paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0105233 | We check REF from REF . Since MATH is weakly embeddable in MATH, by REF we have that MATH and MATH, giving REF . Let MATH, MATH, MATH, MATH, MATH, MATH, with MATH. We want to prove that MATH . Let MATH be such that MATH. Then MATH . Since MATH, MATH, and the amalgam is strongly embeddable in MATH, we know that MATH . Since MATH, its MATH-th power also lies in MATH. This is equal to: MATH so MATH, and symmetrically we also have that MATH. Therefore, in MATH, we have MATH since MATH. Symmetrically, in MATH we have MATH . Therefore, MATH, as needed. So MATH satisfies REF , and thus is strongly embeddable in MATH. |
math/0105233 | If MATH is weakly embeddable in MATH, then trivially it is so in MATH, and the condition on MATH and MATH follows from REF . For the converse, assume that MATH is weakly embeddable in MATH. Let MATH, MATH, MATH, MATH, MATH, with MATH, MATH, MATH with MATH as in the statement of REF . We want to prove that for each MATH, MATH . Write MATH, MATH. Let MATH. Note that, as before, MATH . Also, since MATH, so is MATH and symmetrically for MATH. Since the amalgam is weakly embeddable in MATH, we know that for each MATH, MATH . The MATH-th factor on the left hand side of this equation yields MATH since MATH, and symmetrically for the right hand side. Therefore, for each MATH: MATH giving the conditions for weak embeddability in MATH, as desired. |
math/0105233 | By REF , it suffices to verify that MATH where MATH, since the amalgam is embeddable in MATH. Write MATH, MATH, with MATH. Let MATH. We want to show that it lies in MATH. Since MATH, it lies in MATH, because MATH is embeddable in MATH; likewise, we also have that MATH, and therefore lies in MATH, since the amalgam is also embeddable in MATH. Therefore, MATH has exponent MATH modulo MATH in MATH. Since MATH, this means that it is of exponent MATH modulo MATH, so it lies in MATH, as desired. A symmetric argument shows that MATH is contained in MATH, giving the result. The final statement follows because MATH has the descending chain condition, so any filter is principal. |
math/0105233 | If the amalgam is strongly embeddable, then MATH yielding REF . For REF , note that MATH, since MATH is central in MATH. Conversely, we prove that if the amalgam satisfies REF , then it is strongly embeddable. Note that MATH, since MATH is cocentral in MATH. On the other hand, MATH since MATH, and MATH by REF . Now let MATH, MATH, MATH, MATH with MATH for some MATH, MATH. We want to prove that MATH. Write MATH, MATH, with MATH, MATH, MATH. Then MATH . Therefore, if MATH, then so is MATH. By REF , we must have MATH, and MATH, so MATH as well. Therefore, MATH . On the other hand, MATH so we have that MATH. To see that it lies in MATH, note that it lies in MATH, and MATH. |
math/0105233 | Again note that if the amalgam is embeddable at all, then it is strongly embeddable, since MATH is normal in MATH. If the amalgam is strongly embeddable, then REF certainly holds, and we have MATH. Since MATH is central, the latter commutator is trivial, yielding REF . Conversely, assume the amalgam satisfies REF . We have that MATH and together with REF , we get REF from REF . For REF , let MATH, MATH, MATH, MATH with MATH for some MATH, MATH. We want to prove that MATH. But MATH because MATH is central in MATH; and REF says that MATH. So the amalgam satisfies the conditions of REF , and we are done. |
math/0105233 | The equivalence of MATH and MATH follows from REF . Clearly, MATH. To see that MATH, assume that MATH does not hold. First, we note that MATH. For, given MATH, MATH, then MATH, and also MATH because MATH, and MATH. Therefore, if MATH, then there exists MATH. Let MATH be a MATH-overgroup of MATH such that MATH. That MATH exists follows by REF . Let MATH, where MATH is cyclic of order MATH, infinite cyclic if MATH. Then MATH in MATH by REF . Now consider the amalgam MATH. It cannot be weakly embeddable in MATH, because MATH, but MATH. Now assume instead that there exist MATH, with MATH, and MATH with MATH, but for which there exists a MATH-overgroup MATH of MATH with MATH. Let MATH, with MATH cyclic of order MATH. Note that MATH, because the cartesian is isomorphic to MATH . If the amalgam MATH were embeddable into MATH, then we would have MATH, since MATH, so we would have MATH in MATH, which is a contradiction. Therefore, if MATH fails REF , it is not a weak amalgamation base. Finally, we prove REF . Let MATH be an amalgam of MATH groups, where MATH satisfies the conditions given in REF . We prove MATH satisfies the conditions in REF . Let MATH. Since MATH, MATH. Also, MATH for some MATH, MATH, so MATH. Thus, we have that MATH. Therefore, MATH, and symmetrically for elements of MATH. So MATH satisfies REF . For REF , let MATH, MATH, MATH, MATH, MATH such that MATH. By REF here, there must exist MATH, MATH such that MATH, MATH, since it is clear that MATH is an overgroup of MATH where MATH, and MATH is an overgroup of MATH where MATH. Thus, in MATH we have that: MATH which clearly lies in MATH, and is equal to MATH by symmetry. Therefore MATH is strongly embeddable in MATH, proving MATH. |
math/0105233 | Let MATH be a strong MATH base. We may assume that MATH, for otherwise we would have MATH and the statement is trivial. Thus, it will suffice to check REF for MATH, as per REF . Let MATH, MATH, and let MATH. Since MATH is a strong MATH base, either MATH lies in MATH, or there is no MATH-overgroup MATH of MATH with MATH in MATH. If MATH, we are done. Otherwise, there can be no MATH-overgroup MATH of MATH with MATH in MATH, since there is no MATH-overgroup with the property, and MATH. This proves REF for MATH, and since MATH, this implies REF , giving the result. |
math/0105233 | Since MATH is bounded, we may write MATH as a sum of cyclic groups, MATH, with MATH. Assume first that MATH. If MATH is a sum of cyclic groups of order MATH, then it is easy to verify that MATH. Also, fix MATH with MATH, and let MATH. There is a MATH overgroup K of MATH with MATH if and only if MATH. But this occurs if and only if MATH. Thus, MATH satisfies the conditions for being a strong MATH base. Conversely, note that if some MATH, then MATH includes the entire cyclic summand, which cannot all be MATH-th powers unless the summand is trivial. And if for some MATH we have MATH, then the MATH-th powers are MATH-th powers only if MATH, which occurs only if MATH, which is impossible. So all cyclic summands must be of order MATH for REF to be satisfied, yielding the converse. For REF , assume that MATH. Here we need MATH. If some cyclic summand has MATH, then the entire summand is of exponent MATH, which yields a contradiction unless the summand is trivial. And if MATH, the elements annihilated by MATH are the MATH-th powers; they are MATH-th powers only if MATH. But MATH, so this is impossible. Thus, an abelian group cannot be a strong amalgamation base when MATH, unless it is trivial. |
math/0105233 | Note that for MATH, the claim is that an abelian group is a strong MATH-amalgamation base if and only if MATH, which also follows from the fact that we also need MATH. So we may assume MATH. Let MATH, MATH; the third clause of REF from REF is always false in MATH, since we cannot have MATH for any MATH; and the condition on the order of MATH is trivially false, since MATH. So, if MATH is a strong amalgamation base, then we must have MATH. Setting MATH yields that MATH is MATH-divisible. Conversely, if MATH is MATH-divisible, then it is MATH-divisible for all MATH, giving REF , and from there we get REF as stated in REF . |
math/0105233 | It is easy to verify that, in general, if MATH is an amalgamation base (weak, strong, or special), then so are both MATH and MATH. Conversely, suppose that both MATH and MATH are strong amalgamation bases in MATH. Since MATH, and MATH, it follows that MATH satisfies REF from REF . For REF , assume MATH is a prime power, MATH, and let MATH. If MATH, then MATH, since MATH is MATH-divisible. Suppose that there exist a MATH-overgroup MATH of MATH with MATH. Since MATH, we would have MATH, which shows that MATH. Since MATH is a strong base, and MATH is also an overgroup of MATH, we must have MATH; but if both MATH and MATH, then we have MATH, yielding REF , and we are done. If, on the other hand, MATH, then MATH since MATH is MATH-divisible, and the symmetric argument holds. |
math/0105233 | Use REF , with MATH the set of all primes not dividing the exponent of MATH. |
math/0105233 | By REF , we may assume that MATH a prime, or MATH. Necessity follows from REF , so we only need to prove sufficiency. Assume first that MATH. Let MATH be a prime power dividing MATH. Since MATH is MATH-divisible, for any prime MATH, we may assume that MATH, with MATH. Let MATH, and we want to prove that either MATH, or else no MATH-overgroup MATH of MATH has a MATH-th root for MATH. Note that MATH, so MATH. So, if MATH, then MATH, and we are done. Otherwise, there exists MATH with MATH, but MATH, thus showing that MATH satisfies the last clause of REF . Thus MATH is a strong amalgamation base for MATH. Next, suppose that MATH is of exponent REF. Then MATH. If MATH is odd, then MATH is MATH-divisible for any MATH, so REF always holds and we are done. If MATH, then MATH is trivial and MATH, so the condition on the center stated in the theorem is equivalent to MATH. In particular, MATH. Let MATH be a power of MATH (any other prime power will yield that REF holds), and let MATH. If MATH is central, then MATH and the first clause of REF holds. Otherwise, there is some MATH such that MATH; and since MATH, MATH, so the last clause of REF holds instead. Finally, assume that MATH. Note that we may then assume that MATH. If MATH, we may proceed as above. Otherwise, we must have that MATH, so that MATH. Then MATH is exactly the central elements of exponent MATH. Again, if MATH is not central, we get a MATH which does not commute with MATH, to satisfy the last clause of REF ; if MATH is central of exponent MATH, then it lies in MATH and the first clause of REF is satisfied. And if MATH is of order REF, then MATH, so the second clause is satisfied, and we are done. |
