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math/0101167 | Suppose that MATH and MATH. From the NAME identity it follows that MATH for every MATH. Therefore MATH . Thus MATH. If we let MATH then MATH is MATH - gradable. |
math/0101167 | REF follows from the MATH - derivative property and MATH . For REF we use the same proof as in REF , that is, REF and MATH . REF can be easily checked by using the same proof as in CITE. |
math/0101167 | The first statement is clear. Consider a NAME module MATH and a submodule MATH generated by MATH and MATH, that is, MATH is characterized by a non - split exact sequence MATH . If MATH is generated by its singular vectors then it has to be generated by MATH and some singular vector belonging to MATH. Hence it is of the form MATH, MATH. It is easy to check that any such singular vector is a multiple of MATH (because of MATH). This would imply MATH. Therefore MATH is not generated by its singular vectors. |
math/0101167 | Let MATH. From REF it follows that MATH, MATH, where MATH is a singular vector in MATH. If MATH then this would imply that there are three linearly independent singular vectors of the form MATH, MATH and MATH inside MATH. Therefore MATH should be singular too. On the other hand, the singular space for MATH is at most one - dimensional for every MATH. Hence MATH is proportional to MATH. Therefore MATH is a linear combination of MATH and MATH. |
math/0101167 | We may assume that MATH and MATH are homogeneous, that is, they are contained in some generalized eignespaces. Clearly MATH and MATH for some MATH, MATH and MATH, MATH. We will prove by induction on MATH that MATH which implies the desired result. We already proved that if MATH the statement holds for every MATH. Let MATH. Suppose that the statement holds for every MATH, MATH such that MATH. For every homogeneous MATH there is MATH such that MATH. Then MATH where MATH and MATH . REF can be written as MATH . After expanding REF - by using the binomial theorem - we obtain MATH . If we apply the induction hypothesis, REF reduces to the following differential equation MATH where MATH. Every solution of REF inside MATH is of the form MATH where MATH . |
math/0101167 | Notice that both MATH and MATH satisfy the NAME identity (but the MATH - property does not hold for MATH). To prove that MATH gives a MATH - homomorphism we can use the same proof as in CITE REF , since it uses only the NAME identity. Then MATH for MATH . |
math/0101167 | Assume that MATH for some MATH. Then MATH for every MATH, MATH and MATH. From the commutator formula it follows that MATH and therefore REF holds for every MATH. Hence MATH for every MATH and MATH. Claim: MATH. If not then MATH is a submodule of MATH such that MATH. But, since MATH does not have singular vectors out side MATH the same thing holds for MATH. Therefore MATH has to be zero. Therefore MATH, which implies MATH, by the above argument. |
math/0101167 | From CITE it follows that MATH for some MATH and MATH. On the other hand MATH where an isomorphism is given by MATH. Hence if MATH the space of MATH - homomorphisms is at most two - dimensional. By using the previous lemma we have the proof. |
math/0101167 | We prove first that MATH for every MATH and MATH. Note that MATH . Then by the NAME identity MATH . As in REF , we check that for every MATH and MATH . From this fact it follows that MATH for every MATH. So we have the proof. |
math/0101167 | If REF holds then we have constructed MATH, such that the MATH - property does not hold for MATH. If we let MATH then MATH clearly satisfies the NAME identity. From REF it follows that MATH . Hence MATH is a logarithmic intertwining operator of type MATH . Now in order to project MATH to MATH one has to check that MATH for every MATH, where MATH is the maximal submodule of MATH. Since MATH is generated by the singular vectors and the NAME identity holds, it is enough to check that MATH for MATH and MATH. Write MATH for some MATH. Then MATH for some differential operator MATH (compare REF). This NAME 's differential equation reduces to a differential equation with constant coefficients. Because of REF the corresponding characteristic equation has double roots at MATH, there is a logarithmic solution. Therefore MATH and hence MATH, for every MATH. |
math/0101167 | Directly follows from REF . |
math/0101167 | It is not hard to see that MATH where MATH is a MATH - (bi)module generated by MATH . Hence MATH . |
math/0101167 | Follows directly from REF . |
math/0101167 | Directly follows from the formulas: MATH and MATH . |
math/0101167 | It follows from the proof of the main result in CITE. This proof was used to show that MATH has essentially unique central extension. |
math/0101167 | It is trivial to prove that MATH is a NAME subalgebra. Now, integration by parts implies ``integral" REF . If we take MATH we obtain the NAME subalgebra. If we let MATH then we have MATH . Then MATH is a NAME algebra isomorphism. |
math/0101167 | Let MATH be a REF - cocycle considered in CITE (where was denoted by MATH - which is essentially the NAME - NAME cocycle) and MATH an non - trivial REF - cocycle on MATH induced by the isomorphism MATH. REF implies that MATH . Hence MATH . |
math/0101167 | The proof is straightforward. |
math/0101167 | denote by MATH the projection to a subspace MATH. It is enough to show that MATH for every MATH, MATH. If MATH the result holds. Suppose that REF holds for MATH. Then MATH . |
math/0101171 | Denote the elementary symmetric functions in MATH (respectively, MATH) indeterminates by MATH . It is clear that the MATH depend only on MATH. The polynomial equation which admits MATH as solutions is MATH . Since MATH, the coefficient of MATH is exactly MATH, and all the coefficients of MATH, MATH, are affine functions of MATH, which proves the first assertion. To prove the second assertion of the lemma, it will be enough to show that the MATH are polynomials in MATH and MATH. But MATH with MATH, so the relation MATH allows us to conclude by induction. |
math/0101171 | The product defining MATH vanishes if and only if one its factors does, that is, there exists MATH, MATH, such that MATH . If MATH, this simply means that MATH, and we have REF . If MATH, MATH is not locally invertible at MATH, therefore MATH, and the vanishing of the factor means that MATH which happens exactly when MATH. |
math/0101171 | We have seen that MATH is holomorphic on the whole disc. By the remarks at the beginning of REF, in a neighborhood of the unit circle the maps MATH are all globally defined, holomorphic, with a non-vanishing derivative, it is clear that each function MATH is a MATH function : holomorphic on a neighborhood of the unit circle in MATH, bounded for MATH, and for MATH, MATH tends to MATH in the MATH norm on the unit circle. The MATH-norm of MATH is bounded by MATH. Since MATH is bounded from below in a neighborhood of the unit circle, a similar MATH bound holds for MATH. Now the map MATH, seen as a map from MATH to MATH, is a composition of linear and MATH-linear maps, so to check that it is continuous it is enough to see that MATH which follows from the above estimates and repeated use of NAME 's inequality. For MATH, the required estimate follows immediately from MATH being bounded from below. |
math/0101171 | Suppose first that MATH. Then, using the fact that MATH for any MATH, MATH so MATH, because the last product involves a finite set of powers of MATH and polynomials which are symmetric in the MATH, and therefore can be expressed as polynomials of MATH and bounded holomorphic functions symmetric in MATH, which are then necessarily bounded holomorphic functions of MATH. Suppose MATH in MATH. By the above we have MATH, MATH. Now in the canonical factorization, the harmonic function MATH is the NAME integral of its boundary values, while MATH is the NAME integral of a singular measure MATH, where MATH is a positive measure (carried on a NAME set of NAME measure MATH). Factoring both MATH and MATH, and using the essential uniqueness of the factorization, we see that MATH for all indices MATH (their difference is a positive measure). Since MATH must converge to MATH in MATH, MATH converges weakly to MATH, so MATH, which implies that MATH divides MATH. On the other hand, MATH convergence implies convergence of the function and all its derivatives uniformly on any compactum, and therefore since each MATH has zeroes at the points of MATH with at least the same multiplicity, the same holds true for MATH. Since dividing by MATH only affects the boundary values of MATH on a set of measure MATH, the remaining function is still in MATH, QED . |
math/0101171 | If MATH, then by REF , MATH, which implies that MATH has no inner factor by NAME 's theorem. If MATH has finite codimension MATH, then it must contain a non-zero polynomial MATH of degree MATH. Then MATH is analytic across the unit circle, so cannot admit a singular inner factor. On the other hand, any zero MATH of MATH in MATH provides a continuous functional on MATH which vanishes on MATH : in REF , the functional is MATH, in REF . It is easy to see that if the set of zeros of MATH is infinite, this family of functionals is not of finite dimension, which contradicts the hypothesis. |
math/0101171 | Let MATH be a positive number large enough so that the interior MATH of some component of the lemniscate MATH contains the unit disk, and MATH whenever MATH. Denote by MATH the NAME mapping from MATH onto the unit disk , which maps MATH to the origin. Then MATH is analytic in MATH and the NAME space in MATH, MATH (we equip MATH with harmonic measure), is isometric to MATH in the unit disk: an isometry is being given by MATH . Note that MATH is a finite NAME product and MATH, are analytic in a neighborhood of the unit disk. This implies that MATH have trivial singular parts in their inner-outer factorization for all MATH and their inner parts are finite NAME products. By REF , MATH contains MATH-invariant subspaces generated by these NAME products. In particular this implies that for every fixed MATH, MATH contains all polynomials vanishing with greater or equal multiplicity at all zeros of MATH in MATH. Since the MATH convergence implies MATH-convergence in the unit disk, REF implies that the closure of MATH in MATH contains the MATH-invariant subspace determined by the zeros of MATH which lie in the interior of the unit disk, and, therefore, has finite codimension. Let MATH be the inner part of MATH. Thus, since every MATH-submodule MATH of MATH is a MATH-submodule, the number of generators of MATH as MATH-module does not exceed the number of generators of MATH as MATH-module. By the NAME decomposition theorem this last number does not exceed the order of MATH (see CITE for the details). |
math/0101171 | Let MATH be common zeros in MATH of the polynomials MATH. There are polynomials MATH such that MATH . By REF MATH is in MATH. Since MATH, MATH is outer, and MATH. |
math/0101171 | If MATH, a non-trivial MATH-invariant subspace contained in MATH is given by REF . If MATH and MATH is a nontrivial MATH-invariant subspace contained in MATH, choose MATH. Choose a point MATH such that MATH; then for any MATH such that MATH and MATH (the above Theorem guarantees the existence of such points), we must have MATH, but MATH, which shows that MATH, thus MATH and MATH is not MATH-invariant. |
math/0101171 | Define an integer-valued function in the disk by MATH . The hypothesis says that MATH for any MATH. We claim that MATH is upper-semicontinuous on MATH. Indeed, let MATH, and let MATH a sequence tending to MATH such that MATH. We may assume that MATH is contained in a neighborhood of MATH on which MATH, MATH are all well-defined and distinct. For each MATH, there exists a subset MATH such that MATH if and only if MATH, and MATH. For MATH large enough, this is constant and equal to MATH. Since there is only a finite number of subsets of MATH, we may also assume, by passing to a subsequence again denoted by MATH, that MATH is a constant subset. Therefore for any MATH, MATH . So MATH, as claimed. Let MATH. This makes sense since MATH is integer-valued, and since it is u.s.c. the set MATH is open. We claim that MATH (and therefore MATH) only has isolated points. Indeed, let MATH be the set of non-isolated points within MATH. Since MATH is closed in MATH, so is MATH. On the other hand, suppose MATH. Restrict attention to a neighborhood of MATH where a continuous choice of MATH can be made. Then, reasoning as in the proof of the previous claim, we can find a sequence MATH tending to MATH and a set MATH such that MATH if and only if MATH. Then, for any MATH, the analytic function MATH vanishes on a subset which has a limit point, and therefore is identically MATH in a small disk MATH. This proves that any point in this disk is in MATH, and therefore in MATH. So MATH is also an open set. Since MATH is connected, if MATH, then MATH, which is absurd. Therefore MATH, QED . Now let MATH . Then MATH (and therefore MATH) only has isolated points. Indeed, MATH, if and only if there is some MATH such that MATH, MATH. Take a point MATH, and a compact neighborhood of MATH of MATH; then MATH is a compact set (since a finite NAME product is a proper map from the disk to itself), therefore can only contain a finite number of points from MATH by the above claim, so there can only be a finite number of points in MATH in MATH, QED . For any point MATH, the fiber MATH is made up of exactly MATH points, and each point in this fiber is part of a group of MATH points for which the values of MATH coincide. So the theorem holds with MATH. |
