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math/0104093 | For MATH we have MATH where for MATH . The lemma is now immediate. |
math/0104093 | Suppose MATH is a spectrum for MATH. The orthogonality of MATH is an easy consequence of REF . Let MATH for MATH. To simplify the notation we will write MATH in place of MATH. By orthogonality and linearity MATH extends to an isometry mapping MATH into itself. We must show that the range MATH is all of MATH. Let MATH be the subspace of MATH spanned by the exponentials MATH, MATH with MATH and let MATH be the subspace of MATH spanned by the exponentials MATH, MATH with MATH. Then MATH for all MATH, so MATH. Since MATH preserves orthogonality, MATH. We must show MATH. Since MATH is arbitrary, we also have that the map MATH is an isometry mapping MATH into itself. By construction MATH and MATH for all MATH. It follows that MATH. Hence, MATH as desired. The proof that MATH is a tiling set provided MATH is, follows from the last part of the proof of REF below. |
math/0104093 | Let MATH be a spectrum for MATH. For any partition MATH of MATH, let MATH denote the set of MATH for so that MATH for MATH; MATH for MATH; MATH for MATH and MATH for MATH. Note MATH is empty unless MATH is non-empty. For MATH let MATH, if MATH and let MATH if MATH. Then for MATH, MATH for any MATH in the cube MATH given by MATH if MATH, and MATH if MATH. It follows from REF and disjointness REF of the cubes MATH, MATH that MATH where MATH is the set of MATH for which MATH for MATH, MATH for MATH, MATH for MATH, and MATH for MATH. By definition of MATH we have MATH where MATH is the cardinality of MATH. Since the number of possible partitions MATH only depends on the dimension MATH of MATH, the proof is complete . |
math/0104093 | Suppose MATH is a spectrum for MATH but not a tiling set for MATH. Let MATH. Let MATH. If MATH are so that MATH and MATH intersect MATH then it follows from REF that MATH is an integer, since MATH for MATH because MATH and MATH intersect MATH. So either MATH for all MATH so that MATH intersects MATH or MATH for all MATH so that MATH intersects MATH. In the first case MATH in the second case MATH. |
math/0104093 | Suppose MATH is a spectrum for the unit cube MATH. By REF the pair MATH is non-overlapping. We must show that the union MATH is all of MATH. To get a contradiction suppose MATH. Let MATH be so large that MATH. Let MATH. Let MATH. Pick MATH so that MATH intersects MATH (if such a MATH exists). Use REF with MATH and MATH to conclude MATH has the basis property. It follows from REF that MATH for any MATH so that MATH intersects MATH. Let MATH. Pick MATH so that MATH intersects MATH (if such a MATH exists). Use REF with MATH and MATH if MATH and MATH if MATH to conclude MATH has the basis property. It follows from REF that MATH for any MATH so that MATH intersects MATH. Note we did not move any of the cubes in MATH with MATH, for MATH. Continuing in this manner, we end up with MATH having the basis property so that MATH for any MATH with MATH for MATH. Note MATH for any MATH. So if at some stage MATH is derived from MATH then we have MATH. Repeating this process for each of the other coordinate directions we end up with MATH so that MATH is a subset of the integer lattice MATH, any MATH is obtained from some MATH, and any MATH is translated onto some MATH. In short, we did not move any point in MATH very much. By REF it follows that MATH is non-empty, hence there exists MATH, so that MATH. Replacing MATH by MATH, if necessary, and applying the process described above we may assume MATH. To simplify the notation let MATH. We have MATH . The first sum MATH since MATH and MATH, the second sum is MATH for MATH sufficiently large by REF . This contradiction completes the proof. |
math/0104093 | Let MATH be a tiling set for MATH. Suppose MATH. The proof is by induction on the number of MATH's for which MATH. Suppose that MATH for all but one MATH. Let MATH be the exceptional MATH, then MATH. Fix MATH, MATH so that the line MATH passes through both of the cubes MATH and MATH. Considering the cubes MATH, MATH that intersect MATH it is immediate that MATH. For the inductive step, suppose MATH for MATH values of MATH and MATH for the remaining MATH values of MATH implies MATH for some MATH. Let MATH be so that MATH for MATH values of MATH and MATH for the remaining MATH values of MATH. Interchanging the coordinate axes, if necessary, we may assume MATH . If MATH is an integer, then there we are done. Assume MATH. Let MATH, and for MATH let MATH . In particular, MATH and MATH. We claim the set MATH is a tiling set for MATH. Assuming, for a moment, that the claim is valid, we can easily complete the proof. In fact, MATH and MATH for MATH, so by the inductive hypothesis one of the numbers MATH, MATH is a non-zero integer. It remains to prove that MATH is a tiling set for MATH. We must show that MATH is non-overlapping and that MATH. First we dispense with the non-overlapping part. Let MATH be distinct points in MATH. Suppose MATH is a point in the intersection MATH, then MATH, in particular, MATH and MATH for MATH. It follows that MATH for MATH, so first paragraph of the proof shows that MATH, hence either MATH or MATH. In both cases we get a contradiction to the non-overlapping property of MATH. In fact, if MATH, then MATH. If MATH, then MATH. Let MATH be an arbitrary point, then MATH. If MATH for some MATH with MATH then there is nothing to prove. Assume MATH for some MATH with MATH. The point MATH is in MATH for some MATH. First we show that MATH. Since MATH and MATH we have MATH for MATH; so using MATH, for MATH, it follows that MATH, for MATH; an application for the first paragraph of the proof yields the desired result that MATH. Using MATH we conclude MATH; so using the second half of REF and the definition of MATH we have MATH as needed. |
math/0104093 | This is a direct consequence of NAME 's Theorem and REF . |
math/0104093 | Suppose MATH is a tiling set for MATH. By NAME 's Theorem MATH is an orthogonal set of unit vectors in MATH, so by NAME 's inequality MATH for any MATH. Note that MATH is the NAME transform of the characteristic function of the cube MATH at the point MATH. For any MATH we have MATH where we used NAME 's Theorem, the tiling property, and NAME 's REF . It follows that MATH for almost every MATH in MATH, and since MATH is arbitrary, for almost every MATH in MATH. Hence for almost every MATH the exponential MATH is in the closed span of the MATH, MATH. This completes the proof. |
math/0104098 | Consider any MATH different from MATH. Then MATH has an inversion MATH. So the number of copies of MATH in MATH is the number not containing MATH plus the number which do contain MATH. The permutations in the latter case can not contain MATH. So REF gives an upper bound for the number of copies of MATH which is strict unless MATH has exactly one inversion. The theorem follows. |
math/0104098 | From the previous theorem, we see that the number of zeros directly before MATH is MATH since MATH. |
math/0104098 | We induct on MATH. The result is clearly true if MATH. Assuming it is true for MATH, first consider MATH and let MATH satisfy the lemma. Then the concatenation MATH works for such MATH. On the other hand, if MATH then consider MATH. Pick MATH with MATH copies of MATH and none of MATH. Then MATH is the desired permutation. |
math/0104098 | Let MATH be the number of MATH-copies of MATH that are disjoint from the last layer. The number of MATH-copies of MATH is clearly MATH . So once MATH is chosen, MATH will have the maximum number of copies only if MATH is maximal. |
math/0104098 | It is easy to verify that MATH is NIZ. So, by REF , it suffices to show that if MATH is NIZ then so is MATH. To simplify notation in the two proofs which follow, we will write MATH for MATH, MATH for MATH, and so forth. Now given MATH with MATH we will construct a permutation MATH having MATH copies of REF. Because of REF and MATH we have MATH. So it is possible to write MATH (not necessarily uniquely) as MATH with MATH and MATH. Since MATH is NIZ, there is a permutation MATH with MATH. Also, by REF , there is a permutation in MATH with no copies of REF and MATH copies of REF. Let MATH be the result of adding MATH to every element of that permutation. Then, by construction, MATH and MATH as desired. |
math/0104098 | We prove this theorem by induction on MATH. As previously remarked, it is true if MATH. Now suppose we know the statement for all integers smaller than MATH, and prove it for MATH. If MATH is NIZ, then we are done. If MATH is IZ then, by the proof of REF , MATH is IZ. So MATH and we have MATH by REF . Now take MATH with MATH so that we can write MATH with MATH and MATH. Since the portion of MATH up to MATH has no internal zeros by induction, we can use the same technique as in the previous theorem to construct a permutation MATH with MATH for MATH in the given range. Furthermore, this construction shows that if MATH for MATH or MATH then MATH. This completes the proof. |