math/0105233 | First, assume that for some MATH, all three conditions fail. If MATH, then MATH and MATH. Then we take a sum of two cyclic groups of order MATH, which is a MATH-base (by REF if MATH, or if MATH, because the MATH-parts consist only of abelian groups), but not a MATH base, regardless of MATH. If, on the other hand, MATH, then MATH, and either MATH, or else MATH. If both MATH and MATH are zero, then we have MATH, and the MATH order of MATH is strictly less than the MATH order of MATH, so MATH. We may use REF to find a group MATH which is a MATH-base but not a MATH-base, so not every MATH-base is a MATH-base. If MATH, and both MATH and MATH nonzero, then we may use either REF or REF (depending on whether MATH or not, respectively), to construct a group MATH which is a strong amalgamation base for MATH (and thus also a MATH base), but it is not a strong amalgamation base in MATH, hence also not a MATH amalgamation base. If MATH, then we may use REF to construct a MATH-group which is a strong MATH base (and hence a MATH base, since its MATH-parts are either a strong base or trivial), but not a strong base in MATH, and hence cannot be a MATH-base either. This proves that REF , and REF are necessary. To prove sufficiency, we consider the finite exponent cases first: if both MATH and MATH are nonzero, we may assume they are powers of the same prime MATH; given any prime MATH, REF holds for MATH. If REF also holds for MATH that means that MATH and MATH, so trivially the strong bases in both are the same. If REF holds for MATH, then MATH is the class of all abelian groups of exponent MATH, and MATH of all abelian groups of exponent MATH, and since MATH and all groups are strong bases in both, we again have that every MATH-base is also a MATH-base. Finally, if REF holds for MATH, then we have MATH, and in particular, MATH must be an odd prime. But if MATH is an odd prime and MATH, then every MATH-strong amalgamation base is also a MATH-strong amalgamation base by REF, so it must also be a MATH base by REF . Next assume that MATH. If for any MATH we have that REF holds, then MATH, so MATH, and there is nothing to prove. Otherwise, either REF or REF hold for each prime, so MATH, and again there is nothing to prove. Finally, assume that MATH and MATH. Then REF can never happen, so for each prime either REF or REF hold; and we may assume that both MATH and MATH are power of a prime MATH. For primes other than MATH, REF always holds. For the prime MATH, if REF holds, then we have MATH and MATH, with MATH. The groups in MATH are all abelian, and they are MATH-divisible, so they are MATH-strong bases by REF . And if REF holds, then we again have MATH, and we apply REF to get that every MATH-strong amalgamation base must also by a MATH-amalgamation base. This proves the theorem. |
math/0105233 | First we prove sufficiency: let MATH be a MATH-overgroup of MATH, MATH. We want to prove that MATH. Let MATH, MATH, MATH, such that MATH. We want to prove that MATH. Let MATH, MATH. Clearly, MATH is a MATH-overgroup of MATH where MATH and MATH lie in MATH, so REF cannot hold for MATH, MATH, and MATH. Therefore, REF holds. But then, by REF , we must have, for some MATH, MATH which proves that MATH. Thus, MATH equals its own dominion in MATH, giving sufficiency. For necessity, assume that MATH, and a given MATH, MATH, do not satisfy REF , nor REF . We construct an overgroup MATH, where MATH. For ease, we do this one prime at a time in the case MATH. If MATH, then REF always holds, as mentioned in REF , since the only possible MATH is MATH. If MATH, we replace MATH with MATH, MATH with MATH (MATH, MATH if MATH is odd, MATH if MATH), and replace MATH with MATH, where MATH. We proceed as we did in REF . Let MATH with MATH, MATH infinite cyclic, and let MATH be the least normal subgroup containing MATH and MATH. Since REF fail, we will conclude, as we did in REF , that MATH. We also show that MATH for all MATH. This will show that in MATH, MATH. Proceeding as we did in REF , we set a general element of MATH equal to MATH, and we have MATH, MATH, and MATH. We also get, using REF , and REF in the proof of REF , that they are equal to: MATH . Since we also know that MATH, we must have that MATH or equivalently, that MATH. However, this we know to be impossible, since REF is not satisfied either. Thus, MATH is properly contained in its dominion in MATH. If MATH we are done, by looking at the dominion of MATH in MATH, which contains MATH by construction. If MATH, we adjoin central MATH-th roots MATH and MATH to MATH and MATH, and then convert certain powers of those roots into commutators, as we did in the proof of REF . Neither process will collapse MATH into MATH, since they can all be performed via central amalgams. So again, in the group MATH that we obtain after all the adjunctions are done, we have that MATH lies in the dominion, but its value is the same is MATH. So there is an element in the dominion of MATH which is not in MATH, and so MATH cannot be a special amalgamation base in MATH. For the MATH case, we proceed again as above, and we will always be in the situation analogous to MATH. |
math/0105233 | Let MATH be any overgroup of MATH. If MATH is a special amalgamation base in MATH, then MATH . So MATH is absolutely closed in MATH as well. |
math/0105233 | The result follows because in each case described, MATH will be absolutely closed in MATH: see REF (cyclic groups); REF (squarefree exponent); and REF (abelian group, general case). |
math/0105233 | Let MATH be any MATH overgroup of MATH. Then MATH, MATH, and MATH are all normal in MATH, since they are central subgroups. Let MATH be any of them. Then: MATH since MATH is absolutely closed. But that means that MATH, as desired. |
math/0105233 | Suppose that MATH, and MATH. Let MATH, MATH, and MATH. On the other hand, if MATH, and MATH, then set MATH, MATH, MATH, and MATH. |
math/0105233 | Given MATH, MATH, and MATH, MATH, if either MATH or MATH lies in MATH, then REF is satisfied by REF . If neither lies in MATH, then no MATH-overgroup of MATH has either one as a MATH-th power modulo a commutator, hence it cannot have both of them, proving that either REF or REF holds in REF . |
math/0105233 | Note that if MATH, then so do MATH and MATH. Thus, both commutator brackets lie in the dominion of MATH. Expanding the bracket bilinearly, we have MATH . Since MATH, and MATH, the last three terms on the right hand side lie in MATH. Thus, the left hand side lies in MATH if and only if the MATH lies in MATH, as claimed. |
math/0105233 | In general, if MATH is absolutely closed, then so are MATH and MATH. Conversely, assume both MATH and MATH are absolutely closed in MATH, and let MATH be a MATH-overgroup of MATH. Let MATH, MATH, and MATH, MATH, such that MATH. We want to prove that MATH. We may assume that MATH is a prime power. Write MATH, MATH. If MATH, MATH and MATH both have MATH-th roots in MATH, so there exists MATH such that MATH, MATH. Then MATH so MATH. By the perturbation argument, MATH lies in MATH if and only MATH, so the latter lies in MATH, as desired. A symmetric argument holds if MATH. |
math/0105233 | First, assume that MATH. Let MATH. We want to prove that MATH is absolutely closed in MATH. Denote the generators by MATH for MATH, and MATH for the MATH-th cyclic factor of order MATH. Let MATH be a MATH overgroup of MATH, and assume that MATH, MATH, and some MATH with MATH has MATH. We want to prove that MATH. If MATH, there is nothing to do, since the commutator is trivial, so assume MATH. In particular, MATH. Write MATH, MATH. Their MATH-th powers are trivial, so we must have that MATH, Therefore, each MATH and each MATH are multiples of MATH. Let MATH and MATH be MATH-th roots of those elements (we may assume that MATH and MATH lie in MATH); perturbing MATH and MATH by MATH and MATH, we may assume that MATH and MATH both lie in MATH. But that means that MATH is in the dominion of MATH, which equals MATH (since cyclic groups are absolutely closed), and this proves that the original MATH. Thus MATH is absolutely closed. For the necessity in the case MATH, assume that MATH has two cyclic summands of order less than MATH, and let MATH and MATH be the generators. Since we may adjoin central MATH-th roots to MATH and MATH, it follows that MATH, MATH, and MATH do not satisfy REF or REF . To see they also do not satisfy REF , note that if MATH is congruent, modulo MATH, to a MATH-th power, then MATH must divide both MATH and MATH. So if we could solve the congruence in REF , we would have that MATH divides both MATH and MATH, which makes it impossible for MATH to be congruent to MATH modulo MATH. This proves REF . For REF , note that since MATH is cyclic, it is absolutely closed in MATH, and therefore also in MATH. Conversely, assume that MATH is not cyclic; decompose MATH into a sum of cyclic summands, each of order MATH, MATH, and let MATH and MATH be generators of distinct cyclic summands. Consider MATH, MATH, and MATH. Since MATH, for MATH and MATH to satisfy REF we would need for either of their MATH-th powers to be nontrivial; but since MATH, this is not the case. Since the group is abelian, they also do not satisfy REF . As for REF , once again, if some MATH is congruent to a MATH-th power modulo MATH, then we must have that MATH divides both MATH and MATH; we conclude, as we did in REF , that REF cannot be satisfied, for we would have that MATH and MATH, and yet MATH, which is impossible. This proves necessity. |
math/0105233 | First, assume that there is some prime MATH such that MATH and MATH is not cyclic. Since MATH is the sum of cyclic summands, choose MATH, MATH in MATH which project to generators of distinct cyclic summands. Then note that if for some MATH we have MATH then we must have that both MATH and MATH are multiples of MATH. Since MATH is abelian and MATH, MATH and MATH cannot satisfy REF or REF relative to MATH. But if they satisfied REF , then we would have integers MATH, all multiples of MATH, and MATH. But MATH is a multiple of MATH, as are MATH and MATH, so this is impossible. Therefore, MATH is not absolutely closed in MATH. For sufficiency, let MATH be a prime power, MATH, and MATH. Let MATH be a MATH overgroup of MATH, MATH, MATH, with MATH. We want to prove that MATH. Write MATH, MATH. If MATH is cyclic, then so is MATH (since the latter has the same number of cyclic summands as MATH). Let MATH project to a generator of MATH. Then we may perturb MATH and MATH by elements of MATH, MATH of MATH so that their MATH-th powers lie, modulo MATH, in MATH: write MATH as a MATH for some MATH, and set MATH, and similarly for MATH. Then MATH lies in the dominion of MATH, which is absolutely closed; by the Perturbation Argument, MATH also lies in MATH, and we are done. Finally, assume that MATH, MATH, but MATH does not divide MATH. Then MATH, so there exists MATH such that MATH. In particular, we have MATH . By REF , MATH. However, MATH. Since MATH, MATH, so MATH, and since the latter lies in MATH, we are done. |
