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math/0005119
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Induction on MATH, where MATH for a root MATH. It is crucial that MATH is simply laced.
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math/0005124
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We first compute the shifts MATH for the orbifold MATH associated to a conjugacy class MATH in MATH. Consider the typical class containing MATH where MATH. Recall from the previous section that a fixed point in MATH by the action of MATH is of the form MATH where MATH. Since the calculation can be done locally, we will assume that we take local coordinates MATH near a point MATH such that the action is given by MATH . Equivalently, MATH is locally given by the diagonal matrix MATH where MATH. Then on MATH near MATH, MATH is given by a block diagonal matrix with blocks of the form MATH . The characteristic polynomial of this matrix is MATH, hence it has eigenvalues MATH . Notice that MATH if and only MATH. So the shift for the component of MATH containing MATH is given by MATH . Here we have assumed that MATH lies in the component MATH, and MATH is the shift for the component MATH. For a general conjugacy class containing an element MATH of type MATH where MATH, the description of the fixed-point set MATH given in REF implies that the components for MATH can be listed as MATH where MATH satisfies MATH. Then the shift for the component MATH is given by MATH . By using REF we have MATH . Namely we have proved that MATH which implies immediately the theorem by means of REF .
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math/0005130
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The standard reals MATH are a subset of some hyperfinite set MATH of nonstandard reals. It is a well - known (classical) fact that for every finite set of reals MATH, and every positive MATH there is a natural number MATH such that for all MATH, MATH. So by transfer of the classical fact, setting MATH infinitesimal, we get a MATH such that for all MATH, MATH is infinitesimal.
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math/0005130
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If MATH is a NAME, then so is MATH (since NAME - sequences converge pointwise). Clearly, MATH is a closed subspace of MATH. And any normed (and complete) vectorspace can be factored by a closed subspace, resulting in a normed (and complete, respectively,) vectorspace. If MATH is complete, then clearly so is MATH (otherwise take any non - converging NAME - sequence MATH in MATH, and map it to the sequence MATH in MATH).
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math/0005130
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CASE: closed: Assume MATH such that MATH, and choose a MATH. Let MATH be such that MATH. MATH. If MATH (for a suitable MATH) then MATH and MATH (since MATH), so MATH. CASE: invariant: Assume MATH. Then MATH. CASE: maximal: If MATH is continuous on MATH and MATH, then by REF is an element of MATH.
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math/0005130
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See CITE.
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math/0005130
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The quotient MATH of the NAME MATH is a NAME, and so is its isometric image. The rest is clear.
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math/0005130
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CASE: Assume, MATH, and fix MATH. Then there is a MATH such that for all positive reals MATH, MATH. Assume that MATH is a representant of the quotient MATH, and that MATH. Then MATH, and therefore MATH in MATH, so MATH and MATH. CASE: this follows from REF . CASE: Assume that MATH, and MATH in MATH. Let MATH and MATH be representants of MATH and MATH, respectively, Assume MATH is a positive infinitesimal. Then MATH, which is smaller than every positive standard MATH, since MATH gets arbitrary small, MATH, and MATH is infinitesimal. To show the invariance, assume that MATH, and let MATH be a representant. Let MATH be a positive standard real, and MATH is a positive infinitesimal. MATH, which is infinitesimal. So MATH, and therefore MATH.
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math/0005130
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To see that MATH, it is enough to show that for all positive (standard) reals MATH, MATH is infinitesimal. For any standard reals MATH, MATH. MATH and MATH are finite, and MATH and MATH are infinitesimal. Therefore MATH is infinitesimal. To see that MATH, it is enough to note that MATH is infinitesimal, but MATH is not.
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math/0005130
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This follows from REF , and the fact that MATH for all MATH.
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math/0005130
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By definition, MATH iff for all reals MATH there is a real MATH such that MATH for all reals MATH. Let MATH correspond to MATH. Assume that MATH is an infinitesimal nonstandard real. We want to show that for an arbitrary fixed positive real MATH, MATH. Let MATH be the real corresponding to MATH as above. MATH corresponds to a sequence of reals MATH REF such that MATH is in the ultrafilter. MATH corresponds to the sequence MATH. If MATH is in MATH, then MATH, and MATH is a filter - set, therefore MATH.
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math/0005130
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Define MATH by case distinction: MATH. Then MATH.
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math/0005130
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CASE: By transfer of REF we get: MATH, which is infinitesimal for infinitesimal MATH. CASE: For every standard MATH and infinitesimal MATH, we have MATH. Therefore transfer of REF implies that MATH for infinitesimal MATH. Since MATH was arbitrary, MATH is infinitesimal. CASE: By transfer of the definitions of MATH and MATH, we get the following: For all positive nonstandard reals MATH there is a positive nonstandard real MATH such that MATH for all MATH. Let MATH be a fixed positive infinitesimal, and choose an infinitesimal MATH for the appropriate MATH. Then MATH, and by REF, therefore MATH. CASE: Since MATH, MATH for all infinitesimal MATH, and the same applies to MATH and MATH. So we have MATH. So we just have to show that MATH is infinitesimal for infinitesimal MATH. Transfer of REF shows that MATH for all positive MATH and MATH. Now apply this to MATH.
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math/0005130
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Assume MATH has a representant MATH as in the lemma. We have to show that MATH and MATH, that is, we fix a (standard) real MATH and have to find a MATH such that MATH for all MATH. By REF , MATH is infinitesimal for infinitesimal MATH. Note that MATH is an internal function, since it is element of MATH. Therefore the set MATH is internal as well, and we can apply the spillover principle REF: Assume toward a contradiction that for all standard MATH there was a standard MATH such that MATH, that is, there are arbitrary small standard reals in MATH. Then there is a infinitesimal real in MATH as well, a contradiction. Therefore MATH. Assume on the other hand that MATH, MATH are elements of MATH such that MATH. We want to show that MATH and MATH for some representant MATH of MATH. Let MATH be arbitrary representants of MATH and MATH, respectively, (According to REF , we cannot hope that MATH already has the required properties.) For all nonstandard natural numbers MATH define MATH to be the set if nonstandard natural numbers MATH such that MATH. For all standard natural numbers MATH, the set MATH is not empty, since MATH. The set MATH is internal, and contains all standard natural number, therefore it contains an infinite natural number as well (according to the spillover principle). Let MATH be such an infinite number, and MATH an element of MATH. Now we apply the transfer of REF, setting MATH. So we get a MATH in MATH such that MATH, which is infinitesimal, since MATH and MATH is infinitesimal. Therefore MATH is a representant of MATH as well. Also, MATH, so MATH is infinitesimal, therefore MATH as well.
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math/0005130
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Assume MATH. Since MATH is a closed subspace of MATH and MATH dense in MATH, there must be a MATH which is element of MATH but not MATH. We chose a representant MATH of MATH such that MATH, MATH. Since MATH is an element of the ultraproduct, MATH is an equivalence class of a sequence MATH REF such that for all MATH, MATH. But then MATH (according to REF), so MATH, a contradiction.
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math/0005131
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Any two maximal chains MATH, MATH can be completed to maximal chains in MATH by including the elements of the same maximal chains in MATH and MATH (where we first adjoin a MATH and a MATH to MATH if not already present). Then we use the fact that MATH.
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math/0005131
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Assume that MATH and MATH. Then we claim that MATH and MATH. For, MATH and MATH . Similarly, MATH. It follows that if MATH, then we must have some covering MATH such that MATH and MATH. Therefore, MATH, MATH, MATH, and MATH cannot all be elements of the same chain MATH.
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math/0005131
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Let MATH be a chain, and let coverings MATH be indexed by MATH, such that MATH implies MATH. Then, for any arbitrary MATH, consider the chain MATH and the chain MATH. If we take any refinement of the first of these chains to a maximal chain MATH, we can find a refinement of the second chain to a maximal chain MATH, by letting MATH consist of the elements of MATH less than or equal to MATH, the lattice elements MATH for MATH such that MATH, and the elements of MATH greater than or equal to MATH. By the modular law, the coverings in MATH between MATH and MATH correspond in a one-to-one fashion with the coverings in MATH between MATH and MATH, and corresponding coverings are projectively equivalent. Since MATH, and that number is finite, we must have MATH, proving that MATH is upper regular. Lower regularity is proved similarly.
