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math/0002117
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Immediate from REF since MATH and MATH.
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math/0002117
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Let MATH be the fraction field of the NAME algebra MATH and consider the MATH-representations on MATH and MATH given respectively by MATH and MATH. The NAME transform REF extends uniquely to an algebra isomorphism MATH. Moreover MATH identifies the MATH-finite parts of MATH and MATH. MATH is the MATH-finite part of MATH. In REF we have identified MATH with a big cell in MATH and the map REF embeds MATH into MATH. To see that MATH is all of MATH, we consider the algebra filtration MATH where MATH is the subspace spanned by the quotients MATH where MATH and MATH. (Notice that MATH is MATH-filtered while MATH is MATH-filtered.) Then MATH is commutative and embeds, as a graded NAME algebra, into the function field MATH. This embedding is MATH-linear with respect to the MATH-representation on MATH given by MATH. Clearly MATH lies in MATH and in fact MATH is all of MATH. To see this we recall that, by REF , MATH is a NAME open MATH-orbit in MATH and MATH. So MATH where the third equality is automatic since MATH is a MATH-orbit. Thus our embedding MATH induces an isomorphism on the associated graded rings. In particular then, MATH vanishes for MATH and so MATH if MATH. But MATH and so MATH if MATH. It follows now that MATH for all MATH. Thus MATH. But also MATH. So MATH.
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math/0002117
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Using REF we find the commutative diagram MATH . Then commutativity of the middle square gives MATH.
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math/0002117
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To prove this is it easiest to start with the geometry of MATH and MATH. In REF, we introduced MATH and MATH expressly for the purpose of extracting a square root of the function MATH. In fact, MATH occurs naturally in the geometry of MATH. To begin with, MATH and so MATH is a smooth Lagrangian submanifold of MATH with respect to the KKS symplectic form MATH. Then MATH is simply the restriction of MATH, that is, MATH . The composite map MATH makes MATH into a MATH-equivariant fiber bundle over MATH with typical fiber MATH. Indeed, the cotangent bundle MATH identifies with the contracted product bundle MATH and then MATH identifies with MATH. We will treat the map MATH as an inclusion. The cotangent bundle MATH, and hence the subbundle MATH, trivializes over the big cell MATH. We have identified MATH with MATH. Now we get the following commutative diagram: MATH . Here all maps are birational symplectomorphisms, except the two permutation maps are anti-symplectic. The mathematical content of the left part of REF is that MATH is a trivial bundle over MATH and moreover the standard trivialization MATH induces the trivialization MATH. This is true because MATH is abelian. The bottom row of REF read right to left defines an anti-symplectic NAME open embedding MATH. Using REF we see that the composition of this embedding with the map MATH, MATH, is the moment map MATH. This proves the first statement. The covering MATH induces a covering MATH. This identifies with the covering MATH we constructed in REF (on account of REF for instance) in such a way that we get the commutative square MATH where MATH (compare REF ). Then the inclusion of MATH into MATH is MATH-invariant, the NAME groups MATH and MATH identify naturally and MATH is Lagrangian in MATH. Now let MATH. We can lift the projection MATH to a map MATH in the following way. Notice that if MATH lie above MATH, then MATH breaks into two connected components MATH and MATH which contain MATH and MATH respectively. Then MATH. Now we define MATH by MATH. Then MATH identifies naturally with MATH and the rest of the result follows.
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math/0002117
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The only point that is not immediate is the surjectivity of MATH and MATH in REF . But this follows since, as MATH and MATH are MATH-orbits, MATH and MATH are the MATH-finite parts of the function fields MATH and MATH.
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math/0002117
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Let MATH be a non-zero MATH-sided ideal in MATH. Then MATH is in particular MATH-stable with respect to the representation REF . Let MATH be a subspace carrying a non-zero MATH-irreducible representation. Then MATH lies in MATH or MATH. This follows since, by REF , MATH is isomorphic as a MATH-representation to a subspace of MATH and so, by REF , MATH is multiplicity-free. So MATH lies in MATH which is equal to MATH by REF . So MATH. But MATH is anti-isomorphic to MATH and so is simple by hypothesis. Thus MATH contains MATH and so MATH contains MATH.
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math/0002117
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Define MATH by MATH. Then MATH extends to an algebra anti-involution MATH where MATH and MATH. This follows from the definition of MATH in the proof of REF . Now MATH naturally extends to an anti-involution of MATH and then to a MATH-invariant anti-automorphism MATH of MATH such that MATH. Then MATH preserves the MATH-filtration of MATH. The relation MATH implies MATH. So MATH. The rest is now clear.
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math/0002117
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The equivalence REF is immediate from REF . The implication MATH follows easily from REF . We get REF as follows. Given REF , we know by REF that MATH admits a square root MATH. But the equation MATH in MATH forces MATH. So MATH. So MATH. Finally, the equivalences with REF follow from REF .
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math/0002117
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By decomposing MATH as a MATH-representation, we find that the (unique) MATH-stable direct sum complement to MATH in MATH is MATH. So in particular, MATH is the complement to MATH in MATH. But MATH and so MATH. We find that MATH generates MATH, and so MATH.
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math/0002117
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Since MATH is a lowest weight vector in MATH and MATH is MATH-semi-invariant, it follows that MATH. But MATH.
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math/0002117
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The commutator MATH is a differential operator on MATH of order at most MATH and so we can write it uniquely as the sum of a vector field MATH and a function MATH. It is convenient to compute these parts individually. Using REF we find MATH . The fourth equality follows from NAME identities. Indeed, REF the operator MATH defined by MATH is self-adjoint and REF MATH. Hence MATH. Next using REF and self-adjointness of MATH we find MATH . For the third equality we used the identity MATH, and for the fourth we used MATH (see CITE).
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math/0002117
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Starting from REF we find MATH . We will explain the second equality. We start from the fact that MATH is a primitive idempotent in the NAME algebra MATH. Indeed, MATH is a decomposition of MATH into orthogonal primitive idempotents. For any primitive idempotent MATH, then the map MATH is the orthogonal projection onto MATH and so MATH. Then MATH because of the NAME identity MATH (see CITE). This proves the second equality in REF .
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math/0002117
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If MATH then REF gives MATH and using REF we find MATH. If MATH then REF gives MATH and we find MATH. Thus REF is true when MATH equals MATH or MATH. Now we can prove REF for all values of MATH without further calculation by simply examining the form of MATH. Let MATH be arbitrary. Then MATH where MATH. Since MATH we get MATH . The Right-hand side is the sum of a vector field which in linear in MATH and a function which is quadratic in MATH. But the Right-hand side vanishes for the two distinct values MATH and MATH. So the vector field vanishes identically and the function is of the form MATH where MATH. Comparing coefficients of MATH we find MATH. Using REF and the fact that MATH is a primitive idempotent we find MATH.