math/0105233 | Since MATH is a vector space over MATH, let MATH be elements of MATH which project onto a basis for MATH. Then pick elements MATH whose projections extend MATH into a basis for MATH. Since MATH is of exponent MATH, MATH is a direct summand for MATH. If MATH, then we have a sum of two cyclic groups, both of order less than MATH, so the sum is not absolutely closed, and since MATH has a direct summand which is not absolutely closed, it cannot itself be absolutely closed. This proves necessity. Now assume that MATH. Let MATH be an overgroup of MATH, MATH. If some element MATH of MATH lies in MATH, MATH, then it must be central in MATH, for given any MATH, MATH . So assume that MATH, MATH, and MATH. Then these two elements are central in MATH, so we can perturb MATH and MATH by elements of MATH so that both MATH and MATH are powers of MATH, whose projection generates MATH. Thus, MATH lies in the dominion of MATH, which is absolutely closed. Therefore, MATH, proving sufficiency. |
math/0105233 | We may assume that MATH is a MATH-group, with MATH a prime. If MATH does not divide MATH, then MATH is abelian, and we know that MATH is absolutely closed in MATH if and only if for every prime either MATH or MATH is cyclic or trivial, which is the condition we have above. So we may assume that MATH, in which case MATH. Assume first that MATH satisfies the condition, and let MATH be a prime power dividing MATH. If MATH, then MATH is MATH-divisible, so any pair of elements will satisfy REF . If MATH and MATH, then again, given any pair of elements MATH, find MATH with MATH. Then we have MATH and MATH, which shows that MATH and MATH again satisfy REF . If, on the other hand, MATH, then we know that MATH modulo MATH equals MATH, and is cyclic. So MATH is generated by the image of some element MATH. Since an element of MATH has a MATH-th root modulo a commutator in a MATH-overgroup of MATH if and only if it is central, we either have that MATH and MATH are both central, or else they satisfy REF . If they are both central, we may assume (by perturbing them by commutators) that they are both powers of MATH. But then they satisfy REF as elements of MATH, since a cyclic group is absolutely closed, and therefore they also satisfy REF as elements of MATH. So the condition given is sufficient. Now assume that there is a prime MATH with MATH and MATH neither cyclic nor trivial. First note that if MATH, then MATH, so the quotient would be trivial. Thus, we must have MATH. And since MATH is a MATH-group, the condition given is that MATH is not cyclic nor trivial. Then MATH is not absolutely closed in MATH, since it has a direct summand which is the sum of two cyclic groups, so it cannot be absolutely closed in MATH either. |
math/0105233 | Note that REF from REF do not depend on the value of MATH; so as long as we do not change MATH they remain the same. If MATH lies in MATH, then REF will always be false for any MATH, and any MATH, because the exponent that appears there is always a multiple of MATH, and MATH. So if MATH is absolutely closed in MATH, for any MATH and any pair of elements MATH, either REF or REF hold; and in that case, they also hold in the larger variety MATH. |
math/0105233 | Let MATH be any MATH-overgroup of MATH; let MATH, MATH, MATH be a prime power with MATH, and assume that MATH. We want to prove that MATH as well. If MATH, then note that MATH. Since MATH does not divide MATH, there exist an integer MATH such that MATH. Therefore, MATH. If we raise MATH to the MATH-th power, we are still in MATH; but this equals MATH, which equals MATH, because MATH is of exponent MATH. Therefore, MATH, as desired. If MATH, things are somewhat more complicated. By adding central MATH-th roots MATH and MATH to MATH and MATH, respectively, and replacing MATH with MATH and MATH with MATH, we may assume that MATH. Consider the subgroup MATH generated by MATH, MATH, and MATH. Note that MATH is still in the dominion of MATH in MATH. We claim that MATH. To prove this it suffices to show that MATH is of exponent MATH, as are MATH and MATH for each MATH. To see that MATH is of exponent MATH, note that MATH, since MATH; but also MATH since MATH and MATH-th powers of elements of MATH are central. Therefore, the order of MATH divides MATH. Now let MATH be an arbitrary element. Since MATH, MATH. However, MATH . Thus, MATH is a commutator in MATH of exponent MATH. Since MATH, and MATH, that means that MATH. A symmetric argument shows that MATH as well. Thus, MATH, and it is an overgroup of MATH. Since MATH is absolutely closed in MATH, and MATH lies in the dominion of MATH in MATH, it follows that MATH, which is what we wanted to prove. Therefore, MATH is absolutely closed in MATH, as claimed. |
math/0105233 | We prove necessity first. Assume that there is a prime MATH for which none of them hold. If MATH, we may assume both are prime power of the same prime. Note that for any prime other than that one, REF is satisfied, so MATH and MATH are powers of MATH itself. If MATH, then we must have MATH, and we can use REF to find a MATH-special base, which is not a MATH-special base, hence not a MATH-special base; and the MATH-base is necessarily a MATH-special base as well. If MATH, then MATH, and we may take the sum of two cyclic groups of order MATH. If MATH, then we have that MATH, and either MATH, or MATH. If MATH, then we may use REF if MATH, and REF if MATH, to see that not every MATH-special base is a MATH-special base. If MATH, and MATH, then we use REF or REF to show that not every MATH-absolutely closed group is absolutely closed in MATH. This proves necessity when MATH. If MATH, we have MATH. Since MATH fails all three conditions, MATH, and either MATH does not divide MATH and MATH, in which case we use REF , or else MATH and MATH, in which case we may use REF instead. If MATH and MATH, we may assume that MATH and MATH are prime powers; if they are powers of some prime MATH, then MATH and we use REF again. If MATH, then we proceed as in the case where both MATH and MATH are nonzero. This proves necessity. For sufficiency, let MATH, and assume that for every prime MATH the conditions are all satisfied. If MATH, we may assume they are prime powers. If REF holds, then in both MATH and MATH all groups are absolutely closed and there is nothing to do. If REF holds, then MATH and again the result is trivial. Finally, if REF holds, then we are going from a MATH to a MATH, and everything which is absolutely closed in the former is also absolutely closed in the latter by REF . If MATH, and REF holds for any prime, then MATH and the result is trivial. Otherwise, for each prime either REF or REF holds, so we may write MATH, with MATH, and MATH squarefree. In this case, the result follows from REF . Finally, if MATH and MATH, we may assume MATH and MATH are powers of MATH. Note that REF cannot hold. Given any prime MATH other than MATH, the conditions imply that MATH does not divide MATH. For MATH, if REF holds, then we may write MATH again, with MATH squarefree and relatively prime to MATH, and MATH, The result now follows by applying REF . On the other hand, if REF holds, then all groups in MATH are special amalgamation bases, and since MATH does not divide MATH, every MATH-group in MATH also is a special base, and we are done. |
math/0105233 | One implication is trivial. So assume that MATH is not absolutely closed. Let MATH be an overgroup, and let MATH. We want to show that MATH is not absolutely closed. If MATH there is nothing to do. So assume that MATH. In particular, there exist MATH, MATH, MATH, MATH with MATH, and MATH. Let MATH, MATH. We claim that MATH, MATH, and MATH do not satisfy REF , or REF , relative to MATH and MATH. If REF or REF holds, then no overgroup MATH of MATH has MATH; however, MATH, so neither REF nor REF can hold. That leaves only REF , and we proceed by contradiction. Assume that we have MATH, and integers MATH, with MATH, and such that MATH . By REF , it follows that in MATH we have MATH . Note that since MATH is obtained from MATH by adjoining central elements, all of which are of exponent MATH, MATH and MATH, so the congruences above hold modulo MATH as well. By the description of dominions, we may write MATH for MATH, where MATH, and MATH. Since MATH, we have MATH which means that MATH, which contradicts our choice of MATH and MATH. Therefore, REF cannot be satisfied in MATH either, so MATH is not absolutely closed. |
math/0105235 | Suppose the student starts by picking a set MATH at random, and then keeps picking sets MATH, until there are none left, and making sure never to repeat a set. The sequence MATH is a permutation of the sequence MATH, and it is clear (for reasons of symmetry) that every permutation is equally likely. Since for any MATH, precisely MATH permutations have MATH in the MATH-th position, the lemma is proved. |
math/0105235 | Let MATH be the probability of switching on the MATH-th step or earlier. Then we have the equation: MATH . Since MATH, it is easy to check that MATH. If MATH is the probability of switching on the MATH-th turn, then MATH, and the expected time of switching is MATH the first equality being obtained by telescoping the sum. |
math/0105236 | Note that MATH while MATH . Both probabilities decrease roughly as MATH using the estimates REF . |
math/0105236 | The proof is quite simple. Indeed, since MATH we write MATH . Since MATH we can continue: MATH where we have assumed that MATH is large enough that MATH . |
math/0105236 | If MATH (that is, MATH), then we have MATH . |
