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math/0104120 | Suppose MATH and MATH . We shall show that if MATH is small enough we obtain the conclusion of the lemma. First pick a map MATH from MATH so that for each MATH, MATH and there exists MATH with MATH for MATH . Then, by a simple counting argument we have the existence of MATH so that MATH and if MATH then MATH . We can estimate MATH . Hence for MATH we have MATH . It follows that MATH where MATH. Choosing MATH such that MATH, where MATH is the constant from REF , and applying this theorem, we obtain the existence of MATH, MATH, with desired property. |
math/0104120 | Let MATH and MATH be the smallest integer greater than MATH, where MATH will be chosen later. Suppose first that MATH . Then we can find MATH, MATH, and MATH so that MATH . Let MATH be the space of all MATH-subsets of MATH and let MATH be normalized counting (probability) measure on MATH . If MATH denote the indicator functions MATH if MATH and MATH otherwise then MATH if MATH . Thus MATH and MATH if MATH . Let MATH so that MATH . Then working in the MATH-norm we have MATH . Hence MATH . Since MATH we have MATH . We now suppose that for each MATH we have MATH . Then we get an immediate contradiction. We conclude that for each MATH there exists MATH such that MATH for at least MATH choices of MATH . Let MATH if MATH and MATH otherwise so that MATH . Now suppose MATH is chosen as a function of MATH so that we can apply REF to obtain the existence of a set MATH with MATH and so that MATH where MATH is an absolute constant, and MATH. Then MATH . Recall that MATH so that if we choose MATH such that MATH we have MATH where MATH . Now suppose MATH . Let MATH be defined by MATH if MATH and MATH otherwise, while MATH if MATH and MATH otherwise. Then MATH . Thus MATH . This implies for MATH . Letting MATH and applying REF we obtain MATH . Note that MATH for some MATH so that the result follows. |
math/0104120 | We can assume MATH . Then by REF we can find MATH with MATH so that MATH where MATH . Let MATH be the space of dimension MATH with unit ball MATH . Since MATH is MATH-convex we have REF MATH . Finally observe that for a suitable MATH: MATH . The result then follows. |
math/0104120 | Let MATH be a subspace of MATH of dimension MATH so that MATH . Then if MATH there is a linear operator MATH with MATH and MATH . MATH can then be extended to a norm-one operator on MATH and so MATH has a quotient MATH of dimension MATH so that MATH . It follows immediately from REF with MATH that there is a further quotient MATH of MATH with MATH and for some cube MATH in MATH, and fixed constants MATH and MATH, we have MATH where MATH is the quotient map onto MATH . |
math/0104120 | Let MATH and MATH be the largest integer such that MATH has a subspace MATH of dimension MATH with MATH. Assume first MATH. By REF (and its proof) we have MATH for MATH, where MATH. Then, by REF , we obtain MATH which implies MATH . Therefore MATH. Finally we obtain MATH (of course we may assume that MATH). Suppose now MATH. By above we have MATH. So REF implies the existence of absolute constants MATH, MATH and a rank MATH projection MATH such that MATH for some cube MATH. By REF we obtain MATH so that we have (if MATH), MATH . This implies the theorem. |
math/0104124 | The classical NAME representation formula for minimal surfaces implies that if the properties MATH and MATH hold, the map MATH defined in REF is a pluriminimal immersion. Conversely, let us first prove that each one of the MATH is holomorphic: indeed, we know that MATH is holomorphic on each holomorphic curve MATH (see CITE). Chosen MATH we can find local coordinates MATH such that MATH for every MATH . Since MATH, we can write MATH . Restricting this form to each line MATH, MATH, MATH, we get MATH which clearly implies MATH for any MATH and MATH. Let us now prove that MATH has to be closed: we know that MATH is closed and MATH is holomorphic. Then MATH. Since MATH is of type MATH and MATH is of type MATH we get MATH and MATH which immediately imply MATH. The conformality condition MATH follows directly from the fact that given any vector in complexified tangent space MATH, there exists a complex curve with MATH as tangent vector. On this curve MATH has to be minimal, which, by the classical NAME representation formula, implies MATH. REF follows by contradiction. Indeed, if MATH, where MATH stands for the jacobian, we can take a complex curve MATH in MATH tangent to MATH. By restricting MATH to this NAME surface we get MATH for any MATH, and then MATH is not an immersion. |
math/0104124 | Let us consider complex local coordinates MATH on MATH, and consider the real and imaginary parts as real coordinates MATH. In real notation we can write MATH, and in matrix form MATH . Clearly MATH and MATH for MATH. Therefore, having set MATH, MATH . Thus we can write the matrix associated to MATH as MATH . On the other hand the vanishing of the tensor MATH can be written as MATH, which gives, by REF , the following system of REF . Therefore the matrix associated to MATH is hermitian and its associated form can be written, as in the classical case of minimal surfaces, as MATH which is clearly positive definite and of type MATH. It is also closed since each MATH is closed. |
math/0104124 | Since the induced metric is NAME, at every point of MATH we can choose an orthonormal basis for the tangent space of the form MATH. Since the map restricted to every holomorphic direction has to be minimal, MATH for any MATH. Therefore MATH . Thus, MATH, which implies directly the conclusion. |
math/0104125 | Note that REF are gauge invariant equations. The transformation MATH and MATH are the same gauge change. The choice of MATH is possible because MATH . We simply choose a NAME gauge with MATH . If MATH at infinity, MATH will be unique. |
math/0104125 | MATH and on the other hand MATH . Here we have used MATH. In the first equation, we see the terms MATH and since MATH . Next, we use that MATH and that MATH to write MATH and MATH . Now MATH . Then MATH . Hence, MATH . |
math/0104125 | The proofs of REF are similar, so we show only REF. By REF , MATH . |
math/0104125 | The proof is a reprise of the argument behind Proposition MATH in CITE, so we will just indicate the main points. CASE: If MATH, we imagine that we have the additional restriction MATH in the multiplier MATH, which will be artificial and it will not play any role. Compute MATH. Thus, we might be in any of the cases MATH, MATH, MATH in CITE, but in all of them, we get MATH . CASE: If MATH, we will impose again the additional artificial restriction MATH. Then MATH and the rest follows from REF. |
math/0104125 | Observe that one can assume MATH without loss of generality since all the proofs proceed by putting absolute values inside the integrals anyway. For the case of weight MATH and MATH write MATH . By taking a dyadic decomposition on the space frequency, we have MATH where MATH . An application of NAME and NAME from REF yield MATH . To complete the desired estimate it will suffice to show that MATH . We test MATH against an unimodular MATH function MATH to get the trilinear form MATH applied to the MATH functions MATH. We estimate away the elliptic weights MATH to get MATH for positive MATH. Let MATH, MATH and MATH. An application of REF yields MATH . Thus MATH and therefore REF holds. For the case of weight MATH, MATH we need to consider different pairing of our functions: MATH versus MATH rather than MATH versus MATH as we have just done. One obtains two trilinear forms where the weights have the same signs and we apply REF to each one of them. Then we perform similar and in fact simpler argument to the one presented above. In the case MATH, MATH we once again pair functions with same signs weights and use REF . Finally, in the case MATH, MATH the argument is identical to the one presented above. We omit the details. |
math/0104125 | CASE: First, we note that the bilinear estimate REF implies the embedding REF, if one takes MATH. As it has been already noted, the endpoint MATH is contained in REF, but can also be obtained from REF . For the bilinear estimate REF, one uses as an endpoint MATH result REF. Note that one needs a little bit of extra regularity in MATH REF in order to be able to sum REF in MATH. By complex interpolation, it will suffice to show the other endpoint MATH. The proof for both REF is the same for MATH, so we concentrate on REF. Since MATH, we reduce it to showing MATH . Consider a function MATH, such that MATH . We have MATH . |