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math/0002117
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The result is clear once we prove that MATH where MATH. It suffices to check this for MATH and MATH. Clearly MATH is the identity on MATH. So for MATH we find MATH. Next suppose MATH. Then MATH . The second equality follows by REF because MATH and so MATH.
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math/0002117
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MATH is simple by REF since MATH does not lie in MATH. (Note MATH by REF .) The result follows by REF .
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math/0002117
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MATH follows by REF while MATH follows by REF . MATH is maximal since MATH is simple. The infinitesimal character follows by REF ; the two weights are then NAME group conjugate.
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math/0002117
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Both MATH and MATH are faithful as right or left modules over MATH since the algebra MATH has no zero-divisors. Since MATH identifies with MATH, the left and right annihilators in MATH are MATH and MATH.
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math/0002117
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Let MATH be the algebra generated by MATH and MATH. Then MATH and this is the MATH-grading. Now MATH, regarded as a space of multiplication operators, lies in MATH. Consequently MATH, regarded as a space of functions, lies in MATH. The problem then is to show that MATH is MATH-stable. We know REF that MATH is generated by MATH and MATH, and MATH is generated by MATH, MATH. The multiplication operator MATH certainly preserves MATH, and the operators MATH, MATH, preserve MATH. So we need to check that the operators MATH preserve MATH. For MATH, this is clear. For MATH, this follows using the bracket relations MATH from REF .
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math/0002117
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By REF we know that MATH is a NAME module for MATH with lowest weight vector MATH of weight MATH. Similarly MATH is a NAME module for MATH with lowest weight vector MATH of weight MATH. This follows because for MATH we have MATH, and for MATH we have (by REF ) MATH. The weights MATH exponentiate to characters of MATH and then MATH and MATH are MATH-modules. REF CITE says in particular that MATH is irreducible as a MATH-module if MATH. Our values MATH satisfy this bound.
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math/0002117
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of REF . MATH and MATH are faithful simple MATH-modules by REF . Since MATH and MATH carry different MATH-representations, they are the only non-trivial MATH-submodules of MATH. Neither MATH nor MATH is MATH-stable, since multiplication by MATH moves each into the other. Thus MATH is simple for MATH. Faithfulness is automatic as MATH is a simple ring. Our descriptions of MATH and MATH in REF imply that MATH maps isomorphically onto MATH.
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math/0002117
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Clearly MATH lies in MATH. The converse follows since MATH is a localization of MATH and so any differential operator on MATH extends to one on MATH.
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math/0002117
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The map is injective because any differential operator on MATH is uniquely determined by its values on MATH. This is true since any vector space basis of MATH is a set of local étale coordinates on MATH. To prove surjectivity we need to show that if MATH then MATH extends to a differential operator MATH on MATH. We may write MATH where MATH. (NAME MATH like MATH.) Since MATH is MATH-finite, its components MATH and MATH are each MATH-finite, and so in particular are MATH-finite. Now MATH acts by commutator with multiplication by MATH, that is, MATH. It follows that MATH lies in MATH. But also MATH and so MATH is MATH-finite and thus lies in MATH. Let MATH and MATH be the differential operators on MATH defined by restriction of MATH and MATH respectively. Then MATH is the operator we wanted.
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math/0002117
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The first part is clear. Now MATH is a filtered MATH-invariant algebra automorphism of MATH which is trivial on MATH. By considering the induced action of MATH on MATH, we see easily that MATH lies in MATH. We have MATH and so MATH.
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math/0002117
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Let MATH. Since MATH is multiplicity-free as a MATH-representation, there is a unique MATH-stable complement, call it MATH, to MATH in MATH. Then MATH is a MATH-stable vector space grading and MATH acts on MATH by multiplication by MATH. So MATH . The pairing MATH is MATH-invariant. Suppose MATH. We know by REF that MATH and MATH contain no common MATH-types and all MATH-types appearing are self-dual. It follows by NAME 's Lemma that MATH pairs MATH and MATH trivially. Now suppose MATH. Notice MATH for any MATH. So MATH . This proves MATH is a supertrace.
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math/0002117
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Everything is immediate except non-degeneracy. Now MATH because MATH is simple REF and MATH is a MATH-sided ideal in MATH which does not contain MATH.
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math/0002117
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We need to show that MATH is a simple bimodule over MATH; we already know that MATH is a simple ring. Suppose MATH is a MATH-bisubmodule of MATH and MATH with MATH. Then MATH. Now MATH is non-degenerate on MATH and so there exists MATH such that MATH. Then MATH since MATH is a supertrace. Thus MATH is not MATH-orthogonal to MATH (defined in REF). It follows, since MATH is self-dual and appears only once in MATH, that MATH contains MATH. But MATH generates MATH as a bimodule over MATH by REF . Thus MATH.
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math/0002117
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The map MATH is MATH-invariant, equivariant with respect to MATH and MATH, and also MATH intertwines MATH and MATH. So our results on MATH transfer over to MATH via MATH. This proves the first paragraph. Since MATH is a filtered superalgebra product, we have MATH. Now proving REF reduces to showing that if MATH is not MATH-orthogonal to MATH then MATH. Showing this is easy since the hypothesis means that there exist MATH, MATH and MATH such that MATH. Then MATH has a component in MATH and so MATH. But also MATH (since MATH is a supertrace) and so similarly MATH. Hence MATH. Since MATH is an anti-automorphism we find MATH . Then MATH and this proves REF . Next the relations MATH and MATH follow since MATH is inverse to MATH. But MATH and so MATH. This proves REF .
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math/0002117
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Let MATH. Then we have MATH by REF . Now the result is immediate because of REF .
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math/0002117
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The graded star product is defined by MATH where MATH and MATH are NAME homogeneous. This is strongly MATH-invariant by REF .
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math/0002117
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Define MATH by MATH. Then REF imply REF . We claim MATH. We may assume MATH and MATH. Then REF gives MATH since the grading of MATH is MATH-orthogonal. Now MATH since MATH is a supertrace and MATH and MATH are even. But MATH and MATH by the parity relation REF . This proves our claim. Now REF of MATH follow immediately from the corresponding properties of their MATH-adjoints, the operators MATH. This works because of the properties of MATH.
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math/0002117
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We can easily compute the restriction to MATH of MATH. We find MATH where MATH, MATH and MATH was defined in REF . Now if MATH and MATH then MATH.
