for MATH and MATH. Let MATH and MATH. For all MATH, MATH, we have MATH for large enough MATH and MATH, because MATH. Thus MATH, verifying REF for MATH and, by REF, for MATH. By REF, and REF for MATH, we have REF for MATH and MATH. MATH and MATH generate the same filter by REF, while MATH (and therefore also MATH) satisfies REF by REF .
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math/0005153
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CASE: Let MATH, MATH be such that MATH for every MATH, and let MATH and MATH. For every MATH, MATH for large enough MATH and MATH. Since MATH is arbitrary we have MATH, which implies MATH. Thus, MATH, implying that MATH is NAME. To prove MATH is complete, let MATH be a net in MATH, NAME with respect to MATH. For each MATH, let MATH be a representative of MATH. For any MATH, for large enough MATH, MATH, MATH for large enough MATH and MATH. For, we have MATH for large enough MATH and MATH. Consider MATH as a directed set with ordering given by reverse inclusion, and for each MATH, let MATH, MATH be such that MATH for MATH and large enough MATH, and let MATH. We claim that MATH is a NAME net in MATH with respect to MATH. To prove this, let MATH. Let MATH be such that MATH. Then if MATH, MATH, we have MATH for some MATH such that MATH and MATH, and some MATH, implying that MATH. It follows that MATH is a NAME net. Let MATH be the class of NAME nets equivalent to MATH. We will show MATH, for which it suffices to show that for all MATH, MATH for large enough MATH. Again, let MATH be such that MATH. If MATH and MATH is large enough, then MATH whenever MATH, MATH, and MATH is large enough. That is, MATH for large enough MATH. Thus, MATH is complete with respect to MATH. CASE: Given MATH, let MATH be such that MATH. Now suppose that MATH, that is, that there are sequences MATH and MATH, converging to MATH and MATH respectively, such that MATH for large enough MATH and MATH. Then since MATH and MATH converge to MATH and MATH respectively, we have MATH for large enough MATH and MATH, implying MATH. Thus, MATH. On the other hand, suppose MATH. Then if MATH, MATH converge to MATH, MATH respectively, we have MATH for large enough MATH and MATH, showing that MATH. Thus, MATH. CASE: By REF , MATH, implying MATH is uniformly continuous. CASE: Given MATH, and MATH, let MATH where MATH. Let MATH be such that MATH for MATH, MATH, let MATH, and let MATH be a constant net at MATH. Then MATH for large enough MATH and MATH, or MATH. Thus, MATH is dense in MATH. CASE: By REF , MATH. Thus, by REF , MATH. CASE: We must show that if MATH is uniformly continuous, where MATH is complete and NAME, then there is a unique uniformly continuous MATH such that MATH. If MATH, then MATH is a NAME net in MATH, and we define MATH to be the limit of MATH in MATH. This is well-defined by REF . Clearly MATH. MATH is unique, because MATH, having dense image, is an epimorphism. To show MATH is uniformly continuous, let MATH and let MATH, MATH be such that MATH. Let MATH and MATH be elements of MATH with MATH; then for some MATH, MATH, MATH, MATH we have MATH, whence MATH. Thus, MATH. CASE: Suppose MATH, MATH, and MATH are given as stated. Let MATH. By REF , there is a unique, uniformly continuous function MATH such that MATH. We must show that MATH is a uniform isomorphism. If MATH, then since MATH is dense in MATH, there is a net MATH in MATH such that MATH. MATH is NAME with respect to MATH by REF . MATH represents some MATH, and by the construction of MATH, MATH. Thus, MATH is onto. If MATH, then the closures of MATH and MATH in the topology of MATH coincide. Since the completion is NAME, we must have MATH; thus, MATH is one-one on the image of MATH. Suppose that MATH and MATH are NAME nets in MATH which are not equivalent. Then there is a MATH such that MATH for arbitrarily large values of MATH and MATH. But MATH for some MATH, by the definition of MATH. Thus, we have MATH for arbitrarily large values of MATH and MATH. It follows that MATH and MATH are not equivalent. Thus, by REF , they have different limits in MATH. It follows that MATH is one-one. It remains to show that MATH is uniformly continuous. We have MATH, by REF , and since MATH we have MATH it follows by REF that MATH, whence MATH is uniformly continuous. CASE: Let MATH. Since MATH is dense in MATH, and MATH is dense in MATH, it follows that MATH is dense in MATH. Then by REF , there is an isomorphism MATH, where MATH, such that MATH. However, we have MATH. CASE: We have MATH and also MATH by REF . Then since the image of MATH is dense, we have the desired result by REF .