math/0104125 | By symmetry, it suffices to obtain estimates only for MATH in REF. We first perform a dyadic decomposition on MATH and MATH to get MATH . We split into two pieces, MATH and MATH. For the small frequency case, the argument goes along the lines of the estimate for MATH in REF. Observe that MATH . Therefore, just as in the estimate for MATH . To estimate MATH, one uses NAME embedding with one derivative in the space variable. We get MATH . From the definition of MATH, the boundedness of the NAME transforms and the bilinear estimate REF, we have MATH . Thus MATH and we have shown REF for the quintic nonlinearity in the small frequency case. For the large frequency case, we will have to show just as in the small frequency case MATH . To verify that, following the argument in REF with MATH, we will have to demonstrate some decay in MATH for MATH. More precisely, we need to show MATH for some MATH. By NAME REF reduces to proving MATH . By NAME embedding performed in the spatial variable only, the boundedness of the NAME transforms and the definition of MATH, we get MATH . Thus, MATH and by the bilinear estimate REF, we get MATH . |
math/0104125 | Assume otherwise. Then MATH. Since MATH, it follows that MATH. On the other hand MATH. But MATH. Thus MATH, a contradiction. |
math/0104125 | Observe that the multiplier from REF has the form MATH where MATH are cubes with sidelength MATH. Since all the variables are well localized, we can use REF to estimate the MATH norm of the multiplier in REF. Indeed, let us assume for simplicity that MATH. Then choose MATH in REF so that MATH . For fixed MATH, MATH and MATH span intervals of length MATH and MATH respectively. Therefore, since MATH are both within a ball with radius MATH, we obtain from REF MATH . Based on REF, we have MATH which implies REF. |
math/0104125 | An application of the NAME yields MATH where MATH and MATH . Observe that by REF , we can estimate MATH . By REF we need to compute MATH . Then the estimates are MATH . Combining REF gives REF |
math/0104125 | In that case, we will fully use the quatrilinear form, instead of relying on NAME and then deal with the resulting trilinear forms. We estimate the multiplier in REF by REF. We have an upper bound of MATH where MATH are fixed and MATH . Note that for a fixed spatial variables the time variables span intervals of length MATH and MATH respectively. Also, we have MATH . For a fixed MATH, we have (based on REF) MATH . By the parallelogram law, MATH . Furthermore, since MATH and MATH, we infer MATH. Thus taking into account that MATH, we conclude that MATH and therefore by REF, MATH is contained in an annulus with radius MATH and thickness MATH. Observe that MATH is also in a ball with radius MATH, therefore it belongs to a rectangle with sides MATH and MATH. Finally, since MATH belongs to a ball with radius MATH, one estimates REF by MATH thus implying REF. |
math/0104125 | This case is very similar to REF We estimate the multiplier by MATH where MATH are the sets defined before. We have MATH . For fixed MATH , we have by the parallelogram law MATH . Since MATH and MATH, it follows that MATH. In particular, after taking into account that MATH, we obtain MATH . By REF, one has that MATH is contained in an annulus with radius MATH and thickness MATH. Since MATH is also contained in a ball with radius MATH, we have that MATH is contained in a rectangle with sidelengths MATH and MATH for every fixed MATH. The usual observation that MATH and MATH sweep intervals of length MATH and MATH respectively, leads us to estimate REF by MATH which again implies REF. |
math/0104125 | We estimate the norm of the multiplier by MATH where MATH are the sets defined in REF Like in the previous cases, we have a relation involving some of the variables. Here, we have MATH . We change variables MATH, MATH and we are interested in the measure of the corresponding set in REF. Since the Jacobian of the transformation is two, we pass to the new variables. Fix MATH. Observe also that since MATH, MATH. We have then by the parallelogram law MATH . That implies that for fixed MATH, MATH is contained in an annulus with thickness MATH. On the other hand, MATH is contained in a ball with radius MATH. These estimates, together with the usual observations that MATH are in a intervals of length MATH and MATH respectively, and the fact that MATH sweeps a ball with radius MATH imply the following bound for REF MATH which clearly implies REF. |
math/0104125 | Apply NAME to MATH obtain MATH where MATH and MATH were defined in REF. Compute again MATH. Thus, based on the estimates in REF , we conclude that MATH and by the estimate for MATH in REF MATH . We have the estimate (after quickly going through the appropriate cases - MATH is either MATH or MATH or MATH is either MATH or MATH) MATH . Thus we can sum up the expression in REF as follows MATH . |
math/0104125 | We concentrate on the high frequency case MATH. The case MATH is trivial, because then MATH and one easily estimates (see estimates below). For simplicity, we assume once again that MATH. Observe that MATH appears with the smallest power in the denominator and that should be the worst case for the maximum to occur. Moreover later in the proof regarding that case, we will see that we could perform the same argument with any other configuration of MATH. We use again the quatrilinear form MATH. By REF, we get MATH . We have the relation MATH . There are two distinct possibilities now. Either MATH or MATH (or both). We show the desired estimate, for the case MATH, the other case being similar. Observe that MATH. We introduce again the new variables MATH and we fix MATH. The parallelogram law and REF imply MATH and thus, we have that for fixed MATH, MATH is contained in an annulus with thickness MATH. On the other hand it is contained in a ball with radius MATH. The usual observation that MATH span intervals of length MATH, gives us the following estimate for REF MATH which implies MATH which is the desired estimate REF. |
math/0104125 | We will have to apply NAME 's inequality with a reorganized pairs of functions. We do that in order to take advantage of the disparity in the sizes of MATH and MATH. MATH where MATH and MATH . For MATH, we are in a position to use REF . Since MATH, we have MATH . For MATH, we use REF to infer MATH . For MATH, we combine the estimates for MATH and MATH to show MATH which is the desired REF . For MATH, the estimates above can be improved greatly and thus one estimates in that case as well. |
math/0104125 | For our purposes it will suffice to check the statement of the theorem for some specific sequence MATH, so that MATH, since for every MATH, we will find MATH so that MATH and the solution has a lifespan at least MATH. One could obtain estimates for the indeces in between by the NAME rule for fractional differentiation in the MATH spaces. We do not pursue these however since they are not necessary for our purposes. Due to the nature of the estimates, it will be convenient to take MATH. We will show the theorem for a cubic nonlinearity, since the others are treated in the same way. The common between them is the (anti) linearity structure that all of them exhibit. Take MATH. Since MATH, it is clear that a solution exists and it satisfies the integral REF . Differentiating REF yields MATH . Assume now that MATH for simplicity. Note that by the linearity of MATH and the product rule MATH where MATH or MATH as usual. Thus, by combining estimates REF, we obtain MATH . Thus, taking MATH will allow us to hide MATH and we get (for a time MATH) MATH . We can basically iterate this result in the following manner. For MATH, we proceed as follows. Differentiate REF once more. Then, we have two groups of terms. In the first, the derivatives fall on different MATH, for example MATH, for the second we will have terms like MATH. All the terms are estimated by REF. We get MATH . We further have MATH . Inserting the estimates in REF, we get MATH . Choosing MATH with MATH allows us to hide again. We obtain MATH . In a similar fashion, one obtains the estimates for every MATH. We omit the details. |
math/0104127 | Observe that MATH for any subset MATH. For MATH we have MATH. Therefore it reduces to see that MATH where we have used the fact that MATH. |