math/0104098 | The basic idea behind all four of the inequalities is as follows. Let MATH be the permutation obtained from our MATH- optimal permutation MATH by replacing its last two layers with a last layer of length MATH and a next-to-last layer of length MATH. Then in passing from MATH to MATH we lose some MATH-patterns and gain some. Since MATH was optimal, the number lost must be at least as large as the number gained. And this inequality can be manipulated to give the one desired. For the details, the following chart gives the relevant information to describe MATH for each of the four inequalities. In the second case, the last two layers of MATH are combined into one, so the value of MATH is irrelevant. MATH . Now REF follows easily by cancelling MATH from the inequality in the first row of the table. From the second line of the table, we have MATH and cancelling MATH, which is not zero becase MATH, gives us REF . To prove REF we induct on MATH. If MATH, then we must have MATH, so MATH. Now we assume MATH. If MATH, then the leftmost MATH elements of MATH contain no copies of MATH, so we may replace them with any MATH-permutation and still have MATH optimal. Therefore we may pick MATH and MATH, and thus the second row of the table shows MATH so MATH, as desired. If MATH, recall that from REF , the leftmost MATH elements of MATH form a MATH-optimal permutation, so we may, without loss, choose MATH maximal and thus assume that MATH. From the third line of the chart, we have MATH . Using REF we get that MATH. Substituting this in the previous equation, cancelling MATH, and solving for MATH gives MATH . Since MATH, we have by induction that MATH. Substituting and solving for MATH again and then cancelling MATH, we get MATH. A final substitution of MATH results in REF . For REF , notice that the last row of the table gives MATH so cancelling MATH gives MATH, which can be converted to the desired inequality. |
math/0104098 | It suffices to show that MATH for all MATH. This is equivalent to showing that MATH . However, by definition of MATH, we know that for all MATH, MATH . Subtracting REF from REF , we are reduced to proving MATH. We will induct on MATH. If MATH, then we would like to show that MATH so it suffices to show that MATH, which follows from REF . For MATH we have, by induction, that MATH, so it suffices to show that MATH which simplifies to MATH, and this is is true because MATH. |
math/0104098 | Induct on MATH. The lemma is true for MATH since MATH. Suppose the lemma is true for integers smaller than or equal to MATH, and prove it for MATH. For simplicity, let MATH, MATH, and MATH. Since we have already proved the lower bound, it suffices to show that MATH . Note that we do not have to consider MATH because of REF . We prove REF by induction on MATH. For the base case, MATH, we wish to show MATH . But since MATH is optimal by assumption, we have MATH . Subtracting REF from REF and rearranging terms, it suffices to prove MATH . First, if MATH, then REF is easy to verify using REF and the values MATH, MATH, and MATH for MATH. Therefore we may assume that MATH. Let MATH, MATH, and MATH be layered MATH-optimal permutations having last layer lengths MATH, MATH, and MATH, respectively, as short as possible. Also let MATH, MATH, and MATH. We would like to be able to assume the lemma holds for these permutations, and thus we would like to have MATH. But by REF we have MATH if MATH. Since MATH this holds for MATH and the case MATH is easy to check directly. Therefore we may assume that MATH, MATH, and MATH all satisfy the lemma. If MATH then let MATH be the largest element in the last layer of MATH (namely MATH). Otherwise, MATH and removing the last layer of both MATH and MATH leaves permutations in MATH and MATH, respectively. So we can iterate this process until we find the single layer where MATH and MATH have different lengths (those lengths must differ by REF) and let MATH be the largest element in that layer of MATH. Similarly we can find the element MATH which is largest in the unique layer were MATH and MATH have different lengths. Now let MATH . Note that there is a bijection between the MATH-patterns of MATH not containing MATH and the MATH-patterns of MATH. A similar statement holds for MATH and MATH. So MATH . Note also that MATH because increasing the length of the layer of MATH results in the most number of MATH-patterns being added to MATH. It follows that MATH. By REF , MATH, so to obtain REF it suffices to show that MATH. But MATH is the total number of subsequences of MATH having length MATH and containing MATH and MATH. So the inequality follows. The proof of the induction step is similar. Assume that REF is true for MATH so that MATH where MATH. We wish to prove MATH . Subtracting as usual and simplifying, we need to show MATH . Proceeding exactly as in the base case, we will be done if we can show that MATH . Because MATH we have MATH, so it suffices to show that MATH . This simplifies to showing that MATH, and this is guaranteed by our choice of MATH. |
math/0104098 | For the upper bound, recall that MATH is the total number of subsequences of MATH of length MATH containing MATH and MATH while the double difference just counts those subsequences corresponding to the pattern MATH. For the lower bound, we showed that MATH . Recall that MATH is the total contribution of MATH in MATH, and MATH is the total contribution of MATH in MATH. Therefore MATH, as otherwise one could create a permutation with more MATH-patterns than MATH by inserting a new element in the same layer as MATH . |
math/0104098 | Let MATH. We induct on MATH. It is easy to check the base cases MATH. Note that by REF , either MATH or MATH. If MATH, then we have MATH and maximizing this as a function of MATH gives MATH . If MATH, then we have MATH . By induction, we have MATH, and thus we have that MATH . By REF , this function is to be maximized on the interval MATH and for MATH this maximum occurs at MATH. So MATH as desired. |
math/0104098 | For REF we induct on MATH. The claim is true trivially for MATH since then MATH for all MATH, so we will assume MATH. If MATH then the claim is true by definition. If MATH then MATH by REF and we are able to apply induction. We would like to show that MATH and we know by induction that MATH . Subtracting as usual, we are reduced to showing that MATH. This further reduces to MATH which is true by REF and the fact that MATH. REF is implied by the definition of MATH, so we are left with REF . By the definition of MATH we have that MATH, so it suffices to show that for all MATH, if MATH then MATH. Subtracting in the usual way, we are reduced to showing that MATH . Since we know that MATH by REF , our approach will be to show that MATH for MATH by showing that MATH . Before we prove REF , we will need the following two facts. MATH and MATH. The first fact follows from our proof of REF , in which we showed that MATH for MATH. So to prove the second fact, it suffices to show that MATH implies MATH for MATH. This is proved in exactly the same way as REF with all the inequalities reversed. Now we are ready to prove REF . First we tackle the case where MATH by induction. If MATH then MATH and we are done. So suppose MATH. If MATH, then since MATH and MATH we have MATH as desired. Hence we may assume that MATH. In this case we claim that MATH, which will imply REF by induction. Let MATH. We want to show that MATH and we have MATH . Subtracting, it suffices to show that MATH . By REF , MATH, so it suffices to show that MATH . Since MATH, we have that MATH, and since MATH, we have that MATH, so REF is true, and thus REF holds. For the case where MATH, we examine the quadratics MATH which agree with MATH, wherever both MATH and MATH are defined. We will also need to refer to the roots of MATH, which occur at MATH . REF gives us that MATH so MATH and MATH are real numbers and for MATH, MATH. These roots are important in our situation for the following reasons: MATH . REF is easily verified. Assume to the contrary that the forward direction of REF is not true, and thus MATH but MATH. Let MATH be such that MATH. By REF , we have that MATH, and thus MATH by REF . However because MATH, we have that MATH, a contradiction. To prove the reverse direction of REF , notice that if MATH then by REF and the definition of MATH, we must have that MATH. Therefore by REF , either MATH (as we would like) or MATH, and by REF , it cannot be the case that MATH, as that would imply that MATH if MATH, contradicting REF . To prove REF , note that by REF we cannot have MATH as then we would have MATH, contradicting the definition of MATH. Also, we cannot have MATH as then we would have MATH by REF , again contradicting the definition of MATH. Hence we must have REF . With these tools, REF is easy to prove; we have MATH for MATH, and thus MATH, as desired. It is easily checked that MATH for MATH. |
math/0104098 | By REF it suffices to show the following inequalities: MATH and MATH . REF is clear for MATH because MATH, MATH for all MATH by REF , and MATH. We prove REF by induction on MATH. First, if MATH, then MATH, so it suffices to show that MATH and since MATH, we have MATH . Subtracting that latter from the former, it suffices to show that MATH . So we're done in this case since MATH which follows from MATH and MATH. Now assume that MATH, so we may prove REF by showing the stronger statement that MATH and thus we would like to show that MATH and as MATH, we have MATH . Subtracting as usual, we are reduced to showing MATH . By REF MATH . The upper bound in REF now completes the proof of REF . To prove REF , we want to show MATH and we are given MATH . Subtracting as usual, we are reduced to showing that MATH . Cancelling MATH and simplifying, it suffices to show that MATH . By REF , MATH, so it suffices to show that MATH which is true for MATH. For MATH, note that proving REF reduces to showing MATH which we will prove by induction on MATH. Checking the base cases MATH is easy. Also note that REF holds for MATH if we make the strict inequality weak, so we still can conclude the MATH part of the Lemma. There are now two cases. If MATH then by induction MATH. By REF and the part of the Lemma that we've already proved, the only other possibility is MATH and MATH. But then MATH which is equivalent to the desired inequality. |