math/0105236 | We have MATH . Since MATH, then in the limit MATH this quantity tends to zero. |
math/0105236 | MATH . The quantity MATH is positive because MATH, so we can write MATH where we have assumed that MATH is large enough that MATH . |
math/0105236 | For MATH (or MATH) we would like to reason as follows: MATH . Since the function MATH is unbounded, the weak convergence of the distributions MATH to the stable distribution MATH is not enough to justify the last step equality in the sequence REF above. To justify it we need the following Lemmas: Let MATH be positive independent identically distributed random variables. Let MATH . Then, MATH . Note that MATH . Now, in our case MATH where the inequality follows from REF (and recall that MATH). The probability MATH has the following properties: CASE: MATH for MATH CASE: MATH for MATH CASE: MATH for MATH . MATH for MATH . Follows from the definition of MATH and Property A. There exists a MATH such that MATH for all MATH for some MATH . This follows from Property B, with MATH . If MATH (MATH as in the statement of REF ), then MATH . Follows from REF . MATH . Follows immediately from REF . Now we write: MATH . To analyze the above decomposition, we should first belabor the obvious: MATH . Integration by parts. MATH . The integrand vanishes in the interval by REF . MATH . By REF , MATH . The result follows by integration by parts REF . MATH . Follows from REF . MATH . This follows from the weak convergence of MATH to MATH . The derivation REF is justified. Indeed, if we make the constant MATH above large, we see that the integral of MATH is bounded, hence so is the integral of MATH . Convergence follows from the dominated convergence theorem (or by making MATH small). We have incidentally shown that the density of the stable law decays exponentially as MATH (exact expression can be found in CITE), |
math/0105236 | The proof is immediate. |
math/0105236 | The smallest root MATH satisfies the equation MATH . By REF , MATH . The result follows immediately from REF . |
math/0105236 | This is immediate, since the MATH-th row of MATH is MATH times the MATH-th row of MATH. |
math/0105236 | Immediate, since the bottom eigenvalue of MATH (given in REF ) is zero and the rest are MATH. |
math/0105236 | By inspection - all the MATH minors of MATH are MATH. |
math/0105236 | We know that the MATH from REF . From this the assertion follows immediately. |
math/0105236 | From REF combined with REF , we see that the coefficient of MATH in MATH is given by MATH . The sum is just the MATH-th elementary symmetric function of the MATH which is equal to MATH. The assertion follows. |
math/0105236 | Combining REF we can see that in order for the learner to pick up the correct set with probability MATH, we need to have at least MATH sampling events. Since MATH see REF , we have MATH. Using bounds REF which relate MATH to the harmonic mean, and the weak law of large numbers REF , we obtain estimate REF . This estimate should be understood in the following sense: as MATH, the probability that the ratio MATH deviates from MATH by a constant amount, tends to zero. Therefore, the right hand side of REF behaves like the right hand side of REF with probability which tends to one as MATH tends to infinity. |
math/0105236 | The proof uses the results on the harmonic mean in REF. |
math/0105236 | The proof falls naturally in two parts, both of which require estimates on the characteristic function of MATH. First, we show that we can throw away the tails of characteristic function, and then we estimate convergence in the remaining region. We introduce the following notation: let MATH be the characteristic function of MATH, and let MATH be defined as MATH . In addition, let MATH . Then MATH . REF will be proved in REF. We will also need the following Let MATH be the characteristic function of the stable density, MATH, and MATH . Then MATH . Let us recall that MATH . Now we write MATH . To end the proof of REF we need to estimate how closely MATH is approximated by MATH, for MATH, since by the above estimates we know that MATH (plus lower order terms). It remains only to estimate the difference MATH . From REF we have MATH . |
math/0105236 | By the NAME Lemma, MATH such that MATH for all MATH. Setting MATH, the assertion of the lemma follows immediately. |
math/0105237 | Consider MATH because MATH, due to MATH. Thus it equals (we continue) MATH . |
math/0105237 | As noted above, the vector field MATH corresponds to the NAME - NAME differential for the trivial representation. The NAME derivative MATH corresponds to the NAME - NAME differential for other (nontrivial) representations. Thus the condition that the cobracket on MATH is a cocycle is equivalent to MATH, for the Hamiltonian MATH that defines the corresponding NAME - NAME structure on MATH. On the other hand, we have, by a straightforward calculation, MATH for the bracket MATH defined by MATH. Hence, if MATH, then the derivation property holds. Conversely, if the derivation property holds, then for all MATH, MATH, and this implies MATH by the virtue of the non-degeneracy of the canonical NAME bracket. |
math/0105237 | To prove REF , notice that MATH, thus the equation MATH is equivalent to the MATH-condition (see the proof of REF ). To prove REF , notice that the equation MATH implies the equation MATH for the corresponding Hamiltonian field. Finally, to show that MATH gives exactly the NAME bracket in NAME 's double, we compare the explicit formulas. For MATH we obtain, using the general REF : MATH . Similarly, we obtain MATH . Thus MATH reads MATH and in the case of a purely even MATH it is reduced to MATH which reproduces REF for the NAME bracket. |
math/0105237 | Notice, first, that the bracket of an arbitrary MATH with MATH equals MATH . Now, for an arbitrary MATH we get the function MATH on MATH. Substituting REF into REF, we obtain REF by a direct simplification (which we omit). Notice that the terms containing ``cross" products like MATH remarkably cancel. |
math/0105237 | We have MATH, in the notation of REF . Applying the lemma to the functions MATH and MATH, we obtain the formulas MATH . Here in REF we changed the order and separated the factors MATH. The point is that the correspondence MATH respects the NAME bracket: MATH. In the same way, MATH. Hence MATH, MATH by the virtue of these correspondences, and MATH because MATH and MATH do not contain conjugate variables. Thus, MATH. The identity MATH holds by the construction of MATH. To notice that MATH corresponds to the cobracket in NAME 's double (up to a common sign), it suffices to compare the explicit formula obtained MATH with REF in the even case. |
math/0105237 | Similarly to REF : the condition of a NAME bialgebroid is equivalent to MATH, and MATH. |
math/0105237 | The key statement is that MATH (after the identification of MATH and MATH). The second statement would follow from REF . We have to show that the NAME bracket on MATH induced by the algebroid structure in MATH and specified by MATH is the same as given by the quadratic Hamiltonian REF. To this end, we check the non-vanishing NAME brackets for coordinates. We get MATH and MATH, by the definition of the NAME bracket induced by an algebroid structure and REF . (Here MATH is MATH.) This exactly coincides with the NAME brackets given by the Hamiltonian REF. Hence, MATH. |
math/0105237 | We can lift the scaling transformations to the bundles MATH, MATH, MATH, MATH. If on MATH we have MATH, then for the natural coordinates in the fibers of MATH, MATH, MATH, MATH we obtain, respectively, MATH, MATH, MATH, MATH. It can be checked directly that coordinate changes respect these weights. |
math/0105237 | Consider the Hamiltonian MATH. (Notice that it is odd, as it should be.) We have: MATH (as MATH), MATH (because MATH defines a NAME bracket), and MATH (because MATH is a derivation of the bracket). Thus MATH. Hence, MATH. The part concerning weights follows from the table above. |
math/0105237 | Similar to the proof of REF . Consider the even function MATH. We have: MATH (as MATH), MATH (because MATH defines a NAME bracket), and MATH (because it is equivalent to MATH being a derivation of the bracket MATH, by the virtue of the non-degeneracy of the canonical NAME bracket, compare the proof of REF ). Thus MATH. Hence, MATH. The part concerning weights follows from the table. |
math/0105237 | Consider MATH on the double MATH. REF is clear. REF immediately follows from the coordinate REF . To obtain REF , notice that MATH, hence, because MATH, it follows that MATH, for arbitrary functions MATH. Now suppose that for an arbitrary homological field MATH REF hold. Then, by REF , we conclude that MATH is Hamiltonian, with some odd Hamiltonian MATH, where MATH hence MATH . From REF we conclude that MATH, hence without loss of generality MATH can be set to zero, and that MATH. Taking MATH for arbitrary functions on MATH and applying REF , we deduce that no terms of order MATH can be present in REF and that MATH. This proves the uniqueness of MATH. Similar argument is applicable to an arbitrary homological field MATH with the stated properties (without assuming a given MATH-structure on MATH). We get a field MATH and a bracket MATH. That they make a MATH-structure follows directly. |
math/0105237 | Because MATH is quadratic, MATH is linear and the canonical bracket has degree MATH, the function MATH is quadratic. Now, let us calculate weight. The weight of MATH is the same as that of MATH, MATH, and the weight of MATH is MATH. Hence, because the canonical bracket has zero weight, MATH. |
math/0105237 | The map MATH for an arbitrary NAME superalgebra MATH is defined as follows. If we consider the NAME bracket as the linear map MATH, then the usual adjoint gives the map MATH, and MATH is the composition of MATH with the natural isomorphism MATH and the natural (odd) isomorphism MATH (left multiplication by MATH). In terms of bases we have: MATH hence MATH. (Notice that for the basis MATH right-dual to MATH, MATH .) Now, we have MATH, and we use the canonical identification MATH with MATH. Thus, for the composition, we get MATH. Or: MATH . In other words, MATH restricted to linear functions (with MATH). |