math/0104127 | For MATH, it follows by definition that MATH . CASE: The conjugacy class MATH is even and split. Suppose on the contrary there is a part of even integer in MATH. Without loss of generality we can assume that MATH contains a representative element MATH with the signed cycle decomposition MATH where MATH is even and MATH is empty (we can take all MATH empty corresponding to parts in MATH) and MATH is even. Consider the element MATH where MATH with MATH, for MATH and MATH otherwise. We claim that MATH . In fact the MATH-th component of MATH equals MATH for MATH and it also equals MATH for MATH. Noting that MATH we have MATH . Moreover by REF we have (recall MATH) MATH . Thus MATH is conjugate to MATH in view of REF . Therefore if the even-parity conjugacy class MATH splits then MATH. Now suppose that MATH. Then MATH contains at least two parts since we assume that MATH is even. Without loss of generality we can assume that MATH contains an element MATH such that MATH where MATH. If both MATH and MATH are of cycle length MATH, then MATH. Assume that MATH, so MATH. Consider MATH, where MATH for MATH and MATH otherwise. Then MATH . Subsequently MATH. Hence if MATH of even parity splits then MATH is empty. Together with the above we have shown that split conjugacy class of even parity should have REF . CASE: The conjugacy class MATH is odd and split. If on the contrary MATH, we can assume that MATH contains an element MATH with the signed cycle decomposition MATH where MATH is empty and MATH is odd. Take the element MATH where MATH and MATH for MATH and MATH otherwise. Similarly we can verify that MATH by using REF . In fact MATH since MATH is odd. Hence MATH does not split if MATH is odd and MATH. Next we assume on the contrary that MATH contains two identical parts, then by taking conjugation if necessary we can assume that MATH contains an element MATH such that MATH and MATH for some MATH. Consider the element MATH, where MATH. Then we have MATH which is a contradiction. CASE: Suppose that REF holds. If on the contrary the even conjugacy class MATH does not split, then we can assume that MATH contains an element MATH, where MATH with each MATH being odd cycle, and MATH. Therefore MATH, that is, MATH, which in particular implies that MATH. Then MATH by REF , and so MATH which contradicts with the assumption on MATH. Suppose that REF holds. Assume on the contrary that the odd conjugacy class MATH does not split. Since an identification of two elements in MATH implies that their respective components in MATH are already equal, we can assume that MATH consists of one strict partition MATH for some MATH. Thus MATH contains a non-split element MATH, where MATH is odd and MATH. Let MATH for some element MATH. It follows that MATH commutes with MATH and MATH is a product of disjoint cycles with mutually distinct orders, the permutation MATH equals MATH for MATH. Thus we can write MATH with MATH. As in the proof of REF we have MATH which must equal MATH up to a power of MATH. Set MATH where MATH is MATH or MATH. We claim that MATH is always MATH. Without loss of generality we let MATH, MATH, MATH with MATH, then MATH implies that MATH, which in turn implies the exponent MATH is equal to MATH. Therefore MATH and similarly we have MATH since MATH is odd. This is a contradiction. |
math/0104127 | The associativity follows from the obervation that the two different embeddings are conjugate: MATH . Using the cocycle REF we can check the coassociativity as we did in CITE. Using the cocycle MATH again and super analog of NAME 's theorem we can check that MATH preserves the multiplication structure, for details see CITE in a similar situation (also compare CITE for NAME algebra structures in different but related setups). |
math/0104127 | Let MATH be a character of MATH follows from REF that MATH . It follows from REF that MATH is multiplicative on MATH. Thus given two characters MATH of MATH, we have MATH . Therefore the proposition is proved. |
math/0104127 | First the map MATH is a vector space isomorphism by comparing dimension. The algebra isomorphism follows from the NAME reciprocity. On the other hand, one can check directly that MATH . Since MATH (respectively, MATH) for MATH generate MATH (respectively, MATH) as an algebra, we conclude that MATH is a NAME algebra isomorphism by REF . |
math/0104127 | Let MATH and MATH be any two super class functions in MATH. By definition of the characteristic map REF it follows that MATH where we have used the inner product identity REF . |
math/0104127 | The first identity follows from REF and the second one is obtained by noting that MATH is the adjoint operator of MATH with respect to the bilinear form MATH. |
math/0104127 | All the commutation relations without binomial coefficients are easy consequences of REF by the usual vertex operator calculus in the twisted picture (see CITE). The corresponding relations with binomial coefficients in MATH are equivalent to MATH . This is proved by using REF by a standard method in the theory of vertex algebras (compare for example, CITE). |
math/0104127 | The relations can be established using REF by means of standard techniques in the theory of vertex algebras, see CITE for detail for a similar circumstance. |
math/0104128 | Endow the manifold with a bundle-like metric, and choose any partition of unity subbordinate to the basic cover. Orthogonally project the functions in the partition of unity to the space of basic functions; the smoothness and other desired properties of the resulting functions are guaranteed by the results of CITE. |
math/0104128 | The proof of this theorem is the same as the proof of REF, with forms replaced by basic forms and partitions of unity replaced by basic partitions of unity. |
math/0104128 | First observe that since we have given MATH a bundle-like metric, the tubular neighborhood MATH is a union of leaves, and so the restriction MATH of MATH to MATH makes sense. Choose MATH small enough so that MATH misses the cut locus of MATH; since MATH is compact and MATH is closed, this can always be done. Fix MATH, and let MATH be the exponential image of the ball of radius MATH in the normal space MATH. Then MATH . The holonomy of MATH acts by a finite subgroup MATH of the orthogonal group CITE, so we have MATH and this isomorphism commutes with the exterior derivative. Suppose MATH is closed, and let MATH be the closed form associated to MATH via the isomorphism above. Since MATH is diffeomorphic to Euclidean space, there exists by the ordinary NAME lemma a form MATH such that MATH. Now, MATH may not be MATH-invariant, but the averaged form MATH is, and MATH also equals MATH. Therefore MATH, and hence MATH, is exact. |
math/0104128 | For each MATH, let MATH be the collection of basic MATH-forms restricted to MATH-fold intersections, and define a double complex MATH . The horizontal differential is the map MATH defined above, and the vertical differential is MATH, where MATH is the exterior derivative. We compute the cohomology MATH of this double complex in two ways. First, augment MATH by the column MATH, and then map into MATH by the restriction map MATH. By REF , the rows of the augmented double complex are exact. By CITE, the cohomology MATH of the original double complex is isomorphic to the cohomology of the initial column, which is precisely the basic NAME cohomology MATH of MATH. Second, augment the original double complex with the row MATH. Then the columns of the new double complex are exact, and thus MATH is also isomorphic to MATH. |