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math/0002117
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We have MATH and so MATH is MATH-abelian. Suppose MATH and MATH for all MATH. We can write MATH where MATH and MATH. Since MATH is graded it follows easily that MATH for all MATH. But then MATH since MATH is maximal NAME abelian. It follows by induction on MATH that MATH.
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math/0002118
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CASE: MATH is simple implies MATH since MATH is a MATH-sided ideal in MATH which does not contain MATH. CASE: Let MATH be a non-zero two-sided ideal in MATH. Pick MATH with MATH. Then REF implies there exists MATH such that MATH. It follows that MATH where MATH lies in MATH. So MATH contains MATH. But then MATH contains the MATH-subrepresentation generated by MATH. Since MATH contains no non-zero MATH-invariants it follows (by completely reducibility of MATH as a MATH-representation), that MATH contains both MATH and MATH. Thus MATH is simple.
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math/0002118
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Suppose MATH is simple. Then MATH is non-degenerate on MATH (by REF ) and hence (since MATH is MATH-invariant) MATH is non-degenerate on MATH. Then (by REF again) MATH is simple. Conversely, assume MATH is simple. Let MATH be a non-zero MATH-sided ideal in MATH. To show MATH is simple, it suffices to show that MATH is non-zero. Let MATH. Consider MATH where MATH is some listing of the elements of the NAME group MATH. Clearly MATH. But also we can see using REF that MATH. Indeed, let MATH be the filtration order of MATH in MATH and let MATH be the MATH-symbol of order MATH of MATH. Then MATH lies in MATH and the MATH-symbol of order MATH of MATH is MATH. This product is non-zero and so MATH must be non-zero.
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math/0002118
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The second sentence is clear. Now REF is equivalent to MATH . We have MATH since MATH is a superalgebra. Because of REF , showing REF reduces to showing that MATH is orthogonal to MATH if MATH. So suppose MATH is not orthogonal to MATH. Then there exist MATH, MATH and MATH such that MATH. Then MATH has a component in MATH and so MATH. But also MATH (since MATH is a supertrace) and so similarly MATH. Hence MATH. Now for MATH and MATH we can write MATH where MATH. Now REF implies that MATH and MATH. But also REF implies that the map MATH (which is an algebra automorphism with respect to the ordinary product) is an algebra anti-automorphism with respect to MATH. Thus MATH . Then MATH. This proves REF . Then in particular MATH. So MATH and we get REF .
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math/0002118
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Identical to the proof of CITE.
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math/0002118
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If MATH and MATH then MATH. Then MATH because of the parity REF .
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math/0002118
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Given MATH, we define a product MATH on MATH as follows: MATH is MATH-bilinear and if MATH and MATH with MATH where MATH then MATH. This is the only possible way to extend MATH to a graded star product. The properties of MATH imply that MATH is in fact a perfectly graded, MATH-invariant star product. The converse is clear.
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math/0002118
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Same as the proof of CITE.
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math/0002119
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First suppose that MATH is reachable. Then there is a firing sequence MATH. Therefore, as above, there exist MATH such that MATH. Hence MATH. For the converse, suppose MATH. Then MATH . The proof is by induction on MATH. For the base step put MATH then MATH. The correspondence between markings and their associated polynomials is one-to-one, so here MATH and MATH is clearly reachable. For the induction step we assume that a marking MATH is reachable from MATH if MATH for a fixed MATH. Now suppose MATH is a marking such that MATH . Then for some MATH either MATH or MATH where MATH. In the first case MATH. Observe that MATH enables MATH and define a marking MATH by MATH. Then MATH so, by assumption, MATH is reachable from MATH and so MATH is reachable from MATH. In the second case MATH. There is a marking MATH such that MATH and MATH . Now, MATH is reachable from MATH by assumption and MATH is reachable from MATH by a firing of MATH. By reversibility, therefore, MATH is reachable from MATH and hence MATH is reachable from MATH.
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math/0002119
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Let MATH be a field. First observe that MATH. Let MATH be a NAME basis for MATH. Then MATH if and only if there exists MATH such that MATH and MATH reduce to MATH by MATH.
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math/0002124
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MATH. By choosing MATH small enough as a function of MATH, the NAME fillings contribute negligible volume so this property is retained by MATH.
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math/0002124
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According to CITE a non-oriented minimizer among all nonzero codimension one cycles always exists and is smooth provided the ambient dimension is at most MATH. Let MATH be this minimizer. For a contradiction, assume MATH. The NAME surgeries in section MATH were confined to MATH, so the surfaces MATH, MATH persist as submanifolds of MATH. By NAME 's theorem, for almost all MATH, MATH intersects MATH transversely. Let MATH, MATH denote the intersection. By the co-area formula. MATH . Consequently, for some transverse MATH, MATH . Since both MATH and MATH represent the nonzero element of MATH, the complement MATH can be two colored into black and white regions (change colors when crossing either surface) and the closure MATH of the black points is a piecewise smooth MATH - homology between MATH and MATH. For homological reasons, the reverse NAME surgeries MATH have cores with zero (mod REF) intersection with MATH. This means that the tori MATH each meet MATH in a null homologous, probably disconnected, MATH - manifold MATH. Again, if MATH is a sufficiently small function of MATH, we may MATH-cut off" MATH along these tori to form MATH where MATH denotes a bounding surface for MATH in MATH, with negligible increase in area. In particular, we still have: MATH . More specifically choose MATH to be the MATH-black" piece of MATH, that is, MATH. If we set MATH and recall MATH we see that MATH is a MATH - homology from MATH to MATH. It is time to use REF : MATH separates MATH into two subsurfaces meeting along their boundaries: One subsurface sees black on the positive side, the other on its negative side. REF 's CITE, a converse to the NAME isoperimetric inequality, states that MATH, in the presence of bounded sectional curvatures, yields an upper bound on MATH. Thus, the smaller of these two subsurfaces, call it MATH must satisfy: MATH where MATH is independent of MATH. Combining with line REF , we have: MATH . Now modify MATH to MATH by cutting along MATH and inserting two parallel copies of MATH. This may be done so that the result is disjoint from MATH but bordant to it by a slight modification MATH of MATH, with MATH still disjoint from MATH. See REF . combining REF : MATH . Now reverse the NAME surgeries and consider: MATH . The middle term of line REF is diffeomorphic to MATH, which is a codimension REF submanifold of MATH. This proves that MATH and in particular MATH is orientable. But this looks absurd. Apparently, we have constructed an oriented surface MATH oriented-homologous to the fiber MATH of MATH of smaller area (compare line REF with the first line in the proof of REF ). Let MATH be the divergenceless flow in the interval direction on MATH. Lift MATH to MATH in the infinite cyclic cover MATH and consider the flow through the lift MATH, the lift of MATH. The divergence theorem states that the flux through MATH is equal to the flux through MATH. Since MATH is orthogonal to MATH, MATH completing the contradiction.