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math/0005131
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CASE: MATH. CASE: MATH by REF . If, in addition, MATH, then MATH which implies MATH. Thus, MATH. On the other hand, if MATH, then MATH by REF , and MATH. Thus, MATH.
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math/0005131
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It suffices to prove that if MATH, then for any such MATH and MATH, we have MATH. MATH implies MATH, so we have MATH. Thus, there exists MATH such that MATH. Then we have MATH and MATH, whence MATH.
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math/0005131
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This follows from REF and from the fact that MATH iff MATH.
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math/0005131
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Let MATH. If MATH is principal, say MATH, then MATH is also principal. On the other hand, if MATH is principal, say MATH, then let MATH be such that MATH, and let MATH. We cannot have MATH because both MATH and MATH belong to MATH. Thus, by modularity, MATH. But, this implies that MATH. Thus, the set of atomic maximal based filters is closed under projective equivalence, and by REF , the same is true of the set of atomic filter coverings. If MATH and MATH is strictly meet-irreducible, then MATH. However, MATH is the unique cover of MATH and is principal. It follows that MATH, and MATH is atomic.
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math/0005131
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Let MATH. Then MATH where MATH, because MATH is atomic. The set of elements MATH such that MATH and MATH is closed under joins of chains, by meet-continuity. Then by NAME 's Lemma, MATH is nonempty. Thus, MATH is nonempty. It is easy to see that because MATH is atomic, MATH consists of strictly meet-irreducible elements.
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math/0005131
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Let MATH. If MATH is quasi-atomic, then MATH contains an element MATH such that MATH. Consider the element MATH. If MATH, then MATH, because by modularity, MATH. Thus, MATH, and MATH is quasi-atomic because if it were atomic, then MATH would also be atomic by REF . On the other hand, if MATH is quasi-atomic, then there is an element MATH such that MATH implies MATH. Let MATH be such that MATH, and let MATH. We have MATH, so MATH. If MATH, then by modularity, MATH. But, MATH because if MATH, then MATH. Thus, MATH and MATH. It follows that MATH is atomic or quasi-atomic, but MATH cannot be atomic, because then MATH would also be atomic by REF . Now, if MATH and MATH is meet-irreducible, but not strictly meet-irreducible, we have MATH. However, MATH is the unique cover of MATH in MATH. Thus, MATH, which is quasi-atomic.
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math/0005131
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Similar to the proof of REF . Instead of MATH, we have a MATH such that MATH. The set of elements MATH such that MATH and MATH is again closed under joins of chains by meet-continuity, and nonempty by NAME 's Lemma. Thus MATH is nonempty, and so is MATH by REF .
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math/0005131
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The sets of aomic and quasi-atomic filter coverings are closed under projective equivalence. Since the set of anomalous filter coverings comprises the rest of the filter coverings, it is also closed under projective equivalence. If MATH and MATH, then clearly MATH is meet-irreducible. Thus, if MATH is anomalous, MATH must be empty by REF . Finally, if MATH is finitely decomposable, then MATH where the MATH are meet-irreducible. If MATH, then MATH also belongs to MATH for some MATH, because MATH is closed under finite meets. This would imply that MATH was atomic or quasi-atomic.
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math/0005131
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For all MATH, MATH. It follows that MATH and hence, MATH. We must show that MATH, which we will show by showing that if MATH, then we cannot have MATH, or in other words, we cannot have MATH for any MATH-tuple MATH of elements of MATH, for any cardinal number MATH, where MATH runs through ordinals less than MATH. Assume the contrary, where MATH is the least cardinal possible. By modularity, and the fact that MATH is atomic, we can assume without loss of generality that MATH for each MATH. For each ordinal MATH, let MATH. By the minimality of MATH, we have MATH if MATH, so we must have MATH for MATH. Note MATH cannot be finite, because MATH is a filter. By lower regularity, we have MATH, contradicting the assumption that MATH. We have MATH, MATH, and MATH, whence MATH, proving REF . Also, MATH for any MATH. REF follows by REF . We have MATH. For, let MATH. Then MATH, and we cannot have MATH, because then we would have MATH, contradicting the fact that MATH. MATH because MATH is strictly meet-irreducible. On the other hand, let MATH. Since MATH, MATH. It suffices to show MATH, because, MATH being strictly meet-irreducible, MATH will imply MATH. Let MATH be the least cardinal number such that some MATH-tuple MATH of elements of MATH, where MATH runs through ordinals less than MATH, satisfies MATH. MATH is strictly meet-irreducible because MATH is atomic. Let MATH be the unique cover of MATH, and for each MATH, define MATH. We have MATH if MATH, by the minimality of MATH. Thus, for each MATH, we have by modularity MATH . If MATH, then it is easy to see that MATH . Now, if MATH is infinite, then by the lower regularity of MATH, we must have MATH. However, this is absurd because MATH. Thus, MATH is finite. It follows that MATH, proving REF .
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math/0005131
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It suffices to prove upper regularity, because MATH is coalgebraic, which implies that all coverings are automatically lower regular. Suppose given MATH, indexed by MATH, for some chain MATH, and such that MATH for MATH. We must show that MATH. For each MATH, let MATH. We have MATH for MATH by REF , and MATH. For, if MATH for all MATH, then MATH for all MATH so MATH, and MATH for all MATH, so MATH. Thus, it suffices to show that MATH is nonempty. For each MATH, let MATH and MATH. Since MATH is lower regular, we have MATH and MATH by REF , and for MATH, we have MATH. For, MATH, MATH, and MATH, because for any MATH, we have MATH and consequently, MATH. It follows that MATH. We also have MATH and MATH, whence MATH. Since MATH is upper regular, we have MATH where MATH and MATH. This implies that MATH is nonempty, by REF . Let MATH. Then MATH for all MATH so MATH for all MATH. On the other hand, MATH so there is a MATH such that MATH. If MATH then MATH, so MATH. If MATH, then MATH and MATH for all MATH. Thus, MATH for all MATH. On the other hand, if MATH, then by REF , MATH, implying that MATH in this case as well. It follows that MATH is nonempty. However, by REF , MATH for each MATH. Thus, MATH is nonempty, and MATH is regular.
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math/0005131
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Let MATH be a maximal chain in MATH, and MATH a maximal chain in MATH refining the image of MATH. If we have a principal filter MATH, then we must have MATH. For, if MATH, then there must exist MATH such that MATH and MATH are not comparable. But, MATH, so either MATH, implying MATH, or MATH, implying MATH. If MATH, MATH are principal and such that MATH, say MATH and MATH, then we must have MATH, MATH with MATH, and if MATH then MATH. Let MATH, MATH be such that MATH and it is not true that MATH and MATH are both principal. We will show that MATH is not projectively equivalent to MATH. If MATH is principal and MATH is not, then MATH is not of atomic type, is not projectively equivalent to MATH, and does not count in the multiplicity of MATH. The case MATH principal and MATH non-principal cannot occur, because MATH. The only remaining case is that both MATH and MATH are non-principal. We can assume that MATH is atomic and lower regular, since otherwise MATH is impossible. If we had MATH with MATH, then we would have MATH; thus, we must have MATH in order to have MATH. Denote this set by MATH. Since MATH is atomic, let MATH and MATH with MATH. For each MATH, we have MATH, and for MATH, we have MATH. Since MATH is lower regular in MATH, we would have MATH . Now, if we had MATH, we would have MATH, implying that MATH, which is impossible. The only other possibility is MATH. In this case, we claim that we must have MATH and MATH. For, if MATH, then MATH, implying that MATH. Then because MATH is maximal, we must have MATH and MATH. However, this contradicts the fact that MATH, because MATH. It follows that the case MATH atomic, lower regular, and neither MATH nor MATH principal is impossible.