math/0104128 | Choose coordinates MATH near a point MATH as described in the paragraphs above. As before, we write MATH in the form MATH where MATH is tangent to MATH, MATH, and MATH. Given any MATH such that MATH, we may multiply the tangent component MATH by a radial basic function that is zero in MATH and is MATH outside MATH. In doing so, we preserve the index and orientation line bundle and yet restrict to the case where MATH is a basic normal vector field. We now assume that MATH has already been modified in this way. Next, the basic normal vector field MATH is determined on MATH by its restriction to a normal ball MATH with the MATH coordinates, and we may write MATH on MATH. Again, we may multiply the component MATH by a similar radial basic function if necessary so that this component vanishes on a given MATH such that MATH, without changing the relevant properties of MATH. For the remainder of this proof, assume that we have already modified MATH so that it is in the form MATH restricted to MATH, choosing a slightly smaller MATH if necessary. It follows that MATH on MATH. Given MATH, we write MATH in terms of its polar decomposition MATH. Let MATH be the closed subgroup of isometries on MATH induced from the representation of the holonomy group at MATH on MATH. For every MATH, MATH, whence MATH. This in turn implies that every MATH commutes with MATH and MATH. Let MATH be defined by MATH on the MATH -eigenspace of MATH, where each MATH is any smooth positive function such that MATH and MATH. Since every MATH and MATH have simultaneous eigenspaces, every MATH also commutes with each MATH. Thus, the smooth, one-parameter family of transformations MATH is a deformation of MATH REF to an orthogonal transformation MATH REF that has constant index (that is, the sign of the determinant of the linear transformation does not change). Next, since MATH is orthogonal, there is a complex orthogonal basis MATH of MATH consisting of eigenvectors such that MATH where MATH. If MATH is MATH, then MATH acts by the identity on MATH . If MATH is MATH, then MATH multiplies each vector in MATH by MATH. If MATH and MATH, then MATH acts by a rotation of MATH on MATH, which in this case is necessarily MATH-dimensional. We let MATH and define the transformation MATH for MATH by MATH . The smooth, one-parameter family of transformations MATH is a deformation of the orthogonal transformation MATH REF to a transformation MATH REF that has constant index. Observe that since each MATH commutes with MATH the obvious action of MATH on MATH satisfies MATH for some MATH; it follows that MATH commutes with each MATH. The final transformation MATH may be described in a real orthogonal basis as a diagonal matrix whose diagonal consists of MATH's and MATH's; this transformation is the linear part of a vector field of the form MATH, where the MATH are geodesic normal coordinates on MATH corresponding to that particular basis. We also observe that MATH is then a basic function that is well-defined on a small neighborhood of MATH, and the linear part of MATH at MATH is MATH. Note that we may extend MATH to be a basic function on all of MATH by multiplying by a radial cutoff function and extending by zero. Combining the two deformations described above, we see that MATH may be smoothly deformed to a MATH-equivariant transformation of the form MATH in such a way that the index is unchanged throughout the deformation. The argument that follows is somewhat similar to that found in CITE. Let MATH be a smooth, one-parameter family of MATH-equivariant transformations constructed above such that MATH and MATH-is the linear part of MATH. Thus each MATH is a vector field that is well-defined in a sufficiently small tubular neighborhood of the leaf closure MATH. Next, let MATH,with MATH and MATH as above and MATH a smooth positive function such that MATH for MATH and MATH for MATH. Because of the remarks above, MATH for all MATH. Let MATH be a radial (and therefore basic) cutoff function that is MATH in a MATH-neighborhood of MATH and MATH outside a MATH -neighborhood of MATH (MATH will be chosen shortly). Then for MATH sufficiently small, MATH is a well-defined basic vector field on MATH. Let the basic vector field MATH be defined by MATH so that MATH . Observe that MATH is a smooth, one-parameter family of basic vector fields that agree with MATH outside a MATH-neighborhood of MATH, and MATH inside a MATH-neighborhood of MATH. Let MATH and let MATH . Observe that MATH and MATH. Therefore, if MATH, then MATH for MATH and hence MATH is invertible. Choose MATH so small that MATH for MATH; then the leaf closure MATH is the only critical leaf closure of the basic vector field MATH in the MATH-neighborhood of MATH. We continue by defining for MATH where MATH is MATH in a MATH-neighborhood of MATH and MATH outside a MATH-neighborhood of MATH (where MATH is chosen as above). Eventually, we will find a MATH such that MATH outside a MATH -neighborhood of MATH and is MATH inside a MATH -neighborhood of MATH. Furthermore, MATH is the only critical leaf closure of MATH in a MATH -neighborhood of MATH, and the index of MATH at MATH is the same as the index of MATH at MATH for MATH. Finally, MATH inside a MATH-neighborhood of MATH. Observe that if we let MATH, MATH, and MATH, then the first four properties are satisfied. Moreover, since the orientation line bundles of MATH and MATH at MATH are constant in MATH, the last property is satisfied as well. |
math/0104128 | We use induction on MATH. For MATH, we clearly have that MATH . Similarly, MATH. By ellipticity, there exist MATH and MATH (independent of MATH) such that MATH. Thus, MATH . Assume that the conclusion is true for MATH. Then MATH for positive constants MATH, MATH, and MATH and for MATH, all independent of MATH. Also, it follows from our induction hypothesis that MATH for some MATH. Finally, by the definition of the NAME norm, there exists a positive constant MATH such that MATH whence the result follows. |
math/0104128 | By ellipticity, there exist MATH and MATH such that MATH. By REF , there exists MATH such that MATH. Thus MATH . |
math/0104128 | The standard proof (see CITE) works for the basic case. |
math/0104128 | The standard proof (see CITE) works for the basic case. |
math/0104128 | Let MATH be a family of smooth basic functions approximating (in MATH) the characteristic function of MATH as MATH (see the results of CITE and CITE). Then MATH since MATH is basic CITE. We then apply the dominated convergence theorem. |
math/0104128 | We have MATH where MATH is the set of points of distance MATH from the leaf closure MATH. Observe that MATH since MATH is formally self-adjoint with respect to MATH. By CITE, MATH where MATH is the one-form defined by MATH. Then MATH which follows from REF and the fact that MATH is basic. Furthermore, by the results of CITE, MATH. Since MATH by hypothesis, the divergence theorem yields MATH where MATH is the unit vector field normal to MATH with orientation chosen compatibly with the choice of orientation of MATH. Locally MATH. Since MATH is real, we conclude that MATH . |
math/0104128 | One easily checks that such an operator MATH is formally self-adjoint on the space of basic forms, and MATH; thus, it is sufficient to prove the result for small positive MATH . Since MATH is basic, the support of MATH and its complement are saturated. Since MATH is a Riemannian foliation and since MATH is compact, for every leaf closure MATH, there exists a tubular neighborhood and MATH such that for every leaf closure MATH in that tubular neighborhood, the set MATH of points of distance less than MATH from MATH misses the focal locus and cut locus of MATH. Choose any MATH that is at a distance MATH or more from the support of MATH; let MATH denote the leaf closure containing MATH. Since MATH is Riemannian, the set MATH is also disjoint from the support of MATH. Then MATH by REF . Hence MATH at MATH for MATH. |
math/0104128 | The operator MATH is a bounded, self-adjoint operator on MATH. By REF , MATH maps MATH into MATH for every MATH. By REF , this implies MATH maps MATH to smooth basic forms. Using the fact that MATH is formally self-adjoint on MATH, the standard proofs are easily adapted to show the existence and properties of MATH (see CITE). |
math/0104128 | The proof is identical to the proof of the standard case in CITE, replacing MATH with MATH and the standard NAME theorem with the basic NAME theorem. |