math/0104098 | The inequality follows from the fact that each copy of MATH in MATH gives rise to a copy of MATH in MATH. For the equality, if MATH is layered then it is the unique permutation giving rise to the poset MATH. So every copy of MATH in MATH corresponds to a copy of MATH in MATH and we are done. The only other case we need to consider is if MATH is layered and MATH is not. But then both sides of the equality are zero. |
math/0104098 | We will write MATH for MATH and MATH for MATH. Given MATH, it is easy to construct a poset MATH with MATH. So let MATH be a MATH-optimal poset. Now there must be some MATH with MATH. Now adjoin an element MATH to MATH to form a poset MATH with MATH in MATH if either CASE: MATH with MATH, CASE: MATH, MATH with MATH, or CASE: MATH, MATH with MATH. Then MATH so MATH. |
math/0104098 | As before, we write MATH for MATH. Since MATH it is enough to show that MATH . First, REF is clear since MATH and MATH agree on all subsets not including MATH. Next, notice that MATH and thus to prove REF , it suffices to show that MATH, but this is easy. Let MATH with MATH, MATH, and MATH. Then MATH is an occurrence of MATH in MATH, that is, MATH, so REF is proved. Finally, to prove REF , let MATH be an occurrence of MATH in MATH which contains MATH and MATH, that is, MATH. Then we have that MATH as well. This is because MATH in MATH since MATH are maximal and MATH is LOT. So MATH forms an occurrence of MATH in MATH, and thus REF is proven. |
math/0104098 | Choose MATH with MATH, let MATH and MATH. First consider what happens when MATH. Then REF implies that MATH for all MATH. This forces MATH by REF . Now REF yields MATH, contradicting REF . So we may assume MATH. Note that MATH, and since MATH we get that MATH . Furthermore, since MATH is LOT we get that MATH and MATH . Also, since MATH, we have that MATH . Furthermore, since MATH is LOT, MATH is LOT, so we have MATH . Therefore MATH . Furthermore, by REF we have that MATH. So we must have MATH and, by REF again, MATH. It follows that MATH. Therefore since MATH we have MATH and so MATH. Also, because MATH, we have MATH and thus MATH is LOT, as desired. |
math/0104098 | Notice that the claim is trivial for MATH as all posets on less than MATH elements cannot have any MATH-patterns and thus they have the trivial MATH- decomposition MATH. Assume to the contrary that the claim is not true and let MATH be a MATH- optimal poset of least cardinality that does not have a LOT MATH-decomposition with MATH maximal over all such choices of MATH and MATH. Let MATH be the set from REF , MATH and MATH. First, we claim that MATH is LOT. If not, then there is some element, say MATH. Also let MATH. If MATH, then by REF either MATH or MATH, both contradictions, so MATH and MATH is MATH-optimal. Since MATH, by our choice of MATH we know that MATH has a MATH-decomposition MATH. If MATH, then MATH, so by REF , MATH (because MATH), a contradiction to our choice of MATH. Hence we may assume that MATH, so MATH is LOT. As the only element MATH and MATH disagree on is MATH, we have that MATH is LOT. Hence by REF , MATH is also LOT. Now that we know that MATH is LOT, we get that MATH, so MATH is MATH- optimal. By induction, MATH has a MATH-decomposition MATH and thus MATH is a MATH- decomposition for MATH. |
math/0104098 | Assume that REF does not hold and choose MATH with MATH and MATH maximal over all such choices. Let MATH, MATH and MATH. We must have MATH for all MATH as otherwise by REF we would have either MATH or MATH, a contradiction. Hence we have MATH . First we tackle the easier case, where MATH. Pick two maximal elements of MATH, say MATH, so that MATH. By REF we have that MATH, and thus by REF we know MATH has a MATH-decomposition MATH. Since MATH, we must have MATH, so we also have that MATH and MATH. Therefore MATH and consequently MATH are LOT. Hence by REF , MATH is LOT, a contradiction. Now assume MATH. Let MATH be as in REF , pick MATH (MATH must exist as MATH is not LOT) and MATH. Now MATH and thus MATH by REF . However if MATH then we have contradicted our choice of MATH as MATH. Therefore MATH so by REF , MATH has a MATH-decomposition MATH. By the same reasoning as the previous case, MATH, so again MATH and MATH are both LOT. Although we cannot apply REF in this case, REF still holds for MATH with MATH, so MATH . Therefore we must have MATH. If MATH, this implies that MATH, so REF is true with MATH. If MATH then we must have MATH, so there is precisely one element, say MATH. Since MATH is LOT, MATH must lie in MATH. Let MATH. Then we have MATH . Because MATH is LOT, we have that MATH, and because MATH is LOT we have that MATH. Notice that because MATH is MATH-optimal, by the comment after the proof of REF , MATH is layered, and thus MATH is layered. Since the MATH-patterns in MATH containing both MATH and MATH are formed with exactly one element which lies in MATH, MATH. Finally, MATH. Now combining all these MATH-values with REF gives MATH so MATH. We have by REF that MATH and MATH (this follows from the fact that MATH is layered and MATH-optimal), which forces MATH. This in turn implies MATH. Now it can be checked by direct computation that for MATH in this range either the theorem is true vacuously or one of REF to REF holds. |
math/0104098 | Induct on MATH. If MATH, then MATH, so the theorem is true vacuously. If MATH, then MATH and MATH. Hence we may assume that MATH. If REF is true, let MATH be the poset guaranteed there, MATH and MATH. Then by REF , MATH, so MATH, a case we have already dealt with. It is routine to check that the poset MATH satisfies REF of this theorem if REF is true. Therefore we may assume that REF is true, and thus there is a LOT MATH-poset MATH so that MATH. Since MATH is LOT, MATH. As MATH, we must have MATH. If MATH, then by REF , there is some layered MATH-poset MATH so that MATH, and thus MATH is layered, MATH and MATH for some MATH. If MATH, then by induction, there is some poset MATH, MATH, which satisfies this theorem. So MATH is the desired poset. |
math/0104098 | Assume that the theorem is false. Since MATH for MATH and MATH for MATH, there must be some maximal MATH so that MATH. By REF , there is some MATH so that MATH and MATH. Also note that since MATH, by REF we have MATH, so we may apply REF to see that MATH. By our choice of MATH, MATH, so there is some MATH so that MATH. By REF , MATH, and thus REF produces a poset MATH for some MATH and integer MATH which is positive since MATH. Let MATH. By REF , there is a layered MATH-optimal MATH-poset MATH, and so we must have MATH. Therefore, by REF , we have MATH, and thus the inequality in REF implies that MATH. However, if MATH then we have MATH, contradicting our choice of MATH. |
math/0104100 | CASE: By REF , the symbol of a graph with no roots has polynomial degree MATH. This proves the first assertion. CASE: By REF , any graph MATH is MATH, MATH with no roots, MATH with no wheels. It is easy to see from the definition that the weights are multiplicative : MATH. The only things that remain to be considered are the MATH factors : when one looks at a decomposition MATH, it is unique only up to labelling, which explains the factors MATH. |
math/0104100 | By an argument similar to the one in the proof of REF , one proves that MATH. We now need to prove convergence of the series MATH, and by homogeneity one can set MATH. Without loss of generality, one can assume that the structure constants MATH. Therefore MATH . And more generally MATH with MATH . NAME monomial in MATH of degree MATH and MATH. Let MATH with only one wheel of length MATH. If the absolute values of all components MATH of MATH respectively are less than MATH, MATH . Besides, the following inequalities can be found in CITE : MATH . Finally, the terms of the series in MATH can be bounded by MATH, which proves convergence for MATH small enough. |
math/0104100 | Since for fixed MATH the series for MATH is finite, we can substitute to MATH a real number MATH, and interchange summation with the test function-distribution bracket : MATH and we avoid convergence problems for a fixed MATH by taking a test function MATH with small enough support. We know that MATH, and MATH is scalar valued REF . So MATH . By REF, MATH, and by NAME 's formula for polynomials, MATH . The proposition follows. |
math/0104100 | Let MATH and MATH be the supports of MATH and MATH. Assume that MATH and MATH are included in a ball or radius MATH. Assume that MATH is less than the radius of convergence of MATH. It is clear that REF makes sense. We need to prove that the functional MATH defined therein is a distribution. Since MATH is analytic from MATH to MATH, the pushforward of the compactly supported distribution MATH under MATH is a distribution. But MATH is obtained by multiplying this distribution by the analytic function MATH, so it is a distribution. |
math/0104100 | CASE: Let MATH. The convolution MATH is given by : MATH for MATH a test function on MATH. Since MATH is an algebra homomorphism, by REF we get MATH . Therefore, applying REF , we get for any distributions MATH with support MATH : MATH . Fix a neighborhood of MATH in MATH on which MATH is defined. Assume that MATH on that neighborhood. We have MATH . Since this equality holds for any MATH and MATH supported at MATH, this implies that the two functions MATH and MATH have same derivatives at MATH; since they are analytic, they must be equal. CASE: The second statement follows immediately from REF . |