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math/0005131
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Let MATH be a chain in MATH. Then the image of MATH in MATH can be refined to a maximal chain MATH in MATH, and it is clear that MATH. Now, let MATH where MATH for all MATH. Then MATH, MATH such that MATH and MATH. For each MATH, define MATH, MATH. Then MATH, MATH, MATH, and MATH, implying that MATH and MATH . It follows that MATH, and combined with the fact that MATH, this implies that MATH if MATH is finite or countable, and that MATH is infinite if MATH is.
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math/0005131
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REF is clear. To prove REF , note that we have MATH and MATH, so REF is true for MATH. If REF is true for MATH then for MATH, MATH, and the square root function is also increasing, whence MATH if MATH. Thus, REF follows by induction. To prove REF , we use REF and note that if MATH, then MATH. A computation shows that REF holds for all MATH. Suppose REF holds for MATH, and let us prove it is true for MATH. We have by the induction hypothesis MATH . Squaring, we have MATH . Thus, REF holds for MATH, and by induction, for all MATH.
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math/0005131
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Let MATH be maximal for the property that MATH but MATH. (By meet-continuity, the set of such elements is closed under joins of chains, so a maximal such element exists by NAME 's Lemma.) Then MATH is strictly meet-irreducible, and has a unique cover, MATH. Now, let MATH be maximal for the property that MATH, MATH, and MATH. We cannot have MATH, because MATH. MATH is the unique cover of MATH in the interval MATH, because if MATH and MATH, we must have MATH. Similarly, we successively choose MATH, MATH, MATH such that MATH is maximal among elements MATH such that MATH, MATH, and MATH, and we obtain covers MATH. We have MATH . For each MATH, let MATH be the join of all elements of MATH containing MATH, and MATH the meet of all elements of MATH not containing MATH. We have MATH, MATH because the maximal chain MATH is be closed under joins and meets. Clearly, MATH for all MATH. The mapping MATH sends each MATH to the unique covering in MATH such that MATH, and thus partitions the ordered set MATH into intervals. If MATH is one of these intervals, then we claim that MATH. For, if MATH then MATH is strictly meet-irreducible, and so we must have MATH. If MATH, then we have MATH, so MATH, because, MATH being a chain, MATH. If MATH did not belong to MATH, then there would be an element MATH such that MATH. Then we would have MATH, but MATH because MATH is closed under meets and does not contain MATH. This is impossible, because MATH is meet-irreducible in MATH. Thus, the claim that MATH is proved. It follows that MATH. Clearly, we have MATH, MATH, MATH, MATH, MATH, as well as MATH. Thus, since we have shown that MATH is atomic, MATH, MATH, MATH have unique covers MATH, MATH, MATH. Defining MATH for MATH, MATH, MATH, we obtain MATH and each MATH. Now, either the number of intervals is MATH, or the cardinality of the largest interval is MATH. In the first case, we have MATH by REF . In the second case, by induction on MATH (and noting that recursive application of the construction of this proof will find filter coverings above MATH and therefore distinct from it), we also have MATH.
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math/0005131
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Suppose MATH is a chain in MATH. For each MATH, define MATH and MATH. Then for all MATH, MATH, and if MATH, MATH with MATH, we have MATH. Since MATH is upper regular, MATH. However, MATH. If MATH, then we would have MATH. Thus, MATH.
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math/0005131
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By the Theorem, if MATH is upper regular, then MATH is closed under joins of chains. Then, by NAME 's Lemma, MATH has maximal elements. However, this is impossible for anomalous MATH by REF .
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math/0005131
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Use meet-continuity as in the proof of REF or REF .
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math/0005131
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Let MATH be a chain in MATH, and let MATH. We have MATH for some MATH such that MATH and MATH. For each MATH, MATH such that MATH, we have MATH. Then MATH by the upper regularity of MATH. If we had MATH, then we would have MATH, which is absurd. It follows that MATH.
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math/0005131
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Let MATH be a chain, and MATH be filter coverings projectively equivalent to MATH and such that MATH implies MATH. For each MATH, let MATH, and let MATH. We have MATH for each MATH, so MATH. On the other hand, we have MATH. For, MATH some element of MATH, whence MATH. Also, MATH for MATH, so MATH in that case. If MATH, then MATH. However, MATH. Thus, MATH. So, MATH for all MATH and MATH. Now, MATH is closed under joins of chains, by REF . Thus, MATH for all MATH. It follows that MATH. Thus, MATH, implying that MATH, and that MATH is upper regular, hence regular.
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math/0005131
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If MATH, then MATH is not anomalous, so MATH is nonempty by REF . Furthermore, by distributivity, it has cardinality one, proving that MATH is principally bounded. The Corollary then follows from REF .
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math/0005131
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If MATH were not principally bounded, MATH would contain a sequence of strictly meet-irreducible elements MATH, MATH, MATH such that for all MATH, MATH. For each MATH, let MATH be the unique cover of MATH. Then if MATH is such that MATH, we have MATH. For, we must have MATH, whence MATH. If we had MATH, then the elements MATH, MATH, MATH, MATH, and MATH would form a sublattice isomorphic to the lattice MATH, which cannot happen in a modular lattice. It follows that MATH with MATH, the general covering in the chain, equivalent to MATH for all MATH. This sequence of elements can be refined to a maximal chain MATH such that MATH is infinite. However, this is contrary to the assumption that MATH is finite. Thus, MATH is principally bounded, and regular by REF .
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math/0005131
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If MATH were MATH, there would be MATH such that MATH. Then we would have to have MATH, implying that MATH which is not true. Thus, MATH is an atom, that is, an ultrafilter. The ultrafilters MATH of MATH all form coverings MATH which determine distinct projective equivalence classes, because given two ultrafilters MATH and MATH, if we had MATH, the multiplicity of MATH in MATH would be MATH, and this is impossible.
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math/0005131
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Let MATH, MATH, MATH, MATH be the instances of rules of inference in a proof, in order. Let pretheories MATH, MATH, MATH, MATH be defined by MATH, MATH for MATH, where MATH is the conclusion of MATH. For each MATH, let the set of steps of order MATH in MATH be MATH, and the set of steps of order MATH covered by MATH, by MATH. We have MATH. For, if MATH infers MATH from the finite set of REF, we have MATH where MATH. However, MATH, because, the MATH-tuple MATH being a proof, MATH. Thus, MATH, but the left side is the set of steps of order MATH in the proof of MATH from MATH, and the right side is the set of steps of order MATH covered by the proof.
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math/0005131
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Let MATH be that set of propositios, and we will show that MATH. Since MATH is generated from MATH by MATH, MATH is the intersection (join) of all pretheories MATH such that MATH and MATH imply MATH. Clearly, MATH and MATH imply MATH, because we can construct a proof of MATH from proofs of the elemtns of MATH. Thus, MATH. On the other hand, suppose that MATH is such that MATH and MATH imply MATH, and let MATH. The existence of a proof of MATH from MATH using instances from MATH implies that MATH. Thus, MATH, so MATH. Thus, MATH.
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math/0005131
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If MATH, then the conclusion follows from REF . If MATH, then we have MATH and we have MATH. Thus, the steps in the interval MATH are a subset of the set of steps in the proof of MATH from MATH. By NAME 's Lemma, there is an ideal MATH such that MATH. The covering MATH is an step (of order MATH) in the proof of MATH from MATH that is not covered by MATH.
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math/0005132
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Any map MATH determines a cameral cover MATH of MATH, namely MATH, compare REF. For a NAME bundle, which involves a MATH-equivariant map MATH, the cameral cover MATH is itself MATH-equivariant, so by descent theory, it is pulled back from a unique cameral cover MATH. Clearly, the assignment MATH constructed above is functorial.