math/0104128 | This proof is the same as CITE, but we include some details here because of subtleties. Choose a constant MATH so that MATH for all MATH in the complement of a MATH-neighborhood of MATH. By REF and the remarks following that equation, MATH for every MATH that is supported on the complement of such a neighborhood and for sufficiently large MATH. Let MATH denote the NAME space of MATH basic forms that vanish on a MATH-neighborhood of MATH. Then MATH is a positive symmetric operator on a dense subset of MATH, so it extends to a self-adjoint operator MATH on MATH satisfying the same inequality above. Let MATH be a basic form supported on the complement of a MATH -neighborhood of MATH, and let MATH which is a solution to the generalized wave REF corresponding to the operator MATH with initial conditions MATH, MATH . The family of forms MATH is the unique solution to this generalized wave equation as well, by the statements before and after REF . Note that the formula above implies that MATH is basic. By the unit propagation speed property of the basic wave equation REF , MATH is identically zero on the MATH-neighborhood of MATH if MATH. This implies that MATH for MATH, so that MATH is the unique solution to the system MATH . We may therefore write MATH. Let MATH be a real-valued function with the following properties: CASE: MATH is a positive even NAME function; CASE: MATH; CASE: the NAME transform MATH is supported in the interval MATH. For each nonnegative integer MATH, define MATH by the formula MATH note that each MATH satisfies REF - REF . For a basic form MATH that is supported on the complement of the MATH -neighborhood of Crit-MATH, MATH . The operator MATH is positive and has operator norm is bounded below by MATH for MATH sufficiently large. Thus, the operator norm of MATH (as an operator from MATH to itself) is bounded above by MATH . It is clear that MATH is rapidly decreasing as MATH. By REF , MATH for every basic form MATH supported on the complement of a MATH -neighborhood of MATH. Next, let MATH denote the MATH-norm closure of the space of smooth basic forms, and let MATH denote the closure of the space of such basic forms under the NAME MATH-norm. By the basic elliptic estimates REF , the operator norm of MATH is bounded by a polynomial in MATH. The basic version of the NAME imbedding theorem REF implies that MATH is a bounded map whose operator norm is bounded by a polynomial in MATH if MATH . Using basic duality REF and essential self-adjointness of MATH, we see that MATH is also a bounded map whose operator norm is bounded by a polynomial in MATH whenever MATH. Note that all of the statements above hold for the operator MATH as well as for MATH. Now, given a basic form MATH supported on the complement of a MATH -neighborhood of MATH and MATH, MATH where MATH is a polynomial in MATH. Next, since MATH is rapidly decreasing, MATH has a continuous basic kernel MATH REF , and we have the inequality MATH from the above. Thus, MATH uniformly as MATH. |
math/0104128 | Without loss of generality, we may assume that MATH is a basic normal vector field (otherwise, project to MATH). By REF , we have that MATH, which may be obtained by integrating the traces of the corresponding kernels of MATH and MATH. By REF , the kernels of these operators go to zero uniformly on the complement of a fixed but arbitrarily small neighborhood of the critical leaf closures of MATH. For each critical leaf closure MATH, let MATH be a smooth, radial, basic function that is identically MATH in a tubular neighborhood of radius MATH around MATH and supported within a tubular neighborhood of radius MATH (assume that we have chosen MATH small enough so that this is possible for each MATH). Then we have that MATH . We now use REF to observe that MATH may be deformed without changing MATH or MATH so that within a tubular neighborhood of radius MATH around MATH, MATH for a basic function MATH, such that if MATH then MATH has even NAME index and if MATH then MATH has odd NAME index; again we possibly decrease MATH so that the conclusion of this proposition holds. A unit propagation speed argument shows that the traces are independent of the choice of MATH with those properties, and thus we may calculate the contributions from each tubular neighborhood as if MATH. By the results of CITE, if MATH is the basic Laplacian on MATH with coefficients in MATH , then MATH with the analogous result for the odd case. Thus, MATH . |
math/0104128 | The hypotheses imply MATH . Since MATH, MATH. Since MATH injects into MATH CITE, NAME duality and the Universal Coefficient Theorem imply that MATH . The result follows. |
math/0104130 | First observe that if MATH then MATH must be an isomorphism onto MATH since MATH exceeds the spectral radius of MATH . Furthermore since the spectrum of MATH is invariant under rotations it is clear that MATH cannot be an isomorphism onto MATH . Also note that MATH is always injective and that if MATH is a linear functional annihilating its range then MATH for some constant MATH, that is, MATH. This implies that the closure of the range is either the whole space or the subspace of codimension one when MATH. Here MATH and the formula must be modified if MATH . We next show that if MATH then MATH is an isomorphism onto a closed subspace of codimension one. Next let MATH and MATH . We remark that MATH is easily seen to be closed because MATH is an isomorphism on the unweighted MATH and MATH is bounded for MATH . If we show MATH is closed then we are done, since it is clear this will imply that MATH is closed and this is a subspace of co-dimension one in the range. However MATH so that MATH is an isomorphism. Restricting to MATH this implies MATH and hence MATH is closed. The proof is completed by showing that if MATH then if MATH if has closed range it must satisfy MATH . Note first that it is enough to establish this for MATH since the set of operators with NAME index one is open. Suppose MATH and MATH is closed. Then MATH has a lower estimate MATH for all MATH where MATH . Assume MATH for some MATH and MATH . Then consider MATH . Note that MATH . However MATH . Let MATH . Then we have if MATH . In particular if MATH . Now since MATH we can find MATH so that MATH or MATH . Iterating gives us that MATH is monotone increasing. Now for any large MATH and any MATH we have MATH where MATH, MATH such that MATH and MATH. This gives us MATH . Hence MATH . Letting MATH gives MATH . To show that in fact MATH needs only the observation again that the set of MATH where MATH has NAME index one is open. |
math/0104130 | Let us suppose next that either REF MATH or REF MATH . This implies there exists a constant MATH so that MATH for all MATH; in REF MATH maps onto the subspace of MATH defined by MATH, while in REF MATH is an isomorphism onto the whole space (see REF ). We observe that in REF the linear functional MATH extends to a continuous linear functional on MATH as MATH . Now suppose MATH with MATH with the additional assumption in REF that MATH . Then we may find MATH such that MATH and MATH . Then MATH since MATH . In REF we additionally have MATH . Thus we can find MATH with MATH and MATH . Then we can find MATH such that MATH and MATH . Let MATH . Then MATH . Now MATH and MATH . Hence MATH and so MATH with MATH . From this it follows that in REF we have MATH and in REF MATH . Next we consider the converse directions. Assume either (aa) MATH is continuous on MATH and MATH or (bb) MATH . In either case there is a constant MATH so that if MATH then MATH . Observe that in REF the linear functional MATH is continuous on MATH and so the range of MATH is contained in its kernel; in case (bb) its range is dense. Assume MATH with MATH in case (aa) we also assume MATH . We first find MATH with MATH and so that MATH for MATH . Let MATH so that MATH with MATH . In case (aa) we have additionally that MATH . Now we can find MATH so that MATH and MATH . Now let MATH and MATH . Then MATH . We argue that MATH where MATH . To show REF we note that MATH and (since MATH) MATH . Let MATH . Then MATH and MATH . But now MATH so that MATH is an isomorphism onto the kernel of MATH in case (aa) or onto MATH in case (bb). These two cases combined with the observation that MATH can only be a proper closed subspace of MATH if MATH is continuous on MATH complete the proof of the Theorem. |
math/0104130 | These results follow immediately from the NAME theorem once one observes that MATH if and only if MATH is of the form MATH where MATH . |
math/0104130 | We can then apply REF with MATH and MATH. To estimate MATH we note that if MATH and MATH then MATH and so, for MATH . Hence we can describe the numbers MATH of REF by MATH and MATH . Since MATH and MATH this proves the Proposition. |
math/0104130 | It suffices to consider MATH . Then by REF MATH is bounded and analytic in the upper-half plane and so we have: MATH . Applying the NAME - NAME inequality we prove the lemma. |
math/0104130 | We start with the remark that the functions MATH together with MATH form an orthonormal basis of MATH for MATH . Hence MATH . By REF the last term in REF can be estimated by MATH for MATH . Now if MATH the map MATH defined by MATH satisfies MATH provided MATH . Hence if MATH we have MATH . However using REF gives: MATH . Now by integrating for MATH we obtain REF . |