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math/0005153
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CASE: Let MATH be a MATH-ary term, and let MATH be a MATH-tuple of elements of MATH. Let MATH be representatives of MATH for each MATH. We define MATH to be the NAME net given by MATH and MATH to be the equivalence class of MATH. Since MATH is uniformly continuous, this is independent of the chosen representatives MATH. Thus we have mapped MATH to a MATH-ary operation on MATH. We claim that these mappings, for MATH, constitute a clone homomorphism. If MATH, then MATH; it is straightforward to prove that MATH. Thus, MATH. On the other hand, suppose MATH where MATH is a MATH-ary term and MATH is a MATH-tuple of MATH-ary terms. We have MATH, and MATH, where MATH is the diagonal map; it follows easily from this that MATH, or MATH. Thus, we have a clone homomorphism and MATH is an algebra in MATH. It is clear that MATH is a homomorphism. The uniqueness of the algebra structure follows from the fact that the image of MATH is dense in MATH. CASE: Let MATH be a MATH-ary term for MATH, and let MATH. Then if MATH, there exist MATH, MATH such that MATH for large enough MATH and MATH. Given any MATH, MATH, MATH, with representatives MATH, MATH, MATH respectively, we have MATH for large enough MATH. This implies MATH showing that MATH is uniformly continuous with respect to MATH in the first argument; by REF , MATH is compatible. CASE: If MATH, then by REF , the diagram of uniform spaces MATH commutes. Since MATH, MATH, and MATH are homomorphisms by REF , MATH is a homomorphism on the dense subalgebra MATH of MATH. Since all of the functions in the diagram are uniformly continuous, it follows that MATH is an algebra homomorphism. REF is clear. CASE: Follows from REF : It remains to prove only that if MATH is a uniformly continuous homomorphism and MATH, then the uniformly continuous function MATH, given by the universal property of the completion, is a homomorphism. We have MATH, so the restriction of MATH to the dense subalgebra MATH of MATH is a homomorphism. Since MATH and MATH are uniformly continuous, it follows that MATH is a homomorphism. CASE: The universal property of MATH gives a uniformly continuous homomorphism MATH, which is an isomorphism of uniform spaces by REF . MATH is a homomorphism because its restriction to the dense subspace MATH is a homomorphism.
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math/0005153
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Each element MATH can be seen as a choice of elements MATH for each MATH, such that whenever MATH, MATH. Given MATH, let MATH be some element such that MATH, for each MATH. Then MATH is a NAME net, with respect to MATH, when MATH is viewed as a directed set given by the opposite of the inversely-directed set described above. It thus represents an element MATH, and it is clear that this element is independent of the choices made in the definition of MATH. We define MATH. It is straightforward to prove that MATH is a homomorphism. To prove MATH is onto, let MATH, where MATH. For each MATH, there is a MATH-class MATH such that MATH for large enough MATH. The MATH determine an element MATH of MATH such that MATH.
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math/0005153
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It suffices to verify REF through REF . CASE: Follows from the fact that an isomorphism in MATH is an algebra isomorphism MATH such that MATH and MATH. CASE: Given MATH, let MATH be the closure of the image of MATH in MATH, and MATH the inverse image uniformity MATH for MATH the inclusion. MATH, being the closure of a subalgebra of MATH, is also a subalgebra, and MATH. Then MATH, and MATH, considered as an arrow from MATH to MATH, is an arrow of MATH. CASE: Given a commutative square of the form MATH with MATH and MATH, for any MATH, let MATH be a net in MATH such that MATH. We have MATH. Therefore, MATH. However, MATH; it follows by REF that MATH is a NAME sequence with respect to MATH. Since MATH is complete, it follows that MATH. We define MATH. It is clear that MATH and MATH. To show MATH is uniformly continuous, let MATH. Then MATH for some MATH. Since MATH, MATH.