math/0101171 | Let MATH be a zero of MATH (recall that MATH is the NAME part of the canonical factorization of MATH). CASE: Suppose that MATH, for MATH. Then set MATH. By our hypothesis, MATH, so by REF , MATH is divisible by MATH . We have MATH and MATH for MATH. So MATH which implies MATH. CASE: Suppose that MATH for some MATH. Now MATH is due to the fact that MATH. Then any MATH must satisfy MATH, but if we had MATH, then for MATH we would have MATH, so MATH. |
math/0101171 | Suppose MATH, let MATH for MATH. If a sequence MATH can be found in MATH which approximates MATH pointwise boundedly, then MATH belongs to the weak closure of MATH in MATH. Since the weak closure of MATH coincides with the strong closure, MATH is in the MATH-closure of MATH. The proof of REF therefore reduces to the following Proposition. |
math/0101171 | First we need to introduce notations for two kinds of singularities (in a certain sense) that we will encounter. Recall from REF that MATH is the (finite) set of MATH such that MATH is made up of fewer than MATH points. Larger still is the set MATH of MATH such that MATH is made up of fewer than MATH points. The previous considerations show that MATH is contained in the zero set of MATH, so that if it is not discrete, MATH, and our theorem is vacuously true. Let MATH be any holomorphic function in the disc (later we will choose MATH). Restrict attention for a moment to points MATH. Then there is a unique polynomial MATH of degree not exceeding MATH such that MATH . The polynomial MATH can be written MATH, since its coefficients depend only on the set MATH, and, being symmetric in the elements of that set and given by fractions (NAME 's interpolation formula), are holomorphic in MATH. So we have, at any such point MATH, MATH . Now it suffices to show that when MATH, the MATH turn out to be bounded analytic functions in the unit disk. We write MATH to emphasize the dependency on MATH. In a punctured neighborhood of any point of MATH, MATH is given by the NAME formula, MATH . On the other hand, MATH . When MATH goes through the set MATH, MATH goes through the set MATH, so the above numerators cancel out with the denominators REF . So finally MATH . This formula makes sense whenever MATH, and shows that the coefficients of MATH are holomorphically extendable to MATH. Consider now a neighborhood MATH of the unit circle chosen so that its closure is disjoint from MATH. For MATH, MATH is bounded below. So to prove that the MATH are bounded analytic functions, it will be enough to prove that the singularities for MATH are removable. This will be done by considering several cases in turn for the possible singularity MATH. First, we consider a type of function MATH which allows us to describe explicitly what the interpolating polynomial MATH turns out to be when MATH is a critical point of MATH. CASE: MATH for any MATH, and MATH is one-to-one on the set MATH. Note that this hypothesis is equivalent to the fact that MATH is one-to-one in some neighborhood MATH of MATH. Then MATH is holomorphic on the open set MATH, and MATH is exactly the interpolation polynomial of degree less or equal than MATH assuming the same values as MATH at the points of MATH, for MATH. Choose a smaller neighborhood MATH of MATH, MATH relatively compact in MATH, such that the boundary of MATH is a finite disjoint union of circles. For MATH belonging to a small enough neighborhood of MATH such that MATH, and for MATH, MATH where MATH CITE. This formula depends continuously on MATH in the whole neighborhood MATH, and converges, when MATH converges to MATH, to the solution of an NAME interpolation problem given by MATH where MATH form a partition of MATH such that MATH if and only if MATH and MATH belong to the same set MATH. CASE: For any point MATH such that there exists a MATH with MATH, we have MATH for any such MATH. This hypothesis says that the equation MATH has solutions of multiplicity never exceeding MATH. It does not depend on MATH. Notice that this case is generic: NAME products of degree greater than one always admit critical points inside the disk, and those for which multiplicities of the above solutions can be MATH or more form a small set, in a sense that will be made precise at the beginning of REF . Then the various terms in the sum which represents MATH either converge to a finite limit (when MATH is such that MATH) or can be regrouped in pairs MATH where MATH, MATH for MATH. To simplify notations, let us assume that MATH, MATH. For MATH in a small enough neighborhood of MATH, the sum of the two terms can be written MATH . Define a meromorphic function whose coefficients depend on MATH by MATH . The poles of MATH depend on the parameter MATH and stay away from MATH for MATH in a neighborhood of MATH, and we have MATH where MATH is another meromorphic function (of two variables) with its pole set (which is a hypersurface of MATH) avoiding a neighborhood of MATH. The expression from REF is then equal to MATH . Letting MATH tend to MATH, MATH tends to MATH as well, and MATH tends to MATH, so the coefficients of MATH admit finite limits, and the possible singularity of its coefficients at MATH is actually removable. CASE: General case. The NAME products of degree MATH which do not satisfy the hypotheses of REF form an algebraic subset, therefore of empty interior, since their derivative, which is a rational fraction of fixed degree, must admit at least one multiple root. Given one such exceptional NAME product MATH, let MATH be a sequence of NAME products satisfying the hypotheses of REF and converging to MATH uniformly in a neighborhood the closed unit disk. By continuous dependency of the roots of a polynomial equation upon the coefficients (NAME Theorem), we see that the corresponding functions MATH converge to MATH uniformly in a some neighborhood of the unit circle (where they are all well-defined), and that any symmetric polynomial in the MATH's converges uniformly in the closed disk to the same polynomial in the MATH's. In particular, if we set MATH, then MATH converges to MATH pointwise boundedly on the disk. Given a function MATH, write MATH . From REF and the proof in REF , we see that the MATH are bounded holomorphic functions and MATH. Fix some degree MATH between MATH and MATH. On any circle centered at the origin of radius MATH close enough to MATH, MATH converges uniformly, because MATH is bounded on the circle, all the MATH are well defined and converge uniformly, and all the denominators in the expression are bounded away from MATH. If MATH is furthermore bounded, the limit is a bounded holomorphic function which we denote by MATH. We need to see that for any MATH, MATH such that MATH, then MATH. In that case, there is a MATH such that MATH, and by the convergence of the MATH, we can select a subsequence (which we will denote the same way as the original sequence) and an index MATH such that MATH as MATH. Then MATH where the next to last equality is by the symmetry of the MATH, and the last by uniform convergence on compacta. So we can write MATH, where MATH. Now MATH . |
math/0101171 | The proof proceeds exactly as the proof of REF , except that now we will need to assume MATH, with MATH. This time we need to prove that the functions MATH are continuous up to the boundary (in addition to being holomorphic). For this it will be enough to show that the coeffients of MATH are themselves continuous up to the boundary. This follows from REF , since MATH and MATH are continuous, and the MATH are all distinct near the unit circle. Furthermore, the arguments of REF now show that MATH converges to MATH uniformly on the closed disk (using the uniform continuity of MATH on the closed disk). So the coefficients MATH converge uniformly to functions in the disc algebra, and one proves in the usual way that the limits are functions of MATH alone. |
math/0101171 | CASE: Any zero of MATH in the open disk provides a continuous, non-trivial linear form which vanishes on MATH, by the proof of REF . If MATH has a zero MATH on the boundary, then since MATH never vanishes on the unit circle, there must exist MATH, MATH, such that MATH, MATH. Then MATH provides a non-trivial linear form, continuous on MATH, which vanishes on MATH. The converse is immediate by the remark made before the statement of REF . CASE: As above, an infinite number of zeros of MATH would provide arbitrarily large (finite) sets of linearly independent continuous linear forms vanishing on MATH, so the codimension could not be finite. Suppose on the other hand that the factorization of MATH admit a singular inner factor MATH. It is easy to see that the map MATH is continuous on MATH, so that whenever MATH, MATH is divisible by MATH, therefore by MATH. Arguing as in the proof of REF again, if MATH was of finite codimension, it would contain a polynomial, which would lead to a contradiction. Conversely, if MATH is finite, it will be enough to show that the codimension of the closure of MATH is finite. Let MATH. Then MATH is a closed ideal of MATH, and the main result of CITE shows that MATH, the set of all functions in MATH which vanish on the closed set MATH and which are divisible by the inner function MATH. The hypothesis on MATH imply that MATH is a finite NAME product (because singular inner factors are excluded), and that MATH is a finite subset of the circle. Therefore MATH reduces to those functions in MATH which vanish on MATH to an order greater or equal than the order of vanishing of MATH at those points, and which vanish on MATH (where the notion of ``order of vanishing" makes no sense). Notice that this allows us to compute precisely the codimension of MATH, as the sum of the number of points in MATH, counted in the usual way, and the number of points in MATH, counted with multiplicities. |
math/0101174 | Recall that the coarse graining is initially defined in terms of the mesoscopic NAME MATH which depend only on the bond configurations. The phase label MATH takes also into account the random coloring of good NAME. In fact, MATH and the claim of the lemma follows by the domination by NAME site percolation REF, actually for any choice of MATH. |
math/0101174 | The proof essentially follows REF. The only difference with the periodic case discussed there is the existence of open contours attached to the boundary. But, for any choice of the macroscopic boundary condition MATH (as defined above), these boundary contours are, in the language of CITE, large, that is their diameter exceeds MATH on all sufficiently big MATH renormalization scales. Let us briefly recall the three main steps of the argument: The first step is to check that the volume of MATH-blocks is negligible. This simply follows by REF : MATH . The second and the third steps are devoted to the control of the regions surrounded by contours, that is, by connected surfaces of MATH-blocks. By phase rigidity REF the contours play the role of mesoscopic interfaces separating regions with mesoscopic phase labels of different signs so that they contribute to the total perimeter of the configurations MATH. We distinguish between two types of contours, the small contours with diameter smaller than MATH (where MATH is a given constant) and the remaining contours, namely the large ones. NAME type estimates can be applied to bound the probability of events such that the collection MATH of large contours has a total area larger than MATH . The small contours are not controlled in terms of their total area but of the volume of the regions they surround. By the choice of the macroscopic boundary condition MATH there are no open small contours, and, proceeding exactly as in the proof of REF, one can show that MATH . As a consequence of the previous estimate, the small contours will have no contribution to the MATH-norm. Combining estimates REF, we arrive to the claim of the Theorem. |