math/0104100 | For MATH be two distributions on MATH with small support, and MATH a test function on MATH, let us define MATH . By inspection, MATH is a bi-differential operator with polynomial coefficient acting on distributions. For distributions MATH, we know that MATH . Now bi-differential operators with polynomial coefficients are completely determined by their action on distributions with point support. This proves that MATH . |
math/0104100 | The second statement is easy. We prove the first. Let MATH be any representative of MATH. By definition of MATH, there exists MATH such that MATH . Let MATH be a MATH function supported in MATH which is identically equal to MATH in MATH. Clearly, MATH is a representative of MATH with support included in MATH. |
math/0104100 | We study the restriction of MATH to MATH. The NAME formula implies that there exists a positive number MATH such that the MATH-valued map MATH is analytic for MATH. Furthermore MATH . Therefore, the implicit equation MATH can be solved : there exists a constant MATH and an analytic map MATH, defined for MATH and MATH such that MATH is equivalent to MATH for MATH. Now since MATH and MATH is an open cone, we conclude that there exists MATH such that, for MATH, and any MATH, MATH. Finally, we have proven that, for MATH with norms less than MATH, MATH, MATH implies MATH. Using a straightforward topological argument, we get MATH . Assume that MATH is a test function supported in MATH. Then, by REF , MATH is included in MATH, as we wanted. |
math/0104100 | Let MATH be compatible germs, and MATH two pairs of representatives. By REF , since the multiplication factor MATH does not play a role, it is enough to prove that, for any MATH, the formula MATH provided the supports of MATH are adequate. Assume that MATH (respectively, MATH) are supported in MATH (respectively, MATH). Since MATH, MATH, there exists MATH such that for any test function MATH with support included in the ball MATH, MATH . Using REF we deduce REF for MATH supported in MATH. |
math/0104100 | It is enough to do it for MATH. We consider MATH from REF As before, we chose representatives MATH of MATH supported in MATH and MATH respectively. Since the germs are invariant, there exists a MATH such that, for any test function MATH supported in MATH for all MATH. Applying again REF , consider MATH a test function supported in MATH. We prove that MATH for all MATH. Indeed, since the functions MATH are invariant, it is enough to prove MATH . Using the covariance of MATH under the adjoint action of MATH: MATH writing MATH and differentiating at MATH, we get MATH . Thus we get MATH . By REF MATH . Now we use REF to conclude. |
math/0104100 | The first step is to prove that REF holds under the assumptions of the lemma, provided MATH has small enough support. We proceed as in REF : from the invariance of MATH, we have a constant MATH such that MATH for all MATH and any MATH supported in MATH. We derive MATH from REF . Let MATH have support in MATH. In NAME 's proof which was just outlined above, the distribution MATH is paired with the function MATH and some derivatives of MATH. Since MATH, we can write MATH where MATH is some smooth function. Therefore the support of MATH (and any derivative) is included in the support of MATH. By REF wee see that MATH is included in MATH. Since we are pairing with MATH, with MATH verifying REF , we still conclude as in NAME that MATH is independent of MATH, so that REF still hold. Now we apply REF , but to the algebra MATH (note that this MATH is not ``the same" as the MATH used before !). Using REF , since the MATH function for MATH is given by MATH (see CITE) we derive : MATH . Using the fact that MATH are analytic in a neighborhood of MATH, we conclude from REF that MATH is analytic for MATH. |
math/0104100 | Let MATH be the differential of MATH at MATH of order MATH, evaluated at the MATH-tuple MATH. It follows from REF that the right hand side of the lemma is MATH where MATH . Clearly, this expression is the NAME transform of a bidifferential operator with polynomial coefficients. |
math/0104100 | By symmetry, it is enough to prove that for MATH with MATH an adjoint vector field, MATH, MATH . But MATH . Since the support MATH is included in the support of MATH, we can now apply REF . |
math/0104106 | Consider an entire function on MATH . Define a MATH-linear functional MATH on MATH . Taking MATH and MATH we get MATH by REF. Minimizing the right term, we have MATH . This shows that MATH can be expressed in the form MATH with MATH, MATH with finite NAME norm MATH for any MATH. Therefore we derive MATH . |
math/0104106 | We can prove this Lemma with modifications of the proof of the previous Lemma. Therefore, we omit the proof. |
math/0104106 | First, we will show that for any MATH, MATH, there exist MATH and MATH such that MATH. Let MATH, MATH, be given. Since it has been proved CITE (see also CITE) that every test function in MATH has an analytic extention, there exist MATH and MATH such that for any MATH for any MATH. Hence it is derived by REF that MATH . To prove the converse, The multiple NAME of MATH is given by MATH . Then observe that for MATH . By the condition MATH, MATH for all MATH. Therefore, MATH . By the condition MATH, we have MATH . Thus, it is easy to get MATH where MATH . (Note that finiteness concerning MATH can be shown easily by simple estimation and the NAME theorem CITE.) Then applying REF with REF , we have MATH . We complete the proof. |
math/0104106 | First we shall prove sufficiency. Suppose that MATH is supported in MATH for some MATH and REF holds. Then for any MATH, MATH . With the help of REF , MATH and MATH is a continuous linear functional on MATH. Therefore, MATH is a positive product NAME measure which induces a positive generalized function MATH in MATH. Conversely, suppose that MATH is a positive product NAME measure. Then for all MATH, MATH is a continuous linear functional with respect to MATH by REF . Thus there exist constants MATH such that for all MATH . Let us define an analytic function MATH on MATH by MATH where MATH is the bilinear pairing on MATH. On the other hand, REF implies that MATH . (It is easy to find an increasing function MATH equivalent to a function MATH.) This implies that MATH. Thus, from REF with MATH we obtain that MATH . Due to REF , we have MATH . However REF says that MATH. Therefore, MATH . By choosing an appropriate MATH satisfying MATH, so that MATH. Therefore we get the assertion. |
math/0104109 | By CITE, compare CITE, MATH admits only finitely many tight contact structures next to the MATH, provided that MATH. In the case MATH there is a further infinite family of tight contact structures. |
math/0104109 | The scaling map MATH defines a contactomorphism MATH. Hence, given two contact embeddings MATH, MATH, we can compare either with a third such embedding that maps MATH into the interior of MATH. We may therefore assume that MATH, the contact embedding MATH is the inclusion map MATH, and the contact embedding MATH sends MATH into the interior MATH of MATH. Note that if MATH, then the diffeomorphism MATH may be assumed to have the following effect on MATH and MATH, since MATH is determined up to isotopy by its action on homology, corresponding to an element of MATH: MATH where MATH is some integer. (Different choices of MATH correspond to NAME twists along a meridian of the solid torus that is glued back; these NAME twists extend to diffeomorphisms of the solid torus and hence have no topological effect.) Then MATH sends MATH to MATH. This implies that as we pull back MATH to MATH, we obtain a contact structure on MATH with MATH and MATH. So by REF the extension of MATH to a tight contact structure MATH on the copy of MATH to be glued back is unique. Let MATH be the contact manifold obtained from MATH by contact MATH - surgery along MATH using the inclusion MATH, and let MATH be the contact manifold obtained similarly using the contact embedding MATH. By what we have just observed, the tight contact structure MATH is uniquely determined by the fact that it coincides with MATH near MATH. We also know that MATH coincides with MATH outside MATH, and the manifolds MATH and MATH are diffeomorphic under a diffeomorphism that is the identity near MATH. By the definition of contact MATH - surgery, the contact manifold MATH is tight. It now suffices to show that MATH is tight, because we then know that it is completely determined by its boundary data, which coincide with those of MATH. Recall that if a contact structure MATH on a manifold MATH is written as the kernel of a MATH - form MATH, there is a one-to-one correspondence between contact vector fields MATH and functions on MATH given by MATH, compare CITE. The function MATH is called the Hamiltonian function corresponding to MATH. So the contact vector field MATH for MATH on MATH corresponds in this way to some Hamiltonian function. By multiplying this function with a bump function that is identically MATH on MATH and identically zero near MATH we can construct a contactomorphism MATH that is the identity near MATH and sends MATH into MATH for any given MATH. By precomposing MATH with such a diffeomorphism, we may assume that MATH for a suitable MATH. By multiplying the Hamiltonian function of MATH with a bump function that is identically MATH on MATH and identically MATH outside MATH, we get a Hamiltonian function defined also on MATH whose contact flow will ultimately move MATH into MATH. So this will define a contact embedding MATH . This completes the proof of the proposition. |