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math/0005153
|
Given an arrow MATH, representing an element MATH of the MATH-quotient lattice, let MATH. MATH is clearly well-defined. MATH is onto MATH because if MATH, with MATH, then MATH. MATH is one-one because if MATH and MATH, where MATH and MATH, then both MATH and MATH are isomorphic to MATH by REF , by isomorphisms MATH and MATH satisfyin MATH and MATH, implying that MATH and MATH represent the same element MATH of the MATH-quotient lattice of MATH.
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math/0005154
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We begin by proving the existence of the flat limit MATH. Take a radial gauge MATH for MATH; from the bound on the curvature, we deduce MATH from this we deduce that MATH and MATH have limits MATH and MATH when MATH goes to infinity; moreover, the bound on the curvature implies that for each MATH, the connection MATH is flat on MATH. It remains to see that it is independent of MATH: for this we pick a base point in MATH and prove that the monodromies along the two circles remain conjugate when MATH varies; this is a consequence of the bound on the curvature and the following lemma (see for example CITE): Suppose we have a connection MATH on MATH, and MATH is the monodromy of MATH along the circle MATH; note MATH the parallel transport from the point MATH to the point MATH; then MATH . Therefore we have constructed a flat limit MATH on MATH for the connection MATH. Now pass to the approximation statement. On MATH, there exists a gauge so that MATH, MATH. This statement (a MATH gauge only), can be proven by elementary means and is left to the reader. Now, we extend radially this gauge on MATH to MATH, and the bounds REF imply that MATH with still MATH on MATH; then we choose a cutoff function MATH so that MATH and define a connection MATH by MATH on MATH, the curvature of MATH is MATH and this remains bounded by MATH on MATH, which means that MATH is uniformly bounded by MATH.
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math/0005154
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If MATH is torus invariant, then a torus invariant gauge transformation MATH acts on MATH only by MATH, and the bounds on the curvature immediately imply the required bounds on MATH. Therefore, we are reduced to look at a connection MATH on MATH. Now note that the region MATH is conformally equivalent to the half-cylinder MATH (with coordinate MATH); in the rest of the proof we will use only the flat metric on the cylinder. The bound on the curvature becomes MATH; eventually pulling back MATH using the transformation MATH with MATH sufficiently small, we may suppose that MATH is very small. This means that we are now able to use REF , for some MATH very big, to produce on each MATH a gauge MATH so that MATH . We perform recursively diagonal gauge transformations with coefficients of type MATH (MATH integer) so that we have MATH this is possible because of REF , and the operation does not affect the bound on MATH (but we have only MATH). We want to glue together these local gauges: the transition MATH satisfies MATH the Right-hand side is controled by MATH, and this implies that MATH is very close to some MATH in the kernel of MATH; replacing MATH by MATH, we now may suppose that the transition MATH is close to the identity (in MATH norm), and a standard argument now enables us to glue together all these gauges: for a similar argument, see CITE. If we choose diagonal matrices MATH so that MATH we finally get a gauge MATH, with MATH . NAME embedding implies that MATH is controled as well; translating back these bounds in the metric of MATH, we get the proposition.
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math/0005154
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From REF , we have: MATH for some constant MATH; in particular MATH and we deduce from REF , for MATH big enough, MATH which proves the MATH-estimate of the lemma. The MATH-estimate with weights is proven in the same way. In the integration by parts, new terms appear because of the weight MATH. However, as in the proof of REF , these terms have all a coefficient MATH and therefore are a small perturbation if MATH is large enough (note that we can take the same MATH if the weight remains bounded). Finally, one may deduce the MATH and MATH estimates from the MATH estimates as in REF , since the operator MATH has injective symbol, and the boundary condition MATH is an elliptic boundary condition. The proof is a slightly more complicated, because one has to compose the decomposition REF with a MATH-projection on the image of the operator.