math/0104130 | By NAME 's theorem, REF above, MATH is an unconditional basis for a closed subspace MATH of codimension one which is the kernel of the linear functional MATH . Let MATH be the weight sequence MATH and let MATH. It follows from the basis property that the map MATH defined by MATH is an isomorphism (onto). Clearly MATH is an isomorphism of MATH onto MATH. Hence by interpolation MATH is an isomorphism of MATH onto MATH. In other words, setting MATH, we have MATH and the almost normalized family MATH forms a NAME basis in MATH. Thus, if MATH is a closed subspace in MATH, MATH forms an unconditional basic sequence in MATH also. We next estimate MATH to have the inverse implication. In fact from interpolation between MATH and MATH we have MATH where MATH depends only on MATH . Similarly if we define MATH then the norm of MATH in MATH can be estimated by MATH . From the other hand, MATH, what gives MATH. Therefore the norms MATH and MATH are both equivalent to MATH. Therefore the assumption that MATH is an unconditional basic sequence leads to equivalence of metrics MATH and MATH. |
math/0104130 | CASE: Suppose MATH are distinct and MATH . Then MATH while MATH . Hence MATH which contradicts REF . For the second part note that if MATH then MATH so that MATH . REF is immediate from the fact that MATH . CASE: If MATH then, in view of REF , MATH and MATH . Let MATH. Then MATH . Since MATH, we have REF . CASE: Let MATH satisfy MATH . If MATH and MATH have opposite signs or if MATH, then MATH and so that MATH . If they have the same sign define MATH and note that MATH . Since MATH for MATH real, we obtain REF from REF . |
math/0104130 | Let us apply REF to MATH. Then MATH . Now for each MATH there is a choice of sign so that: MATH and hence MATH . This yields REF for MATH and this extends to MATH in view of REF and the fact that MATH . |
math/0104130 | We observe that for each MATH we have (taking MATH) MATH . Hence MATH . Along with a similar estimate for MATH and setting MATH we get: MATH . Now let MATH where MATH are the separation constants of MATH and MATH respectively. We next make the remark that there is a constant MATH so that if MATH and MATH then MATH . Let us fix MATH . Suppose that MATH and let MATH be chosen so that MATH and MATH . It may happen that MATH . Note that we have an automatic estimate, MATH . Then if MATH and MATH we have MATH and so MATH . By REF we have MATH and MATH . Hence we have by REF MATH . Next suppose MATH and MATH . Then MATH . We also have MATH and so by REF MATH . Combining and summing over all MATH we have MATH where MATH and MATH if MATH and MATH if MATH . To conclude we need only consider the case MATH . In this case MATH. We also have MATH by REF and so by REF MATH . Hence MATH . This concludes the proof. |
math/0104130 | This follows from REF taking MATH in view of REF . |
math/0104130 | The left-hand inequality in REF is an immediate consequence of REF . We turn to the right-hand inequality. We first use REF . There are constants MATH so that if MATH we have MATH, MATH and for every MATH . Choose MATH so that MATH for MATH . Let MATH be the orthogonal projection with respect to MATH of MATH onto the subspace MATH of all MATH so that MATH and let MATH . Then MATH . The orthogonal complement of MATH (with respect to MATH) is a REF-dimensional space with orthonormal basis MATH . Hence MATH and MATH . Since MATH, we deduce MATH and a suitable constant MATH . It remains therefore only to estimate MATH . We first argue that MATH . Since MATH for MATH we deduce that MATH . Now if we use REF we get an estimate valid for MATH . Since MATH we then have MATH and we can then rewrite the above estimate as MATH . Now MATH . We next use the sampling REF : MATH . However we can combine with REF to deduce that MATH for MATH which gives the conclusion. |
math/0104130 | In this case we compare the carrier function MATH for the basis MATH with the carrier function MATH for the basis MATH . Clearly MATH . We can next use REF to estimate MATH and then the theorem follows directly from REF together with REF . |
math/0104130 | Combining REF and the boundedness of MATH gives us REF . Note that if MATH . Now MATH where MATH for a suitable constant MATH . By REF and the assumption of the theorem, this implies that MATH and yields the Theorem. |
math/0104130 | We will use the connections between NAME basis property and sampling/interpolation in the spaces of entire functions of exponential type. These connections in the case of MATH and the NAME - NAME space may be found in CITE. Let MATH consisting of all entire functions of exponential type at most MATH and satisfying MATH . (Note that the NAME transform of MATH is the set of all distributions from MATH supported on MATH . ) Now the formal adjoint of the map from MATH to MATH defined by MATH is the map from MATH to MATH given by MATH . Hence MATH is an unconditional basic sequence (respectively, unconditional basis) if and only if MATH is an interpolating sequence (respectively, complete interpolating sequence) in MATH . Note that if MATH is interpolating for MATH then it is interpolating for MATH by the simple device of considering functions of the form MATH where MATH and MATH . It follows that our result can be proved by showing that MATH is a complete interpolating sequence for MATH for all MATH for some MATH . To do this we use the results of CITE that this is equivalent to requiring that MATH is a MATH-weight for MATH . Now MATH is a MATH-weight (CITE or CITE) and so there exists MATH so that MATH is a MATH-weight (compare CITE p. REF ). Hence the NAME transform is bounded on both MATH on MATH for MATH . It then follows by complex interpolation that MATH is a MATH-weight when MATH . |
math/0104140 | By definition of the counting function, any point MATH can be covered by a sufficiently small cube MATH such that the number of isolated roots of MATH in this cube is no greater than the number MATH finite by the assumptions of the theorem. It remains only to choose a finite subcover of the compact MATH and add together the corresponding cyclicities. |
math/0104140 | Consider the leaf (analytic curve) of the foliation MATH passing through the point MATH on the transversal, and denote by MATH the (oriented) segment of this leaf between the the initial point and the next intersection with MATH. By definition of the chart MATH, the displacement function measured in the chart MATH, is the difference of MATH between the endpoints of MATH, hence MATH (the equality is exact). The form MATH vanishes on MATH, therefore the integral above is equal to the integral of MATH along MATH and, since MATH converges uniformly to the closed curve MATH as MATH, we conclude that MATH with MATH uniform and analytic in MATH and MATH. This yields REF for the derivative. |
math/0104140 | Consider an arbitrary vector MATH and embed it into a vector field MATH on MATH. We claim that in a sufficiently small neighborhood of the preimage MATH one can construct a smooth vector field MATH such that MATH, that is, MATH and MATH are MATH-related. Such a field obviously exists near each point MATH, since MATH is surjective (and takes the form of a parallel projection in suitably chosen local coordinates by virtue of the theorem on rank). Now it remains to choose a finite covering of MATH by these neighborhoods and patch together the corresponding vector fields, using the appropriate partition of unity. To conclude the proof, notice that the (local) flows of MATH and MATH are conjugate by MATH (by construction), hence the local flow of MATH, defined in some neighborhood of MATH, takes the latter preimage into preimage of the corresponding point MATH on the flow curve of MATH. Since the initial vector MATH can be chosen pointing to any direction, this proves that all sufficiently close preimages are diffeomorphic to each other. |
math/0104140 | It will be shown below in REF that if MATH of degree MATH is transversal to infinity, then there exist two polynomials MATH of degree MATH such that MATH where the dots stand for a bivariate polynomial of degree MATH. One can easily check that the rational vector field MATH-related to the field MATH, in the chart MATH is regular (smooth) on the infinite line MATH (more precisely, on its affine part covered by this chart). The other affine part is covered by the field that is obtained in a similar way from solution of the equation MATH. Thus near each point of the compactified level curve MATH one has a smooth vector field MATH-related to MATH in the finite part MATH (in particular, this implies that this field vanishes at all points of indeterminacy of MATH on MATH). The rest of the proof is the same as in REF . |