math/0101174 | Let us briefly work out the MATH case: Recall that MATH . On the other hand, any phase label MATH which is compatible with MATH is, after the repainting of the MATH-cluster into MATH, automatically compatible with the boundary condition MATH and, therefore, satisfies the conditions of REF under the measure MATH. Thus, the claim of the proposition follows from the general tightness REF and, in the case of negative magnetic fields MATH, by the representation formula for the boundary surface tension REF. |
math/0101174 | It is a standard result of the theory of convex bodies (see for example, REF ) that for every MATH there exists a convex polyhedral set MATH such that the NAME distance MATH. Of course, the support function MATH of such MATH satisfies MATH . Define now MATH . A comparison with REF reveals that MATH where MATH and MATH are the support functions of MATH and MATH respectively. Now, for every bounded convex MATH the volume MATH can be written in terms of its support function MATH as MATH . Consequently, REF implies that MATH . As we have already mentioned in the end of Subsection REF, MATH . On the other hand, for MATH, the support function MATH equals to MATH. Moreover, for every MATH any supporting hyperplane to MATH at MATH is also a supporting hyperplane to MATH. It follows that the support function MATH on MATH. Thus, as it follows now from REF, MATH . Since by REF, MATH, and both MATH and MATH are bounded above by some finite constant MATH, we finally obtain the following estimate MATH and the claim of the theorem follows once we redefine MATH. |
math/0101174 | We shall give the proof only in the more difficult case of negative boundary fields MATH. Also according to REF , it will be sufficient to derive the proposition with MATH replaced by MATH (for MATH small enough). Starting from the approximate shapes MATH let us use the transformation REF to define the scaled polyhedral approximation MATH of MATH. By the approximation result of REF it will be enough to prove REF with MATH instead of MATH. Define MATH. Since MATH is polyhedral and convex it is an easy matter to show that there exists a side length MATH and a finite number MATH of disjoint parallelepipeds MATH with bases MATH included in MATH of side length MATH and height MATH such that: I REF The sets MATH cover MATH up to a set of measure less than MATH denoted by MATH and they satisfy MATH where MATH is the unit normal to MATH (and, hence, to the corresponding facet of MATH). I REF For any MATH the base MATH divides MATH into two parallelepipeds MATH and MATH, such that in the case of MATH, MATH and MATH, whereas in the case of MATH the corresponding base MATH, while MATH and MATH. In order to enforce a microscopic interface close to the polyhedral set MATH, we define MATH . Let us also introduce the set MATH chosen such that for any configuration MATH in MATH the bonds inside and outside MATH are decoupled. For this, it is enough to close the bonds which are in a neighborhood of MATH (the microscopic counterpart of MATH) MATH . We get MATH . We first check that MATH . From the argument developed in the proof of REF , we are going to check that in each of the two connected components of MATH the phase labels MATH are, with a uniformly positive probability, close to the equilibrium values (for the MATH-norm). Since MATH decouples the connected components, it is enough to consider the mesoscopic phase labels in the interior of the regions decoupled by MATH. As the boundary of MATH is regular, these regions are fairly exhausted by the mesoscopic boxes for any scale MATH. Using the terminology of REF , we assert that with large probability, the mesoscopic label configurations are almost uniformly constant and contain only contours which have no contribution in terms of the MATH-norm. According to estimate REF, the volume of the regions surrounded by the small contours is arbitrarily small in the thermodynamic limit. Furthermore, from the usual NAME estimates, the volume of the large contours is negligible as well. This implies REF. Finally as MATH, we see by applying REF that MATH . The denominator can be easily estimated by REF MATH . On the other hand, as we are trying to bound probabilities from below, even under the conditioning on MATH one can still apply FKG arguments (non-crossing is a decreasing event) and use REF to decouple between different MATH-events which constitute MATH in REF. Therefore, since, MATH, we are entitled to conclude that MATH . Combining REF with the estimates REF; the approximation REF ; the choice of the polyhedral set MATH through REF and the representation REF of MATH, we arrive to the claim of REF . |
math/0101176 | we consider MATH for MATH and MATH. We fix MATH for MATH. We consider MATH for MATH and MATH. Then we have MATH. Then we have MATH and MATH. Then we have MATH . Then MATH . Hence MATH . |
math/0101176 | Let MATH be the strict transform of MATH, MATH the strict transform of MATH for MATH, and MATH such that MATH. In MATH case, we consider that, MATH . As in REF , we have MATH. In MATH case, we consider that, MATH . As in REF , MATH. In MATH case, we consider that, MATH . As in REF , MATH. In MATH case, we consider that MATH . We fix MATH for MATH. We consider MATH. We have MATH or MATH. Then MATH or MATH. Then we have MATH and MATH. Then MATH . Hence we have MATH. |
math/0101176 | Case-MATH . Let MATH for MATH. We consider that, for MATH, MATH . Then, as REF , MATH. Case-MATH. Let MATH be MATH. We consider that, for MATH, MATH . We fixed MATH and MATH for MATH such that MATH and MATH. Then we have MATH or MATH . Then MATH. We have MATH. Then we have MATH . Hence, we have MATH. CASE: Let MATH be MATH where MATH for MATH. Then we consider that, for MATH, MATH . Then we have MATH . Hence, we have MATH. The proofs of cases MATH and MATH are the same as the proof of case MATH. Case-MATH. Let MATH be MATH or MATH, MATH or MATH, MATH. We consider that, for MATH, MATH . As in case MATH, we have MATH. |
math/0101176 | Let MATH the strict transform of MATH, MATH the strict transform of MATH, and MATH such that MATH. In MATH case, we consider that, MATH . As in REF , MATH. In the MATH case, we consider that, MATH . As in REF , MATH. In the MATH case, we consider that MATH . As in REF , MATH. In the MATH case, we consider that MATH . As MATH case in REF , MATH. In the MATH case, we consider that, MATH . As MATH case in REF , MATH. |
math/0101176 | We change the multiplicity of subvariety at the point with the weighted multiplicity of subvariety at the point for MATH in [KREF ( compare [KaREF])]. The proof is the same as [KREF ( compare [KaREF])]. |
math/0101176 | (compare [KaREF]) We have MATH by REF . Let MATH = sup MATH; there exists an effective MATH-divisor MATH such that MATH and MATH . Let us define a function MATH for MATH with MATH to be the largest real number such that MATH whenever MATH, MATH and MATH. Then MATH is a convex function [KaREF lREF - lREF ]. Since MATH, there exists a real number MATH such that MATH and MATH for any MATH. Let MATH be a large and sufficiently divisible integer and MATH the evaluation homomorphism. We consider subspaces MATH of MATH for integers MATH such that MATH. First, we have MATH . Let MATH be a member corresponding to MATH for some MATH. Since we have MATH, the number of conditions in order for MATH is at most ( the number of homogeneous polynomials of order MATH in MATH variables ) MATH, that is, MATH . Therefore, we have MATH because MATH . By [KaREF last part], we have that MATH. |
math/0101176 | Let MATH. Let MATH be the weighted blow up of MATH at MATH such that MATH with the exceptional divisor MATH of MATH such that MATH. Hence MATH by REF . Let MATH be a rational number such that MATH. Since MATH, we can take MATH. Let MATH be a rational number such that MATH for MATH. By REF , there exists an effective MATH-Cartier divisor MATH such that MATH and MATH. Hence MATH. Let MATH be the log canonical threshold of MATH at MATH: MATH . Then MATH. Let MATH be the minimal element of MATH. If MATH, then MATH is free at MATH by REF , since MATH. CASE: We consider the case in which MATH is a curve. By REF , MATH is normal at MATH, that is, smooth at MATH. Let MATH be a neighborhood at MATH. Then MATH. Hence we have MATH by REF . Since MATH, there exists a rational number MATH with MATH and an effective MATH-Cartier divisor MATH on MATH such that MATH and MATH. As in [KaREF StepREF], there exists an effective MATH-Cartier divisor MATH on MATH such that MATH and MATH. Let MATH be a general effective MATH-Cartier divisor on an affine neighborhood MATH of MATH in MATH such that MATH and MATH. Then we have MATH, hence MATH. Let MATH . Since MATH is chosen to be general, we have MATH. We have an element MATH such that MATH and MATH. By REF , MATH has an element which is properly contained in MATH. By REF , we conclude that MATH is MATH at MATH, and MATH has an element which is properly contained in MATH, that is, MATH. CASE: We consider the case in which MATH is a surface. Let MATH be a neighborhood at MATH and MATH its closure of local universal cover, that is, MATH and MATH. Let MATH be MATH and MATH. Since MATH is a MATH point [KaREF] and MATH is ramified only over MATH, MATH is also irreducible and MATH is at most MATH. Moreover MATH is a hypersurface, MATH is also NAME. Therefore, MATH is a rational NAME point, that is, MATH is a smooth point or a rational double point. If MATH is a smooth point of MATH, then MATH for MATH. By REF , we have MATH. If MATH is a rational double point of MATH, then MATH is A type or D type or E type, that is, MATH and MATH is invariant by the action of MATH for MATH and MATH. We have MATH by REF . Let MATH or MATH. CASE: We assume first that MATH. As in REF , there exists a rational number MATH with MATH. By REF and MATH, there exist an effective MATH-Cartier divisor MATH on MATH and a positive number MATH such that MATH, MATH, MATH is MATH at MATH, and that the minimal element MATH of MATH is properly contained in MATH. Thus we have the theorem when MATH. We consider the case in which MATH is a curve. Since MATH, we have MATH. As in REF , we take a rational number MATH, an effective MATH-Cartier divisor MATH on MATH, and a positive number MATH such that MATH, MATH, MATH, MATH is MATH at MATH and that the minimal element MATH of MATH is properly contained in MATH, that is, MATH. CASE: We assume that MATH. As in REF , we take a rational number MATH with MATH and an effective MATH-Cartier divisor MATH on MATH with MATH and MATH. Here we need the factor MATH because MATH has the weighted multiplicity MATH at MATH. Then we take MATH such that MATH is MATH and MATH has an element which is properly contained in MATH. We shall prove that we may assume MATH. Then we can take MATH as in REF , and the rest of the proof is the same. For this purpose, we apply REF . In argument of REF through REF, the number MATH was chosen under the only condition that MATH. So we can take MATH, where the MATH for MATH will stand for very small positive rational numbers. Then MATH and MATH. This means the following: for any effective MATH, if MATH, then MATH. We look for MATH so that there exists an effective MATH-Cartier divisor MATH with MATH and MATH. The equation for MATH becomes MATH . We have MATH, MATH, and in particular, MATH. By [KaREF StepREF], we obtain a desired MATH, and can choose a new MATH such that MATH. Then we repeat the preceding argument from REF . If we arrive at REF again, then we have MATH and MATH. |