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math/0005154
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Again the NAME formula MATH gives the MATH-estimate (for forms vanishing on the boundary) MATH from which the MATH-statement (without weight) follows immediately. One can then deduce weighted statements as in the proofs of REF .
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math/0005154
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First, note that: MATH . Therefore, using the decomposition in REF , we have: MATH from which the the estimates below follow: MATH . Using MATH and the estimate in REF , we get: MATH from the third hypothesis, we have MATH if MATH is small enough, these two inequalities give the required estimate.
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math/0005154
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We give a concise proof, since this is parallel to CITE. Remark that MATH now the problem to be solved is MATH that is, using MATH, MATH this is a MATH-problem on small disks near infinity; for the model REF the NAME formula gives us an explicit solution; in general, with the small perturbation MATH, the solution is produced by a fixed point theorem, and we even have an estimate MATH one can then deduce the regularity statement on MATH.
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math/0005154
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For a contradiction, let MATH be an instanton with MATH and MATH, and consider the extended holomorphic bundle MATH given by REF . The restriction of MATH to the elliptic fibres MATH must be semistable for all MATH (see CITE). Moreover, MATH cannot be generically the nontrivial extension of MATH by itself, since this would give a non-constant map from MATH to MATH (which parametrises the extensions of MATH by itself). Therefore, as shown in CITE, index theory tells us that for each MATH: MATH . But if MATH, then MATH, thus contradicting the assumption that MATH.
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math/0005154
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The lemma is a consequence of the NAME transform of doubly-periodic instantons defined in CITE, more exactly of its holomorphic aspects; we anticipate a bit here, but see the introduction to REF for a summary of the construction. Again for a contradiction, let MATH be an instanton with MATH and asymptotic state MATH not of order two. The corresponding NAME transformed NAME field MATH has simple poles at MATH; its residues have rank one. However, as we shall see in the proof of REF , the non-zero eigenvalues of the residues of MATH are exactly MATH, and more generally, the eigenvalues of MATH at MATH are the MATH such that MATH; hence, the vanishing of MATH implies that the eigenvalues of MATH remain bounded when MATH goes to MATH. Now if MATH then MATH remains isomorphic to some MATH on each torus near infinity. It is then clear (again, see the proof of REF ) that the eigenvalues of MATH must go to infinity and we get a contradiction.
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math/0005154
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We analyze the situation locally near infinity; in the decomposition REF , the metric is approximately MATH and we will simplify the problem by using this metric to make the calculations (the correction term can be easily bounded); at a point on MATH where MATH is a subbundle, we suppose for example that MATH is not contained in MATH; choose a local flat section MATH for MATH, and note MATH the dual flat section of MATH; extend MATH near MATH, keeping it parallel on MATH (this is possible with our approximation for the metric); locally, MATH is generated by MATH, where MATH is holomorphic, and an orthogonal section is given by MATH, and MATH from which we deduce MATH and finally, since our choice of MATH satisfies MATH, and MATH is holomorphic, MATH in order for MATH to be in MATH, it is necessary that MATH on MATH, and therefore MATH is constant. Now restrict to the case of nontrivial decomposition MATH (the other cases are similar); therefore we may suppose that MATH on MATH; if the first order term of MATH does not vanish, then MATH this still is not in MATH if MATH (but it is in MATH if MATH, which corresponds to the case MATH); this means that we need MATH to vanish up to first order. Concerning MATH, it is easy to verify that the MATH-condition is always satisfied.