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math/0002124
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We actually show that any homotopically essential loop obeys this estimate. The long collar condition MATH REF implies that any arc in MATH with end points on MATH can be replaced with a shorter arc with the same end points lying entirely within MATH. It follows that any essential loop in MATH can be homotoped to a shorter loop lying in the complement of the NAME surgeries. Thus, it is sufficient to show that any homotopically essential loop MATH in MATH has MATH. For a contradiction, suppose the opposite. Since the bundle projection MATH is length nonincreasing, degree MATH. Lift MATH. to an arc MATH in MATH. The lift MATH joins some point MATH to MATH where MATH degree MATH. Since MATH and since REF requires order-MATH, we see that MATH and MATH differ by a non-trivial covering translation of the cover MATH. Nevertheless, any non-trivial covering translation moves each point of the total space at least twice the injectivity radius of the base, a quantity guaranteed by REF to be MATH. Now using that the projection MATH is also length nonincreasing, we see that MATH. Since MATH, the same estimate MATH applies to MATH.
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math/0002126
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Consider the derivation MATH of degree MATH given by MATH. On the polilinear form of weight MATH acts by MATH. Define the homotopy MATH to be MATH on the forms of weight MATH and MATH on the forms of weight MATH. We define map of complexes MATH by MATH where MATH . This is a map of (total) complexes. Indeed, using identities MATH and MATH we have: MATH . It is clear that MATH . As for MATH we have MATH where MATH - total differential in the complex MATH, MATH being MATH on MATH, and the homotopy MATH is given by the formula MATH . This equality is also verified by direct computation. Indeed, we have, for actions on MATH and adding these equalities we get the desired result.
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math/0002126
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We check all the required properties. First MATH . Then MATH . Also MATH and MATH .
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math/0002126
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Let MATH be the differential graded algebra of MATH matrices over the algebra MATH can define an action MATH of MATH on MATH by MATH, where MATH, MATH, MATH. Put now MATH . It is easy to see that MATH, MATH satisfy REF, and hence we can twist the action MATH by MATH, MATH to define a new action MATH, as in REF. Consider now the linear functional MATH on MATH defined by MATH . Then MATH is a closed graded MATH-invariant MATH-trace on MATH with respect to the action MATH. Hence we can define the characteristic map MATH . Consider now two imbeddings MATH, MATH defined by MATH . It is easy to see that MATH and MATH. Now to finish the proof it is enough to recall the well-known fact that MATH and MATH induce the same map in cyclic cohomology. Since we will later need an explicit homotopy between MATH and MATH we give the proof below. Put MATH . Put MATH. Notice that MATH, MATH. Consider the family of maps MATH. Since we have MATH, where MATH these maps satisfy MATH, where MATH is the operator defined by MATH . Define also an operator MATH by MATH . Then it is easy to verify that MATH, MATH and MATH. Hence MATH, MATH. We conclude that MATH, where the homotopy MATH is given by MATH. Hence MATH where MATH .
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math/0002126
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We use the notations of the proof of REF . There we established that MATH. We need only to verify that MATH is well defined on the quotient complex MATH. But since MATH, and MATH is easily seen to be MATH on MATH, the result follows.
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math/0002126
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We have: MATH . Next, if we write MATH we have: MATH . Also, if MATH is compact the algebra MATH has a unit given by the function MATH . Then MATH . Finally, we have MATH .
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math/0002126
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The first identity is clear, the second follows from the fact that MATH is a constant map, taking the value REF.
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math/0002126
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Let us chose another trivialization of the bundle MATH, and let MATH be a transition matrix between the two bases of the fiber MATH. Then we have a new map MATH, related to MATH by MATH . Let MATH denote the action corresponding to the map MATH. Consider now the pull-back MATH as a map MATH, where we consider forms on MATH as the form on MATH which is MATH outside the space of units. When MATH is not compact, we obtain not an element in algebra, but rather a multiplier. Hence we see that if we define MATH and MATH we will have MATH .
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math/0002126
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Since MATH is horizontal, it can be written as a sum of the expressions of the form MATH, where MATH is a function, MATH - projection, and MATH is a form on MATH of the type MATH. Then MATH is also of the type MATH.
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math/0002126
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The curvature of the connection MATH is a horizontal form MATH on MATH, given by MATH . Hence MATH has only components of the type MATH and MATH. The statement of the lemma will then follow from the fact that MATH also has only components of the type MATH and MATH. But this follows from REF .
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math/0002127
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By definition, MATH if and only if MATH; this fact combined with the commutation relation MATH establishes the equivalence of MATH and MATH. Furthermore, these two facts show that if MATH is invariant under the action of MATH, then so is MATH, whence MATH is as well. Finally, suppose that MATH is invariant under MATH. Then for MATH, MATH is invariant under MATH, whence MATH is invariant under MATH. It follows that MATH is invariant under MATH.
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math/0002127
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We prove by contradiction. Let MATH. Hence, MATH for some MATH. Let MATH, for MATH, MATH. Suppose that MATH is not divisible by MATH, and MATH is a set of non-zero measure such that both MATH and MATH are subsets of MATH. Let MATH; we have, MATH a contradiction. If. It suffices to show that MATH for some MATH. Again, let MATH. Let MATH be such that MATH is a bijection. (It can be easily shown that MATH is a surjection.) The injective property of MATH can be assured in the following manner: for each MATH, define the set MATH, then for MATH choose MATH to be REF if MATH, if not, choose MATH, else choose MATH. Let MATH. Note that by construction, MATH is MATH translation congruent to MATH. Hence, MATH where MATH and is MATH periodic. Thus, for MATH, MATH . For almost any MATH, there exists a MATH and an integer MATH such that MATH. Moreover, by hypothesis, MATH is a multiple of MATH, since MATH. Since MATH is MATH periodic, we have that for MATH, MATH . This completes the proof.