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math/0005132
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Let us observe first that, since we are using only roots rather than arbitrary weights, it is sufficient to consider the case when MATH is simply-connected. We have MATH, hence, by REF MATH . However, by REF, so we obtain that MATH. By restricting to MATH, we obtain MATH. Since MATH is simply-connected, every coroot is primitive. Therefore, there exists a weight MATH, such that MATH. By applying MATH to the above isomorphism MATH, we obtain an isomorphism MATH. Now it only remains to check that this isomorphism is independent of the choice of MATH. However, since the MATH's are locally pull-backs of the corresponding MATH-bundles on MATH, it suffices to consider the universal situation, namely the case MATH. In the latter case, the MATH-bundle MATH itself is trivialized over an open dense part of MATH, namely over MATH. This is because MATH is not among the set of roots which become negative under the action of MATH. In particular, we obtain an isomorphism MATH over MATH. Moreover, it is easy to see that for any MATH as above, the isomorphisms MATH and MATH coincide. In particular, MATH is independent of MATH over MATH and hence over the whole of MATH, which is what we need. MATH .
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math/0005132
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Let us consider first the universal situation: MATH, MATH, where the MATH's satisfy MATH and MATH, where MATH satisfies MATH . The natural maps MATH and MATH are a cameral and a spectral cover, respectively and it is easy to see that in this case MATH and MATH. This proves the first assertion of the proposition. Indeed, any cameral (respectively, spectral) cover is locally induced from MATH (respectively, MATH). For a spectral cover MATH there is a natural map MATH, that attaches to a map MATH given by a MATH-tuple MATH of maps MATH the composition MATH. The resulting map MATH is an isomorphism, because this is so in the universal situation, that is, for MATH. Similarly, we have MATH maps MATH which correspond to the natural map MATH. We claim that they define an isomorphism MATH. Indeed, both the fact that these maps satisfy the condition on the characteristic polynomial and that the resulting map is an isomorphism follow from the corresponding facts for MATH.
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math/0005132
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First we need to show that the map MATH is well-defined, which is equivalent to saying that the projection MATH is an isomorphism. Since the latter projection is proper and MATH is reduced, it is enough to show that the scheme-theoretic preimage in MATH of every MATH is isomorphic to MATH. This is clear on the level of MATH points, since by definition of regular elements, the only abelian MATH-dimensional subalgebra in MATH that contains MATH is its own centralizer. For MATH, the tangent space MATH can be identified with the space of maps MATH that satisfy: MATH . We claim that the tangent space to MATH at MATH is zero. Indeed, this is the space of maps MATH as above, for which, moreover MATH. However, since MATH, any such MATH is identically zero. This implies that MATH which means that MATH is reduced, that is, MATH. Now let us show that MATH is smooth. Let MATH be equal to MATH. Using the above description of the tangent space to MATH it is easy to see that MATH sends an element MATH to the unique map MATH that satisfies: MATH . Consider now the map MATH given by MATH. The above description of MATH implies that the composition MATH coincides with the tautological projection MATH. However, since MATH is regular, the fact that MATH implies that MATH is an injection. We conclude that MATH is an isomorphism, hence MATH is contained in the smooth locus of MATH. Furthermore, MATH is surjective, so MATH is smooth as claimed.
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math/0005132
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Consider the fibered product MATH. By definition of MATH, there is a closed embedding MATH that sends a triple MATH to MATH. We claim that this embedding is in fact an isomorphism. Indeed, the statement is obvious over the preimage in MATH of the regular semisimple locus of MATH. Therefore, the two schemes coincide at the generic point of MATH. This implies what we need, since MATH is reduced.
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math/0005132
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The map MATH is smooth, since it is a base change of a smooth map. Hence, the fact that MATH is smooth and connected implies that MATH has the same properties. A well-known theorem of NAME (compare CITE or CITE, p. REF) says that the restriction of the NAME map MATH to MATH is smooth and that it gives rise to a Cartesian square: MATH . Therefore, the natural action of MATH on the preimage in MATH of the regular semisimple locus in MATH extends to the whole of MATH. The same is true for MATH, because the map MATH is flat and surjective. The étale local isomorphism follows from comparison of our Cartesian square with that of REF.
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math/0005132
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Let MATH be as in the formulation of the proposition. Consider an element MATH such that MATH for MATH and MATH for MATH. In this case MATH is a NAME subalgebra of MATH. Let MATH be the corresponding NAME subgroup. It is well-known that MATH is a NAME subalgebra in MATH. Let MATH be an element in the unipotent radical of MATH, which is regular with respect to MATH. We then see that MATH is a regular element in MATH, since MATH. It is known that if a NAME subalgebra contains a regular element, then it also contains its centralizer (compare REF). Therefore, MATH. Moreover, it is easy to see that every pair MATH is MATH-conjugate to one of the above form. To conclude the proof, it remains to show that MATH. For that, it suffices to show that the image of MATH as above under MATH belongs to the corresponding locus of MATH. However, the above image is just MATH, which makes the assertion obvious.
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math/0005132
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First, we have a natural closed embedding MATH . Its image consists of pairs MATH such that MATH and MATH. This map is compatible with the MATH-action. Hence, it extends to a finite map MATH . Since both varieties are smooth, in order to prove that MATH is an isomorphism, it suffices to do so over the open part, that is, over MATH. However, in the latter case, the assertion becomes obvious.
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math/0005132
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Let MATH be the group-subscheme of MATH over MATH defined by the condition: MATH . First, let us show that MATH is commutative and smooth over MATH. Let MATH be a MATH-point of MATH. The tangent space to MATH at MATH consists of pairs MATH such that MATH. The differential of the map MATH sends MATH to MATH. We claim that it is surjective. It is known that if MATH is of adjoint type, then the centralizer of every regular element is connected. (In particular, each MATH is commutative; this holds even if MATH is not of adjoint type.). Therefore, MATH. However, the latter, as we saw in the proof of REF, coincides with MATH, since MATH is regular. To prove that MATH is smooth over MATH, it remains to observe that the fibers of MATH are smooth (since they are algebraic groups in char. MATH) and all have dimension MATH, by the definition of MATH. The fact that MATH is commutative was established in the course of the above argument. Now let us prove the assertion for MATH. We have a natural closed embedding MATH, which is an isomorphism over the regular semisimple locus of MATH. Hence, it is an isomorphism, because MATH is reduced. Therefore, since the map MATH is flat and surjective, this shows that MATH is commutative and smooth over MATH. It is irreducible, because this is obviously true over MATH.
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math/0005132
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REF implies that the map MATH is smooth and surjective. Therefore, all we have to show is that if MATH is a map such that the induced automorphism of MATH is trivial, then MATH maps to MATH. Observe that MATH acts on MATH via its action on MATH. Since the isomorphism MATH is MATH-equivariant, we obtain a commutative diagram of actions: MATH where the top horizontal arrow is the adjoint action. Therefore, if a map MATH induces the trivial automorphism of MATH, its adjoint action on MATH is trivial too. But this means that it factors through MATH.
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math/0005132
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The map MATH is an isomorphism because it is a closed embedding and at the same time an isomorphism over the generic point of MATH. Commutativity of the diagram can be checked over the preimage of MATH, in which case it becomes obvious.
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math/0005132
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The proof will consist of two steps. The first step will be a reduction to the case of MATH and the second one will be a proof of the assertion for MATH. CASE: Both MATH and MATH are vector bundles of rank MATH over MATH and the map MATH is clearly an isomorphism over MATH. Since the variety MATH is smooth, it remains to show that MATH is an isomorphism on an open subset of MATH whose complement has codimension at least MATH. It follows from REF that such an open subset is formed by the union of the MATH-orbits of the images of MATH, where MATH for all simple roots MATH. Therefore, by MATH-equivariance and by REF, it suffices to show that the map MATH is an isomorphism. This reduces us to the case when MATH is a reductive group of semi-simple rank MATH. Moreover, the statement is clearly invariant under isogenies, so we may replace MATH by MATH. Clearly, the assertion in such a case is equivalent to the one for MATH, which in turn can be replaced by MATH. CASE: For MATH, the variety MATH can be identified with MATH in such a way that the sheaf MATH goes over to MATH. Moreover, MATH can be identified with the MATH-cover MATH. To prove the assertion, it is enough to show that MATH has degree MATH, since any non-zero map between two line bundles of the same degree is automatically an isomorphism. By definition, MATH is the MATH-module of anti-invariants of MATH in MATH. Therefore, MATH .