math/0104109 | By the preceding proposition it suffices to consider the following situation: Let MATH be the manifold obtained from MATH by contact MATH - surgery along MATH, using the inclusion MATH. Let MATH be the manifold obtained from MATH by contact MATH - surgery along a spine of the solid torus MATH that was attached to MATH to form MATH. We want to show that MATH is contactomorphic to MATH. We can obtain MATH by gluing MATH to MATH using the attaching map MATH described by MATH and then extending MATH over MATH to a unique tight contact structure MATH. We observed in the proof of the preceding proposition that the torus MATH in MATH is a convex surface with MATH and MATH. By REF and arguments similar to those in the preceding proof, we can find a contact embedding MATH isotopic to the identity and sending MATH into MATH. Now perform the MATH - contact surgery on MATH using this embedding (composed with MATH), and call the resulting contact manifold MATH. The gluing for this surgery may be described by MATH . It is a straightforward check that the topological effect of this second surgery is to cancel the first surgery, because the composition of these maps sends MATH to MATH (in fact, it is the identity map). A further application of REF shows that MATH is indeed contactomorphic to MATH. |
math/0104109 | For MATH, that is, MATH, this is the result of NAME mentioned in the introduction, which holds true even for MATH. By the preceding proposition, MATH, MATH, is obtained from MATH by contact MATH - surgery. The result now follows from REF . |
math/0104109 | For MATH this is well-known, see CITE. For positive MATH it is a consequence of REF (which holds true also for MATH and MATH). For negative MATH we use a construction analogous to CITE. Let MATH be a smooth function with strictly positive derivative, MATH for all MATH, and MATH. Notice that MATH then implies MATH for all MATH. So we may define MATH as kernel of the contact form MATH (defined on MATH). Projection onto the MATH - and MATH - coordinate gives MATH the structure of a principal MATH - bundle MATH. Let MATH be the associated complex line bundle MATH, and write MATH for its zero section. Write MATH for the angular coordinate and MATH for the radial coordinate in the MATH - fibre, so that MATH on MATH. The vector fields MATH and MATH are defined on MATH, and MATH extends to a MATH - invariant MATH - form on MATH satisfying MATH and MATH. We then have MATH . Set MATH . It is a straightforward check that MATH is a symplectic form defined on all of MATH, and that MATH is a NAME vector field for MATH defined on MATH, and MATH. So the unit disc bundle MATH gives a strong symplectic filling of MATH for MATH. |
math/0104109 | It is well-known (and easy to prove) that MATH is generated by MATH and MATH. Moreover, the product MATH is of order MATH in MATH, which implies that MATH and MATH can be expressed as a product in MATH and MATH. Thus, with MATH given, we may write it in the form MATH with MATH and MATH. Set MATH and MATH for MATH, so that MATH. If MATH, then by REF we know that MATH is obtained from MATH by contact MATH - surgery, where MATH is chosen suitably. If MATH, we observe that with MATH we can write MATH . Since conjugate matrices MATH and MATH give rise to contactomorphic torus bundles MATH and MATH, we conclude once again that MATH is obtained from MATH by contact MATH - surgery for a suitable MATH. By induction, there exists MATH such that MATH is obtained from MATH by MATH times contact MATH - surgery. Thus, NAME 's theorem and REF imply that MATH is not strongly symplectically fillable for MATH. |
math/0104109 | Given MATH, we write it in the form MATH with each MATH equal to MATH or MATH. Let MATH be a nodal elliptic surface with a section, without multiple fibres, and with NAME number (or number of singular fibres) equal to MATH. For the existence of such a surface see CITE. This surface is algebraic CITE and thus NAME; in particular we find a symplectic from MATH on MATH that restricts to an area form on each nonsingular fibre (since these are complex submanifolds). By the arguments in REF we find a simple closed loop MATH in MATH along which the monodromy of the fibration MATH equals MATH. Let MATH be the disc whose oriented boundary is MATH. Then MATH is the desired symplectic manifold. Here is an alternative and slightly more direct argument: Observe that MATH and MATH correspond to positive NAME twists of MATH. This implies that there is an orientable NAME fibration MATH with generic fibre a torus, MATH singular fibres, and monodromy along MATH equal to MATH, compare CITE. Such a NAME fibration admits a symplectic form MATH with the described properties, see CITE. Since the base of the fibration is MATH, the second homology group of the total space is generated by the fundamental class of the fibre (this remains true in the presence of singular fibres). So the homological condition in the cited theorem, necessary to apply NAME 's symplectic fibration construction, is trivially satisfied. |
math/0104109 | Represent MATH by MATH, that is, MATH where MATH is as described in the introduction. The properties of MATH imply that we can find a smooth function MATH such that the contact MATH - form MATH is invariant under the transformation MATH and thus descends to a contact form (which we continue to denote MATH) on MATH representing MATH. Observe that the MATH - form MATH is a contact form for any MATH, and in view of the well-known NAME stability theorem CITE it defines a contact structure equivalent to MATH. For MATH the contact planes MATH approach the tangent spaces along the fibres of MATH. Hence, the symplectic form MATH on MATH constructed in the preceding proposition will have the property that MATH is nondegenerate for MATH sufficiently small. |
math/0104109 | It is possible to choose the values of MATH equal to those of MATH at MATH and MATH and still satisfy the appropriate equivariance condition because MATH is an eigenvector of MATH with eigenvalue MATH. CASE: First consider the case that MATH is an eigenvector of MATH with positive eigenvalue. This is equivalent to saying that MATH is of type MATH. A straightforward analysis shows that in this case MATH . The same analysis applies to MATH. That is, the function MATH with the described properties has twisting MATH determined by MATH . Since MATH by assumption, we have MATH for MATH, and MATH for MATH. CASE: Now assume that MATH is not of type MATH. Then MATH, and one verifies that the twisting MATH of MATH is determined by MATH compare CITE. Let MATH (with MATH) be the smooth function defined by MATH, and let MATH be the lift of MATH with MATH. One checks that MATH is strictly increasing and MATH for all MATH, with equality MATH for MATH. The required function MATH can be defined by smoothing the function MATH at MATH, and then extending it to all MATH by imposing the appropriate equivariance property. Since the smoothing is done at MATH, and MATH is not an eigenvector of MATH with positive eigenvalue, we can ensure that this does not lead to a twisting MATH larger than MATH. The properties of MATH imply that this twisting MATH, determined by MATH is equal to MATH or MATH. |
math/0104109 | The contact plane MATH is spanned by MATH and MATH. We may choose MATH of the form MATH where MATH is a smooth convex curve in the MATH - plane, close to MATH. Moreover, we may assume that MATH is tangent to MATH only at the two points on MATH with MATH. The assumption MATH guarantees that the singular set of the characteristic foliation MATH consists of the two circles MATH. Furthermore, the vector spanning MATH away from its singular points always has a non-zero MATH - component, and the coefficient functions of this vector field may be chosen not to depend on the MATH - coordinate. The two circles MATH divide this singular foliation. Now apply REF . |
math/0104109 | (for MATH). Let MATH be as in REF . Write MATH respectively, MATH for the contact structures on MATH respectively, MATH defined by these functions. Fix a positive real number MATH. Let MATH be the smooth function satisfying MATH for all MATH. Observe that MATH and MATH, and MATH is strictly monotone increasing. With MATH as above, set MATH . We continue to write MATH, MATH for the lift of those contact structures from MATH respectively, MATH to MATH. Define contact embeddings MATH, MATH as follows: MATH . Notice that MATH is the composition of contactomorphisms MATH . Fix a positive real number MATH. Choose MATH sufficiently small such that MATH and MATH . Let MATH be a smooth, strictly monotone increasing function such that MATH . Let MATH be a smooth, monotone increasing function such that MATH . It is easy to see that MATH can be chosen in such a way that MATH . For MATH set MATH . Define MATH by MATH . Define MATH . Notice that MATH coincides with MATH for MATH and with MATH for MATH. Moreover, one easily verifies that MATH is an injective immersion. We compute MATH . It follows that the singular foliation MATH is represented by the vector field MATH. The singular foliation MATH, on the other hand, is represented by MATH. We claim that these two singular foliations are identical as smooth foliations. Indeed, the two functions MATH and MATH vanish only at MATH and have positive derivative there. It follows that either of them can be written as MATH with MATH a smooth, nowhere zero function on MATH, so MATH is smooth and non-zero on all of MATH. By REF there exists a neighbourhood MATH of MATH in MATH and a contact embedding MATH that coincides with MATH respectively, MATH on the common domain of definition, and with MATH on MATH for MATH or MATH. By REF , and with MATH as in that lemma (which holds true for the contact structure MATH in place of MATH), this MATH extends to a contact embedding MATH . Let MATH be the Legendrian circle in MATH defined by MATH . Then MATH induces a contact embedding MATH where we may think of the tubular neighbourhood MATH of MATH as MATH . Again by REF (adapted suitably), MATH is obtained from the manifold MATH by contact MATH - surgery on MATH. To verify the sign of this surgery we need to make the following observations. Let MATH be a meridian of MATH defined by MATH, say, and let MATH be a longitude of MATH defined by MATH, MATH. We take MATH to be oriented in positive MATH - direction, and MATH to be oriented in counterclockwise direction with respect to the oriented basis MATH of the MATH - plane. This is consistent with our orientation assumptions in the definition of contact surgery. Moreover, it is this choice of longitude that gives MATH, so the surgery coefficient MATH is determined by expressing the attaching map in terms of MATH and MATH. The effect of the map MATH (up to isotopy) is to send MATH to MATH, and MATH to MATH, as can be checked from our explicit formulae. So MATH maps to MATH, which shows that it is this curve MATH on MATH which becomes homologically trivial when we glue in a solid torus in place of MATH to obtain MATH. |