math/0101176 | Let MATH the weighted blow up with the weights MATH with the exceptional divisor MATH of MATH such that MATH. Then MATH by REF . Since MATH and REF , as in REF of the proof of REF , there exist a rational number MATH and an effective MATH-Cartier divisor MATH such that MATH, MATH, and MATH. Let MATH be the log canonical threshold of MATH at MATH: MATH . Then MATH. Let MATH be the minimal element of MATH. If MATH, then MATH is free at MATH by REF , since MATH. Let MATH be a neighborhood at MATH, and MATH its closure of local universal cover, that is, MATH and MATH, MATH and MATH, MATH and MATH, MATH and MATH, MATH and MATH, MATH and MATH. CASE: We consider the case in which MATH is a curve. By REF , MATH is smooth at MATH. Since MATH, MATH, MATH, MATH, MATH, and MATH are points, then MATH. Hence by REF , we have MATH. Since MATH, as in REF of the proof of REF , we take a rational number MATH such that MATH, and an effective MATH-Cartier divisor MATH on MATH and a positive number MATH such that MATH, MATH, MATH is MATH at MATH, and the minimal element MATH of MATH is properly contained in MATH, that is, MATH. CASE: We consider the case in which MATH is a surface. Let MATH be MATH, and MATH be MATH. Since MATH is MATH-factorial NAME terminal at MATH, MATH is factorial at MATH([KaREF ]). In particular, MATH is a NAME divisor at MATH. Therefore, MATH is a rational double point. Then MATH is A type or D type or E type, that is, MATH and MATH is invariant by the action of MATH for MATH or MATH. By REF , We have MATH. CASE: As in REF of the proof of REF , there exists a rational number MATH with MATH. By REF and MATH, there exist an effective MATH-Cartier divisor MATH on MATH and a positive number MATH such that MATH, MATH, MATH is MATH at MATH, and that the minimal element MATH of MATH is properly contained in MATH. Thus we have the theorem when MATH. We consider the case in which MATH is a curve. Since MATH, we have MATH. As in REF , we take a rational number MATH, an effective MATH-Cartier divisor MATH on MATH and a positive number MATH such that MATH, MATH, MATH, MATH is MATH at MATH and that the minimal element MATH of MATH is properly contained in MATH, that is, MATH. |
math/0101176 | Let MATH and MATH be the weighted blow up of MATH at MATH such that MATH with the exceptional divisor MATH of MATH such that MATH. By REF , MATH. CASE: Since MATH and REF , as in REF of the proof of REF , there exists a rational number MATH and an effective MATH-Cartier divisor MATH such that MATH, MATH, and MATH. Let MATH be the log canonical threshold of MATH at MATH: MATH . Then MATH. Let MATH be the minimal element of MATH. If MATH, then MATH is free at MATH by REF , since MATH. CASE: We consider the case in which MATH is a curve. By REF , MATH is smooth at MATH. Let MATH be a neighborhood at MATH. Then MATH. Hence we have MATH. Since MATH, as in REF , we take a rational number MATH, an effective MATH-Cartier MATH on MATH, and a positive number MATH such that MATH, MATH, MATH, MATH is MATH at MATH, and that the minimal element MATH of MATH is properly contained in MATH, that is, MATH. |
math/0101184 | Here, MATH, is the even NAME character as defined in CITE, mapping the even NAME of MATH into even periodic cyclic cohomology, and MATH denotes the pairing between NAME and periodic cyclic cohomology defined in CITE. We have MATH where (for each MATH) MATH is the cyclic MATH-cocycle defined by MATH . Since MATH it follows that MATH . Now, MATH hence MATH. Therefore MATH as claimed. |
math/0101184 | Again, MATH, is the odd NAME character as defined in CITE, mapping the odd NAME of MATH into odd periodic cyclic cohomology, and MATH denotes the pairing between NAME and periodic cyclic cohomology CITE, pREF. It is straightforward to calculate this pairing directly, as in the previous example, but it is quicker to appeal to NAME 's index theorem CITE, pREF, which states that MATH where MATH is the natural orthogonal projection MATH. We have MATH hence the result. This shows that MATH is a nontrivial NAME module, and also that MATH. |
math/0101184 | We use the universal coefficient theorem to calculate the NAME. Since the NAME are free abelian, it follows from REF that MATH, MATH, as claimed. We will exhibit the generating even NAME modules. Since MATH, we start by noting the NAME of MATH. Both the even and odd NAME of MATH are isomorphic to MATH. We exhibit the generating NAME modules. This follows from REF , since the NAME of MATH are both isomorphic to MATH. The generator of MATH is the even canonical NAME module MATH REF corresponding to the *-homomorphism MATH, MATH. Explicitly, MATH . Since MATH is a crossed product MATH (via the trivial action) the generator of MATH is the odd NAME module MATH, REF where MATH, and MATH. We note that under the NAME assembly map CITE MATH we have MATH, MATH. We modify these NAME modules to give an even NAME module MATH over MATH, modelled on the corresponding MATH for MATH, and two even NAME modules MATH and MATH over MATH induced from the odd NAME module MATH over MATH. The first generator of the even NAME is given by the canonical even NAME module MATH REF corresponding to the *-homomorphism MATH, defined on generators by MATH. Then MATH . Let MATH be any of the projections MATH, MATH, MATH. Then it is immediate that MATH, and MATH . Hence MATH, for each of MATH, MATH, MATH. Now we define the NAME module MATH. First of all, consider MATH acting on MATH, with orthonormal basis MATH, via MATH . Define the diagonal operator MATH by MATH. Let MATH. We define a *-homomorphism MATH by MATH . Define MATH . It is easy to verify that MATH and MATH are both rank-one operators. Hence MATH is compact for all MATH in the dense subalgebra of elements of the form MATH, where MATH, the NAME functions on MATH, and so for all MATH. Hence MATH is an even NAME module over MATH. We calculate the pairing of the NAME character of MATH in even periodic cyclic cohomology with the elements of MATH represented by the projections MATH, MATH, MATH. We have MATH where we denote by MATH the rank one operator MATH. Hence MATH, so MATH . Therefore MATH. It follows that MATH is a nontrivial element of MATH. The same calculation for MATH gives us that MATH since MATH. Hence MATH. Since MATH it is obvious that MATH. We obtain the third NAME module MATH from MATH as follows. We pull back MATH via the automorphism MATH, defined by MATH . Then MATH is as follows. We again have MATH acting on MATH, this time via MATH . Take MATH, MATH and MATH as before, and define the *-homomorphism MATH by MATH that is, the same as MATH, except the representation on MATH is different. Then MATH is an even NAME module over MATH, as can be verified exactly as before. To distinguish MATH from our previous example, we calculate the pairing of its NAME character with the projections MATH, MATH and MATH. We note that MATH . Hence MATH . We summarize the results of all these calculations in the following table. The pairings of the generating NAME modules with NAME are given by : MATH . Since MATH pairs nontrivially with MATH, we can see that MATH, MATH and MATH in fact generate MATH as an abelian group - since their pairings with each of the projections are either REF or REF, by NAME 's index theorem CITE, pREF, they are generators, rather than just generating a copy of MATH inside MATH. This completes the proof of REF . |
math/0101184 | We know CITE, pREF that for finite groups MATH and MATH, and any left MATH-module MATH, that for MATH the group cohomology of the free product MATH is given by MATH . Since MATH, and MATH it follows that MATH for all MATH. By definition, MATH since MATH is a trivial MATH-module. To find MATH we use the NAME spectral sequence. Recall that, for an exact sequence of groups MATH and some MATH-module MATH, MATH is a MATH-module, and MATH converges to MATH. In our case we have an exact sequence MATH so MATH and MATH . So MATH for MATH. Now, MATH with the action of MATH on MATH being by inner automorphisms (inversion) so MATH. Finally, MATH is the only non-zero MATH. So everything is over at the first step of the spectral sequence. |
math/0101184 | Given a REF-cocycle MATH, we show that we always have MATH, for some linear map MATH, with MATH, MATH. By our previous work we have MATH . Now define MATH . So we have shown that every REF-cocycle is in fact a coboundary, so MATH. |
math/0101184 | From our previous work we see immediately that MATH, while MATH. The three distinguished REF-cocycles are ``dual" to the generators of MATH, and NAME 's periodicity operator MATH maps them to cohomologically nontrivial even cocycles of higher degree. We see that for each MATH we have MATH, with generators MATH, MATH and MATH. |
math/0101184 | By REF. Let MATH be a linear functional, with MATH. Suppose that MATH . Obviously MATH, and MATH. We want to solve MATH . CASE: MATH, MATH, MATH. This gives MATH which has the solution MATH REF MATH, MATH, MATH. This gives MATH which has the solution MATH, for all MATH, MATH. CASE: MATH, MATH, MATH. This gives MATH . Hence MATH . This has the solution MATH where MATH is a scalar. Hence we have shown that for any of REF-cocycles MATH, with MATH, the corresponding REF-cocycle MATH is a coboundary. So for each such MATH, and any MATH, MATH. |
math/0101184 | Both these statements are immediate, the second because MATH. |
math/0101184 | We see that MATH is a degenerate NAME module, since MATH, and hence is zero in MATH. |
math/0101184 | This follows immediately from REF . |
math/0101184 | We have MATH where MATH, and MATH is as above. We construct a homotopy of NAME modules from MATH to a degenerate NAME module. For MATH, we define MATH where MATH is given by MATH . Then MATH, and MATH is a degenerate NAME module, since MATH. Hence MATH is a trivial element of NAME. |
math/0101185 | Treating MATH as a family of skew - symmetric matrices parameterized by MATH, it is easy to check that the coefficient of the linear term in the characteristic polynomial of MATH is MATH . Clearly, MATH is maximally nondegenerate if and only if this coefficient is never zero. By REF , the first term only vanishes at MATH. At these points the second term becomes MATH which does not equal zero by REF . Thus, MATH is maximally nondegenerate. Near the boundary of MATH we have MATH . Since MATH is nondegenerate on MATH, we see that MATH spans MATH near MATH. Hence, MATH has REF . In order to check REF , we consider a nonvanishing vector field MATH on MATH such that MATH spans MATH. We choose MATH so that it equals MATH near MATH and we recall that any characteristic of MATH is equivalent (after some reparameterization) to a trajectory of MATH. It is straightforward to check that MATH where MATH is the standard volume form with respect to our coordinates on MATH. Since MATH, it follows that the MATH-component of MATH, MATH, satisfies MATH . This equation implies that MATH vanishes only when MATH does. Hence, by REF , MATH equals zero along the two-dimensional tori MATH defined by MATH, respectively. By continuity and our choice of MATH near MATH, we also see that MATH is strictly positive away from these tori. On MATH, MATH . Hence, MATH is tangent to both tori and lies in the kernel of MATH on each. This means that the flows of MATH on MATH are both conjugate to the irrational flow for the angle MATH. Let MATH be the negative limit set of MATH under the flow of MATH. The previous observations then imply that all the points in MATH asymptotically approach the aperiodic flow on MATH . Consequently, the trajectories of MATH through all of these points are trapped in MATH and REF holds. At this point, it is also easy to see that none of the trajectories of MATH are closed and hence REF is satisfied. Since MATH is nonnegative, any periodic orbit of MATH would have to lie on either MATH or MATH. However, this is impossible since the flow of MATH on both these sets is aperiodic. Now, let MATH be the map which sends MATH and acts as the identity on the MATH-component. By REF , it is possible to choose MATH such that MATH . Consider a trajectory of MATH (a characteristic of MATH) which starts at some point MATH and exits MATH through MATH. The MATH-component MATH can not vanish along this trajectory since it would then be trapped in one of the internal tori MATH and could never exit MATH. Consequently, the (anti-)symmetry of MATH described above implies that as the trajectory progresses from MATH to MATH it retraces, in reverse, the progress it made along the MATH-component from MATH to MATH. Hence, the trajectory must exit at MATH and REF has been verified. |