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math/0005154
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We will give two different ideas to prove the proposition, but we will not give the proofs, because they follow essentially well known arguments. The first idea is direct construction: construct a NAME metric on MATH (so that the NAME connection is anti-self-dual); for this, one has first to build a metric MATH on MATH which gives asymptotically at infinity an instanton: this is possible because MATH and the behavior of MATH near infinity (see REF ) give all the parameters at infinity of the instanton; then one wants to deform MATH to a solution MATH of the NAME equation, mutually bounded with MATH; NAME 's method CITE cannot be used, because MATH has infinite volume, but one can apply the method in CITE, using precise analysis at infinity, which will be explained in the next section for the study of the moduli space. The second idea, giving a different proof, consists in using the NAME transform of instantons. Recall that our instantons are in correspondence with NAME bundles with singularities on the dual torus MATH, with a harmonic metric. Actually, the correspondence has a purely holomorphic interpretation, and this is an occurrence of the so-called NAME transform. Stability is ususally preserved by such a correspondence, so that a MATH-stable bundle on MATH would transform into a stable parabolic NAME bundle on MATH; then one can apply NAME 's theorem CITE to construct a harmonic metric, whose inverse NAME transform provides an instanton with quadratic curvature decay, and by REF this instanton has exactly the desired behavior at infinity.
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math/0005154
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First, we have to understand the behavior of the laplacian MATH acting on sections of MATH. We want to prove that it is NAME. This property is not changed by a perturbation in MATH (this adds to MATH a compact operator), and we can therefore restrict to the case when MATH on MATH. On this domain MATH, the laplacian preserves the decomposition MATH. The case of MATH is easier: since we have seen that MATH controls MATH, it follows that MATH, which is MATH, is small if MATH is big enough; therefore REF proves that MATH is an isomorphism on MATH for the NAME boundary condition (the same is true for NAME boundary condition). The case of MATH is more complicated, but can be reduced to standard theory: recall that MATH is torus invariant, so that the operator now reduces to an operator on MATH; the action of MATH on off-diagonal coefficients (which exist only when MATH is trivial) is by MATH and the action on diagonal coefficients is the standard laplacian on MATH (that we obtain by making MATH in the previous formula); now MATH becomes the translation invariant laplacian MATH on the conformal cylinder MATH, so that standard theory CITE now applies: such operator (say, with NAME boundary condition on MATH) is NAME for all weights, except a discrete set of critical weights MATH (they are characterized by the existence at infinity of solutions of type MATH); moreover, as the operator is self-adjoint, its index is REF at the weight REF if it is noncritical, or MATH for small positive weights if REF is critical; in our situation, MATH corresponds to the decay MATH, and this becomes exactly the weight MATH on the cylinder; there are two cases: if MATH or MATH is non zero (off-diagonal coefficients), then the weight REF is not critical, and the operator remains NAME for nearby MATH, with index REF: actually is is an isomorphism, because it easy to verify that is has no kernel; if MATH and MATH are zero, then the laplacian has index MATH for small weights MATH, so that it becomes an isomorphism if we add the possibility to consider solutions MATH of MATH with MATH having some nonzero limit at infinity (and this is exactly our definition of MATH). All these results can also be checked by direct calculation, after decomposing MATH into NAME series along each circle. Finally, we deduce from these considerations that the laplacian MATH is an isomorphism MATH for the NAME boundary conditions on MATH, and gluing this isomorphism with a parametrix on the compact part, it follows that MATH is NAME on MATH. In order to calculate the index, if MATH is nontrivial, we have seen that the index is not changed if we modify MATH so that MATH near infinity; the index of a self-adjoint operator on a compact manifold is zero; by an excision principle, this has the consequence that the index comes only from the contribution at infinity; therefore, it is equal to the index of the operator MATH acting on the trivial bundle MATH; now this operator is completely explicit: on the MATH component, it is an isomorphism, and on the MATH component (that is, diagonal, torus invariant, matrices), it is simply the standard laplacian in MATH, and its index between the spaces that we have defined is again REF, with REF-dimensional kernel and cokernel equal to constant diagonal matrices. If MATH is trivial, we cannot reduce to the operator of flat space, but we can reduce to MATH, with MATH the diagonal connection MATH where MATH is a cutoff function which equals REF for MATH and REF for MATH; then, as above, it is not difficult to prove that MATH is an isomorphism on non-diagonal components (and the operator on the diagonal components is the same as above). Finally, the operator MATH has no kernel in MATH, since an element in the kernel would decompose MATH, which is impossible. This finishes the proof of the first part of the proposition. If MATH is an instanton, observe that the operator MATH acting on self-dual REF-forms, by the NAME formula, equals the laplacian MATH; this means that the above results remain true for MATH, and we deduce that the operator MATH is surjective; its kernel equals the kernel of the laplacian MATH; again one can prove (in particular using REF ) that this operator is NAME (for the weight MATH); remark that the MATH condition corresponds to a critical weight (on diagonal components, where the operator is asymptotically the standard laplacian of MATH), when MATH corresponds to a slightly greater weight; nevertheless, it remains true that the MATH-kernel equals the kernel for slightly greater weights (the possible new solutions in the kernel at the critical weight are never MATH).