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math/0005132
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The fact that MATH is an immediate consequence of the fact that in a group of adjoint type centralizers of regular elements are connected. To prove that MATH, let us observe that if MATH is a regular nilpotent element, MATH is a local non-reduced scheme. Its closed point, viewed as a point of MATH, belongs to the intersection of all the MATH's. Let MATH be the sub-group of MATH which corresponds to maps MATH that send the closed point of MATH to the identity in MATH. Clearly, MATH is unipotent and MATH is the NAME decomposition of MATH. The proof is concluded by the observation that MATH which is exactly REF condition.
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math/0005132
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First, one easily reduces the assertion to the case when MATH is the centralizer of a regular nilpotent element, which we will assume. In this case, MATH entirely consists of nilpotents elements. Let MATH be the normalizer of MATH. Since the nilpotent locus in MATH is a single MATH-orbit, we obtain that MATH is a single MATH-orbit. Thus, let MATH be an embedding. To show that MATH is regular, it is enough to show that its centralizer in MATH coincides with MATH. By REF, the quotient MATH maps isomorphically to the group of MATH-equivariant automorphisms of MATH. If for some MATH we have MATH, then MATH acts trivially on MATH, since MATH is an embedding. Hence, MATH. To prove the implication in the other direction, let us observe that MATH is the only MATH-invariant open subset of MATH consisting of regular elements only. However, the locus of MATH that are embeddings is clearly such a subset.
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math/0005132
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Both MATH and MATH are MATH-equivariant MATH-bundles on MATH. To prove that they are isomorphic, we must show that the two homomorphisms MATH corresponding to the base point MATH coincide. However, this follows from the fact that MATH, which is true since MATH is minimal.
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math/0005132
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Given REF, it remains to show that the data REF of a MATH-equivariant ``value" map MATH is the same as the data of a section MATH of MATH. And indeed, giving such a section MATH is equivalent to giving a MATH-equivariant section MATH of the pullback MATH over MATH, compare REF above. By REF , this is the same as a MATH-equivariant section MATH. Now by the definition of MATH (compare subsection REF), MATH. Here MATH is the MATH-equivariant cameral cover of MATH associated to the NAME bundle on MATH which is MATH of our given NAME bundle MATH on MATH. The section MATH, and hence also our original section MATH, are therefore equivalent to a MATH-equivariant map of MATH-schemes MATH which is also MATH-invariant. But this is the same as a MATH-equivariant map of MATH-schemes MATH, as claimed.
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math/0005132
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We identify MATH and MATH using REF . There is a natural map MATH, sending a MATH-bundle on MATH to its classifying morphism MATH. This map MATH is clearly injective, and its image is contained in the Right-hand side We still have to prove the surjectivity of MATH, that is, to show that a morphism MATH in the Right-hand side satisfies the two compatibility conditions between MATH's and MATH's stated in REF. It suffices to do so locally, and then we may assume that MATH has a section. In this case, we can identify MATH with the sheaf of MATH-bundles on MATH satisfying the two compatibility conditions between MATH's and MATH's and which additionally are trivialized along the section MATH. Similarly, we can identify MATH with MATH-bundles on MATH which are trivialized along the section. Each of the compatibility conditions requires the equality of two given trivializations of some (MATH- or MATH-) bundle over MATH. Now our assumption, MATH, together with the assumed trivialization of all objects along the section, guarantees that these equalities hold over the section. The difference between the two trivializations is therefore a global automorphism which equals the identity along the section, so it is the identity everywhere since the fibers of MATH are integral and proper.
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math/0005132
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Let us fix a cameral cover MATH and consider regularized MATH-bundles on MATH corresponding to this fixed MATH as a sheaf of categories over MATH, denoted by MATH. By REF, MATH is a gerbe over the sheaf of NAME categories MATH. This gerbe is induced from the MATH-gerbe MATH by the homomorphism MATH, compare REF. Thus, according to REF, we have a functor MATH, and for a given object MATH the category-fiber of the above functor is a MATH-gerbe, which can be canonically identified with MATH. Finally, according to REF, an object MATH is equivalent to data REF above.
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math/0005134
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Each MATH is clearly one-one and onto, and MATH is an homomorphism of pointed MATH-overalgebras because MATH, and for each MATH, each MATH, and each MATH, we have MATH .
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math/0005134
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Let MATH, and let MATH. By the properties of MATH and REF , we have MATH .
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math/0005134
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MATH. Thus, MATH.
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math/0005134
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In that case, MATH for all MATH and MATH.
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math/0005134
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Let MATH, MATH, and MATH. We have MATH . Thus, if the section MATH is given, MATH and MATH determine each other, because the mappings MATH together make up an isomorphism of MATH-sets from MATH to MATH.
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math/0005134
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Given a factor set MATH for MATH and MATH, we define MATH, and for each MATH, the MATH-ary operation MATH, by the equation MATH . The fact that MATH is a factor set makes the mapping MATH a clone homomorphism, that is, makes MATH an algebra of MATH. For, let MATH for MATH, let MATH be a MATH-tuple of elements of MATH, and let MATH. We have MATH while MATH and the desired equality of these two elements follows from the analogous facts for MATH and MATH. Note MATH is a homomorphism with respect to the algebra MATH just defined. Let us denote MATH by MATH, and MATH by MATH. We have MATH iff MATH. We define MATH by the equation MATH . Each MATH preserves the distinguished element because MATH . Also, MATH preserves the MATH-operations because for each MATH, MATH for MATH, and MATH for MATH, we have MATH while on the other hand, MATH . Thus, MATH is a homomorphism of pointed MATH-overalgebras, and it is clear that each MATH is REF and onto. It follows that MATH is an isomorphism of pointed MATH-overalgebras. Now, let MATH (so that MATH) and let us show that the factor set we obtain is MATH. We have MATH so that the corresponding factor set is given by MATH since by definition, MATH .
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math/0005134
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Using equation MATH for MATH to expand MATH, we have MATH . But, MATH by REF ; combining these two results, we obtain MATH as was to be proved.
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math/0005134
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If MATH, MATH, and MATH are extensions in MATH of MATH by MATH, and MATH, MATH are equivalences, then we first observe that MATH . Then, we have MATH we also have MATH, whence MATH.
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math/0005134
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MATH and MATH are isomorphisms, whence MATH is also by REF . Thus, MATH maps each MATH-class onto a MATH-class, in a one-one fashion. Since each MATH-class is the image of a unique MATH-class by REF , these mappings paste together to give MATH as an isomorphism. MATH is unique, because it is determined by MATH, which is determined by REF .
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math/0005134
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For each MATH, we have MATH .
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math/0005134
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Suppose MATH and MATH are equivalent via an equivalence MATH. By REF , MATH. On the other hand, MATH by REF . The desired result follows. Conversely, if the factor set difference MATH for MATH some MATH-function, then we can produce a new section MATH for MATH, namely MATH and we have MATH. For, MATH as desired. Now we will construct an isomorphism MATH, which is an equivalence from MATH to MATH. We define the one-one and onto function MATH by MATH . We have MATH, because MATH maps each MATH to MATH, and we have MATH, because of REF . If we are given MATH, then letting MATH, we have for each MATH, MATH whence REF and onto function MATH is an isomorphism. Finally, for each MATH, with MATH, and all MATH, we have MATH thus, MATH, so that REF is satisfied, and MATH.
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math/0005134
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If MATH is an extension in MATH of MATH by MATH, and MATH is a splitting, then MATH by REF . If MATH is another section of MATH, then MATH for some MATH-function MATH, by REF . Thus, every factor set of a split extension has the form MATH. It follows by the theorem that all split extensions are equivalent.
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math/0005134
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Choosing a split extension MATH, and a splitting MATH as the section, we obtain MATH, by REF .