math/0104140 | If MATH is a homogeneous polynomial of degree MATH without multiple linear factors, then the map between subspaces of homogeneous forms, MATH is an isomorphism. Indeed, in the bases consisting of all monomial forms of the given degrees, the matrix of MATH is the NAME matrix whose determinant is the resultant of the two partial derivatives MATH and MATH. The assumption on MATH implies that this resultant is nonzero. Therefore any homogeneous form MATH of degree exactly MATH is divisible by MATH. Any monomial form of degree greater than MATH can be represented as a monomial REF-form of degree MATH times a monomial function and hence is divisible by MATH without remainder with the same relation between the degrees, MATH . Applying this observation to all homogeneous components of a REF-form MATH, we prove the assertion of the Lemma for homogeneous polynomials. To divide a REF-form MATH of degree MATH by a nonhomogeneous differential MATH, where MATH is the principal homogeneous part of MATH, MATH, we divide it by MATH first, and then transform the result as follows, MATH reducing division of MATH by MATH to division of another form MATH of strictly inferior degree. Iterating this step, we prove the Lemma in the general case. Notice that this is essentially the algorithm of division with remainder of univariate polynomials. |
math/0104140 | This is an immediate corollary to the NAME theorem applied to the system of algebraic equations MATH . The same proof actually works for a more general case of functions MATH defined by a system of algebraic equations MATH, MATH, involving all functions simultaneously. |
math/0104140 | First we show that under the assumptions of the Theorem, the number of isolated zeros of any linear combination MATH with complex constant coefficients MATH, is bounded in any compact MATH uniformly over all such linear combinations. Indeed, without loss of generality we may assume that the functions MATH are linear independent - this does not affect the supply of all linear combinations. Next, it is sufficient to consider only combinations with coefficients on the unit sphere, satisfying the equality MATH. The functions MATH for such MATH are all different from identical zero, hence each of them has only a finite number of isolated zeros in the compact MATH (accumulation of roots to the boundary of MATH is forbidden). Now the standard semicontinuity arguments using compactness of the unit sphere, prove that the number of zeros of all MATH is uniformly bounded. To deal with arbitrarily polynomial combinations, we can treat them as linear combinations of monomials MATH, MATH, reducing the general case to the already studied one. |
math/0104140 | This holds since: CASE: any irreducible component of MATH should belong to an irreducible component of MATH, and CASE: if MATH is a pair of irreducible varieties, then MATH and in the case of equal dimensions necessarily MATH. In other words, when passing from MATH to MATH each irreducible component either completely survives, or is split into a number of other components of lower dimensions. |
math/0104140 | For MATH the claim is obvious, since any decreasing sequence of nonnegative integers must be finite. For an arbitrary MATH, the first ``letters" MATH, MATH, form a non-increasing sequence (it must not necessarily be strictly decreasing). However, no more than a finite number of values is taken. Along any interval of constancy of the first letter, the tails MATH also form a lexicographically strictly decreasing sequence in MATH. Hence the length of each such segment is finite by the induction assumption, and the length of the whole chain is finite as the sum of finitely many finite numbers. |
math/0104140 | The field MATH is independent of the choice of the fundamental solution, hence for each semiaxis MATH one can choose a solution MATH of the NAME system MATH that is real on that semiaxis (recall that the matrix MATH is real so MATH is real-valued on MATH). Then it remains only to define MATH using the identities MATH and similarly for the imaginary part (with obvious modifications for the field of fractions MATH). |
math/0104140 | If MATH, then the imaginary part MATH must vanish somewhere between MATH and MATH on MATH. |
math/0104140 | Draw disjoint circles of radius MATH around each singularity and of radius MATH around the origin (this circle bounds a neighborhood of infinity on MATH). Restricted on each circle, the field is quasialgebraic by REF . On the complement there are no singularities, so after triangulation of this multiply connected domain one can apply REF . |
math/0104143 | It is easy to find from relation REF that MATH where MATH. Therefore we can apply the argument given in the proof of REF. |
math/0104143 | We start with REF . It is easy to see that MATH . Consequently relying on REF we have MATH . Using the inequality (see, for example, CITE) MATH we obtain REF . In a similar way we can establish the estimates MATH which easily imply REF . |
math/0104143 | Multiplying REF by MATH, respectively, and then adding together, we obtain MATH . We have by REF MATH . A simple calculation shows that MATH. We now estimate the right hand side terms. On account of REF , the bilinearity of MATH and REF we can estimate MATH . Similarly, due to REF MATH . By the NAME inequality MATH . Using REF once more MATH . Adding all these inequalities together we obtain the conclusion. |
math/0104143 | Using REF and relation REF we have MATH . Let MATH and MATH be the solution of one dimensional random affine equation MATH . A comparison argument gives that MATH . Here MATH is the dynamical system introduced in REF: MATH, where MATH is the solution to REF with the initial data MATH. REF has the stationary solution given by MATH: MATH . This can be checked by the variation of constants formula. This solution is exponentially attracting which follows from the variation of constants formula again: MATH which tends to zero exponentially fast. Indeed, it follows from REF that for a sufficient small MATH, MATH for large MATH under REF . |
math/0104143 | Since MATH is completely continuous for MATH (see the regularity assertion of REF ) the sets MATH are compact. Since MATH is a random variable the ball MATH is a random set. The continuity of MATH allows us to conclude that MATH is a random set. The construction of MATH ensures that that set in absorbing and forward invariant. The temperedness of REF can be proved in the same way as in CITE for the MATH . NAME equations. |
math/0104143 | We integrate REF . The conditions of REF are satisfied if MATH. MATH can be estimated by MATH . On account of REF MATH has the bound MATH since MATH is forward invariant. Note that the expectation of MATH can be made arbitrarily small if MATH is large. Calculation the expectation of the last expression and choosing MATH sufficiently large yields the conclusion. |
math/0104145 | We have: MATH with equality only when there exists MATH such that MATH for all MATH with MATH nonzero. Dividing two consecutive such equations, we obtain MATH as required. |
math/0104145 | Setting MATH, we have: MATH . Now, MATH is positive on MATH, so we may freely multiply by MATH; we thus need to show that MATH . Now, the left-hand-side has a NAME series convergent in the original annulus, namely MATH . Since this MATH, and has at least one nonzero coefficient, the desired inequality follows. |
math/0104145 | It will suffice to show that MATH with error uniform on any interval MATH with MATH, since then on MATH, MATH is bounded away from REF, while the error converges uniformly to REF. We split into two cases: MATH and MATH. In the first region, we claim that for all sufficiently large MATH, and for MATH, MATH with error uniform in MATH. By NAME 's residue theorem, MATH . Now, MATH, so we have the uniform estimates MATH and thus MATH . Now, MATH so it remains to show that MATH is bounded. But, setting MATH and rescaling MATH, this is MATH by the known asymptotics of NAME functions. We now consider the case MATH. Here, we claim MATH again with uniform error. Again, NAME 's integral gives MATH . Since MATH is continuous and negative for MATH, MATH, there exists a positive constant MATH such that MATH in that region. In particular, when MATH, we find MATH, and thus MATH . Since MATH is bounded, we find that the contribution of this region to the integral is negligible. Now, consider the region MATH. Since MATH is continuous for MATH, MATH, we find the uniform estimate MATH or upon exponentiation, MATH . Similarly, MATH and thus MATH . Since MATH the claim follows. |