math/0105237 | Immediately follows from REF . For an odd bialgebra, to obtain REF for the odd cobracket in MATH we simply use the remark at the end of proof for REF together with the above formula for the vector field MATH. (Notice that MATH.) |
math/0105237 | Consider MATH and MATH . Their NAME bracket equals MATH . After exchanging the indices MATH and MATH in the second and the fourth terms of the last expression, changing order of factors and calculating signs, it turns out that these terms cancel with the first and the second terms respectively. In the same way, the factors in the fifth term can be rearranged so that it completely cancels signs. Finally: MATH which differs from the required formula only by the notation for the indices. |
math/0105237 | The two terms in REF have the zero canonical NAME bracket because they do not contain conjugate variables. Each term is a NAME tensor by itself. Indeed, the second term is the NAME tensor on MATH. The first term equals MATH, the NAME tensor that gives the NAME - NAME bracket MATH on MATH generated by the NAME bracket in MATH. |
math/0105237 | The explicit formulas defining the odd bialgebra structure in MATH are given above: REF . We shall check uniqueness and the statement concerning the cobracket. Recall that an odd symmetric inner product has the property MATH . In MATH the nonvanishing inner products of the basis vectors are MATH. The MATH-invariance condition for an odd inner product means that MATH for all MATH, or MATH . Suppose MATH with indeterminate coefficients MATH, MATH. Then MATH, MATH. From the MATH-invariance we get: MATH, and also MATH, where we used REF that follow from REF . Hence MATH, MATH, and we recover REF . The uniqueness is proved. Consider now the element MATH. Its coboundary in the NAME algebra cochain complex is the function MATH. Hence, MATH where we omitted a straightforward simplification. Similarly, MATH . Thus, MATH. |
math/0105237 | It is possible to pick bases MATH and MATH, MATH, MATH such that MATH, MATH. By an argument as in the proof of the previous theorem, we immediately recover the commutators of the basis elements in the form REF . Because we are given that MATH is a NAME superalgebra, the vector field MATH defining the brackets REF is homological. Hence, its coefficients obey the relations that can be written as MATH, MATH, MATH, and we arrive at an odd bialgebra structure in MATH. The rest is obvious. Notice that MATH and MATH can be interchanged. |
math/0105237 | Let us denote by MATH local coordinates on MATH, by MATH coordinates in the fiber of MATH, by MATH the corresponding coordinates in the fiber of the dual bundle MATH. Then a change of coordinates in MATH has the form MATH (Here MATH.) Denote the conjugate momenta in MATH and MATH by MATH and MATH, respectively. For MATH we obtain MATH and for MATH we obtain MATH . (We assume that MATH, MATH.) Notice that MATH and MATH hence MATH . We see that it is possible to rewrite the change of coordinates on MATH as MATH which is identical with that for MATH if we substitute MATH, MATH. Hence, we define the desired diffeomorphism MATH as MATH . Consider the symplectic form MATH on MATH. Substituting REF, we get MATH, which is the symplectic form on MATH. Thus, the diffeomorphism REF is a canonical transformation. Notice, finally, that REF define the transformation MATH for all vector bundles. One only has to be careful with the distinction of left and right coordinates: our formulas make use of left coordinates in MATH and the corresponding right coordinates in MATH. (If we want to change, say, a left coordinate MATH into the right coordinate, we have to multiply both MATH and the corresponding momentum by MATH.) Rewritten in left coordinates only, the formula for MATH becomes MATH where bar is used to denote the left coordinates in MATH and the respective momenta. (The natural pairing of MATH and MATH is MATH.) Hence, for MATH we will get MATH . At the right hand side MATH stand for (left) coordinates in MATH and MATH for the corresponding momenta. Notice that the natural isomorphism between MATH and MATH is, in left coordinates, MATH. Hence MATH . Taking composition with MATH and using REF we finally obtain: MATH which concludes the proof, if we identify MATH with MATH with the help of MATH. |
math/0105237 | Let coordinates in MATH be denoted, as above, by MATH and MATH. Let MATH and MATH be the corresponding odd momenta. Then, similarly to REF, we have MATH . Consider now the antidual bundle MATH. Let MATH denote the fiber coordinates in MATH that are right-contragredient to MATH, that is, the form MATH gives the invariant canonical odd pairing of MATH and MATH. Notice that MATH. Then MATH and we get MATH for the changes of coordinates in MATH. Notice that MATH. Using REF and moving MATH to the left, we can rewrite it as MATH . Hence, we can define the desired diffeomorphism between MATH and MATH by the formula MATH . Check the symplectic form: on MATH it is MATH, and substituting MATH, MATH we get MATH, which is exactly the odd symplectic form on MATH. We omit the direct check for MATH, which is completely similar to that in the proof of REF . |
math/0105238 | By REF, the structure MATH is stable and the formula MATH is one-based. Suppose that MATH is not uniformly of NAME type witnessed by some family MATH of subvarieties of MATH. Then for any finite sequence MATH of subgroups of MATH there is some MATH for which the formula MATH holds. By compactness, we can find some elementary extensions MATH and a point MATH satisfying all of these formulas for all choices of finite sequences of subgroups of MATH. As MATH is one-based as a definable group in MATH, MATH is also one-based. Thus, every (parametrically) definable (in MATH) subset of MATH is a finite Boolean combination of cosets of MATH-definable subgroups. Thus, MATH, being a definable subset of MATH, is a finite Boolean combination of cosets of MATH-definable subgroups of MATH. As MATH is NAME closed, this combination is necessarily a finite union. Thus, there are subgroups MATH of MATH and points MATH such that MATH. This contradicts the choice of MATH. Therefore, MATH is uniformly of NAME type. |
math/0105242 | Consider the subset MATH of the space MATH of REF of maps from MATH to MATH which consists of jets MATH (MATH, MATH, MATH) such that MATH is orthogonal to MATH in the sense of the quadratic form MATH on MATH. MATH is a submanifold of MATH of codimension MATH. For an immersion MATH the intersection points of the image of the jet extension MATH of the map MATH with MATH are just those points where REF-form corresponding to the vector field MATH vanishes on MATH. The NAME Transversality Theorem implies that, for a generic change of coordinates, the image of MATH intersects MATH (transversally) at isolated points. Now the fact that after the change of coordinates REF-form corresponding to the vector field under consideration has an algebraically isolated singular point at the origin follows from the Curve Selection Lemma. |
math/0105242 | Let MATH, , MATH be the zeros of the form MATH on MATH. In local coordinates MATH, , MATH, MATH, , MATH centred at the point MATH such that MATH (one can take MATH for MATH) let MATH. Let MATH be a small open ball centred at the point MATH. The degree of the map MATH is equal to the degree of the map MATH given by MATH and thus is equal to the multiplicity MATH of the zero MATH (see, for example, CITE). Now consider the manifold MATH . Since REF-form MATH has no zeros on MATH, the degree of the mapping MATH is equal to zero. This implies MATH . If MATH is a perturbation of REF-form MATH with only nondegenerate zeros on MATH then it has just MATH nondegenerate zeros on MATH. |
math/0105242 | The proof follows the same lines as CITE. Let MATH and MATH. The minors of the matrix MATH vanish at a point MATH if and only if the rank of the matrix MATH is not maximal. The set of MATH matrices with complex entries with rank MATH is an affine algebraic variety of codimension MATH inside the set of all MATH matrices with complex entries. Hence MATH has dimension MATH. This implies that MATH. By CITE this implies that the homological dimension MATH of MATH is equal to MATH. By REF - NAME (see, for example, CITE) we have MATH . Since also MATH, it follows that MATH is a NAME ring. Since MATH is a finite mapping, MATH is a finite MATH-module. Therefore MATH is a NAME module over MATH and MATH . Since MATH is a MATH-dimensional regular ring, it follows from the NAME - NAME formula that the homological dimension of MATH as a MATH-module is REF. This means that there is an exact sequence of MATH-modules MATH . Moreover, MATH must be equal to MATH and REF-th NAME ideal of MATH viewed as a MATH-module is generated by the determinant of MATH. Hence MATH is a hypersurface and MATH is a NAME ring, too. Since MATH is a finite mapping, MATH is also a finite MATH-module. But a finitely generated MATH-module is flat if and only if it is free (see for example . CITE). Therefore it suffices to show that MATH is a free MATH-module. By the NAME - NAME formula we have MATH . Since MATH, it follows that MATH. But this means that MATH is a free MATH-module. |
math/0105242 | Consider again the mapping MATH and let MATH. For each MATH, MATH is a finite dimensional vector space over MATH. Denote its dimension by MATH. Define MATH . By REF and CITE, MATH is a locally constant function of MATH. Now for a point MATH where MATH is a regular value of MATH, MATH by REF . On the other hand MATH. |