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math/0005154
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It remains only to calculate the dimension, which, by REF , is the index of the operator MATH. If the limit flat connection MATH is non trivial, this is simple to calculate by comparison to the same operator for MATH: actually, by the excision principle, MATH now for the flat connection MATH, the operator MATH has no kernel (by the NAME formula), but its cokernel equals the cokernel of the operator MATH acting on MATH; we have seen above that the cokernel of this operator on MATH is the MATH-orthogonal of constant, diagonal matrices. This proves the formula for the index. If MATH is trivial, the same result holds, but one must compare with the operator MATH defined in REF .
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math/0005154
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Consider the NAME expansion MATH. Then on the torus MATH, we have: MATH . However, under the hypothesis above, MATH for all MATH, which proves the lemma.
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math/0005154
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By the previous lemma, the estimate holds away from the region where MATH is small, that is: MATH . Actually, we claim that if the estimate of the lemma is satisfied outside this region, then it must be satisfied everywhere. Indeed, one has the inequality for any function MATH, and a constant MATH independent of MATH, MATH and the lemma follows by applying REF to MATH and MATH. The proof of REF is left to the reader.
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math/0005154
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First, note that: MATH . Near infinity, this a consequence of REF and of the fact that MATH. Globally, the estimate follows from the NAME inequality: MATH . To prove the lemma itself, we have that: MATH by REF . Thus, we conclude that MATH, and again by REF we have MATH. The second estimate is obtained in a similar way.
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math/0005154
|
See REF.
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math/0005158
|
From the NAME theorem, one sees that, for an ample divisor MATH, MATH where MATH is the corresponding line bundle and MATH is the corresponding endomorphism. This, coupled with the formula for the degree in terms of the norm MATH where MATH is the reduced norm, shows that MATH . The sign can be determined from the fact that MATH where MATH represents the class of the principal polarization.
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math/0005158
|
In a neighborhood of the point MATH the boundary of MATH is of the form MATH where MATH and MATH are in MATH and MATH denotes the closure of the horizontal sections of MATH. Locally, we can always find a decomposable element whose boundary is of the form MATH so subtracting this element from NAME 's element allows us to assume that the boundary is of the form MATH . To show that the boundary is not simply the restriction of the closure of a divisor in the generic fibre amounts to showing that MATH. To do this we intersect with MATH and use the fact that a function restricts to a divisor of degree REF in a cycle not contained in its divisor. Intersecting with MATH gives MATH as by assumption, only MATH lies on MATH so MATH A similar computation for MATH shows that MATH where here MATH are translates of the MATH and MATH which split MATH and MATH is the closure of MATH. The cocycle condition MATH, gives MATH, as MATH and MATH are the closures of these divisors. Adding the two and observing that MATH is homologous to zero in the fibre (in fact it is torsion) we get our result.
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math/0005158
|
Let MATH be the maps from the closure of the MATH to the base MATH. Let MATH be divisors in MATH be such that MATH. Such MATH exist as one can always find an orthonormal basis for the rational NAME. Intersecting MATH with MATH gives a relation in MATH of the form MATH . The direct image MATH is a rational equivalence of points on MATH so is the divisor of a function, MATH. These functions MATH combined with the elements MATH give the required decomposable elements. Subtracting these elements from NAME 's element gives a relation of the form MATH .
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