math/0104145 | Note that MATH since MATH, we conclude that MATH converges for MATH for some MATH independent of MATH, and satisfies MATH. By NAME 's theorem, MATH . Now, as before, we can restrict the integral to any uniform neighborhood of REF, with exponentially small error. In particular, we may restrict to a neighborhood MATH affording the change of variable MATH. The integral thus becomes MATH . Now, we have the uniform estimate MATH since MATH the contribution of the error term is as required. Once we remove this term, the integral can be extended to MATH, again giving uniform exponentially small error. Evaluating the resulting Gaussian integral gives the desired result. |
math/0104145 | Since the main term of the estimate has order MATH, we can tolerate a multiplicative error of order MATH. In particular, we may replace MATH by its first-order estimate. We thus need to estimate REF. Now, MATH . Thus we obtain MATH . Now, for MATH, MATH and for MATH, we have: MATH . In particular, the MATH term gives a contribution of order MATH. For fixed MATH, the contributions get smaller as MATH increases. We thus find that the terms with MATH are dominated by the terms with MATH, of order MATH . It remains to consider the terms with MATH, that is, MATH . If we replace MATH by MATH, MATH by MATH and MATH by MATH, the resulting error is again MATH. We thus obtain MATH as required. |
math/0104145 | We first note that if MATH does not contain REF, then MATH is bounded below REF on MATH, while MATH is bounded; the result follows immediately. If MATH is the single point MATH, then MATH by assumption, so MATH for sufficiently small positive MATH; we may thus enlarge MATH while maintaining the hypotheses. We may thus take MATH of the form MATH with MATH; moreover, if the theorem is true for MATH with MATH, it is true for MATH. Since MATH and MATH are both positive, we may choose MATH so that MATH and MATH are positive on MATH. In particular, the first function is thus positive on MATH for all MATH; it remains to consider the second function. Now, clearly MATH is analytic on MATH, while MATH is real analytic and positive on MATH. In particular, there exist positive constants MATH and MATH such that MATH . We thus find: MATH which is clearly positive for sufficiently large MATH. |
math/0104145 | We write MATH and thus MATH the formula follows immediately from the lemma. |
math/0104145 | Since MATH, we conclude that MATH is strictly convex on MATH. Now, MATH, and thus MATH is decreasing in a neighborhood of MATH. It thus remains only to show that MATH is increasing near MATH. We compute: MATH . At MATH, this becomes MATH. |
math/0104145 | Uniqueness follows from strict convexity, so it suffices to show existence. But MATH is continuous and decreasing on MATH, and converges to MATH at MATH, so attains every value greater than MATH, in particular MATH. |
math/0104145 | Choose MATH; we will show that for all sufficiently large MATH, MATH. As this will hold for all choices of MATH, the theorem will follow from the continuity of MATH. We also choose MATH. We consider the linear combination MATH, as a vanishing linear combination of the coefficients of MATH. In particular, if we let MATH denote the coefficient of MATH in this linear combination, it will suffice to show that for all sufficiently large MATH, MATH for MATH and MATH. Indeed, if a choice of MATH existed with MATH, this would give a positive linear combination of nonnegative quantities (at least one of which is positive), equal to MATH, a contradiction. We compute: MATH where MATH. We thus need to show that for all sufficiently large MATH, MATH for MATH and MATH, where MATH . We note the following properties: CASE: MATH and MATH are power series with radius of convergence MATH. CASE: MATH. CASE: MATH. Indeed, the first two factors are clearly positive, while the second factor is positive since MATH . Similarly, the third factor is positive, since MATH. CASE: MATH is positive on MATH. This time, the second and third factors are negative; we note the limit MATH. In other words, the hypotheses of REF apply, taking MATH for appropriate MATH. It follows that for all sufficiently large MATH, MATH when MATH. For sufficiently large MATH, MATH, so we are done. |
math/0104145 | Given such a code MATH, of length MATH, let MATH be its weight enumerator, and define a power series MATH. Then from the various NAME theorems CITE, we conclude MATH for appropriate coefficients MATH. In particular, we are in the scenario of REF , with MATH. In each case, MATH, MATH and MATH are all clearly polynomials; since MATH, has radius of convergence MATH, and MATH, the hypotheses of the theorem are satisfied; it remains to compute MATH. Since MATH we find MATH. To compute MATH from MATH, we proceed as follows. Define new series MATH and MATH. From the NAME identity and the fact that MATH and MATH are linear combinations of power series coming from weight enumerators, we find that there exists an integer MATH such that: MATH . Dividing these equations, we find MATH . In terms of MATH and MATH, this becomes: MATH . We thus conclude that MATH since we readily verify MATH. Similarly, to compute MATH, we differentiate the functional equation for MATH at MATH, obtaining: MATH since by REF. Since MATH and MATH, we mutiply this by MATH to obtain the desired bound. |
math/0104145 | Let MATH be an even, strongly MATH-modular lattice, with theta series MATH. Then (CITE,CITE) MATH can be written as a weighted-homogeneous polynomial in MATH and MATH, where MATH is the lowest-dimensional even MATH-modular lattice. Clearly MATH; using the product formula for MATH, we also conclude that MATH. Thus REF applies, and it remains only to compute MATH. We first compute, with MATH, MATH and thus MATH. We have the transformation laws: MATH where MATH . We thus conclude that MATH, and that MATH . Multiplying by MATH (since the lattices are even) and dividing by MATH (since MATH) gives the required bound. |
math/0104145 | Consider a monomial MATH of MATH; note that MATH must be a multiple of REF, and in particular MATH is a multiple nof REF. Under the specified substitution, this yields MATH a polynomial with nonnegative coefficients and with valuation MATH; the first two claims are immediate. Now, the NAME theorem for Type II codes over MATH states that MATH is a (weighted homogeneous) polynomial in MATH . Under the substituion, we have: MATH . The remaining claim follows. |
math/0104145 | We recall that a self-dual binary code MATH has associated to it two enumerators: its weight enumerator MATH and its ``shadow" enumerator MATH CITE, CITE. For appropriate coefficients MATH, we have: MATH . Here MATH can be characterized (up to a multiplicative constant) as the unique linear combination of the first MATH nontrivial coefficients of MATH that is also a linear combination of the first MATH nontrivial coefficients of MATH. We define power series MATH and MATH by MATH . If we let MATH denote the coefficient of MATH in MATH, and let MATH denote the coefficient of MATH in MATH, NAME tells us: MATH . Set MATH, MATH, and let MATH be an even integer of the form MATH. Now, consider the coefficient MATH . This is manifestly a negative linear combination of MATH for MATH (so the same will be true for small perturbations). It thus suffices to consider the coefficients of MATH. By the remark above and REF , this reduces to showing that MATH is positive for MATH, and can be perturbed to be positive at MATH as well; this is clearly the case. |
math/0104145 | As above; the case MATH odd is analogous to the self-dual binary code case, while the case MATH even is analogous to the formally self-dual binary code case. The only respect in which the proof is not straightforward (using the formulae of CITE) is in dealing with the ``other" genera (not covered by REF). In each case, direct summing by a suitable MATH-modular lattice places us into the ``good" genera, and we can proceed from there; the only effect is to multiply the power series in question by a theta series, which clearly has no effect on positivity. |
math/0104145 | We define MATH and observe that MATH and MATH are polynomials with nonnegative coefficients. Since MATH we can proceed as in the case of self-dual binary codes. |
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