math/0101186 | Write MATH as MATH, where MATH and MATH is a discrete MATH-submodule of MATH of rank MATH. Since MATH does not have any nonzero translations, MATH has a fixed point: MATH has a fixed point, that is, there exists MATH such that MATH. Since MATH, in suitable coordinates, we can write MATH . Since MATH is of finite order, we may take MATH as a diagonal matrix MATH. It suffices to show that MATH. We derive a contradition from the assumption MATH. Recall that MATH. Then there exists MATH such that MATH. We can write MATH . Let MATH. Then MATH leads to MATH. Thus the curve MATH in MATH is contained in MATH. This contradicts the fact that MATH is a finite set. Since MATH is an Abelian surface, there exist global coordinates MATH of MATH centered at a fixed point of MATH such that MATH. Recall that MATH, where MATH and MATH is a finite set. Then we have MATH for some integers MATH and MATH where each MATH is coprime to MATH. Let MATH be the group algebra of the cyclic group MATH over MATH. Clearly, MATH. The action of MATH on MATH defines a natural MATH-module structure on MATH: MATH. Write MATH for the MATH-th cyclotomic polynomial. Since MATH and MATH is a torsion free MATH-module. This also holds for MATH. NAME with MATH the tower of modules MATH, gives the tower of vector spaces MATH. Then we get MATH, where MATH is the NAME function. Thus, MATH is either REF or REF. Combining this with the fact MATH is not divisible by REF CITE, we get MATH is either REF or REF. For the case of MATH, we have MATH. Therefore MATH. We also see that MATH and MATH must be roots of unity. Recall that neither MATH nor MATH is MATH. Thus, we conclude that MATH. Changing the generator by MATH to MATH, we obtain MATH. Since MATH is the direct sum of some eigenspaces corresponding to the indicated eigenvalues of the action of MATH on the space MATH, MATH must be a free MATH-module of rank MATH. This gives MATH. Next we consider the eigenspace decomposition: MATH where MATH denotes the eigenspace of the eigenvalue MATH. In our case, we obtain the following MATH-equivariant eigenspace decompositions: MATH . Therefore we have MATH. In particular, MATH. If MATH, then the same argument gives MATH, MATH or MATH. Changing the generator again, we obtain MATH, MATH. However the first case cannot happen because MATH. Therefore, MATH. Then in a similar way we have MATH. Since MATH and MATH, we have MATH and MATH. This gives the result. |
math/0101186 | REF are clear. The last REF comes from the fact that MATH and MATH is a finite group. We need to show that MATH, that is, if MATH then MATH. Since MATH, MATH. This is a NAME isometry of MATH. Set MATH. Then MATH preserves the NAME structure and satisfies MATH for MATH. Therefore MATH is a NAME isometry of MATH. Moreover, since MATH on MATH, and MATH on MATH, we have MATH on MATH. This shows MATH. |
math/0101186 | To show that MATH is a normal subgroup of MATH, we check that for each MATH and MATH, there exists an element MATH such that MATH. We have MATH for some MATH then we have: MATH . In addition, MATH and MATH. Therefore, we may take MATH. Assume that MATH for some MATH and MATH. Take MATH. Since MATH, we have MATH. This shows that MATH. To show that MATH, we take MATH, MATH and set MATH. By REF , there exists MATH such that MATH. Since MATH is an even hyperbolic lattice by REF , MATH. Then there exists a NAME isometry MATH of MATH such that MATH and MATH for all MATH. Recall that MATH is the image of the homomorphism MATH. From REF , there exists MATH such that MATH. Moreover MATH for all MATH by REF again. Hence MATH and MATH. Clearly, MATH. So we have MATH. Then we get the result. |
math/0101186 | To prove the first part of the statement, we have only to do with the same line of REF . It is clear that MATH. To show MATH, take the element MATH corresponds to the element of MATH and this leads to an element of MATH by taking restriction. So, applying REF for the pairs MATH and MATH, we have the element of MATH which comes from the element of MATH. Then we conclude that MATH by REF . |
math/0101192 | For the assertion MATH, see CITE or CITE. Let MATH and MATH. We have MATH, because otherwise MATH. Thus MATH. Then, MATH because MATH. Finally, the inclusion MATH follows in a similar way. |
math/0101192 | Suppose that the cubes MATH have already been chosen and the properties stated in REF hold. Let us see how we can choose cubes MATH, substitutes of MATH, such that MATH is a NAME function on MATH. For a fixed MATH, we set MATH . It is easily seen that this is a non negative NAME function, with constant independent of MATH. Then, we denote by MATH the cube centered at MATH with side length MATH. We have to show that MATH is a good choice as a cube of the scale MATH. Indeed, by REF it is clear that MATH. Thus MATH . Take now MATH such that MATH . This inequality implies MATH, and also MATH. Thus MATH and MATH. The inclusions MATH imply MATH, with MATH depending only on MATH, MATH, MATH. One can verfy that the properties in REF still hold, assuming that the constant MATH is ``absorbed" by the ``error" MATH in REF . |
math/0101192 | For fixed MATH and MATH, we have MATH, with MATH and MATH. Assume MATH. We have MATH, and so MATH . If MATH or MATH, we also have MATH by an approximation argument, and so MATH . Let us see the converse inequality. Given MATH, let MATH be the least integer such that MATH. Now let MATH be the least positive integer such that MATH. Then we have MATH . Also, it is easily checked that MATH . Therefore, MATH . |
math/0101192 | We denote MATH, MATH, and MATH. Recall that, for fixed MATH, MATH, we have defined MATH, where MATH is non negative, smooth, radial, and non increasing. Then we write MATH . We denote MATH and MATH . Thus, MATH . Let us see that MATH is ``big". Recall that MATH for MATH. Then, for MATH, we have MATH. Therefore, MATH . Since MATH and also MATH, for MATH big enough we obtain MATH using REF . If we denote the measure MATH by MATH, we get MATH . Thus, MATH . Now we will deal with the doubling property. If MATH is not MATH-doubling, we write MATH. We have MATH . Therefore, MATH . So if we take MATH big enough, there exists some MATH such that MATH is MATH-doubling and MATH. |
math/0101192 | Take MATH and MATH satisfying REF . By the assumption above, the fact that MATH for MATH, and NAME 's inequality, we get MATH and so MATH. |
math/0101192 | For a fixed MATH, we denote MATH. Observe that MATH . We have MATH where MATH is the least integer such that MATH. If MATH, then MATH. Therefore, MATH . Thus, MATH . If MATH is small enough, we have MATH for all MATH. Therefore, REF holds. Finally, REF follows easily from REF . |
math/0101192 | Given MATH, we set MATH and MATH . So we have to see that there exists some MATH such that, for all MATH and MATH, MATH if we choose MATH small enough. Since MATH is open, we can consider a NAME decomposition of it. That is, we set MATH, so that the cubes MATH have disjoint interiors, MATH for each MATH, and the cubes MATH have finite overlap. Take a cube MATH such that there exists some MATH with MATH. By standard arguments, one can check that for any MATH, MATH where MATH is the centered radial maximal NAME operator: MATH . Since MATH, we get MATH if MATH. For MATH small enough, this implies MATH for all MATH. So we have MATH . Let MATH be such that MATH. Let us check that MATH . For a fixed MATH, let MATH be the least non negative integer such that there exists some MATH. Let us denote MATH for MATH, and MATH. Then we have MATH where MATH is the least integer such that MATH. Let MATH be the MATH-neighborhood of MATH. It is easily checked that MATH for all MATH. Therefore, if MATH is small enough, for MATH we must have MATH. Then, by the weak MATH boundedness of MATH with respect to MATH, we get MATH . Using the finite overlap of the neighborhoods MATH, MATH which proves REF . Since MATH for all MATH, we get MATH, if MATH is small enough. By the MATH condition, MATH . Therefore, by the finite overlap of the cubes MATH, MATH . Thus, MATH. Now we only have to take MATH (which does not depend on MATH, MATH or MATH), and REF follows. |
math/0101192 | We have defined the kernels MATH so that they are CZ kernels uniformly on MATH. By REF we know that they are of weak type MATH with respect to MATH. We only have to check that this holds uniformly on MATH. Indeed, if this is not the case, for each MATH we take MATH such that MATH. Then we define MATH. Since MATH, MATH is a CZO (using also the uniform properties of the the operators MATH). On the other hand, we have MATH for each MATH, because MATH are integral operators with non negative kernel. Thus MATH, which contradicts REF . |
math/0101192 | Since the operators MATH are of weak type MATH with respect to MATH, from REF it follows that their duals are also of weak type MATH with respect to MATH, uniformly on MATH. Then, REF is a consequence of duality in NAME spaces. We only have to argue as in CITE, for example: MATH . |
math/0101192 | Let MATH be some fixed point. First we will show that, for some MATH, MATH if we choose MATH big enough. Let MATH be some ball containing MATH such that MATH . Notice that if MATH, then MATH. Since MATH, we get MATH . Recall also that MATH is formed by MATH ``layers" of NAME cubes, and so we have MATH . We distinguish two cases: CASE: Assume MATH, that is MATH is small compared to MATH. We choose MATH such that MATH. Then there exists some ball MATH containing MATH and MATH with radius MATH. Therefore, MATH and REF holds. CASE: Suppose that MATH. Then there exists some NAME cube MATH in MATH such that MATH and MATH. Otherwise, by REF , MATH which contradicts our previous assumption. Since MATH, we can find MATH such that MATH. Thus, MATH if MATH is chosen big enough. By REF , we obtain MATH. Then there exists some ball MATH containing MATH and MATH with radius MATH . Arguing as in REF , we obtain REF . Now we have to deal with the term MATH. Let MATH be some constant whose precise value will be fixed below. Consider the ball MATH, and assume that MATH is big enough so that MATH, MATH, where MATH is the point chosen in REF or REF above. Let us remark that in both cases we have MATH, where MATH may depend on MATH. We have MATH . If we take MATH big enough (so that MATH), for each MATH we have MATH where MATH may depend on MATH. Thus MATH. We also have MATH where MATH depends on MATH. Therefore, MATH . If we take MATH such that MATH, by REF , we get MATH . |
math/0101192 | We will show that for some MATH, the operator MATH is bounded on MATH. The precise value of MATH will be fixed below. Without loss of generality, we take MATH non negative. Given any MATH, we denote MATH . We will show that there exists some constant MATH, with MATH, such that for all MATH where MATH is some constant depending on MATH but not on MATH. It is straightforward to check that REF implies that MATH is of weak type MATH with respect to MATH for MATH small enough. As in REF , we consider the NAME decomposition MATH, where MATH are dyadic cubes with disjoint interiors (the NAME cubes). Take some cube MATH. Suppose that MATH and MATH are chosen in REF so that the maximum principle REF holds with MATH instead of MATH. To simplify notation, we will write MATH instead of MATH. Then, for MATH, we have MATH, and so MATH . Let MATH be such that MATH. If for all MATH with MATH we have MATH, where MATH are positive constants which we will fix below, then we write MATH (that is, MATH is a ``bad point") and, otherwise, MATH. Notice that MATH. We will see that MATH is quite small. Indeed, we will prove that MATH for some positive constant MATH independent of MATH and MATH. Assume that REF holds for the moment, and let us estimate MATH. For MATH, we have MATH . Using the inequality MATH, for MATH, we get MATH . It is not difficult to check that the family of sets MATH has bounded overlap (depending on MATH). Then, summing over all the indices MATH, we obtain MATH . By REF , if we choose MATH, we get MATH which is REF with MATH. Now we have to show that REF holds. We intend to use the MATH property. Let us see that MATH for all MATH. By REF , if MATH, then MATH, where MATH is some positive constant depending on MATH. Then we have MATH for some MATH. If moreover MATH, then this inequality holds for some MATH, assuming that we take MATH. Suppose that MATH. Let MATH be the least integer such there exists some MATH, and let MATH be the least integer such that MATH. Then we have MATH . It is not difficult to check that if MATH and MATH, then MATH. Therefore, MATH. Then, by the weak MATH boundedness of MATH, we have MATH . Thus, by the finite overlap of the sets MATH, and since MATH, MATH . Notice that MATH depends on MATH, but not on MATH. If MATH is small enough, we obtain MATH. Now, since MATH, we have MATH and REF holds. By the MATH property, we get MATH, and because of the finite overlap of the cubes MATH, MATH which implies REF . |