math/0105242 | It is convenient to define the family MATH for a larger space of parameters. Let MATH be a real (that is, invariant with respect to the complex conjugation) versal deformation of the MATH, MATH, MATH. Here MATH, , MATH are the coordinates on MATH, MATH; let MATH. Let MATH be the MATH-dimensional affine space with the coordinates MATH, , MATH, let MATH (MATH). Let us denote by the same letter, MATH, the trivial extension MATH of the chosen versal deformation: MATH. Let MATH be a REF-form on MATH defined by MATH . Again by the same symbols, MATH and MATH, we shall denote the complex and the real bifurcation sets of the map MATH (the real one being the intersection of the complex one with the real space MATH), MATH, for MATH real, MATH (in fact MATH is a subvariety of the MATH-dimensional affine space MATH). For a point MATH let MATH where MATH REF are the components of the map MATH: MATH. If MATH and MATH, the variables MATH, , MATH are local coordinates on the (MATH-dimensional) variety MATH in a neighbourhood of the point MATH. For a generic MATH, for all zeros MATH of the restriction of REF-form MATH to the level manifold MATH, one has MATH. Moreover, for a generic MATH which does not possess this property such MATH's form a subset of codimension REF, all zeros MATH of the restriction of REF-form MATH to the level manifold MATH are simple (that is, nondegenerate). For a local system of coordinates MATH, , MATH on a manifold MATH the Hessian MATH of a REF-form MATH equal to MATH in these coordinates is defined as the determinant MATH . (The Hessian is a function on the manifold MATH; it depends on the choice of the coordinates.) For MATH, let MATH . For a point MATH with MATH, let MATH be the Hessian of REF-form MATH in the (local) coordinates MATH, , MATH. The Hessian MATH of REF-form MATH in the coordinates MATH, , MATH is given by the formula: MATH . In the coordinates MATH, , MATH REF-form MATH can be expressed as MATH . One has MATH . By NAME 's rule MATH . Substitution in the form MATH yields MATH . By some abuse of notation we shall denote partial derivatives in the coordinates MATH, , MATH of functions on the manifold MATH by MATH (MATH) (in order to distinguish them from partial derivatives of the corresponding functions on MATH). Therefore the Hessian MATH is the determinant of the matrix MATH . For MATH and MATH we have MATH . Therefore we get MATH . REF now follows from the following two lemmas. Let MATH be an invertible MATH-matrix, MATH a MATH-matrix, MATH a MATH-matrix, and MATH a MATH-matrix. Then MATH . Let MATH denote the MATH identity matrix. Then we have MATH . Let MATH be an invertible MATH-matrix, MATH a MATH-matrix, MATH a MATH-matrix, and MATH a MATH-matrix. Let MATH denote the MATH-th row of MATH and MATH the MATH-th column of MATH. Let MATH be the matrix MATH . Then MATH . By REF MATH . Another application of REF yields MATH . For any MATH and for any singular point MATH of REF-form MATH on the (possibly singular) level variety MATH (that is, for a singular point of MATH or for a zero of REF-form on its smooth part), let MATH be the ideal of MATH generated by the functions MATH REF and by the MATH-minors of the matrix MATH . Let MATH. From the proof of REF it follows that: CASE: MATH; REF if MATH, , MATH are real elements of the ring MATH which are representatives of a basis of the factor-algebra MATH (considered as analytic functions defined in a common neighbourhood of the origin in MATH), then, for MATH small enough, the multigerms of the functions MATH, , MATH are representatives of a basis of the algebra MATH. Thus the algebras MATH form a trivial vector bundle over a neighbourhood of the origin in MATH and the choice of (real) representatives of the elements of the basis of the algebra MATH fixes a trivialization of this bundle compatible with the real structure. As it is usual in similar situations, a quadratic form MATH on the algebra MATH is defined by the formula MATH where MATH is a linear function on the vector space MATH, MATH means the product in the algebra MATH. Let MATH be such that MATH for all singular points MATH of REF-form MATH. In particular this means that MATH is not contained in the discriminant MATH of the deformation and the Hessian MATH is defined in a neighbourhood of each singular point MATH. For a singular point MATH of REF-form MATH, let MATH. For MATH, define MATH as MATH where MATH for MATH from the summand MATH for MATH, MATH for MATH, MATH, the sum is over all simple (that is, nondegenerate) zeros of REF-form MATH which emerge from the (generally speaking, degenerate) zero MATH for generic MATH. The expression under the limit sign is defined for a generic MATH (that is, for a generic MATH the form MATH has only nondegenerate zeros). The facts that: it has a finite limit as MATH tends to zero (as an element of MATH), this limit depends analytically on MATH, the corresponding quadratic form is nondegenerate, it becomes real for MATH real, and in the last case the signature of the corresponding real quadratic form is equal to MATH - follow from the know properties of the similar objects for the smooth case see, for example, REF. At the moment the required linear functions MATH (and thus the quadratic forms MATH) are defined for MATH outside of the discriminant MATH and of the set MATH of those MATH for which on MATH there exists a zero MATH of the form MATH with MATH (both MATH and MATH are hypersurfaces in MATH). We want to show that in fact this family of linear functions has an analytic extension to these two subsets as well. Moreover we should control the extension to the last subset so that the quadratic form MATH does not degenerate there. To show that the constructed family of linear functions MATH has an analytic extension to the set MATH of those MATH for which there exists a zero MATH of the form MATH with MATH, it is sufficient to prove that, for a nondegenerate zero MATH of the form MATH with MATH, the linear function MATH has a finite limit different from zero as MATH tends to MATH, where MATH is a singular point of REF-form MATH, MATH. In this case MATH, the vector spaces MATH and MATH are generated by one element MATH, MATH, and it is sufficient to show that MATH has a finite nonzero limit as MATH. Let MATH be a permutation of the indices MATH, , MATH such that the Jacobian MATH of the functions MATH, , MATH with respect to the variables MATH, , MATH at the point MATH (and thus at all points MATH close to MATH) is different from zero. In this case MATH, , MATH are local coordinates on the manifold MATH at the point MATH and thus at all points MATH close to MATH (on the corresponding level manifold). At a point MATH close to MATH and such that MATH as well (and thus where MATH, , MATH are also local coordinates) the Jacobian of the coordinate change MATH, , MATH, , MATH is equal to MATH. Thus the value of the Hessian of the restriction of the form MATH to the corresponding level manifold in the coordinates MATH, , MATH at the point MATH is equal to MATH and therefore differs from MATH by a nonzero analytic factor. This finishes the proof in this case. To show that the constructed family of linear functions MATH has an analytic extension to the discriminant MATH it is sufficient to prove the following. Let MATH be a point at which the corresponding level variety MATH has a singularity of type MATH, REF-form MATH (as a REF-form on MATH) does not tend to zero, and the zero hyperplane of it is in general position with respect to the tangent cone of the variety MATH at the point MATH. In this case MATH. Let MATH from a neighbourhood of MATH be of the form MATH (that is, MATH differs from MATH only by the first MATH coordinates: the values of the functions MATH, , MATH) and such that MATH. REF-form MATH has two nondegenerate zeros MATH and MATH. It is sufficient to show that the linear function MATH has a finite limit as MATH. Without loss of generality one can suppose that MATH, MATH, MATH is the origin in MATH, MATH, MATH, MATH. The last equation means that MATH, where MATH, that is, MATH. Let MATH. As a basis of the algebra MATH (as a vector space) and thus of the algebra MATH one can take MATH and MATH. For the coordinates of the points MATH and MATH one has MATH, MATH for MATH. Here and further on all series are power series in MATH and thus, for example, MATH means MATH. From REF it follows that MATH and thus MATH. Therefore MATH, MATH. Both expressions have finite limits as MATH tends to zero (for MATH, the terms MATH and MATH sum up to zero). |
math/0105244 | The desired filtration clearly exists and is unique over MATH, and in fact all of the eigenvectors of MATH are defined over MATH or MATH. It suffices to show that for MATH or MATH, the eigenvectors of slope MATH are defined over MATH. (In the case MATH, the space they span is then defined over MATH.) Choose MATH such that MATH; we will suppress MATH in the notation of the semi-valuation MATH. Choose an isobasis MATH of MATH over MATH or MATH with MATH for some MATH, such that MATH. Let MATH be the largest integer such that MATH; we must show that MATH are overconvergent. Pick MATH, so that MATH is an eigenvector of maximum slope. Now choose an isobasis MATH of MATH over MATH on which MATH acts via a matrix MATH for which MATH is integral, and write MATH. By rescaling MATH, we may assume that the MATH are integral. Because MATH is defined over an overconvergent ring, we may choose MATH such that MATH for all MATH. We will show by induction on MATH that MATH. Choose MATH to minimize MATH; the equality MATH implies the equation MATH and hence the inequality MATH . If the minimum on the right side is achieved with MATH, we may replace the term MATH by MATH while preserving the inequality, leading to the conclusion MATH. Otherwise, we have by induction MATH . In either case, the induction follows, and we conclude that the MATH are convergent. Therefore the submodule spanned by MATH is indeed defined over MATH and so is the desired MATH. |
math/0105244 | Let MATH be the distinct slopes of MATH. It suffices to show that for MATH, the eigenvectors of MATH over MATH of slope MATH are defined over MATH. Pick MATH with MATH, choose an isobasis MATH of MATH over MATH on which MATH acts by a matrix MATH such that MATH is integral and congruent to the projection onto the span of MATH modulo MATH. Let MATH be a vector over MATH such that MATH. We show by induction on MATH that MATH is congruent to a vector over MATH modulo MATH, starting with MATH. Suppose that MATH with MATH defined over MATH. From MATH we get MATH . Reducing modulo MATH, we obtain equations of the form MATH for MATH and MATH for MATH. Given that MATH, we conclude that MATH, so that MATH is congruent to a vector over MATH modulo MATH. The induction complete, we conclude that MATH is defined over MATH, as desired. |