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math/0005134
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MATH is also an isomorphism, and the formula for factor sets shows that MATH for any section MATH.
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math/0005134
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If MATH is an element of MATH, then the elements of MATH are the elements of MATH of the form MATH, and such a pair will belong to MATH iff MATH. On the other hand, MATH is the set of MATH such that MATH. However, we have MATH, because MATH. Thus, the mapping MATH is one-one and onto. Since this applies to each MATH, MATH is an isomorphism.
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math/0005134
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Let MATH and MATH as defined above. We have MATH .
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math/0005134
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We have MATH .
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math/0005134
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MATH for some MATH-function MATH. Then, MATH. However, MATH. For, MATH . Thus, MATH by REF .
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math/0005134
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Follows from the definition of MATH being a homomorphism of abelian group objects in the category MATH.
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math/0005134
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We will show that representative factor sets for these two extension classes are equal. Let MATH be a factor set representing the homology class corresponding to the extension class of MATH. Then MATH represents MATH. A factor set MATH for MATH can then be defined by MATH . On the other hand, a factor set MATH for MATH can be defined by MATH and then a factor set for MATH can be given by MATH. For each MATH, we have MATH .
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math/0005134
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We have MATH .
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math/0005134
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Let sections MATH, MATH of the extensions MATH be chosen. The factor set MATH is a factor set of MATH. On the other hand, a factor set for MATH can be computed from the factor set MATH computed in REF . It is easy to see that these factor sets are equal.
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math/0005134
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Let MATH be an abelian group MATH-overalgebra totally in MATH, and let MATH be a universal arrow to the functor MATH. That is, MATH is an induced abelian group MATH-overalgebra in MATH of MATH along MATH. By CITE, since MATH is onto, we can equally well consider MATH to be an induced pointed MATH-overalgebra of MATH along MATH. Thus, CITE applies, and the diagram MATH is a pushout diagram, where MATH is the mashing homomorphism defined by MATH. However, MATH, whence MATH. Similarly, MATH. It follows easily that MATH, and that MATH.
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math/0005134
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By CITE restriction along MATH is a full functor from MATH to MATH. Since MATH and MATH are isomorphic to restrictions along MATH, MATH, where MATH, MATH, and MATH. It follows that MATH is determined by any of its components MATH.
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math/0005134
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Given an extension MATH of MATH by MATH, let MATH, and view MATH as a homomorphism from MATH to MATH. Then we have a module extension MATH of MATH by MATH. On the other hand, given MATH, MATH, and MATH, we define MATH. This gives a MATH-function which is clearly one-one and onto, and is a MATH-overalgebra homomorphism from MATH to MATH, where MATH, because MATH thus, MATH is an isomorphism, and MATH is an extension in MATH of MATH by MATH, that is, an element of MATH. The first construction, of MATH from MATH, is one-one by REF , but it is also onto, because if we start with MATH and construct MATH, then we get MATH back as MATH. Thus, the two mappings are inverses to each other. If two extensions MATH and MATH in MATH of MATH by MATH are given, and MATH, then from MATH we obtain that MATH, whence we have a commutative diagram MATH showing that the constructed module extensions remain equivalent, by the customary definition. On the other hand, given a commutative diagram MATH then let MATH and MATH be constructed from MATH and MATH: we have MATH and for all MATH, and MATH, MATH whence MATH. Thus, the constructed elements of MATH are equivalent by our definition. We omit the verifications that this isomorphism between the bifunctors MATH and MATH of MATH and MATH is natural in both arguments, and that it preserves the abelian group operations.
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math/0005134
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The first and second functors are naturally isomorphic, because of an adjunction between the functor of restriction of MATH-sets along MATH and the functor from MATH-sets to MATH-sets given by sending a MATH-set MATH to the MATH-set MATH. The second and third functors are naturally isomorphic, because of the adjunction between the free functor from MATH to MATH, and the corresponding forgetful functor.
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math/0005136
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Let MATH denote the items. If the items are selected without replacement, let MATH be a uniformly random permutation on the numbers MATH. If the items are selected with replacement, let MATH be an i.i.d. sequence of uniformly random integers in the range MATH. Let MATH denote the sequence consisting of the first MATH items with respect to MATH, that is, MATH; MATH is a uniformly random sequence of MATH items chosen without (respectively, with) replacement. Let MATH be the number of bystander items, and MATH the number of relevant items of MATH. Say that MATH is a positively (respectively, negatively) critical integer if the set of the first MATH relevant items of MATH is proper (respectively, improper) but the first MATH relevant items is improper (respectively, proper). Pick a uniformly random permutation MATH on the numbers MATH (independent of MATH). Use MATH to tag a random set of MATH items from MATH, that is . MATH; the tagged items form a random set of MATH items chosen without (respectively, with) replacement, since we could have picked MATH first and then MATH. Suppose that MATH of the tagged items are relevant. Since our sequence of REF is already in a random order, we may instead pick and keep the first MATH relevant items of MATH and the first MATH bystander items of MATH. The resulting set MATH of kept items is a uniformly random set of MATH items chosen without (respectively, with) replacement, and whether or not MATH has the property is determined by MATH. We can write MATH . Now we use the fact that the negatively critical integers and the positively critical integers are interleaved, and that MATH is unimodal in MATH (from REF ). Let the critical integers be MATH, and suppose that of these, MATH maximizes MATH. Then we can write MATH and MATH . Thus MATH where the MATH term becomes small if MATH gets large while MATH and MATH remain fixed. Thus if both MATH and MATH are between MATH and MATH we can write MATH .
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math/0005136
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For MATH-SAT, when MATH is large, we know that MATH and MATH are both close to the critical ratio MATH, where MATH. (Details of the MATH upper bound are not available at this time, so here we use the established bound of MATH CITE.) More generally for MATH-SAT, we know that for MATH or MATH, MATH, where MATH and MATH are lower and upper bounds on the critical MATH-SAT ratio MATH (which could conceivably be a function of MATH). Let MATH, where MATH is a positive constant that we will choose in a moment. By REF , the fraction of clauses which are partially-free is w.h.p. MATH. The value MATH for REF is MATH. Note that MATH is monotone decreasing in MATH, and that MATH is monotone increasing in MATH, so that our bound from REF is at least as good as MATH . For MATH-SAT we take MATH to get the above-stated constant of MATH.
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math/0005136
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For convenience let MATH be the number of green balls, MATH be the number of red balls, and MATH be the number of balls selected. The precise probability that the MATH-th ball is the MATH-th red ball is MATH . The ratio of successive terms is MATH which is monotone in MATH and implies the unimodality claim. Next we identify the mode: MATH so that the optimal MATH is given by MATH, which is within MATH of MATH. We next approximate the maximum value of this probability. For convenience let MATH. Recall NAME 's formula: MATH where MATH. MATH . Consider the MATH error term arising from the MATH portion of NAME 's formula. Since MATH, MATH, MATH, and MATH, the error term will be MATH, so we may drop it to get an upper bound. If the second term on the right were larger than MATH, then we could increase MATH while keeping the ratios MATH, MATH, and MATH fixed, and thereby make the probability as large as we like, and in particular larger than MATH. Thus we can drop this term as well: MATH where the MATH vanishes when MATH, MATH, MATH, and MATH.