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math/0002127
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Now we shall construct a second wavelet set to form an interpolation pair. Take the same multiplicity function, MATH and construct a MATH similar to MATH, except we will take the interval MATH and shift it right MATH. The result is MATH . This new MATH also satisfies REF 's theorem, so there is a wavelet set MATH: MATH . We claim that the wavelet sets MATH and MATH form an interpolation pair. Then MATH is defined as follows: MATH . We need to check that MATH is involutive (see CITE). Notice that MATH and that MATH. For MATH. This yields the following: MATH . Similarly, for MATH, we get that MATH, whence MATH is involutive. Finally, to interpolate between these two wavelets, define the functions MATH on MATH and MATH on MATH as follows: MATH and extend via REF-dilation periodicity. It is also routine to verify that these functions satisfy the matrix REF . Let MATH, MATH, MATH and MATH be as above. Then the wavelet defined by MATH is an element of MATH . As defined above, MATH is a wavelet. Note that MATH above yields that MATH, whence the claim is established by REF . Consequently, there exist wavelets in all MATH's, for dilation by REF.
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math/0002127
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Fix MATH and MATH. We shall first give a brief outline to show that MATH satisfies the consistency REF . For a complete proof of this fact see CITE. Let MATH, then MATH where MATH and MATH, and MATH are in either MATH or MATH. Consistency equation: MATH. A similar argument is used for MATH to prove that MATH satisfies the consistency equation. Here is how to construct the set MATH. Let MATH = support of MATH so MATH . Now take the set where MATH and shift it to the right by MATH, call this set MATH. Thus, MATH and let MATH . It is easy to show that MATH so by NAME 's theorem, MATH is the desired wavelet set.
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math/0002127
|
The above is one way of constructing MATH and MATH. Now, we will construct a second generalized scaling set MATH and a second wavelet set MATH that have the same multiplicity function by moving the interval MATH to the right MATH. MATH . Consider the wavelet sets MATH and MATH, then we have the following: MATH . This gives us the following map MATH: MATH . As with the case of dilation by two, it is routine to check that MATH is involutive. Additionally, we need to define functions MATH and MATH for the interpolation; define the functions MATH on MATH and MATH on MATH as follows: MATH and extend via d-dilation periodicity. Let MATH and MATH, MATH and MATH be as above. Then the wavelet defined by MATH is an element of MATH . As defined above, MATH is a wavelet. Note that MATH above yields that MATH, whence the claim is established by REF . Hence, there exist wavelets in all MATH's, for dilation by MATH. We have now completed the proof of REF .
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math/0002127
|
We shall prove by contrapositive. Suppose that MATH, then there exists a MATH that is not divisible by MATH. We shall show that there exists a positive integer MATH such that for MATH, MATH, by showing that MATH. Note that MATH converges to MATH in measure. Let MATH be such that MATH. Choose MATH, a set of non-zero measure, such that for MATH, MATH for some MATH. Let MATH be the measure of MATH. Choose MATH. Let MATH be such that for MATH, MATH. Let MATH; note that MATH. Likewise, define MATH; MATH. We have that MATH and MATH, and both sets have measure greater than half of MATH, whence they must intersect. Hence, if MATH and MATH are subsets of MATH, then MATH. For MATH, MATH therefore, MATH, as required.
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math/0002129
|
Let MATH be a compact NAME. To show that MATH is hereditarily indecomposable we take disjoint closed sets MATH and MATH and open sets MATH and MATH such that MATH and MATH. We must exhibit three closed sets MATH, MATH and MATH such that MATH, MATH, MATH, MATH, MATH and MATH. Choose a continuous function MATH such that MATH, MATH, MATH and MATH. Using MATH we define three continuous functions, MATH, MATH and MATH, as follows: first MATH second MATH and third MATH (At this point the reader may find it instructive to draw the graphs of MATH, MATH and MATH in case MATH and MATH. The zig-zag that appears when one follows the graph of MATH left-to-right until it meets the graph of MATH then follows the graph of MATH right-to-left until it meets the graph of MATH and finally the graph of MATH left-to-right until the end is characteristic of hereditarily indecomposable spaces.) Note that MATH. Consider the polynomial MATH defined by MATH. Then MATH, for one readily checks that CASE: MATH if MATH; CASE: MATH and MATH if MATH and CASE: MATH if MATH. An application of the Intermediate Value Theorem gives us a continuous function MATH such that MATH. Let MATH, MATH and MATH. We check that these sets have all the required properties. CASE: MATH because MATH; CASE: MATH because if MATH then MATH, hence MATH; CASE: MATH because if MATH then MATH, hence MATH; CASE: MATH because if MATH then MATH hence MATH and MATH; CASE: MATH because if MATH then MATH hence MATH and MATH and CASE: MATH because MATH. We conclude that MATH is indeed hereditarily indecomposable.
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math/0002129
|
For each MATH define MATH by MATH . Observe that MATH and that, for each individual MATH, the sets of values MATH and MATH are equal. It follows from this that the coefficients of MATH, MATH, , MATH in MATH and MATH are the same and hence that the polynomials are the same.
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math/0002129
|
Let MATH be such a polynomial and assume that there are MATH and MATH such that MATH. Completing the square gives us MATH. Now because MATH we know that MATH so that we can write MATH for some nonnegative MATH. We find that MATH; write MATH and MATH. Observe that for each MATH either MATH or MATH and that MATH. We cover our space by three closed sets: MATH, MATH and MATH. We now note that MATH on MATH (because MATH whenever MATH) and that MATH on MATH. We define MATH as the combination MATH . Note that MATH is well-defined because, by continuity, MATH on MATH and MATH on MATH. Also MATH for all MATH; this is clear on MATH and on MATH it holds because MATH and MATH. Finally, MATH is continuous because it is the combination of continuous functions defined on closed subsets.
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math/0002132
|
Let MATH be a lifting of MATH. Ad-MATH is an automorphism of MATH preserving the invariant scalar product and sending MATH to MATH for all MATH. Thus, Ad-MATH for suitable numbers MATH and MATH.
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math/0002132
|
It is known that MATH are distinct positive roots and MATH. Hence, the collection MATH does not depend on the reduced presentation. The vector MATH is a singular vector in MATH. If MATH is another reduced presentation, then the vectors MATH and MATH are proportional. Since MATH is a free MATH-module, we have MATH in MATH for a suitable MATH. MATH since the monomials are equal when projected to the commutative polynomial algebra generated by MATH.
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math/0002132
|
MATH is a singular vector in MATH. It has to have weight components of the form MATH for suitable MATH and MATH. Since MATH is generic, we have MATH and MATH is of the required form for a suitable MATH.
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math/0002132
|
Since MATH, we have that Ad-MATH and Ad-MATH for any MATH orthogonal to MATH. Hence MATH and Ad-MATH.
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math/0002132
|
See REF .