math/0101192 | We denote MATH. Observe that, by the definition of the kernels MATH, we have MATH . Let MATH be such that MATH. If MATH, then MATH by REF , and then MATH. Assume now MATH. Since MATH, we have MATH. If MATH is big enough (depending on MATH), we deduce MATH by REF . Then we get MATH, and so MATH if MATH. We also deduce MATH. By REF , if MATH, we obtain MATH . Suppose finally that MATH. As above, we have MATH, and since MATH, MATH. Then we get MATH, and so MATH if MATH. On the other hand, MATH, because MATH. Thus MATH with MATH depending on MATH. Then, if MATH, by REF we get MATH . |
math/0101192 | The constant MATH will be chosen below. For the moment, let us say that MATH. Let MATH be the least integer such that MATH. For each MATH, we denote MATH. We have MATH. We will show that some cube MATH, with MATH, is doubling with respect to MATH and MATH. Suppose that all the cubes MATH, MATH, are either non MATH-doubling, or non MATH-doubling (for simplicity, we will show the existence of a cube MATH which is MATH-doubling, instead of MATH-doubling). For each MATH, let MATH be the number of non MATH-doubling cubes of the form MATH, MATH and let MATH the number of non MATH-doubling cubes of the form MATH, MATH. From our assumption we deduce MATH . We have MATH. Thus, MATH . Let MATH be the biggest non MATH-doubling cube of the form MATH, MATH. Then, for MATH, we have MATH if MATH is big enough. Let MATH be such that MATH. We denote MATH. From the properties of the kernels MATH, it is easily seen that, for MATH and MATH, we have MATH. Then, using the boundedness of MATH in MATH, we obtain MATH if MATH. Thus, MATH . By REF , MATH. Then, from REF , we get MATH . Therefore, MATH if MATH. Thus MATH if MATH is big enough, which is a contradiction. |
math/0101192 | We denote MATH. Then, we have MATH, for all MATH, with MATH. Therefore, MATH . Thus, MATH. We only have to set MATH. |
math/0101192 | We will construct inductively the set MATH. Let MATH. If MATH, the lemma is straightforward. Otherwise, we take MATH such that MATH. Assume that MATH have been chosen. We set MATH and we choose MATH such that MATH and MATH. By construction, MATH, and also MATH for MATH. We have MATH for each MATH, because otherwise MATH, and then MATH. Finally we show that the third property holds. Suppose that MATH and MATH. If MATH, it is easily seen that MATH. Because of the definition of MATH, we must have MATH (otherwise MATH, which is not possible). However the last inclusions imply MATH, which is a contradiction. |
math/0101192 | Let MATH be some fixed MATH-doubling cube, with MATH. We also denote MATH. We will show that REF holds for MATH. To this end, by an inductive argument, for each MATH we will construct a sequence of MATH-doubling cubes MATH. First we will show how the cubes MATH are obtained. Let MATH where MATH is some big positive constant which will be fixed below. By REF , this set is open. We consider some NAME decomposition MATH, where the cubes MATH are dyadic cubes with disjoint interiors. Let us check that MATH. If MATH, then for all MATH there exists some cube MATH centered at MATH, with MATH with MATH (where MATH is some fixed constant). Since MATH, we have MATH. By NAME 's Covering Theorem, there exists some covering MATH with finite overlap. Using that MATH is MATH-doubling, we obtain MATH . Therefore, MATH, which is a contradiction if MATH is big enough. Since MATH, by the properties of the NAME covering, we have MATH for any NAME cube MATH such that MATH. Moreover, subdividing the NAME cubes if necessary, we may assume that MATH. Let MATH be such that MATH. Observe that if MATH, then MATH, and so MATH. For MATH, we consider some MATH-MATH-doubling cube MATH, with MATH given by REF . Now we take a NAME 's covering of MATH with this type of cubes: MATH, and we define MATH. Notice that, for each MATH, MATH, because all the NAME cubes intersecting MATH have side length MATH. In particular, we have MATH. For each MATH, let MATH be such that MATH. If MATH is centered at some point in MATH, then MATH. This finishes the step MATH of the construction. Suppose now that the cubes MATH (which are MATH-MATH-doubling, with MATH, and have finite overlap) have already been constructed. Let us see how the cubes MATH are obtained. For each fixed cube MATH we repeat the arguments applied to MATH. We denote MATH and let MATH be such that MATH. We consider the open set MATH, and a decomposition of it into NAME dyadic cubes with disjoint interiors: MATH. Arguing as in the case of MATH, we deduce MATH, and if MATH, then MATH. Given MATH such that MATH, for MATH, we consider some MATH-MATH-doubling cube MATH. It may happen that the union MATH is not pairwise disjoint, and so for a fixed MATH there are several indices MATH such that MATH is defined. In any case, for each MATH we choose MATH with MATH so that MATH (no matter which MATH). Now we take a NAME covering of MATH with cubes of the type MATH. So we have MATH, and the cubes MATH have bounded overlap. Moreover, for each MATH there exists some MATH such that MATH. We define MATH, and we denote by MATH the integer such that MATH. The first step to estimate MATH. We want to show that given any MATH, if MATH is big enough, we have MATH . We will prove this estimate inductively. First we deal with the case MATH. We have MATH . Given some small constant MATH, let MATH. Let us see that MATH is small. By REF , for all MATH there exists some MATH-doubling cube MATH centered at MATH such that MATH. We consider a NAME 's covering of MATH with this type of cubes. That is, MATH, with MATH. We have MATH . In particular, we deduce MATH. Then we obtain MATH . Now we have to estimate MATH. Given MATH, let MATH be such that MATH. From REF , the following maximum principle follows: MATH where MATH is some fixed constant depending on MATH. Let us see that if MATH, then MATH . Indeed, we have MATH (with equality if MATH). If MATH, then REF follows from the fact that MATH. If MATH for some MATH with MATH, then MATH, and so MATH by REF . Thus, MATH . Moreover, it is easily checked that, for MATH (and MATH), we have MATH. Therefore, REF holds in any case. We denote MATH. Notice that MATH. We have MATH where we have used that MATH if MATH and MATH. Now we write MATH . First we will estimate MATH. For MATH, we have MATH . Therefore, MATH where we have used that MATH. It remains to estimate MATH. Given MATH, by REF we have MATH . Then we get MATH . Given MATH, for MATH, we denote MATH. It is easily seen that if MATH, then MATH. Then, summing over all the cubes MATH such that MATH, due to the finite overlap of the cubes MATH, we obtain MATH . So we have shown that MATH with MATH, but not MATH, depending on MATH and MATH. The MATH-th step to estimate MATH. Now we will show that for any MATH, MATH with MATH, but not MATH, depending on MATH and MATH. The arguments to prove REF are similar to the ones we have used to obtain REF , although a little more involved because the cubes MATH are non pairwise disjoint. For each MATH we define MATH. Arguing as in REF , we deduce MATH . We denote MATH. Using the definition of MATH, we obtain MATH . To estimate MATH, we need to introduce some additional notation. Assume MATH. We denote MATH. We set MATH . It is easily checked that the sets MATH, MATH, are pairwise disjoint, that MATH, and moreover that if MATH, then MATH. We have MATH . Now we set MATH, and MATH. For MATH, we have MATH . Therefore, operating as in the case MATH, we get MATH . Summing over MATH, we obtain MATH . Finally we deal with MATH. Let MATH be the collection of all the NAME cubes (originated from all the sets MATH) such that if MATH comes from MATH, then MATH. Assume MATH. We denote MATH. We set MATH . The sets MATH, MATH, are pairwise disjoint and MATH . Moreover, it easily seen that if MATH, then MATH, and so MATH . If MATH is originated by MATH, arguing as in the case MATH, we deduce MATH . Therefore, MATH . In the following estimates the notation MATH means that MATH is a NAME cube of MATH: MATH . By REF follows. From REF , we get MATH . This is the same as REF , except for the last term on right hand side. However, we will see below that this term equals MATH. The estimate of MATH. We are going to prove that MATH . We denote MATH. It is easily seen that MATH for all MATH (this is the main advantage of MATH over MATH). We will show that there exists some positive constant MATH such that MATH for all MATH. This implies REF , because MATH and MATH is MATH-doubling. For a fixed MATH, by the covering REF , there exists some subfamily MATH such that CASE: MATH. CASE: MATH if MATH. CASE: If MATH, MATH, and MATH, then MATH. First, we will see that MATH for some fixed constant MATH. By REF it is enough to show that, for each MATH, there exists some cube MATH centered at MATH such that MATH, with MATH sufficiently small. Let MATH some cube which forms MATH such that MATH. By our construction, there exists some cube MATH such that MATH, so that MATH comes from MATH. Because of REF of the covering, we have MATH. Therefore, MATH, which implies MATH and MATH. Let MATH be some MATH-doubling cube whose center is in MATH (which implies MATH). Let MATH be the set of indices MATH such that MATH. We have MATH for MATH, because MATH (since MATH). Thus, MATH. From our construction, we deduce MATH . Since MATH for MATH, by the weak MATH boundedness of MATH, and by the MATH-doubling property of MATH, we obtain MATH . Thus, by the finite overlap of the cubes MATH and the fact that MATH is MATH-doubling, MATH . Since we may choose MATH as big as we want, MATH can be taken arbitrarily small, and REF holds. Let us see that REF follows from REF . We denote MATH. Since the cubes MATH, MATH, are disjoint, REF implies MATH. By REF of the covering and the fact that MATH is MATH-doubling, we have MATH . Then, MATH . Therefore, MATH . The end of the proof. We only need to prove the lemma for MATH. For each MATH we have MATH . From REF it follows that MATH as MATH, and then the integral on the left hand side above tends to MATH as MATH. Now the lemma follows from REF . |
math/0101192 | Let MATH be some cube with MATH and MATH. We write MATH . First we will estimate the integral MATH. For each MATH, let MATH be some MATH-MATH-doubling cube with MATH. Notice that for each MATH and MATH, we have MATH. Thus by REF , if we denote MATH, we get MATH . By NAME 's Covering Theorem, there exists some subfamily of cubes MATH which covers MATH with finite overlap. Since MATH, we have MATH. Then we obtain MATH . It is easily seen that, for all MATH, MATH. Therefore, MATH . Now we turn our attention to the integral MATH. For MATH, MATH . Thus, MATH . Finally we deal with MATH. For MATH and MATH, we have MATH . Thus, MATH . From REF , we deduce MATH for some fixed constant MATH with MATH. Therefore, MATH . So we have MATH . Choosing MATH and taking the supremum over all the cubes MATH, we get MATH. Thus MATH if MATH. One way to avoid the assumption MATH would be to work with ``truncated" operators of type MATH in REF , instead of MATH; and also to consider a truncated version of MATH in REF , etc. The technical details are left for the reader. |