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math/0005136
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Let MATH if the items we are interested in are clauses of MATH . Boolean variables, and let MATH if the items are edges of a graph REF or MATH-uniform hypergraph (MATH). Recall that MATH is the number of variables or vertices. In each of these cases, the number MATH of possible items is MATH. Recall our assumption that MATH is fixed and that we are looking at sets of MATH items, so that functions of MATH and MATH may be written as MATH. A more detailed analysis could determine what happens when for example . MATH with MATH, but we do not attempt this. Say that an item is MATH-free (MATH), with respect to a set of items, if the first MATH variables (if the item is a clause) or the first MATH vertices (if the item is an edge or hyperedge) do not occur in any other items in the set. The probability that an item is MATH-free when the MATH items are randomly selected without replacement is easily seen to be MATH . We use the identity MATH to estimate each term in the product: MATH (When sampling is done with replacement, the MATH error term does not appear.) From this we see that the probability that an item in the set is MATH-free is MATH. Thus the expected number of MATH-free items is MATH. We wish to show that the actual number of MATH-free items will likely be close to its expected value, so we bound the variance. Let MATH be the indicator random variable for item MATH being MATH-free. For MATH, when sampling is done without replacement we have MATH . In the same manner as above we estimate each term in the product: MATH . (As before, when sampling with replacement, the MATH error term does not appear.) Thus MATH yielding the variance in the number of MATH-free items MATH . Using NAME 's inequality, it follows that the actual number of MATH-free items will with high probability be within MATH of its expected value MATH. Recall that an item is partially free if at least one of its variables/vertices is not contained in any of the other items. We use inclusion-exclusion to estimate the number of partially free items. Let MATH denote the event that REF is partially free, and MATH denote the event that REF is free in positions MATH. Then MATH . Since there are MATH possible values of MATH, and for each one MATH is with high probability within MATH of MATH, the number of partially free clauses is with high probability within MATH of MATH .
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math/0005137
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Let MATH be the complex calculating MATH, and let MATH be the subcomplex calculating MATH, that is, the subcomplex of chains whose support is disjoint from MATH. Of course, one has MATH, which justifies the notation. Write MATH. There is an evident map of complexes MATH which must be shown to be a quasi-isomorphism. Let MATH . Obviously the map MATH is an inclusion. We claim first that MATH is a quasi-isomorphism. To see this, note that all these complexes admit subdivision maps MATH which are homotopic to the identity. Given a chain MATH, there exists a MATH such that MATH. Taking MATH with MATH, it follows that MATH surjects onto MATH. If MATH, we choose MATH such that MATH. Since MATH is injective and commutes with MATH, it follows that MATH is injective on homology as well, so MATH is a quasi-isomorphism. It remains to show the surjective map of complexes MATH is a quasi-isomorphism. The kernel of MATH is MATH which is acyclic as MATH admits an evident homotopy retract. The next point is to show MATH . The assertion for MATH is easy because any point MATH in MATH can be attached to MATH by a radial path MATH not passing through MATH. Then MATH extends uniquely to MATH on MATH and MATH. Vanishing in REF when MATH will be proved in a sequence of lemmas. For convenience we drop the subscript MATH and replace MATH with MATH. Let MATH be a radial line meeting MATH at MATH. Let MATH be the space of sections of the local system along MATH with rapid decay at MATH. Then MATH . Let MATH be the complex of chains calculating this homology, and let MATH be the subcomplex of chains not meeting MATH. Then MATH is contractible, and MATH where MATH denotes classical topological chains. The result follows. One knows from the theory of irregular connections in MATH CITE that MATH can be covered by open sectors MATH such than MATH where MATH is rank MATH and MATH has a regular singular point. Let MATH be a smaller closed sector with outer boundary MATH and radial sides MATH. Recall the NAME lines are radial lines where the horizontal sections of the MATH shift from rapid decay to rapid growth. We assume MATH contains at most one NAME line, and that MATH are not NAME lines. Writing MATH, where MATH are even smaller sectors, each of which containing the NAME line if there is one, one may think of the following lemma as a NAME sequence. With notation as above, Let MATH be a basepoint in the interior of MATH. then MATH . One has MATH and of course the assertion of the lemma is that this coincides with MATH. To check this, by REF one is reduced to the case MATH where MATH has rank MATH and MATH has regular singular points. If MATH does not contain a NAME line for MATH then MATH, and the argument is exactly as in REF . Suppose MATH contains a NAME line for MATH. Then (say) MATH and MATH. Let MATH be the complex of chains calculating the desired homology, and let MATH be the chains not meeting MATH. As in the previous lemma, MATH is acyclic. We claim the map MATH is a quasi-isomorphism. If we choose an angular coordinate MATH such that MATH then rotation MATH provides a homotopy contraction of the inclusion of MATH. This homotopy contraction preserves the condition of rapid decay, proving the lemma. Let MATH be the ramified cover of degree MATH obtained by taking the MATH-th root of a parameter at MATH. By the theory of formal connections CITE, one has, for suitable MATH, a decomposition as in REF for the formal completion of the pullback MATH. Let MATH be the degree of the pole of the connection on MATH when we identify MATH, that is, MATH for a local parameter MATH, and MATH is the order of pole of MATH. We have MATH . Assume first that we have a decomposition of the type REF on MATH itself, that is, that no pullback MATH is necessary. We write MATH as a union of closed sectors MATH where MATH has radial boundary lines MATH and MATH. We assume each MATH has at most one NAME line. Using excision together with the previous lemmas we get MATH . By REF , the map MATH above is given by MATH . An element in the kernel of MATH is thus a section MATH of MATH which has rapid decay along each MATH. Since each MATH contains at most one NAME line, such a MATH would necessarily have rapid decay on every sector and thus would be trivial. This proves vanishing for MATH. Finally, to compute the dimension of MATH, note that if MATH has a connection with pole of order MATH, then it has a horizontal section of the form MATH, where MATH has a pole of order MATH. (The connection is MATH.) Suppose MATH. NAME lines for this factor are radial lines where MATH is pure imaginary. Thus, there are MATH . NAME lines for this factor. Consider one of the NAME lines, and suppose it lies in MATH. If the real part of MATH changes from negative to positive as we rotate clockwise through this line, say we are in case MATH, otherwise we are in case MATH. We have MATH since the two cases alternate, we get a contribution of MATH. If MATH there are no rapidly decaying sections, so that case can be ignored. Summing over MATH with MATH gives the desired result. Finally, we must consider the general case when the decomposition REF is only available on MATH for some MATH. By a trace argument, vanishing of the homology upstairs, that is, for MATH, in degrees MATH implies vanishing downstairs. Since MATH is unramified, an NAME characteristic argument (or, more concretely, just cutting into small sectors over which the covering splits) shows that the NAME characteristic multiplies by MATH under pullback, proving the lemma. In particular, we have now completed the proof of REF .