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math/0002132
|
The first equation holds since MATH commutes with the comultiplication. Now MATH .
|
math/0002132
|
It is sufficient to check the equation for the residues of both sides at MATH, and for the limit of both sides as MATH. The residue equation MATH is true since the NAME operator commutes with the comultiplication. The limit equation MATH is a corollary of REF .
|
math/0002132
|
The Corollary follows from the following simple Lemma. For MATH, the operator MATH is the product in a suitable order of all operators MATH with MATH and MATH.
|
math/0002132
|
If MATH is a reduced presentation, then MATH and MATH .
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math/0002132
|
For MATH we have MATH, and REF is equivalent to REF .
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math/0002132
|
We have MATH. Hence MATH . The last equality follows from REF .
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math/0002132
|
Let MATH be a minuscle dual fundamental weight. We have MATH and MATH according to REF-cocycle property. Now MATH .
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math/0002136
|
Due to the symmetric and dual nature of the NAME diagrams, both arguments will be similar, and I only need discuss one. So without loss of generality, let MATH denote the MATH . NAME diagram, consisting of REF row with MATH boxes. We can use the NAME diagram (see REF ) to count the dimension of each of the representations, as it will be the sum of the dimensions of the representations on the previous level that this one is connected to. Since MATH has fewer than MATH boxes, this module is part of MATH, and using the inductive construction of the NAME diagram, we can see that all the connections leading into this representation come from connections leading out of the representation MATH (and, in fact, go to/from the same representations on the MATH level). These connections are of two types: those connecting MATH to modules in MATH, and those connecting it to modules in MATH. The connections to MATH are easy to see, as they result from the standard NAME 's lattice. There is one representation with a rectangular NAME diagram of shape MATH, of dimension REF, and one representation whose NAME diagram has a single box in the second row. By the hook length formula, this representation has dimension MATH. Whatever connections exist between MATH and MATH again come from reflections of connections from MATH to MATH. Of these, there is only one: MATH only connects to MATH, and since MATH has shape MATH, it only connects to the representation labeled by the NAME diagram MATH of shape MATH. So our third and last downward connection from MATH (and by reflection, upward connection from MATH) is to MATH. Notice that this is precisely the MATH version of the representation we are investigating! Thus by induction, we may assume that this representation has dimension MATH. (One may verify that the claimed formula holds for a sufficiently low base case.) The dimension of the representation MATH is therefore MATH .
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math/0002136
|
Both expressions are simplifications of MATH.
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math/0002136
|
Both expressions can simplify to MATH.
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math/0002136
|
From REF , it is clear that MATH implies MATH. The cited proposition, applied to the present situation, states that if MATH is a minimal idempotent, then MATH is a minimal idempotent of MATH. For the present purposes, it is not necessary to deal with minimal idempotents, but notice that since MATH is REF-dimensional, the minimal idempotent MATH is a scalar multiple of MATH. Therefore, the cited proposition guarantees that a certain scalar multiple of MATH is a minimal idempotent of MATH, which is a stronger result than we require. (Notice also that since MATH is written using only the generators MATH, it commutes with MATH.)
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math/0002136
|
CASE: MATH B. MATH C. MATH D. If MATH, MATH . If MATH, MATH .
|
math/0002138
|
This proof is adapted from CITE. Let MATH be a continuously differentiable function with compact support such that MATH . Set MATH where MATH where MATH is a infinitely differentiable function with MATH=REF when MATH and MATH when MATH and MATH is a positive real number. The properties of MATH and MATH are the same as in CITE for MATH small. So MATH as MATH. For MATH and MATH let MATH be the set of NAME functions MATH with NAME constant MATH, MATH for MATH and MATH. With the supremum norm MATH . , MATH is a complete space. For MATH and MATH, let MATH be the solution of MATH . The bounds on MATH and MATH ensure that the solutions of the above equation exists for all time MATH. Define an operator MATH on MATH by MATH . We know that MATH is a contraction mapping and the centre manifold MATH is a fixed point of MATH for MATH, MATH and MATH small enough from the proof of the existence theorem CITE. Define MATH for MATH such that MATH. The domain of MATH is a closed subset of MATH since MATH. Since MATH, thus MATH is also a contraction mapping. For MATH let MATH where MATH is a neighbourhood of the origin in MATH. Since MATH as MATH, then MATH where MATH is a constant. Thus MATH is not empty because MATH by defining MATH such that MATH. If we can find a constant MATH such that MATH maps MATH into MATH, then MATH is a fixed point of MATH, and MATH that is, MATH is a centre manifold of REF , let MATH, MATH . To finish the proof define MATH . Then MATH . From the properties of MATH and MATH on pREF in CITE and MATH, we have MATH . Using REF , MATH . Using the same calculations as REF on pREF CITE, for each MATH, there is a constant MATH such that MATH where MATH and MATH is the solution of MATH . Using REF on pREF on pREF in CITE, and REF , if MATH provided MATH and MATH small enough so that MATH. Choose MATH and MATH small enough and MATH large enough such that MATH. Therefore MATH. Hence REF holds.
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math/0002139
|
For notational simplicity, we will do the case of pair correlation (MATH). Our test function MATH can then be written as MATH for some MATH, say MATH supported inside MATH. Let MATH be smooth, compactly supported and such that MATH is constant on MATH, where it equals MATH. Set MATH. For further notational simplicity also set MATH and MATH. By the mean value theorem we have MATH where MATH lies between MATH and MATH. For the difference MATH to contain a nonzero contribution from the term indexed by the pair MATH, we must have at least one of MATH or MATH lying in MATH. Now MATH is within MATH of both MATH and MATH, which implies that both lie in MATH, as does MATH if MATH is sufficiently large so that MATH. Since MATH is constant on MATH we find that MATH. Thus we get MATH as required.
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math/0002139
|
Write MATH for MATH, MATH, MATH. Then MATH . Since MATH, we may count only the contribution of those MATH (MATH distinct) for which all the components MATH are divisible by MATH. There are MATH such MATH-tuples. If MATH then since MATH we have MATH and so we find MATH . Since MATH and MATH with MATH, this gives MATH with MATH which is positive if MATH. Thus for MATH sufficiently large, MATH will diverge in these ranges.
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math/0002139
|
Without loss of generality we will assume that MATH. For each pair of coprime integers MATH with MATH, denote by MATH the interval MATH . Then MATH is MATH-approximable if and only if it lies in infinitely many of the intervals MATH with MATH. That is for all MATH, MATH lies in MATH . Thus we need to compute the measure of MATH. Since MATH, we have MATH (allowing overlap of the intervals). Since MATH, the above limit is zero.
|
math/0002147
|
Let MATH, MATH. It is clear that MATH and MATH are disjoint, and MATH has two connected components, one of them contains zero and the other one is not bounded. Thanks to NAME 's theorem, for any MATH there exists a holomorphic function, MATH, (with pole in zero), such that: CASE: MATH on MATH, CASE: MATH on MATH, where MATH. We define MATH, for MATH small enough.