math/0104114 | For MATH and a morphism MATH we need to check that kernel and cokernel of MATH exist, and that MATH (the other axioms are clear). Let MATH denote MATH considered as an element of MATH. Then there are unique morphism MATH and MATH which make the obvious diagrams commutative; moreover, MATH, MATH are easily seen to be a kernel, and a cokernel of MATH (here the diagrams involving MATH are commutative because they inject into the corresponding diagrams for MATH; while in the ones involving MATH one only needs to verify equalities of elements in MATH for various objects MATH, thus it is enough to verify the equality of their compositions with a surjective arrow MATH). This shows existence of MATH, MATH; since an arrow in MATH is an isomorphism if and only if the forgetful functor to MATH sends it into an isomorphism, REF follows. Exactness of the forgetful functor is clear from the explicit description of MATH, MATH. |
math/0104114 | MATH is faithful and exact, because such is its composition with the faithful exact forgetful functor MATH. Let us check that MATH is a full imbedding, that is, MATH . First we claim that REF holds if MATH; moreover, this is true for any pair of adjoint functors MATH, MATH (not necessarily between additive categories). Indeed, using REF we get MATH and it is immediate to see that the resulting isomorphism coincides with the map induced by MATH. Thus to check REF it suffices to see that any MATH is a kernel of an arrow MATH for some MATH. It is enough to find an injection MATH (then apply the same construction to its cokernel); but MATH being faithful implies that the adjunction arrow MATH is an injection. To see that MATH is surjective on isomorphism classes of objects it sufficies to prove that any object MATH is a subobject in MATH for some MATH (then it is also a kernel of a morphism MATH for some MATH; since we know already that MATH for some MATH we conclude that MATH by exactness of MATH). Now for MATH consider the adjunction arrow MATH (coming from REF). We claim it is injective; indeed, since the forgetful functor is exact and faithful, it is enough to see that the corresponding arrow MATH in MATH is injective; it is in fact a split injection, because MATH by the definition of a coalgebra over a comonad. |
math/0104114 | Let MATH denote the group MATH-points of a split real form of MATH. Let also MATH be the group of points of a maximal unipotent subgroup of MATH defined over MATH. Then the manifold MATH admits a unique up to a constant MATH-invariant measure which has unique smooth extension to every MATH. Let MATH denote the space of MATH-functions on MATH with respect to this measure. Since for every MATH as above we have MATH it follows that we have well defined unitary operators MATH acting on MATH (NAME transform along the fibers of MATH). We claim now that these operators MATH define an action of MATH on MATH. Indeed, this is proved in CITE when MATH is replaced by a non-archimedian local field and in the archimedian case is essentially a word-by-word repetition. For every MATH we denote by MATH the corresponding unitary automorphism of MATH. The operators MATH commute with the natural action of MATH on MATH. Set now MATH (here by MATH we mean the space of complex valued MATH-functions). It is clear that MATH is a dense subspace of MATH (since it contains the dense subspace of MATH-functions with compact support). Let us show that MATH is invariant with respect to the operators MATH. Indeed, every MATH is a MATH-vector in the MATH representation MATH. Therefore, since every MATH commutes with MATH it follows that MATH is again a MATH-vector with respect to MATH. Hence for every MATH the function MATH makes sense. Moreover, it is easy to see that for every MATH and for every simple root MATH of MATH we have MATH . Hence MATH which implies that MATH. Hence MATH is invariant with respect to MATH's and therefore it is also invariant with respect to all MATH. It is clear that MATH is a faithful module over MATH. Moreover REF implies that in the space MATH we have the equality MATH . Clearly, this implies our claim. |
math/0104114 | REF shows that categories MATH, MATH for MATH; and functors MATH; MATH form a faithful localization data. Thus the statement follows from REF . |
math/0104114 | It is obvious that MATH. Let us prove the inverse inclusion. Let MATH. Since MATH is irreducible for any MATH, we have MATH for some MATH. Let us prove that the function MATH, MATH is polynomial. For any MATH consider the operator MATH acting on the functions on MATH which is defined by: MATH. Then for any set of functions MATH, where MATH, we have: MATH . In particular, if MATH is a differential operator of order MATH, then MATH for any MATH. But it is well known that the latter property implies that the function MATH is polynomial. It is clear that for every polynomial function MATH there exists an element MATH such that MATH. Let us take MATH. Then MATH for any MATH. Hence MATH which finishes the proof. |
math/0104114 | Let MATH as before be a symplectic vector bundle. Let MATH. Suppose that MATH transforms by a character MATH under the natural action of the group MATH on MATH (coming from the action of MATH on MATH by dilatations). Then it is easy to check that MATH . For MATH we know that MATH . So dilatations in the fibers of the vector bundle MATH act on MATH by the character MATH. By induction on the length of MATH we deduce that MATH. In particular, for any MATH and MATH we have: MATH. The lemma follows. |
math/0104114 | Let us choose a simple coroot MATH and MATH such that MATH. By REF the statement of REF is equivalent to saying that MATH divides MATH. We are going to show that for MATH such that MATH we have MATH. (Recall that we have identified MATH with the algebra of polynomial functions on MATH). This is enough, since the set MATH is a NAME dense subset of the hyperplane MATH. The latter statement is equivalent to saying that MATH for MATH. Let us prove this. We have MATH where MATH, MATH are the dual bases of MATH and MATH respectively. So MATH. We claim that MATH for MATH. Indeed, MATH, hence MATH since MATH is not dominant. (Recall that MATH by the assumption). This proves the lemma. |
math/0104115 | We may assume that the MATH are nonzero. Let MATH be the divisors MATH, MATH. These MATH are degree-MATH divisors whose positive and negative parts MATH each have degree MATH. Set MATH, a nonzero rational function on MATH. If MATH agree on the MATH-th coordinate then MATH is either a pole of both MATH and MATH or a zero of MATH. Let MATH and MATH. Then the negative part of the degree-zero divisor MATH is bounded above by MATH, and thus has degree at most MATH. Thus the positive part of MATH also has degree at most MATH. Hence there are at most MATH choices of MATH for which MATH. Since there are MATH common poles, we deduce that the words associated to MATH have at most MATH common coordinates, as claimed. |
math/0104115 | Let MATH be any other choice, and set MATH. Then MATH is a rational function on MATH with neither pole nor zero at MATH. Thus using MATH instead of MATH in REF multiplies the MATH-th coordinate of every word by the nonzero scalar MATH, for each MATH. Since each coordinate is changed by a permutation of the alphabet MATH, an equivalent code results. |
math/0104115 | Let MATH be the divisor of the function MATH. Then MATH is a rational section of degree MATH of MATH if and only if MATH is a rational section of degree MATH of MATH. This identifies MATH and MATH as sets. Having chosen MATH for MATH, we may choose MATH for MATH. Then REF gives the same coordinates for MATH as an element of MATH that MATH has as an element of MATH. This identifies MATH and MATH as error-correcting codes. |