math/0101192 | We will show that for some MATH, the operator MATH is bounded on MATH. Without loss of generality we take MATH non negative. Given some constant MATH close to MATH, for each MATH, we denote MATH . The precise value of MATH and MATH will be fixed below. As in REF , we consider the NAME decomposition MATH, where MATH are dyadic cubes with disjoint interiors (the NAME cubes). Take some cube MATH and MATH. Suppose that MATH and MATH are chosen in REF so that the maximum principle REF holds with MATH. Then, we have MATH and so MATH . Let MATH be such that MATH. If for all MATH with MATH (where MATH is some positive big integer which will be chosen later) we have MATH where MATH is another constant which we will choose below, then we write MATH (that is, MATH is a ``bad point"). Notice that MATH. We will see that the set of bad points is quite small. Indeed, we will prove that MATH where MATH is some constant which depends on MATH (from the MATH property), MATH, MATH, but not on MATH, MATH, MATH, MATH. We defer the proof of REF , which is one of the key points of our argument, until REF below. Let us denote MATH. Now we have MATH . From REF we get MATH . From calculations similar to the ones in REF , it follows MATH . If we take MATH such that MATH, then the right hand side of REF is bounded above by MATH, and so MATH . Summing by parts we get MATH . Observe that if we assume MATH, then MATH which implies that our summation by parts is right. Since MATH, we have MATH . Thus, MATH . We define MATH for all MATH. To simplify notation, we also set MATH. Given MATH such that MATH, we consider the operator MATH . Since MATH, we obtain MATH . From REF we get MATH where MATH and where MATH is some constant with MATH which will be chosen later. The term MATH is easy to estimate: MATH . Let us consider the term MATH now. By NAME 's inequality and REF , we obtain MATH . It is easy to check that the family of sets MATH has bounded overlap, with some constant which depends on MATH. This fact implies MATH . Therefore, MATH. Finally we have to deal with the term MATH. Notice that, since MATH, for MATH, by REF we have MATH for all MATH. We set MATH. By REF , for MATH, we have MATH . We set MATH, and then we obtain MATH . We denote MATH and MATH. Then we have MATH . We will show that MATH for any MATH and MATH, which implies REF . For the rest of the proof we follow the convention that all indices MATH are restricted to MATH and MATH. Now we will introduce ``principal" cubes as in CITE or CITE. Let MATH be the set of indices MATH such that MATH is maximal. Assuming MATH already defined, MATH consists of those MATH for which there is MATH with MATH and CASE: MATH, CASE: MATH if MATH. We denote MATH, and for each MATH, we define MATH as the smallest cube MATH containing MATH with MATH. Then we have CASE: MATH implies MATH. CASE: MATH and MATH implies MATH. By REF and the fact that MATH, we get MATH . Let us estimate the term MATH first. Notice that if MATH, then MATH. As a consequence, MATH, and MATH. Taking into account this fact, the finite overlap of the cubes MATH (for a fixed MATH), and REF , for any MATH we get MATH . Thus, MATH . Let us estimate the term MATH. By NAME 's inequality and REF , for a fixed MATH, MATH . Summing over MATH, since any cube MATH occurs at most MATH times in the resulting sum, we get MATH . Notice that for each MATH we can write MATH . We have obtained MATH . Notice that for any fixed MATH we have MATH . Therefore, MATH which yields REF . Thus, by REF , MATH . We only have to choose MATH small enough, and we are done. |
math/0101192 | We use the same notation as in the proof of the preceding lemma. Let MATH and take MATH containing MATH recall MATH, with MATH. By REF , we have MATH for some MATH depending on MATH. It is easily seen that this implies that MATH for MATH, where MATH is some positive constant which depends on MATH. Also, MATH for MATH. So, if we choose MATH and MATH, then MATH . We denote MATH. For a fixed MATH, let MATH be the least integer such that MATH and MATH. There exists some cube MATH containing MATH, with MATH for some MATH. Since MATH, by REF there exists some doubling cube MATH centered at MATH such that MATH . Now, by NAME 's Covering Theorem, we can find some family of cubes MATH (with MATH) which covers MATH with finite overlap. Moreover, we assume that the covering has been chosen so that the REF holds. Given MATH with MATH, we will show that if MATH is small enough, then MATH for all MATH. Let MATH some fixed cube from the family MATH, and let MATH be the least integer such that MATH. First we will see that MATH . If MATH for some MATH and MATH, then by REF we have MATH where MATH is so that MATH. Let us denote by MATH the NAME cube of MATH containing MATH, with MATH. Since MATH, we have MATH, and so MATH. In fact, if MATH, which depends on MATH, is very big, then maybe we should write MATH, where MATH is some positive integer big enough, depending on MATH. The details of the required modifications in this case are left to the reader. From REF , we get MATH . For MATH and MATH, we have MATH, because MATH. Thus REF implies MATH, and then, from the weak MATH boundedness of MATH, by REF , and because MATH is doubling, we obtain MATH . So REF holds if MATH is sufficiently small. Now we have to estimate MATH. If MATH, then MATH, by REF . Recall also that MATH. As a consequence, we deduce MATH. Moreover, we have MATH, and so MATH, with MATH. Since by REF we have MATH we obtain MATH assuming MATH, where MATH is some positive integer which depends on MATH and MATH. Recall also that the fact that MATH yields MATH where MATH is given by MATH. If we choose MATH small enough, then MATH and, for MATH, REF implies MATH . On the other hand, if MATH, then by REF we have MATH with MATH. From the fact that MATH we deduce MATH, and so MATH for MATH. Thus, from REF we get MATH for any MATH with MATH. If we take MATH small enough, operating as in REF , we obtain MATH which together with REF implies REF . By REF , using the MATH condition for MATH, we get MATH for each NAME cube MATH. By the finite overlap of the cubes MATH, we obtain MATH . Therefore, MATH . |
math/0101193 | Let MATH. If MATH belongs to MATH, then MATH. Hence MATH which implies that MATH. The same argument applies to MATH and REF follows. For REF, again let MATH. Thus MATH. If MATH, then MATH. Hence MATH . It follows that MATH, MATH and MATH. The same argument applies to MATH and we have proved that MATH is a MATH-skew subdiagram of MATH. |
math/0101193 | Since MATH commutes with the MATH's, it is clear that MATH is nilpotent and commutes also with the MATH's and so the first statement is clear. The last statement is also clear since we are taking connected components. |
math/0101193 | By REF , MATH is a MATH-skew subdiagram of MATH. Since MATH is centrally symmetric, MATH is a skew subdiagram of MATH. Now let MATH and denote by MATH, the MATH-invariant non-degenerate bilinear form corresponding to the type of MATH as explained in REF. We have MATH . Thus MATH and MATH. To finish the proof, it suffices to show that MATH is a connected component of MATH. Let MATH be such that MATH. We have MATH. By central symmetry, this means that MATH. This in turn implies that MATH since MATH and MATH is a skew subdiagram. Thus MATH. The same argument applies for MATH and this proves that MATH is a connected component of MATH. |
math/0101193 | It is clear that MATH commutes with the MATH's and a direct computation shows that MATH. The second statement follows from REF (which is also true if we exchange MATH et MATH) since we are considering connected components. |
math/0101193 | We shall first show that if MATH and MATH (respectively, MATH), then the element MATH (respectively, MATH) as defined in REF (respectively, REF ) is uniquely determined by MATH (respectively, MATH) up to a scalar. Let MATH, MATH and if appropriate, we let MATH, MATH. Let MATH be in MATH with the appropriate property. For MATH, MATH. Now let us fix MATH and MATH and let MATH. If MATH, then MATH implies that MATH . Similarly, if MATH, then MATH . In the case of type A and C, MATH and we have MATH . In the case of type B and D, MATH and we have MATH . Also, we have for types B, C and D, MATH which implies that MATH . When MATH, we have MATH. If MATH is even, then we must have MATH because from the classification of REF, MATH for types B, D and MATH for type C (one should treat the case where MATH separately). Since MATH are fixed and MATH is connected, we observe that the restriction of MATH on MATH and MATH are uniquely determined by the value of MATH (MATH are fixed here). It follows that MATH (respectively, MATH) is uniquely determined by MATH (respectively, MATH) up to a scalar. By REF , we have that MATH is spanned by a set of elements indexed by a subset of MATH. To finish the proof of this theorem, il suffices to construct an element of MATH for each element of MATH and show that they form a basis of MATH. Recall that MATH is non zero only if MATH and MATH. Type A. Let MATH be a skew diagram with barycentre MATH and MATH. Define MATH by MATH . Then a simple verification shows that MATH commutes with MATH and MATH. Further MATH is nilpotent since MATH and therefore MATH. Types B and D. Let MATH, MATH be centrally symmetric skew diagrams. Let MATH be such that MATH or MATH. Define MATH by MATH if MATH is even; and if MATH is odd, we set MATH . Again a simple verification shows that MATH commutes with MATH and MATH, and that MATH is nilpotent since MATH. Finally, we verify easily that for MATH, MATH . Therefore MATH. Now suppose that MATH and MATH is odd. Let MATH be such that MATH. Define MATH by MATH . The same verification works and therefore MATH. Type C. Let MATH, MATH be centrally symmetric skew diagrams. Let MATH be such that MATH or MATH. Define MATH by MATH . Again a simple verification shows that MATH commutes with MATH and MATH, and that MATH is nilpotent since MATH. Finally, we verify easily that for MATH, MATH . Therefore MATH. Now suppose that MATH and MATH is odd. Let MATH be such that MATH. Define MATH by MATH . The same verification works and therefore MATH. Thus we have constructed a set of elements of MATH indexed by the set MATH. Finally, since they are defined on connected components or pairs of connected components, they are clearly linearly independent. The proof is now complete. |
math/0101193 | Since MATH, it is clear that a connected component MATH of MATH is a skew subdiagram of MATH and that MATH is a MATH-skew subdiagram of MATH. |
math/0101193 | First observe that MATH has cardinal MATH. By the previous lemma, we are reduced to counting the number of connected components MATH of MATH. We shall proceed by induction on the cardinality of MATH. The result is trivial if the cardinality of MATH is MATH. Now let MATH be the cardinality of MATH. Let us first suppose that there exist a northeast corner MATH such that MATH is again a skew diagram. We shall compare the connected components of MATH and MATH. Note that the former is obtained from the latter by removing MATH set-theoretically. It follows by an easy observation that, MATH if MATH is a southwest angle of MATH, which is also a southwest angle of MATH. MATH if MATH is a southwest corner of MATH, which is also a southwest corner of MATH. MATH otherwise. For example to illustrate REF, let MATH be the following skew diagram, MATH . If we remove MATH and take MATH, then MATH looks like MATH where MATH is a southwest angle of MATH. So when we remove this box, we recover MATH and the number of connected components is increased by MATH. We are therefore reduced to counting southwest corners and southwest angles of MATH southwest of MATH. Let MATH be the number of southwest corners MATH in MATH such that MATH and MATH, and MATH the number of southwest angles MATH in MATH such that MATH and MATH. Then it follows from the above that MATH . Now MATH is a skew diagram, so each southwest angle is above a southwest corner. Further, the leftmost southwest corner is not below any southwest angle. We conclude by induction that MATH . We are left with the case where removing a northeast corner in MATH gives MATH skew diagrams. This means that MATH is of the form: MATH . Let MATH be the leftmost, that is, MATH minimal, southwest corner of MATH, and let MATH. Then it is clear that MATH if MATH. Let MATH be the leftmost northeast corner of MATH. Then MATH . It follows that MATH. |
math/0101193 | First of all, we note that MATH is a MATH-skew subdiagram of MATH, and since MATH is centrally symmetric, MATH is a skew subdiagram of MATH. Thus MATH. Let us suppose that MATH is non empty and MATH is not a subset of MATH. Then there exists MATH such that MATH or MATH is in MATH. This is not possible because MATH, while MATH or MATH is in MATH which, as noted above, is a MATH-skew subdiagram. The lemma now follows. |