math/0105244 | REF allows us to construct an isobasis MATH of MATH such that MATH for some MATH with MATH and MATH. Put MATH; we may assume that MATH for MATH. Now suppose MATH is an eigenvector of MATH of maximum slope. Equating the coefficients of MATH in the equation MATH gives MATH . Let MATH be the smallest integer such that MATH; by the hypothesis that MATH is not isoclinic, we know that MATH. We wish to show that MATH for MATH by descending induction on MATH. For MATH, we have MATH, so MATH. For MATH, we have MATH by descending induction, since MATH for MATH. Now suppose MATH; assuming MATH, we have that MATH . Let MATH, so that MATH. For MATH and MATH a nonnegative integer, let MATH be the smallest real number MATH such that there exists MATH with MATH. Now MATH if and only if MATH for MATH. Thus we have that MATH for MATH and we wish to show that MATH for MATH, by induction on MATH. For MATH, we have MATH and so MATH, yielding MATH. As for MATH, from REF , we have MATH . We conclude that MATH. |
math/0105244 | By REF , the map MATH (respectively, MATH) induced by MATH is also injective. Let MATH be the first term of the descending slope filtration of MATH (respectively, the ascending slope filtration of MATH). We know MATH is spanned by vectors MATH with MATH. On the other hand, MATH; since we cannot have MATH by injectivity, we must have MATH and hence MATH. If MATH has rank greater than REF, then some linear combination of eigenvectors goes to REF, which is forbidden, so MATH has rank REF and we may choose a generator MATH of MATH with MATH. All that remains to be shown is that MATH is nonempty. Pick an isobasis MATH of MATH over MATH, and write MATH. Now MATH, so by REF again, MATH. In particular, there must exist a nonzero linear combination of the MATH over MATH which sums to REF. If MATH, then MATH, and the latter is nonempty, as desired. |
math/0105244 | Suppose MATH is such that MATH, with MATH. Write MATH with MATH; then MATH. If MATH, then this equation implies MATH; otherwise, we must have MATH. |
math/0105244 | By replacing MATH with a suitable exterior power, we may reduce to the case where MATH has rank REF. Let MATH be a generator of MATH, so that MATH for some MATH. By NAME 's trick CITE, we may (after enlarging MATH) choose a basis MATH for MATH over MATH such that MATH with MATH. Write MATH with MATH; then we have MATH for MATH. In particular, MATH whenever MATH. From this we deduce that for any MATH such that MATH, MATH (since the equation MATH has no solutions in MATH for MATH). Moreover, any two nonzero MATH are multiples of one another by elements of MATH (since these are the only solutions of MATH in MATH). Therefore we can write MATH, where MATH is an eigenvector of MATH over MATH and MATH. By CITE, we may write MATH with MATH and MATH; we may shift factors of MATH to or from MATH to ensure that MATH. Now MATH is defined over MATH and is a direct summand of MATH over MATH, hence also over MATH. Additionally, MATH with MATH. Finally, note that since the span of MATH is a direct summand over MATH up to isogeny, MATH acts on it through a matrix invertible over MATH. Thus this span is actually a subcrystal, as desired. |
math/0105244 | The fact that MATH is semistable follows from REF . Now regarding MATH and MATH as actually being defined over MATH, we can construct a basis for MATH which extends to a basis for MATH (since the action of MATH on MATH is invertible up to scalars). The quotient is defined over MATH and is isomorphic to MATH. |
math/0105244 | By taking a suitable exterior power, performing an isogeny and dividing the action of MATH by a scalar, we reduce to the case where MATH has dimension REF. Then the statement follows immediately from the previous lemma and from CITE. |
math/0105244 | Equivalently, we show that for MATH a MATH-crystal over MATH, and MATH with MATH for some MATH, that MATH. Form the dual MATH for some large integer MATH; from MATH we obtain a map MATH such that MATH for all MATH. By REF , there exists a MATH-stable submodule MATH of MATH whose extension to MATH is the kernel of the map MATH to MATH. Let MATH. By REF , the highest slope of MATH is MATH, has multiplicity REF, and has eigenspace MATH. By REF , MATH is semistable, and the slopes of MATH are all less than MATH, so REF gives a direct sum decomposition MATH. However, REF forces MATH to map to REF under MATH. Since MATH is injective on MATH, MATH. Thus MATH is one-dimensional and constant over MATH, and MATH, as desired. |
math/0105244 | We first show that if MATH is a MATH-crystal all of whose slopes are equal to one of MATH, where MATH, and MATH is an eigenvector MATH over MATH with slope MATH and MATH, then MATH is defined over MATH. Construct the dual MATH of MATH, with slopes MATH. Let MATH be the map induced by MATH, let MATH be the kernel of MATH, let MATH, and let MATH be the top eigenspace of MATH, which has rank REF. Then MATH is isoclinic of slope MATH, and so is potentially semistable, in fact potentially constant, by CITE. Take a finite extension MATH of MATH over which MATH is constant. Take MATH with MATH for some MATH; then we can extend MATH to a basis MATH of MATH with MATH for some MATH with MATH. Applying MATH to this equation gives MATH . This equation uniquely determines MATH: it must have the form MATH, where MATH satisfies and MATH. On the other hand, the equation MATH has a unique solution with MATH, and differentiating gives MATH. Thus we must have MATH and so MATH, whence MATH is an eigenvector of MATH. The upshot is that MATH has a complement in MATH over MATH, so MATH maps into MATH and MATH is defined over MATH. Suppose we are given a morphism MATH over MATH between two MATH-crystals MATH and MATH over MATH with at most two distinct slopes each. If the slopes of MATH and MATH are the same, then MATH has three distinct slopes and MATH corresponds to an eigenvector of the middle slope; by the result of the previous paragraph, MATH is then defined over MATH. Now suppose the slopes of MATH are not the same as the slopes of MATH. Then the image of MATH must be an isoclinic sub-MATH-crystal of MATH; again by CITE, the eigenvectors in the subcrystal are defined over a finite separable extension MATH of MATH. Thus over MATH, MATH maps MATH to a direct sum of trivial crystals. Composing with the projection onto one factor gives a map from MATH to one trivial crystal, which corresponds to an eigenvector of MATH over MATH. By the first claim above, this eigenvector is defined over MATH; after repeating for each factor in the direct sum, we conclude that MATH is defined over MATH. |
math/0105244 | We induct on the dimension of MATH. Suppose MATH becomes constant over MATH. Let MATH be the highest slope of MATH, and let MATH be a dual of MATH. Then the eigenvectors of MATH of lowest slope MATH are defined over MATH. Let MATH be one of these eigenvectors, with MATH; then MATH corresponds to a map MATH such that MATH. By REF , MATH has highest slope MATH with multiplicity REF and eigenvector defined over MATH. By the induction hypothesis, MATH is unipotent over MATH, as then is MATH. |
math/0105244 | Define the pushforward map MATH by applying NAME duality MATH to the transpose of MATH. Let MATH denote the fundamental class on MATH. For arbitrary classes MATH and MATH on MATH of complementary dimension, we have MATH . On the other hand, MATH, the fundamental class on MATH. On the other hand, MATH is NAME dual to the cycle class of a point, whose pullback is NAME dual to the class of the preimage of that point (assuming the point is chosen in the dense open set on which MATH is finite). Thus MATH, where MATH is the generic degree of MATH. In particular, MATH, so MATH is injective, as then must be MATH, and the desired projector is MATH. (For the relevant inputs into this proof related to crystalline cohomology, see CITE, specifically REF for cycle classes and REF for NAME duality.) |
math/0105244 | We first consider the case where MATH is projective. In this case, we may extend MATH to a scheme MATH of finite type over MATH by choosing an embedding of MATH in a projective space over MATH and taking its NAME closure in MATH. Now let MATH be as in REF . Then MATH is isomorphic over MATH to the log-crystal MATH, hence is semistable. Now the surjection MATH induces a map MATH of crystals over MATH, which gives an injection MATH of MATH-crystals over MATH by the previous lemma. Moreover, the projector constructed in the proof of the lemma is an endomorphism of MATH over MATH; by REF , this endomorphism is actually defined over MATH, as then is its image MATH. Thus MATH is semistable, and in particular is also overconvergent; since MATH is a finite extension of MATH, we conclude that MATH is also overconvergent, and is potentially semistable. To handle the case of MATH arbitrary, recall that by NAME 's Lemma, there exists MATH projective such that MATH is a surjective birational morphism. By the same argument as in the previous paragraph, but applied to the map MATH, we conclude that the latter being potentially semistable implies the same for the former. |
math/0105245 | The existence of a MATH-Brinkhuis triple pair implies the inequality MATH for any MATH, because each square-free word of length MATH yields MATH different square-free words of length MATH. This means MATH for any MATH, and hence MATH establishing the lower bound. |
math/0105245 | The proof proceeds as in REF above, with MATH replaced by MATH. |
math/0105245 | As MATH and MATH, the words MATH and MATH start with letters MATH and MATH, respectively. If MATH, then MATH contains the square MATH, so MATH. NAME of the composed words MATH and MATH, MATH, implies that the words MATH have to end on MATH, because MATH is the only word in MATH such that MATH and MATH are both square-free. This in turn implies that all words in MATH and MATH end on MATH and MATH, respectively. Now, square-freeness of the composed words MATH and MATH implies that the first three letter of MATH, for any MATH, have to be MATH, because this is the only word MATH in such that MATH and MATH are both square-free. For MATH and MATH, no such words exist, and the only possibility for MATH would be MATH which is not square-free. For MATH, the square-free word MATH starts with MATH and ends on MATH, but MATH contains the square of MATH. |