math/0104115 | We estimate the error in REF using contour integration. By REF and the discussion around REF we have MATH for any MATH. (In fact we obtain REF for all MATH, but we shall soon need to assume MATH.) On the circle MATH we have MATH by REF . We estimate MATH by using another contour integral to express MATH in terms of MATH: For all MATH with MATH we have MATH . Consider first MATH with MATH. For such MATH we obtain MATH by integrating termwise the product of the absolutely convergent series REF for MATH and MATH. For any MATH other than MATH, the integrand extends to a meromorphic function on MATH with simple poles at MATH and a multiple pole at MATH. The contour in REF encloses the poles MATH but not the poles MATH. Thus analytic continuation gives MATH for all MATH, for any contour that encloses MATH but not MATH. Now when MATH the contour in REF encloses MATH but not MATH. Thus we can evaluate the contour integral in REF by starting from REF , adding the residue at MATH, and subtracting the residue at MATH. The former residue is MATH, and the latter is MATH. This proves REF . Thus REF is MATH . We use REF to estimate both parts of this. For the single integral, we find MATH . Thus the single integral is MATH. We shall show that the double integral is exponentially smaller than MATH; this will prove REF . To estimate the integrand, let MATH, so MATH and MATH . Here MATH, so MATH . Thus our proof of REF will be complete once we show MATH or equivalently MATH and this follows from the observation that MATH . It remains to prove REF and to show that the ``main term" in REF is indeed exponentially larger than the ``error term" as long as MATH. By REF , the main term is MATH . Thus strict inequality in the upper bound REF is what we need to show that REF exceeds the ``error term"MATH. The ratio between MATH and the claimed upper bound is MATH . Trying MATH we find that MATH so the upper bound holds for all MATH. Moreover the bound is strict if MATH is a decreasing function of MATH at MATH. We calculate that the logarithmic derivative of MATH at MATH is MATH . This is negative once MATH, so REF is proved. MATH . |
math/0104115 | Estimate REF follows from REF and the bound REF on each term with MATH nontrivial, together with the facts MATH and MATH (see REF ). For the remainder term to be exponentially smaller we must have MATH (from REF ) and MATH . The ratio between the two sides is MATH where again MATH. For all MATH, the product REF exceeds MATH. For MATH the product clearly falls below MATH once MATH is large enough. Thus REF equals MATH for some MATH, with the minimum attained at some MATH; since MATH is a decreasing function of MATH for that MATH, REF holds for all MATH. It is not hard to check that MATH - even the lower bound MATH on REF suffices for this. REF now follows from REF . |
math/0104116 | Set MATH. Let MATH, and let MATH denote the normal closure in MATH of the subgroup generated by the elements MATH with MATH. We begin by proving the claim that MATH . To see this, note that by REF we have MATH . In MATH, the terms of REF commute, and the right-hand side of REF equals MATH which proves the claim. Let MATH denote the MATH-th term in the lower central series of MATH. We now show inductively that for MATH and MATH we have MATH . Note that MATH, and assume that MATH for each MATH and some MATH. Our earlier claim implies that the element MATH is contained in the normal closure MATH in MATH of the group generated by the commutators MATH with MATH. Hence MATH for each MATH and therefore MATH. Similarly, we have that MATH . As MATH, we have that MATH is well-defined by REF . Furthermore, REF holds by REF with MATH. |
math/0104116 | For MATH, the statement is trivially true. Note that MATH defines an idempotent endomorphism of MATH. Also, MATH commutes with conjugation by MATH on MATH. Hence we have MATH . The statement now follows immediately by induction. |
math/0104116 | We first show that MATH fixes MATH inductively. Set MATH for MATH. Assume that MATH fixes MATH. Once we show that MATH fixes MATH, then MATH will fix MATH, as MATH has NAME twist MATH. We prove the claim that if MATH fixes MATH then MATH fixes MATH. Note first that MATH fixes MATH, so we assume inductively that it fixes MATH with MATH. As MATH is abelian over MATH and MATH has NAME twist MATH, we see that MATH . From REF , we therefore have that MATH will fix MATH, since MATH. We conclude that MATH fixes MATH. The first statement of the proposition now follows. Note that MATH for any MATH. Therefore, we see that MATH for any MATH. The last statement of the proposition now follows from the recursive REF of MATH and REF . By REF , the element MATH has the same image under MATH as MATH. Let MATH denote the abelian subextension of MATH generated by roots of cyclotomic MATH-units, and note that MATH can be considered as a homomorphism of MATH CITE. To prove the second statement of the proposition, it suffices to show that MATH generates MATH. Let MATH denote the unique integer with MATH and MATH. Let MATH, MATH, and MATH. Let MATH denote an element of MATH restricting to a generator of the procyclic group MATH CITE. Then MATH topologically generates MATH as a normal subgroup of MATH. Let MATH denote the restriction of an element MATH to MATH. We claim that MATH is also the normal closure in MATH of the procyclic subgroup generated by MATH. If not, then since MATH, we must have MATH for some MATH. Clearly, we would then have MATH contradicting the definition of MATH. Let MATH denote the largest normal subgroup of MATH fixing MATH. We have MATH, and so we need only show that MATH is the normal closure in MATH of the procyclic subgroup generated by MATH. Inductively assuming this is true for MATH, we prove it for MATH. If MATH, then since MATH is abelian, MATH is a product of conjugates of MATH by powers of MATH. From REF , we see that MATH if and only if MATH . Observing REF , we see that any such element is a product of conjugates of MATH by powers of MATH. |
math/0104116 | Since MATH, this follows directly from REF , the surjectivity of MATH under NAME 's conjecture CITE and the last statement of REF . |
math/0104116 | To prove REF , we need only show that, for each MATH, the element MATH is contained in the pro-MATH subgroup generated by the elements MATH. This follows easily from the fact that the group generated by MATH and MATH contains MATH. For any pro-MATH group MATH, let MATH denote the MATH-th term in its descending central MATH-series, and set MATH. Now assume MATH is free pro-MATH on MATH and the MATH. Note that MATH is a free pro-MATH group as a closed subgroup of a free pro-MATH group. Hence it is free on the MATH if and only if the images of the MATH form a minimal generating set of MATH. Fix MATH, and let MATH denote the free pro-MATH subgroup of MATH generated by MATH and MATH. Let MATH denote the normal closure in MATH of the pro-MATH subgroup generated by MATH. By freeness of MATH, we have MATH . Hence we are reduced to showing that the images of the MATH form a minimal generating set of MATH. As MATH is elementary abelian, we have an injection MATH . Finally, as MATH, we obtain minimality, proving REF . As for REF , consider a free presentation of MATH on generators MATH and MATH mapping to MATH and MATH: MATH . By REF , the group MATH has a free subgroup MATH on elements MATH defined as in REF . Since MATH is free on the MATH, the map MATH is an isomorphism. Hence MATH. Thus MATH is isomorphic to MATH, which is a subgroup of the group MATH generated by the image of MATH. Hence, if MATH is nontrivial, then MATH for some MATH. Therefore we have MATH, a contradiction. Hence MATH is trivial, proving REF . |
math/0104116 | For MATH regular, it is well-known that MATH is free pro-MATH on the generators MATH and MATH with MATH and MATH odd. By the construction REF of the MATH and REF , it follows that MATH is freely generated as a pro-MATH group by the MATH with MATH odd MATH. As MATH is pronilpotent, and since by REF the elements MATH and MATH have the same image on the NAME group of the maximal abelian subextension of MATH in MATH, we have that MATH is also free on the MATH. As MATH is a quotient of MATH, the map MATH is surjective. (Note that the surjectivity of MATH does not depend on the choice of elements MATH.) Assuming NAME 's conjecture, we have that MATH, and hence MATH, is injective. Therefore, MATH is an isomorphism and MATH. Finally, the injectivity of MATH plus the surjectivity of MATH yield the surjectivity of MATH CITE. |
math/0104116 | We begin by assuming merely that MATH is injective and MATH is not. Then MATH for some nonzero MATH and MATH. Choose a lift MATH of MATH. Since MATH is injective, the image element MATH is nontrivial. Let MATH be maximal such that MATH fixes MATH. Let MATH be the image of MATH in MATH. By CITE, we have that MATH is surjective, so there exists MATH such that MATH for some minimum possible MATH. Set MATH for MATH lifting MATH. By construction and the injectivity of MATH, there exists MATH maximal such that MATH fixes MATH. By induction, we obtain a sequence of elements MATH, each of which restricts to some multiple of MATH, and corresponding sequences of elements MATH and exponents MATH such that MATH. If MATH is surjective, then each MATH is zero. Since the sequence of numbers MATH is increasing, the sequence MATH has a limit MATH which restricts to MATH. We see that MATH, as MATH will fix MATH for every MATH. This is a contradiction, proving the theorem. Note that, removing the assumption of the surjectivity of MATH, this argument yields that infinitely many of the MATH are non-zero. |
math/0104116 | If MATH, then since MATH is regular, REF b implies that MATH is free on the generators MATH with MATH odd MATH and hence is free on the MATH by REF . That is, MATH is an isomorphism. Since by REF is surjective, REF implies that MATH is an isomorphism. |
math/0104116 | For each odd MATH, let MATH be an element of MATH with maximal possible image under MATH. Let MATH be a homomorphism satisfying MATH for each odd MATH. Assume that MATH is an isomorphism. The surjectivity of MATH forces that MATH for some MATH and MATH for odd MATH. Setting MATH for MATH, we find that the matrix formed by the MATH with MATH and MATH odd MATH is upper triangular and invertible. Then the MATH form a minimal generating set of MATH, and hence the MATH freely generate MATH, as MATH is a free pro-MATH group. |