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math/0002147
|
To prove REF notice that REF implies: MATH . Taking into account MATH and MATH, MATH, we have MATH . From MATH and MATH, MATH, we obtain MATH . Using the above three inequalities and REF in page REF we conclude the proof of the first assertion in this proposition. To obtain REF , consider MATH. From REF , there is MATH a curve with origin MATH and ending at MATH that verifies MATH and MATH. As MATH (if MATH is large enough), then we can apply MATH, MATH, to obtain MATH. Bearing in mind REF , we get MATH, and then MATH . Therefore, MATH . Now we are going to prove REF . First observe that, if MATH is large enough and MATH is a set in the labyrinth MATH, then it is possible to find a positive constant MATH, only depending on MATH, such that: for all MATH there exists a curve MATH in MATH from MATH to MATH satisfying MATH. This comes from the fact that the Euclidean diameter of MATH is uniformly bounded. Using the former, we obtain MATH which proves REF . From REF , it is not hard to deduce REF . Concerning REF , we are only going to construct the polygon MATH. The other polygon MATH can be constructed in a similar way. Let MATH . MATH is a not empty open subset of MATH. For MATH satisfying MATH, consider MATH . Since MATH is a compact subset of MATH, then there are, MATH, closed balls of MATH such that MATH. Note that MATH and MATH are in disjoint arc-connected components of MATH. Then, we can construct a polygonal line MATH in MATH such that MATH. As MATH is MATH-type, we have MATH. This means that MATH, MATH. Therefore MATH can be chosen in such a way that MATH, because MATH and MATH. As a consequence of REF , we obtain MATH, MATH, MATH and MATH. And so, we have that MATH and MATH, which concludes REF . Finally, we prove REF . Thanks to Maximum Modulus Theorem, we only need to check that MATH . Let MATH. If MATH, we have: MATH . On the other hand, if MATH, MATH, the reasoning is slightly more complicated. From REF , it is possible to find a curve MATH such that MATH and MATH. We define MATH . For a MATH large enough, one has MATH, and so MATH. Therefore, MATH is left divided in three disjoint pieces: MATH from MATH to MATH, MATH from MATH to MATH, and MATH from MATH to MATH (see REF ). To continue, we need to demonstrate the existence of a constant MATH, that does not depend on MATH, such that MATH . Indeed, MATH . Taking into account that MATH, we reason as in REF and obtain MATH . Therefore, using REF , we have: MATH by REF in the hypotheses of REF , we get: MATH . Thus, REF holds for MATH. At this point, we distinguish two cases: CASE: If MATH then: MATH for a MATH large enough. CASE: If MATH then: From MATH, we have, in the set of Cartesian coordinates given by MATH, MATH . Using REF , the fact that MATH, REF , and Property MATH one has MATH where MATH. By NAME theorem, MATH for a MATH large enough.
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math/0002152
|
REF are immediate. Let us see REF. For MATH, we have MATH for MATH. Thus MATH for MATH. Therefore, MATH . Let us check the fourth property. For MATH, we have MATH .
|
math/0002152
|
REF are easy to check. The third property is also easy to prove with the aid of REF below. The fourth property follows from the third.
|
math/0002152
|
Let MATH be some fixed constants. Obviously, MATH. So we only have to show MATH. It is enough to prove that for a fixed collection of numbers MATH satisfying MATH and MATH we have MATH . For any MATH, let MATH be a cube centered at MATH with side length MATH. Then MATH and so MATH. By NAME 's covering theorem, there exists a family of points MATH such that the cubes MATH form an almost disjoint covering of MATH. Since MATH and MATH have comparable sizes, MATH with MATH depending on MATH and MATH. Therefore, MATH . Then we get MATH . Since MATH for all MATH, we obtain MATH where MATH is the number of cubes of the NAME covering. Now it is easy to check that MATH is bounded some constant depending only on MATH, MATH and MATH: If MATH is the NAME measure on MATH and MATH is the NAME constant in MATH, we have MATH . Thus MATH and REF holds.
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math/0002152
|
Let MATH. To see that MATH we set MATH for all cubes MATH. Then REF holds with MATH. Let us check that the second REF is also satisfied. We have to prove that for any two cubes MATH, MATH . Notice that if MATH, then MATH because MATH are doubling. So REF follows if MATH. However, in general, MATH does not imply MATH, and so we have to modify the argument. Suppose first that MATH. Then MATH. We denote MATH. Then we have MATH . Using the properties of REF repeatedly, we get MATH . Since MATH and they are doubling cubes, we have MATH . Now we are left with the second term on the right hand side of REF . We have MATH . Due to the fact that MATH are doubling cubes, MATH and, by REF , we get that REF holds in this case. Assume now MATH. Then MATH. There exists some MATH such that MATH and MATH. Since MATH and MATH have comparable sizes, we have MATH. Then, if we denote MATH, we get MATH . Also, MATH . Therefore, MATH . Now we have to check that MATH. If MATH is a doubling cube, since REF holds with MATH (by REF ), we have MATH . Therefore, for any cube MATH (non doubling, in general), using MATH we get MATH . Thus MATH . It only remains to show that REF also holds with MATH instead of MATH. This follows easily. Indeed, if MATH are doubling cubes, we have MATH .
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math/0002152
|
Assume MATH for simplicity. First we show REF. If MATH, then REF holds with MATH. Moreover, for any cube MATH we have MATH . Therefore, MATH . The second term on the right hand side is estimated as REF in the preceeding lemma. For the first and third terms on the right, we apply REF . So we get, MATH . Thus MATH satisfies REF too. The implication MATH is easier: One only has to consider doubling cubes in REF. Let us see now REF. Let MATH be some cube, non doubling in general. We only have to show that REF holds. We know that for MATH-almost all MATH there exists some doubling cube centered at MATH with sidelength MATH, for some MATH. We denote by MATH the biggest cube satisfying these properties. Observe that MATH, and then MATH . By NAME 's covering theorem, there are points MATH such that MATH-almost all MATH is covered by a family of cubes MATH with bounded overlap. By REF , using that MATH, we get MATH .
