paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0005119 | Induction on MATH, where MATH for a root MATH. It is crucial that MATH is simply laced. |
math/0005124 | 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... |
math/0005130 | 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... |
math/0005130 | 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 (other... |
math/0005130 | 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. |
math/0005130 | See CITE. |
math/0005130 | The quotient MATH of the NAME MATH is a NAME, and so is its isometric image. The rest is clear. |
math/0005130 | 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 b... |
math/0005130 | 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... |
math/0005130 | This follows from REF , and the fact that MATH for all MATH. |
math/0005130 | 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 correspon... |
math/0005130 | Define MATH by case distinction: MATH. Then MATH. |
math/0005130 | 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/0005130 | 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 s... |
math/0005130 | 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, MA... |
math/0005131 | 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. |
math/0005131 | 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. |
math/0005131 | 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 ... |
math/0005131 | 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. |
math/0005131 | 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. |
math/0005131 | This follows from REF and from the fact that MATH iff MATH. |
math/0005131 | 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 ... |
math/0005131 | 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 element... |
math/0005131 | 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... |
math/0005131 | 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 . |
math/0005131 | 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 m... |
math/0005131 | 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. Assum... |
math/0005131 | 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 M... |
math/0005131 | 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, MA... |
math/0005131 | 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 follo... |
math/0005131 | 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 R... |
math/0005131 | 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/0005131 | 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. |
math/0005131 | 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 . |
math/0005131 | Use meet-continuity as in the proof of REF or REF . |
math/0005131 | 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. |
math/0005131 | 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,... |
math/0005131 | 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 . |
math/0005131 | 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 elemen... |
math/0005131 | 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... |
math/0005131 | 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, b... |
math/0005131 | 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, MA... |
math/0005131 | 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 fr... |
math/0005132 | 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 fu... |
math/0005132 | 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... |
math/0005132 | 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... |
math/0005132 | 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/0005132 | 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 ... |
math/0005132 | 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 Cartesi... |
math/0005132 | 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 ... |
math/0005132 | 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 i... |
math/0005132 | 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 th... |
math/0005132 | 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 di... |
math/0005132 | 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. |
math/0005132 | 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 ... |
math/0005132 | 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 inter... |
math/0005132 | 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 singl... |
math/0005132 | 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. |
math/0005132 | 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... |
math/0005132 | 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-han... |
math/0005132 | 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, ... |
math/0005134 | 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 . |
math/0005134 | Let MATH, and let MATH. By the properties of MATH and REF , we have MATH . |
math/0005134 | MATH. Thus, MATH. |
math/0005134 | In that case, MATH for all MATH and MATH. |
math/0005134 | 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. |
math/0005134 | 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, ... |
math/0005134 | 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. |
math/0005134 | 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. |
math/0005134 | 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 i... |
math/0005134 | For each MATH, we have MATH . |
math/0005134 | 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... |
math/0005134 | 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. |
math/0005134 | Choosing a split extension MATH, and a splitting MATH as the section, we obtain MATH, by REF . |
math/0005134 | MATH is also an isomorphism, and the formula for factor sets shows that MATH for any section MATH. |
math/0005134 | 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, M... |
math/0005134 | Let MATH and MATH as defined above. We have MATH . |
math/0005134 | We have MATH . |
math/0005134 | MATH for some MATH-function MATH. Then, MATH. However, MATH. For, MATH . Thus, MATH by REF . |
math/0005134 | Follows from the definition of MATH being a homomorphism of abelian group objects in the category MATH. |
math/0005134 | 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... |
math/0005134 | We have MATH . |
math/0005134 | 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. |
math/0005134 | 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 al... |
math/0005134 | 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. |
math/0005134 | 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 ... |
math/0005134 | 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... |
math/0005136 | 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 i... |
math/0005136 | 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 a... |
math/0005136 | 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 ide... |
math/0005136 | 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/0005137 | 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. Le... |
math/0005137 | 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/0005139 | 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/0005139 | 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 s... |
math/0005139 | 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 li... |
math/0005139 | 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 intersecti... |
math/0005139 | 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 van... |
math/0005139 | 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 propor... |
math/0005149 | 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 ... |
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