paper stringlengths 9 16 | proof stringlengths 0 131k |
|---|---|
math/0005150 | The standard matrix units basis MATH of MATH also satisfies the hypotheses of REF with MATH. |
math/0005150 | 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 MA... |
math/0005150 | 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 M... |
math/0005150 | 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 MA... |
math/0005150 | 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 ... |
math/0005150 | 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 tha... |
math/0005150 | 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 ... |
math/0005150 | 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-uncondit... |
math/0005150 | 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 MA... |
math/0005150 | 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 an... |
math/0005150 | 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. |
math/0005150 | 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 s... |
math/0005150 | 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 hypot... |
math/0005150 | 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 MA... |
math/0005150 | 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)... |
math/0005150 | 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 deno... |
math/0005150 | 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... |
math/0005150 | 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 . |
math/0005150 | 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. |
math/0005150 | 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 . |
math/0005150 | 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, ... |
math/0005150 | 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/0005150 | 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 isometri... |
math/0005150 | 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 almos... |
math/0005150 | 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... |
math/0005150 | 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. |
math/0005150 | 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 clai... |
math/0005150 | 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 ... |
math/0005150 | 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/0005150 | 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 isomor... |
math/0005150 | 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 an... |
math/0005150 | 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".) I... |
math/0005150 | 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 M... |
math/0005150 | 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 ... |
math/0005150 | 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 R... |
math/0005150 | 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/0005150 | 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 MA... |
math/0005151 | 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-... |
math/0005151 | 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 se... |
math/0005151 | 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. |
math/0005151 | 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 i... |
math/0005151 | 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 fo... |
math/0005151 | 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. |
math/0005151 | 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/0005151 | 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 ... |
math/0005151 | 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, th... |
math/0005151 | (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... |
math/0005151 | 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 t... |
math/0005153 | We have MATH . |
math/0005153 | 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 f... |
math/0005153 | 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,... |
math/0005153 | 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/0005153 | 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... |
math/0005153 | Since MATH, MATH. By REF, this implies MATH. |
math/0005153 | 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, ... |
math/0005153 | 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 argum... |
math/0005153 | Follows from REF . |
math/0005153 | MATH is a base for MATH. Thus, MATH iff MATH for some MATH, iff MATH. |
math/0005153 | 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. |
math/0005153 | 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/0005153 | 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. |
math/0005153 | 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/0005153 | 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/0005153 | 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. |
math/0005153 | 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. |
math/0005153 | 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 sat... |
math/0005153 | 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 ... |
math/0005153 | Follows from REF . |
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. |
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. |
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. |
math/0005153 | 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,... |
math/0005153 | 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 le... |
math/0005153 | Follows from REF. |
math/0005153 | Every NAME net in MATH can be shown equivalent to a net MATH where the directed set MATH is MATH, ordered by reverse inclusion. |
math/0005153 | 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 ... |
math/0005153 | 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 M... |
math/0005153 | 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... |
math/0005153 | 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 ... |
math/0005153 | 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 ... |
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/0005154 | 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 i... |
math/0005154 | 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-cylin... |
math/0005154 | 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... |
math/0005154 | 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 . |
math/0005154 | 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. |
math/0005154 | 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 pr... |
math/0005154 | 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 w... |
math/0005154 | 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 no... |
math/0005154 | 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 co... |
math/0005154 | 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... |
math/0005154 | 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 pre... |
math/0005154 | 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 k... |
math/0005154 | 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. |
math/0005154 | 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/0005154 | 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... |
math/0005154 | See REF. |
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... |
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 t... |
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 o... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.