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math/0002152
|
First we will show that if MATH for some MATH, then MATH . We will use the characterization of MATH given by REF . REF follows by standard methods. The same proof that shows that MATH when MATH is a doubling measure works here. We omit the details. Let us see how REF follows. For simplicity, we assume MATH. We have to show that if MATH, then MATH . Recall that MATH is the first integer MATH such that MATH. We denote MATH. Then, for MATH and MATH, we set MATH . Since MATH we get MATH . Now we take the mean over MATH for MATH, and over MATH for MATH. Using the MATH boundedness of MATH, we obtain MATH . For MATH we write MATH . The estimate for the first term on the right hand side is similar to the previous estimate for MATH: MATH . On the other hand, since MATH, we have MATH. Therefore, MATH . So we have proved that REF holds for MATH. If MATH for all MATH, then the integral MATH may be not convergent. The operator MATH can be extended to the whole space MATH following the usual arguments: Given a cube MATH centered at the origin with side length MATH, we write MATH, with MATH. For MATH, we define MATH . Now both integrals in this equation are convergent and with this definition one can check that REF holds too, with arguments similar to the case MATH.
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math/0002152
|
Let MATH be the smallest cube concentric with MATH containing MATH and MATH. Since MATH, we have MATH and MATH. Then, MATH .
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math/0002152
|
For any function MATH, we set MATH, with MATH and MATH. By REF , MATH. Since MATH and MATH, we have MATH. Thus MATH.
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math/0002152
|
We will prove REF for MATH. The proof for other values of MATH is similar. Recall that if REF are satisfied, then REF is also satisfied subsituting MATH by MATH, for any MATH (see REF ). Let MATH. Assume first that MATH is bounded. Let MATH be some fixed cube in MATH. We write MATH. Let MATH be some positive constant which will be fixed later. By REF , for MATH-almost any MATH such that MATH, there exists some doubling cube MATH centered at MATH satisfying MATH . Moreover, we may assume that MATH is the biggest doubling cube satisfying REF with side length MATH for some integer MATH, with MATH . By NAME 's covering theorem, there exists an almost disjoint subfamily MATH of the cubes MATH such that MATH . Then, since MATH and MATH, we have MATH . Since REF is satisfied if we change MATH by MATH, we have MATH and, by REF , MATH . So if we choose MATH big enough, MATH . Now we want to see that for each MATH we have MATH . We consider the cube MATH. If MATH, then there exists some cube MATH, MATH, containing MATH and such that MATH. Thus MATH . The first and third sums on the right hand side are bounded by MATH because MATH and MATH on the one hand and MATH and MATH on the other hand have comparable sizes. The second sum is also bounded by MATH due to the fact that there are no doubling cubes of the form MATH between MATH and MATH, and then MATH. Assume now MATH. Then MATH . Since MATH and MATH have comparable sizes, by REF we have MATH. And since MATH, we also have MATH. So REF holds in this case too. If MATH, then, by the choice of MATH, we have MATH which implies MATH . Thus MATH . As above, the term MATH is bounded by MATH. The last one equals MATH, which is estimated in REF . For the second one, since MATH is doubling, we have MATH . So REF holds in any case. Now we consider the function MATH . Since we are assuming that MATH is bounded, MATH. By REF we have MATH . By REF and taking into account that MATH, we get MATH . Thus MATH . Then, for MATH small enough, MATH with MATH depending on MATH, MATH and MATH. Now the theorem is almost proved for MATH bounded. We have MATH which is equivalent to REF . When MATH is not bounded, we consider the function MATH of REF . By this Lemma and the subsequent remark we know that MATH . Since MATH as MATH, REF holds in this case too.
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math/0002152
|
REF coincide. So we only have to compare REF . For any MATH, the inequality MATH follows from NAME 's inequality. To obtain the converse inequality we will apply NAME. If MATH, then MATH and so MATH.
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math/0002152
|
The proofs of REF are similar to the usual proofs for MATH. Let us sketch the proof of the third property, we can follow an argument similar to the one of REF . Given MATH, it is obvious that MATH and MATH. For the converse inequality, given an atomic block MATH with MATH, it is not difficult to see that each function MATH can be decomposed in a finite fixed number of functions MATH such that MATH for all MATH, with MATH, where MATH are cubes such that MATH and MATH, etc. Then, we will have MATH, which yields MATH .
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math/0002152
|
By standard arguments, it is enough to show that MATH for any atomic block MATH with MATH, MATH, where the MATH's are functions satisfying REF of atomic block. We write MATH . To estimate the first integral on the right hand side, we take into account that MATH, and by usual arguments we get MATH . On the other hand, for the last integral in REF , we have MATH . By REF , for each MATH we have MATH . Also, MATH . Thus MATH and then, taking into account REF , MATH . By REF , we are done.
|
math/0002152
|
Following some standard arguments (see CITE, for example), we only need to show that if MATH is an atomic block and MATH, then MATH . Suppose MATH, MATH, where the MATH's are functions satisfying REF of atomic blocks. Then, using MATH, MATH . Since MATH, we have MATH . From REF we get MATH .
|
math/0002152
|
We have to prove that MATH . We will show that there exists some function MATH such that MATH . Let MATH be some small constant which will be fixed later. There are two possibilities: CASE: There exists some doubling cube MATH such that MATH . CASE: For any doubling cube MATH, REF does not hold. If REF holds and MATH is doubling and satisfies REF , then we take MATH such that MATH if MATH, MATH if MATH, and MATH if MATH, so that MATH (this is possible because of REF ). Then, MATH . Since MATH is an atomic block and MATH is doubling, MATH. Therefore MATH . In REF there are again two possibilities: CASE: For any two doubling cubes MATH. MATH . CASE: There are doubling cubes MATH such that MATH . Assume first that REF holds. By the definition of MATH there exists some cube MATH such that MATH . We consider the following atomic block supported on MATH: We set MATH, where MATH and MATH is supported on MATH, constant on this cube, and such that MATH. Let us estimate MATH. We have MATH . Then, since MATH is doubling and MATH, MATH . Now we have MATH . By the definition of MATH, MATH . On the other hand, by the computation about MATH and since REF does not hold for MATH, MATH . By REF , if MATH has been chosen small enough, MATH . Now we consider REF . Let MATH be doubling cubes such that MATH . We take MATH . So MATH, and MATH is an atomic block. Since MATH and MATH are doubling, MATH . We have MATH . Since we are in REF , the terms MATH are bounded by MATH. By REF , if MATH is chosen MATH, then MATH .
|
math/0002152
|
This lemma is very similar to REF . We only need to show that if MATH is a MATH-atomic block and MATH, then MATH . Suppose MATH, MATH, where the MATH's are functions satisfying REF of MATH-atomic block. Since MATH, MATH where MATH. As MATH, we have MATH . From REF we get MATH .
|
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