math/0101193 | First observe that MATH implies that MATH. But MATH is even, so MATH. We proceed by induction on MATH as in REF . For MATH, the result is clear. Let MATH and let MATH be the northeast corner of MATH such that MATH is minimal. Let MATH denote the cardinality of MATH. Let us first take care of the case where MATH is rectangular, that is, MATH has exactly one southwest corner since MATH is centrally symmetric. Then every MATH is rectangular and MATH. So we have only to count the number of MATH's such that MATH is odd. This is clearly MATH. So let us suppose that MATH is not rectangular. Let MATH and let us suppose that MATH is again a skew diagram. We shall use a similar argument as in REF , but since we are dealing with centrally symmetric skew diagrams, we need to have a detailed analysis on the induction process. Denote by MATH and MATH. Note that MATH. Let MATH be a connected component of MATH, then we have the following REF lemmas: We have MATH if and only if MATH is a connected component of MATH. Let MATH and MATH, then we have MATH, MATH and MATH if and only if MATH is a southwest corner of MATH southwest of MATH. MATH is a connected component of MATH if and only if MATH is neither a southwest corner nor a southwest angle of MATH southwest of MATH. In particular, MATH. MATH is the disjoint union of two distinct connected components of MATH if and only if MATH is a southwest angle of MATH southwest of MATH. In particular, MATH, MATH and MATH. Let MATH, then MATH and MATH if and only if either MATH or MATH and MATH is not a southwest angle of MATH southwest of MATH. MATH is the disjoint union of two distinct connected components of MATH such that MATH if and only if either MATH or MATH and MATH is a southwest angle of MATH southwest of MATH. MATH is the disjoint union of three distinct connected components of MATH such that MATH and MATH if and only if MATH is a southwest angle of MATH southwest of MATH such that MATH and MATH. Otherwise MATH is a connected component of MATH such that MATH. All three lemmas are easy consequences of REF , of the central symmetry of MATH and the fact that MATH is chosen to be the northeast corner with MATH minimal. It follows from these three lemmas that MATH and MATH are equal unless MATH satisfies one of the following conditions: MATH is a southwest corner of MATH southwest of MATH, in which case MATH by REF. MATH is a southwest angle of MATH southwest of MATH such that MATH, in which case MATH by REF. MATH is a southwest angle of MATH southwest of MATH such that MATH and MATH, in which case MATH by REF. MATH with MATH, in which case MATH if MATH is even by REF; MATH+REF if MATH is odd by REF. MATH with MATH, in which case MATH if MATH is even by REF; MATH if MATH is odd by REF. MATH with MATH, in which case MATH if MATH is even by REF; MATH if MATH is odd by REF. MATH is a southwest angle of MATH southwest of MATH, in which case MATH if MATH is even by REF; MATH if MATH is odd by REF. MATH is not a southwest angle of MATH southwest of MATH and MATH, in which case MATH if MATH is even by REF; MATH if MATH is odd by REF. Now if MATH, then only REF apply, thus we deduce as in REF that MATH where MATH (respectively, MATH) is the number of southwest corners (respectively, angles) southwest of MATH. Now if MATH, then REF do not apply. We have therefore MATH . There are REF cases: MATH is even or odd; MATH is or is not a southwest angle of MATH. By applying REF in the appropriate case, we have that: MATH since MATH is not rectangular, so there is always a southwest angle above MATH. It follows that we have again MATH . Finally if MATH, then REF apply also. Again there are the same REF cases to consider, and we deduce that MATH since for these MATH, MATH unless REF or REF is satisfied (note that MATH is the first southwest corner of MATH southwest of MATH because MATH and MATH is chosen to be the leftmost northeast corner of MATH). Hence, we have again MATH . By induction, we have our result. We are therefore left with the case where MATH is not connected. In this case, let MATH be the leftmost southwest corner of MATH. Then MATH is of the following form: MATH A similar argument as in REF reduces MATH to a horizontal diagram which is a simple verification. |
math/0101193 | Let us first suppose that MATH. Then MATH has at least a northeast corner MATH such that MATH, MATH since MATH is a skew diagram of cardinal MATH containing MATH. Thus the pair MATH, MATH is in MATH. We can therefore suppose that MATH and MATH are skew diagrams of cardinal MATH and MATH. There exist MATH such that, exchanging the roles of MATH and MATH if necessary, MATH be a southwest corner of MATH and MATH be a northeast corner of MATH. By central symmetry, MATH and the pair MATH is in MATH. So the first part of the theorem is proved. If MATH, then it is clear that equality holds if and only if REF is satisfied. So we can assume that the cardinality of MATH and MATH are MATH. If they are both rectangular, then one is horizontal and the other vertical. It is clear that equality holds. Now let MATH be non rectangular and that MATH contains MATH (the case MATH contains MATH is similar). Then MATH contains MATH. Let MATH be a northeast corner of MATH, then MATH and MATH. Let MATH be a southwest corner of MATH with MATH. Then either MATH is not in MATH or MATH has a southwest corner MATH with MATH. In the first case, the pair MATH, MATH is in MATH. In the second case, the pair MATH is in MATH. Since the pair MATH is in MATH in both cases, we are done. |
math/0101193 | A centrally symmetric skew diagram has always a southwest corner MATH such that MATH. Let MATH (respectively, MATH) be such a southwest corner in MATH (respectively, MATH). Then it is clear that the pair MATH is in MATH and we are done. |
math/0101193 | First note that MATH has exactly one southwest corner is equivalent to MATH has exactly one northeast corner, and that if MATH, then MATH. So we can suppose that MATH has exactly one southwest corner MATH. Now any skew subdiagram of MATH contains MATH. So MATH is empty if MATH or MATH, and the ``if" part follows. Now suppose that MATH does not satisfy REF , and therefore there are at least MATH southwest corners and at least MATH northeast corners. Let the bottom row of MATH be MATH and the top row of MATH be MATH where MATH. Then MATH does not satisfy REF implies that MATH, MATH and MATH. Now let MATH, then MATH. It follows that the skew diagram MATH is in MATH. So REF follows since MATH. Finally, if MATH does not satisfy REF , then the above can also be applied to the leftmost column and the rightmost column. Thus REF follows. |
math/0101193 | Let us suppose that MATH satisfies REF . If MATH is rectangular, then we have our result by REF . So let us suppose that MATH is near rectangular with southwest corners MATH, MATH with MATH. If MATH satisfies REF of REF , then MATH and MATH. We observe that we have only to find a necessary and sufficient condition for MATH to be in MATH or MATH to be in MATH. This is so if and only if MATH or MATH is odd. But this is equivalent to saying that MATH is not non integral. If MATH satisfies REF of REF , then a similar argument gives our result. Now if MATH satisfies REF of REF , then MATH. We now have to find a necessary and sufficient condition for MATH to be in MATH or MATH to be in MATH. This is so if and only if MATH or MATH is odd. Again this is equivalent to saying that MATH is not non integral. We have therefore proved the ``if" part. To prove the ``only if" part, it suffices therefore to prove that if MATH, then MATH is either rectangular or near rectangular. Let MATH be the top row of MATH with MATH, then MATH is the bottom row of MATH. Note that MATH since MATH is a skew diagram. If MATH, then either MATH is in MATH or MATH is in MATH since either MATH or MATH is odd. So it follows that if MATH, then MATH is non empty for some MATH with MATH, MATH if MATH, or equivalently, MATH. Now let MATH be the leftmost column of MATH with MATH, then MATH is the rightmost column of MATH. Note that MATH since MATH is a skew diagram. A similar argument shows that if MATH, then MATH is non empty for some MATH with MATH, MATH if MATH. We are therefore left with four possibilities: MATH and MATH; MATH and MATH; MATH and MATH; MATH and MATH. In the first case, we have that MATH which is rectangular and the result follows. In the second case, we have that MATH. Since MATH, we have either MATH or MATH. If MATH, then MATH is just a vertical diagram which is rectangular. If MATH, then MATH is near rectangular satisfying REF of REF . The third case is similar to the second yielding REF of REF . The last case implies that either MATH, MATH or MATH, MATH. In the former, MATH is rectangular. In the latter, MATH is near rectangular satisfying REF of REF . Our proposition now follows. |
math/0101193 | The ``if" part is just a direct verification using the computations in the proof of REF . Let us show why type REF is not allowed for MATH integral and leave the other verifications to the reader. Let MATH be integral and near rectangular and MATH, MATH be the two distinct southwest corners with MATH. Then MATH and MATH are both odd, so by the computations in the proof of REF , there are REF elements of MATH which are not in MATH. Note that if MATH is near rectangular and non integral, then MATH. To prove the ``only if" part, it suffices to prove that if MATH, then MATH is near rectangular. Let us suppose that MATH is not near rectangular. Let MATH be the top row of MATH with MATH, then MATH is the bottom row of MATH. Note that MATH since MATH is a skew diagram. Let MATH be the leftmost column of MATH with MATH, MATH, then MATH is the rightmost column of MATH. Note that MATH since MATH is a skew diagram. Recall from the proof of REF that if MATH and MATH, then MATH is non empty for some MATH with MATH, MATH; and if MATH and MATH, then MATH is non empty for some MATH with MATH, MATH. As in REF , we have four cases to consider: MATH; MATH, MATH; MATH, MATH; MATH. In the first case, MATH is rectangular. So MATH. In the second case, if MATH, then MATH is rectangular and MATH. Since MATH is not near rectangular, MATH and MATH. So MATH is non empty for some MATH with MATH and MATH if MATH. Let MATH be a southwest corner distinct from MATH and MATH. Then clearly, MATH. It follows that MATH is a pair in MATH. Therefore MATH. So let us assume that MATH and MATH are the only southwest corners of MATH. Then since MATH, MATH and by central symmetry MATH; otherwise, MATH is a southwest corner distinct from MATH and MATH. It follows that either MATH is in MATH or MATH is in MATH, where MATH. Thus MATH. The third case is similar to the second. Finally, in the last case, we have either MATH, MATH or MATH, MATH. Without loss of generality, we can assume that MATH, MATH. Then MATH or MATH. Since MATH, MATH is not rectangular and therefore we can not have MATH. So MATH and MATH has exactly two southwest corners MATH and MATH. Now MATH is non empty for some MATH with MATH and MATH since MATH. Further, MATH implies that MATH, therefore the rectangular diagram formed by the bottom two rows MATH is a skew subdiagram of MATH. It follows that either the bottom row MATH is in MATH or MATH is in MATH where MATH. Again we conclude that MATH. We have therefore proved that if MATH is not near rectangular, then we have MATH. Our proposition now follows. |
math/0101193 | This is a direct consequence of REF . |
math/0101193 | It suffices to observe that REF does not apply here since we are only interested in MATH. |
math/0101193 | For type A, this is a direct consequence of REF . So let us suppose that MATH is of type D, and let MATH be the corresponding triple of centrally symmetric skew diagram. If they are all non empty, then by REF , MATH and MATH are not empty where MATH. So MATH. If MATH, then since MATH is non integral, MATH by REF . Finally, suppose that MATH. Then the condition MATH would imply that by MATH where MATH. So by REF , we deduce that either MATH or MATH does not satisfy REF . Without loss of generality, we can suppose that MATH does not satisfy REF . Thus we have by REF that MATH . Since MATH is integral, this means that it has at least two distinct southwest corners MATH. Let MATH be a southwest corner of MATH. Then the pairs MATH and MATH are in MATH. Hence, MATH which contradicts MATH. |
math/0101193 | Let MATH be an almost principal nilpotent pair, and let MATH be the associated pair of skew diagrams as given in REF . We have REF cases: MATH or MATH is empty; MATH and MATH are both non-empty. In the first case, by REF , we have condition REFi. Let us show that in the second case, we have REFii. So let us assume that MATH and MATH are both non-empty. By REF , we must have MATH and both MATH and MATH must be rectangular for otherwise one of them will have two distinct southwest corners and one can use the argument in the proof of REF for type D to obtain a contradiction. Now if MATH (respectively, MATH) is a skew subdiagram of MATH (respectively, MATH) such that MATH and MATH are of the same shape, then one sees easily from the fact that the MATH's are centrally symmetric that the pair MATH is in MATH. It follows that if MATH, then there is only one such pair. Now MATH is non integral, therefore MATH contains the square MATH, hence MATH and MATH must satisfy condition REFii. Finally, given MATH satisfying REF. It is clear that the corresponding distinguished nilpotent pair is almost principal. |
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