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math/0005137
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To establish the existence of a pairing, note that if MATH is a rapidly decaying chain and MATH is a form of the same degree with moderate growth, then elementary estimates show MATH is well defined. Suppose MATH and write MATH where MATH denotes MATH. Let MATH and suppose MATH where MATH has moderate growth also. Then MATH . Note MATH may include simplices mapping to MATH. Our REF of MATH factors these chains out. Thus, we do get a pairing of complexes. Of course, chains away from MATH integrate with forms with possible essential singularities on MATH. To complete the description of the pairing, we must indicate a pairing MATH . To simplify notation we will drop the subscript MATH and take MATH. An element in MATH can be represented in the form MATH where MATH is a radial path. Let MATH. Given MATH, choose a sector MATH containing MATH on which MATH has a basis MATH. By assumption, we can represent MATH with MATH analytic on the open sector, such that MATH where MATH from a basis of MATH in a neighborhood of MATH and MATH are meromorphic MATH-forms at MATH. then by REF . The pairing is taken to be trivial on chains which do not contain MATH. If MATH is a MATH-chain bounding two radial segments MATH and MATH and a path along MATH from MATH to MATH. Then NAME 's theorem (together with a limiting argument at MATH) gives MATH . Similar arguments show the pairing independent of the choice of the radius of the disk. Also, if MATH with MATH meromorphic at MATH, then MATH . It follows that the pairing is well defined. The diagrams MATH and MATH commute. Consider the top square. The top arrow is excision, replacing a chain with the part of it lying in the disks MATH. The bottom arrow maps a MATH as above in some MATH to MATH. Along MATH outside the disks MATH is exact; its integral along the chain is a sum of terms of the form MATH where MATH. For the part of the chain inside the MATH of course we must take MATH. Combining these terms with appropriate signs yields the desired compatibility. For the bottom square, the top arrow associates to a relative chain on MATH its boundary on MATH. The bottom arrow associates to a horizontal section MATH on MATH the corresponding element in MATH. Note here the MATH will be constant so in the pairing with MATH only the term MATH survives. The assertion of the lemma follows. Returning to the proof of the theorem, we see it reduces to a purely local statement for a connection on a disk. In the following lemma, we modify notation, writing MATH to denote the corresponding group for a connection on a disk MATH with a meromorphic singularity at MATH. The pairing MATH is nondegenerate on the left, that is, MATH for all relative MATH-cycles implies MATH. We work in a sector and we suppose the basis MATH taken in the usual way compatible (in the sector) with the decomposition into a direct sum of rank MATH irregular connections tensor regular singular point connections. Let MATH be the dual basis. Fix a MATH and suppose first MATH and MATH both have moderate growth. We claim MATH has moderate growth. For this it suffices to show MATH has moderate growth. But MATH . This has moderate growth because, MATH all do. Now assume MATH for all MATH. Fix a MATH and assume MATH is rapidly decreasing in our sector. Let MATH be a radius in the sector with endpoint MATH. We can find (compare REF-REF) a basis MATH of MATH on the sector with moderate growth and such that MATH, so MATH. We are interested in the growth of MATH along MATH. We have MATH . Asymptotically, taking MATH the parameter along MATH, MATH as MATH for some MATH and some MATH. We need to know the integral MATH has moderate growth as MATH. Changing variables, so MATH, this becomes MATH where MATH is a sum of monomials in MATH and MATH with positive coefficients. Clearly this has at worst polynomial growth as MATH as desired. Finally, assume MATH is rapidly increasing and MATH is rapidly decreasing. We have as above MATH . In particular, MATH has moderate growth. This implies MATH has moderate growth as well. Indeed, changing notation, this amounts to the assertion that if MATH is rapidly decreasing and MATH has moderate growth, then MATH has moderate growth. Fix a point MATH with MATH. the mean value theorem says there exists a MATH with MATH such that MATH . Suppose MATH. We get MATH proving moderate growth. We conclude that our representation for MATH has moderate growth, and hence it is zero in MATH. It follows that the pairing MATH is nondegenerate on the left. Returning to the global situation, we have now MATH and to finish the proof of the theorem, it will suffice to show these dimensions are equal. With notation as above, MATH. It will suffice to compute the difference of the two NAME characteristics MATH . It is straightforward to show this difference is invariant if MATH is replaced by a smaller NAME open set, and that the NAME characteristics are multiplied by the degree in a finite étale covering MATH. Using REF , we reduce to the case where formally locally at each MATH we have MATH with MATH rank MATH and MATH at worst regular singular. (Here MATH is the NAME power series field at MATH). Let MATH be the degree of the pole for the connection on MATH. Then one can find coherent sheaves MATH such that MATH . It follows that, writing MATH . Since MATH (which is proven algebraically as above, replacing MATH by the regular connection associated to MATH) it follows that MATH . Referring to REF , we see that this is the desired formula. This completes the proof of the theorem.
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math/0005139
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Given MATH, choose a positive NAME function MATH with the following properties: MATH is radial, that is, MATH depends only on MATH; and the NAME transform MATH, defined for MATH by MATH satisfies MATH for all MATH such that MATH. For instance, we may take MATH where MATH because the NAME transform of a function REF is MATH for any MATH and MATH. By NAME summation, MATH . Under the hypothesis on MATH, the only positive term in the sum over MATH is MATH. The sum over MATH is bounded from below by the sum over MATH of length MATH, which is at least the number of such vectors times MATH. It follows that MATH has at most MATH vectors of length MATH, as claimed.
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math/0005139
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Except for the last phrase, this follows from the previous Lemma by taking MATH and MATH, since then MATH must have a nonzero MATH of length at most MATH, and any vector of MATH of length MATH must be orthogonal to MATH. To assure that MATH can be computed in polynomial time, we take MATH for a positive constant MATH small enough that if MATH has a nonzero vector of length at most MATH then the LLL algorithm will find a (possibly different) nonzero vector of length at most MATH. Our Corollary now holds with MATH.
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math/0005139
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A segment of MATH of length MATH is contained in a box whose MATH-th side is MATH (MATH). The rational points of height at most MATH in this box come from points of MATH contained in a box MATH whose MATH-th side is MATH REF and thus has volume MATH. It takes time MATH to apply lattice reduction and, by REF , either list MATH or find a hyperplane containing MATH. In the former case, we test whether each of the resulting MATH points lies in MATH. In the latter case, we map this hyperplane to MATH and intersect it with MATH, finding at most MATH rational points. Thus in either case we find all rational points of height MATH on our segment in time MATH. Since it takes only MATH segments to cover MATH, we are done.
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math/0005139
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For each positive integer MATH consider MATH. Since MATH is transcendental, MATH is an analytic arc MATH contained in no hyperplane of MATH. Now apply the argument for REF with MATH. As noted in the remarks following the statement of that theorem, the curve need not be algebraic as long as it is MATH and its intersection with any hyperplane is of bounded size. (Here we need not compute this intersection numerically, since we are only bounding the number of rational points of small height on MATH, not computing them efficiently.) The differentiability is clear since MATH is analytic, and the boundedness is proved in the next lemma. We conclude that the number of points of height MATH on MATH is MATH. Since MATH can be taken arbitrarily large, our theorem follows.
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math/0005139
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Fix MATH. We shall say that a compact MATH is ``good" if its boundary MATH is rectifiable and its interior MATH contains MATH. Choose a good MATH, and define a norm on MATH by MATH. Let MATH be the unit ball MATH. It is sufficient to prove the lemma for MATH. For each MATH choose a good MATH such that MATH does not vanish on MATH. Let MATH, and let MATH be the number of zeros of MATH in MATH. By NAME 's theorem, if MATH with MATH then MATH has at most MATH zeros in MATH, and a fortiori in MATH. Now MATH is compact and is covered by the open balls MATH of radius MATH about MATH. Thus there is a finite subcover MATH. Then MATH is an upper bound for the number of zeros in MATH of any MATH, and thus of any nonzero MATH.
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math/0005139
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Let MATH be the distinct rational numbers MATH. Expand MATH at infinity: MATH . For this to be of the form MATH we must have MATH . The discriminant of this quadratic equation in MATH is MATH times a square; thus REF has nonzero rational solutions if and only if MATH is a square. Explicitly we find that MATH are proportional to MATH. Conversely, suppose MATH for some integer MATH. Let MATH . Then MATH . To make this MATH with MATH we now need only choose nonzero MATH so that MATH is a MATH-th power (for example, take MATH), and then choose MATH so that MATH. Specializing MATH to sufficiently large integers in the resulting MATH yields infinitely many integer triples MATH with MATH such that MATH, as claimed.
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math/0005149
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First we prove that MATH contains a maximal torus MATH of the group MATH. For, let MATH where MATH is a reductive group. Let us choose two NAME subgroups MATH and MATH in MATH satisfying the following conditions: MATH and MATH are NAME subgroups of MATH, and MATH, where MATH is a maximal torus of MATH. Let MATH be the unipotent part of MATH. Then MATH, and by lemma, MATH. But, it is well - known (CITE, MATH REF) that MATH where MATH is a maximal torus of MATH and MATH is a normal subgroup of MATH, whence MATH. Assume now that MATH, MATH is the center of MATH, MATH, and MATH and MATH are the system of roots and the system of positive roots of the group MATH with respect to MATH, respectively. The group MATH is contained in the NAME group MATH of MATH, and therefore acts on MATH. Since MATH, MATH is generated by the root subgroups MATH, which it contains (CITE, exp REF, t. REF). If MATH then MATH for any-MATH (since MATH). Hence MATH for any MATH. Let MATH be a group generated by those MATH, for which MATH for any MATH. Then MATH, MATH and therefore MATH. It is known REF that the group MATH is parabolic and its nilpotent radical is MATH (here MATH is the centralizer of MATH in MATH). This completes the proof of the Theorem.
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