math/0104116 | The idea of the proof is to show that if MATH or MATH is an isomorphism then MATH is free, contradicting REF 's conjecture that MATH has no free pro-MATH quotient of rank MATH. The state of MATH or MATH being an isomorphism does not depend on the choice of generators MATH by CITE and REF , so we use our previously defined generators from REF (which is possible by REF ). If MATH is an isomorphism, then MATH is as well CITE. Assume that MATH is an isomorphism. This means that the MATH freely generate MATH, so REF implies that the MATH do as well. We clearly have that MATH is generated by MATH and MATH with MATH odd and MATH. As MATH for any MATH, it follows by REF c that MATH is freely generated by these elements, finishing the proof. |
math/0104116 | We remark that the surjectivity, or lack thereof, of MATH and MATH is independent of the choice of the elements MATH. Thus, REF again allows us to use our choice of MATH from REF . Set MATH if MATH is even, and let MATH denote the kernel of MATH on MATH if MATH is odd. Set MATH. Similarly to the proof of REF-REF, we have an isomorphism of abelian groups MATH . We claim that MATH is a MATH-submodule of MATH. (NAME points out that a similar group need not be MATH-stable.) In fact, we have by definition that MATH and MATH . This proves the claim, so MATH is a direct summand of MATH as a MATH-submodule. Therefore, MATH gives rise to an abelian pro-MATH extension MATH of MATH unramified outside MATH with NAME twist MATH and satisfying MATH. The maximal elementary abelian subextension of MATH descends to the desired elementary abelian MATH-extension of MATH. |
math/0104119 | This can be proved directly or from REF . Let MATH and MATH . Then, by the NAME Theorem, the weak-MATH-closure MATH of MATH is the set MATH. By an application of the NAME and NAME 's Theorem REF , p. CASE: MATH meets the weak-MATH-closure of MATH so that we can apply REF . |
math/0104119 | REF implies REF . Clearly MATH is the only weak-MATH-cluster point of MATH in MATH and so MATH weakly. REF implies REF . Assume for some MATH with MATH, we have MATH . Pick MATH with MATH . Then for each MATH the sets MATH, MATH and MATH satisfy the conditions of REF , so we pick MATH with MATH, MATH and MATH contradicting REF . |
math/0104119 | We only need prove REF implies REF . Pick MATH with MATH . Let MATH . Then there exists MATH with MATH . We now argue that since MATH is in the weak-MATH-closure of both MATH and MATH, then there exists MATH with MATH and MATH . Thus MATH . Letting MATH we are done. |
math/0104119 | We start by proving the Theorem in the special case when MATH . Suppose the conclusion is false. Using REF we produce a bounded sequence MATH with MATH but MATH . We can regard MATH as a subspace of MATH. Now let MATH . For fixed MATH, if MATH, then MATH is in the weak-MATH-closure of both MATH and MATH for some MATH . Hence there is MATH such that MATH by an application of REF and so MATH . We conclude that MATH for each MATH . Let us denote by MATH the MATH-projection of MATH onto MATH and let MATH . Let MATH and MATH . Then MATH and MATH . Let MATH and MATH. We shall define inductively a sequence MATH in MATH and a sequence MATH in MATH such that: MATH . Let us suppose MATH and MATH and MATH have been determined and satisfy REF ; if MATH these sets are empty of course. We shall determine MATH and MATH. Let MATH and MATH . If MATH we define MATH by taking MATH to be a norm-preserving extension of MATH if MATH we simply let MATH . Then MATH . Let MATH where MATH and MATH . We next define MATH to the extension of MATH such that MATH . We claim that MATH . In fact if MATH then we can write MATH where MATH and MATH. Then MATH . Now by REF we can define MATH with MATH and MATH for MATH . Let MATH be the restriction of MATH to MATH. Now consider the sets MATH and MATH . Clearly MATH belongs to the weak-MATH-closure of each set. By REF we can find MATH with MATH and so that MATH . It is now clear that REF hold. For REF note that if MATH we have MATH while MATH . Now the proof is completed (for the special case MATH) by observing if MATH is any weak-MATH-cluster point of the sequence MATH then MATH for all MATH . Since MATH this contradicts REF , since MATH does not converge weakly. To treat the general case suppose MATH . Then MATH is reflexive. Consider the induced map MATH; clearly MATH . We next note that MATH embeds into MATH and MATH is CITE p. REF. Hence MATH satisfies a lower bound on MATH . However it is easily seen that MATH coincides with MATH and MATH . |
math/0104119 | Suppose MATH is a bounded sequence such that MATH . Let MATH be any weak-MATH-cluster point of MATH. Then MATH so that MATH . Therefore MATH weak-MATH. But since MATH is a NAME space this implies MATH weakly and we can apply REF . |
math/0104119 | Suppose MATH is any bounded operator. Then we may find a NAME space MATH with MATH and an extension MATH . We claim MATH is an NAME space. Indeed if MATH is a separable subspace of MATH then let MATH be the closure of MATH which is also separable. Then MATH is separable and reflexive so that since MATH is separable, MATH has separable dual. Now it follows from a deep result of CITE (see also CITE) that MATH is weak-MATH-sequentially compact. Hence if MATH is any sequence in MATH there is a subsequence MATH so that MATH is weak-MATH and hence weakly convergent in MATH . Thus MATH is weakly compact by NAME 's theorem (see REF p. REF) and in particular MATH is weakly compact. |
math/0104119 | Let MATH . Then by REF MATH is also a NAME space. We write MATH where MATH is the inclusion map and MATH. Clearly MATH is reflexive. We observe that MATH is one-one and by REF we obtain that MATH is also one-one. Now, since MATH is reflexive, this implies MATH as required. |
math/0104119 | Let MATH. Then since MATH is non-reflexive, the operator MATH has non-empty spectrum and furthermore for any MATH in the boundary MATH there is a sequence MATH with MATH so that MATH . This implies that for MATH we have MATH is not contained in MATH by REF . Then we apply REF and deduce that MATH is non-reflexive. By REF we have that MATH. |
math/0104119 | Suppose MATH fails to be an isomorphism on any subspace isomorphic to MATH . Let MATH be any maximal Abelian subalgebra of MATH . Then it follows from classical results of CITE that MATH is weakly compact on MATH; by NAME 's Theorem REF (see also REF ), MATH is weakly compact. |
math/0104119 | The dual of any MATH-algebra is MATH-embedded (CITE, CITE) and so it follows from the work of CITE that MATH satisfies the hypotheses of REF implies that if MATH is non-reflexive then it contains a complemented isomorphic copy of MATH. Since MATH, there exists in MATH an isomorphic copy of MATH. |
math/0104119 | If MATH and MATH is a corresponding eigenvector then one of the sets MATH or MATH is an invariant set with non-empty interior. Then use REF . |
math/0104119 | Just apply REF . |
math/0104119 | We note that any MATH is in the approximate point spectrum of MATH, that is, there exists a sequence MATH in MATH with MATH and MATH . Now let MATH be defined by MATH . Then MATH so we can apply REF . |
math/0104120 | This follows easily from: MATH which in turns from the estimate MATH . |
math/0104120 | Note that MATH . Hence MATH . Now observe MATH . This completes the proof. |
math/0104120 | Suppose MATH and suppose MATH . Then MATH where MATH . Hence there is a choice of signs MATH with MATH and MATH . Let MATH . Then MATH. Hence MATH . Iterating we get MATH which leads to MATH which implies MATH . |
math/0104120 | CASE: Observe that MATH. Hence MATH . Let MATH be the largest integer so that the right-hand side is at most MATH. Applying REF with MATH we obtain MATH . The result follows, since MATH for MATH. In REF we note first by a result of CITE (see also CITE for a sharper version) there exist universal constants MATH and MATH so that MATH for some MATH . Recall simple properties of the numbers MATH. Clearly, for every MATH, MATH one has MATH and MATH. Thus if MATH then MATH for every MATH. Therefore we may suppose that MATH is a power of two, say MATH, MATH, and MATH. Since MATH for every MATH, we get MATH for every integers MATH, MATH. Then, taking MATH for some MATH we have MATH provided MATH with appropriate absolute constant MATH. Now take MATH to be smallest integer larger than MATH. Then by REF we obtain MATH for MATH and the result follows. |
math/0104120 | Let MATH and let MATH be as in REF . Then by the proposition we have MATH with MATH . Thus by REF we obtain MATH that is, MATH . That implies the result. |
math/0104120 | By the sharp form of NAME 's Theorem REF there is a projection MATH of rank at least MATH so that MATH . Let MATH and introduce an inner-product norm MATH on MATH so that MATH where MATH . If MATH with MATH there exists MATH with MATH . Since MATH we obtain MATH . Hence MATH which implies, for any MATH, MATH . Hence MATH which proves the theorem. |
math/0104120 | We will follow NAME 's argument in CITE, which is itself a refinement of CITE. NAME shows that for a suitable choice of MATH, if MATH then one can find an increasing sequence of subsets MATH so that MATH and if MATH for MATH then there exists MATH so that MATH . It follows that if MATH there exists MATH with MATH and MATH . We now argue by induction that MATH where MATH and MATH. This clearly holds if MATH . Assume it is true for MATH where MATH . Then if MATH we can observe that there exists MATH with MATH . Clearly, MATH . Hence there exists MATH with MATH and MATH . Finally pick MATH so that MATH and MATH . Then MATH and MATH . This establishes the induction. We finally conclude that MATH and this gives the result, as the case of general MATH follows easily. |
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