math/0105245 | From REF , we know that MATH and MATH starts with MATH and ends on MATH. There are now two choices for the forth letter. Let us consider the case that MATH starts with MATH. Then MATH starts with MATH. Now, from REF , MATH ends on MATH, and square-freeness of MATH implies that MATH starts with MATH, and hence with MATH. Now MATH starts with MATH and MATH with MATH. From square-freeness of MATH and MATH, we can rule out MATH ends on MATH, because both possible extension MATH and MATH result in squares. Hence MATH ends on MATH and, from square-freeness of MATH, it has to end on MATH, and thus on MATH. Consider now the second possibility, that is, MATH starts with MATH. Necessarily, it then starts with MATH. As MATH ends on MATH, square-freeness of MATH implies that MATH starts with MATH. Then MATH starts with MATH and MATH with MATH. NAME of MATH and MATH rules out an ending MATH for MATH, as the only possible extension MATH and MATH both result in squares. Hence MATH ends on MATH, and, from square-freeness of MATH, actually has to end on MATH. Now, in both cases it is obviously impossible to find square-free words of length MATH that satisfy these conditions. For the first case, the one choice left for MATH is MATH, which contains the square of MATH. In the second case, the only word for MATH that satisfies the conditions is MATH. In this case, MATH contains the square of MATH. |
math/0105245 | The proof that these are indeed special NAME triples consist of checking the conditions of REF explicitly. This has to be done by computer, as the number of symmetry-inequivalent composed words of length MATH that have to be checked for square-freeness is MATH, which gives MATH words of length MATH for MATH. CITE program brinkhuistriples.m that performs these checks accompanies this paper. This check is independent of the construction algorithm used to find the optimal triples. In order to show that these triples are indeed optimal, one has to go through the algorithm outlined above. This has been done, giving the results of REF . |
math/0105245 | This can be checked by computer. The number of square-free words of length MATH is MATH. However, we do not need to check all MATH possibilities. Without loss of generality, we may restrict one of the four words to start with the letters MATH, leaving only MATH choices for this word. Furthermore, the two words in MATH may be interchanged, as well as the other two words; so it is sufficient to consider one order of words in both cases. No MATH-Brinkhuis MATH-triple was found for MATH. |
math/0105246 | The proof is almost identical to the proof of the special case in CITE, so we will omit some details. For any random variable MATH, we abuse notation slightly and denote by MATH the random variable MATH where MATH, MATH, and MATH are independent, with MATH and MATH; thus MATH is a random variable with the distribution MATH. By conditioning on MATH, we obtain the fundamental relation MATH and thus the estimate MATH . To complete the proof, we give a series of lemmas. For any real numbers MATH and MATH, the random variable MATH defined by REF satisfies MATH . This follows by REF, using little more than the fact that MATH is convex with MATH on MATH. For any random variable MATH and real MATH, we have MATH. REF yields MATH . Returning to our sequence MATH, the preceding lemma applies to all elements except MATH, that is, MATH which yields the case MATH of REF . We improve the exponent by induction, using REF. Let MATH. If MATH, MATH, then MATH . By REF and the hypothesis, MATH and the result follows by evaluating the beta integral. In particular, using REF yields MATH . This proves REF for MATH, with MATH. Since MATH, for any MATH we trivially have MATH, which by REF establishes REF for all MATH with MATH; applying REF again, we obtain REF for all MATH. Somewhat better numerical bounds are obtained for MATH by taking a geometric average between the cases MATH and MATH; this yields MATH, MATH. In particular, we have MATH, and thus, by REF , MATH. Let MATH. If MATH, MATH, then MATH . This is similar to the proof of REF , substituting the hypothesis (and the trivial MATH) into REF, but the estimate of the integral is slightly more complicated; for details see CITE. REF completes, by induction, the proof of REF and the estimate REF. The bound for MATH obtained above and REF now yield (using Maple) first MATH and then MATH. These bounds and REF further yield MATH for integers MATH; again see CITE for details. To obtain REF if MATH, let MATH. Then, by REF, provided MATH, MATH . The case MATH follows similarly from REF, which completes the proof of REF . |
math/0105246 | REF shows, in particular, that as soon as MATH, MATH and thus MATH is integrable. This implies, as is well-known (see for example . CITE) that MATH has a bounded continuous density MATH given by the NAME inversion formula MATH . Moreover, using REF with MATH, we see that MATH is also integrable when MATH, which by a standard argument shows that MATH is MATH times differentiable, with MATH and thus MATH where the latter integral can be estimated using REF with MATH. The argument above yields the bound MATH . To obtain better numerical bounds we combine REF for MATH, MATH, MATH, MATH, MATH, MATH and REF (for MATH in different intervals; see CITE for details); this yields, provided MATH, MATH; similarly, invoking also REF with MATH, MATH for MATH. |
math/0105246 | By the NAME inversion REF , MATH . In order to estimate the right hand side, note that for any random variables MATH and MATH, MATH since the characteristic functions here depend on the marginal distributions only, this and REF yield MATH . In particular, with MATH, MATH . Further, for any MATH and MATH, REF yields the estimate MATH . Consequently, for any MATH, MATH . For given MATH and MATH, the optimal choice here is MATH, giving the bound MATH . With REF and the estimate REF, this yields REF. Choosing MATH and evaluating the constants numerically, using MATH, we obtain REF. To obtain the final estimate, we use REF and observe that, for MATH, MATH which by REF yields that for MATH, MATH . Choosing the optimal MATH, we find REF [with the constant MATH multiplying MATH], at least when MATH. For MATH, REF follows trivially from REF, since the right hand side of REF then is larger than MATH. |
math/0105246 | For any MATH, MATH where MATH is estimated in REF . The final integral can be estimated by NAME 's inequality: for any MATH where MATH is the moment generating function of MATH. CITE proved that MATH is finite for all MATH; thus MATH for every MATH, and the first claim follows by REF . For REF we choose MATH, for which it will be shown in REF below that MATH, and thus REF implies MATH. Denoting the right hand side of REF by MATH, we thus obtain from REF, observing that MATH, MATH . We optimize by taking MATH and obtain the following bound (for MATH, so that MATH; smaller MATH are trivial since MATH): MATH . |
math/0105246 | If MATH for MATH, then by REF, for MATH, MATH . Hence, REF holds with MATH if (and only if) MATH . Similarly, REF holds with MATH for MATH (respectively, for MATH) if REF holds for MATH (respectively, for MATH). Clearly, MATH decreases as MATH increases, and thus if some MATH satisfies REF, then so does any larger MATH. Following NAME, we argue differently for small and large MATH in order to find a MATH satisfying REF. For small MATH we use a NAME expansion. By straightforward differentiations, MATH . We write the last formula as MATH and note that MATH and MATH, where MATH . Consider first MATH. By NAME 's formula, for MATH, MATH so REF is satisfied for MATH provided MATH . If MATH, we find MATH while if MATH, we find MATH . For MATH, in either case, because MATH, MATH and thus MATH . It is readily checked that this is less than MATH so that REF holds, for MATH and MATH. For larger MATH, we begin by another crude estimate. Let MATH be uniformly distributed on MATH. Then, by MATH and symmetry, MATH . Note that that MATH, too, decreases if MATH is increased. Taking the logarithmic derivative, we find for MATH, MATH . For MATH, this is evidently positive, and thus MATH then is increasing. Hence, if MATH, then MATH . For smaller MATH, we take MATH, and check numerically that MATH. Moreover, MATH and further, if MATH, MATH and thus MATH . Hence, REF shows that MATH is decreasing on MATH, and thus MATH . Finally, MATH . Combining these estimates we find that if MATH, then MATH whenever MATH, while MATH for MATH, and thus REF holds when MATH. We have also shown that MATH will do for MATH and MATH; since MATH is increasing for MATH, and thus less than MATH for MATH but larger than MATH for MATH, REF for MATH follows. For MATH, we again use NAME 's formula for small MATH; arguing as above we see that REF holds for MATH provided MATH . It is easily checked numerically that MATH. It follows that MATH . Hence, REF holds and REF is satisfied for MATH provided MATH . It is readily checked that this holds for MATH and MATH. For larger MATH we argue as follows. The function MATH satisfies MATH . Hence MATH is convex, and since MATH, MATH . Consequently, if MATH and MATH, then MATH and thus MATH. Choosing MATH, this shows MATH for MATH, while MATH for MATH by the preceding case. This completes the proof of both theorems. |
math/0105246 | Since a moment generating function is convex and MATH, MATH is increasing on MATH. Moreover, MATH is decreasing on MATH. Hence, if MATH, the integrand in REF with MATH is for MATH at least MATH and the same holds for MATH by symmetry. Consequently, MATH . Let MATH be a constant to be determined later and choose MATH. Then MATH and thus by REF, for MATH, MATH . If MATH, there thus exists MATH such that for MATH, MATH . Given MATH, let MATH and define inductively MATH, MATH. Let MATH be the smallest integer with MATH. Then MATH, MATH, and thus MATH . It remains to estimate MATH from below. Since MATH is increasing, MATH and thus MATH . Consequently, MATH . We choose MATH, which maximizes MATH. Then MATH, we may choose MATH so small that MATH, and the result follows. |
math/0105246 | For MATH, by REF , MATH and the result follows by taking MATH. |
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