paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0007186 | The integrability of MATH follows from the NAME theorem. The exactness of the NAME complex for MATH is a theorem due to NAME REF . |
math/0007186 | For any MATH, consider the closed form MATH. By REF , there is a function MATH such that MATH. By REF , MATH satisfies the lemma. |
math/0007186 | Let us choose MATH such that the NAME complex and the usual MATH-NAME complex MATH are exact. On the NAME complex MATH consider the NAME filtration generated by MATH. Considering the ``stupid" filtration on MATH we have that the natural inclusion MATH is a filtered quasiisomorphism, that is, induces a quasiisomorphism ... |
math/0007186 | By REF one can choose MATH such that MATH. For any functions MATH from MATH the functions MATH also satisfy REF . Let us find MATH such that MATH. We find MATH from the equation MATH . Note that MATH is a closed REF-form of the MATH-NAME complex and the equation can be rewritten as MATH where MATH. By REF , such MATH e... |
math/0007186 | REF follows from the existence of functions MATH satisfying REF . CASE: One needs to construct, locally, a NAME algebra morphism MATH such that MATH. Let us choose functions MATH and MATH, MATH, as in REF and assign MATH. Since the elements MATH form a local basis in MATH, this defines a map of MATH modules, MATH. One ... |
math/0007186 | For any point of MATH there is an open neighborhood, MATH, such that on it MATH, where MATH is an REF-form. Let MATH be the vector field on MATH corresponding to MATH by the isomorphism MATH determined by MATH. Then, MATH is a formal automorphism of MATH which transforms MATH to MATH. Hence, one may suppose in the prop... |
math/0007186 | The operators MATH are pairwise commuting derivations of MATH restricted to an open set MATH. Let MATH be the free MATH module over MATH spanned on MATH. Denote by MATH the basis dual to MATH. Let us define in the obvious way the complex MATH, where MATH . Since MATH and MATH, this complex is a deformation of the NAME ... |
math/0007186 | The same as of REF . |
math/0007186 | The same as of REF . |
math/0007186 | By REF one can suppose that MATH and MATH. So, MATH, where MATH is the algebra of functions constant along MATH. Let MATH and MATH be a sufficiently small neighborhood of MATH. Let MATH be functions on MATH such that MATH form a local basis in MATH and MATH be functions such that MATH, where MATH is the NAME bracket in... |
math/0007186 | Follows from the previous proposition. |
math/0007186 | Choose an open covering MATH of MATH such that, for each open set MATH there exist flat connections MATH on MATH and MATH on MATH. Let MATH denote the splitting of the extension MATH given by the negative of the NAME derivative. This splitting is a NAME algebra homomorphism, but is not MATH-linear. Instead, for MATH an... |
math/0007186 | By REF , there is an open covering MATH of MATH such that there exist isomorphisms of polarized quantizations MATH . These isomorphisms induce isomorphisms MATH over each MATH. We may suppose that both MATH and MATH admit on each MATH flat connections MATH, MATH, such that MATH. Since MATH mod MATH, we may choose the c... |
math/0007186 | First, we assume that MATH corresponds to the trivial deformation MATH. By REF , there is an open covering MATH of MATH such that MATH are isomorphic to NAME polarized quantizations. Let MATH be MATH linear derivations of MATH such that for MATH one has MATH. Such MATH exist because they obviously exist for the NAME st... |
math/0007186 | Follows from the NAME construction, see also CITE. |
math/0007186 | By REF we have the equalities MATH . By REF we have MATH and MATH. Combining these with the preceeding identity we obtain MATH . By REF both differences in the left hand side of the last identity are equal to the differences of the respective constant terms, hence coincide (since the constant terms are not affected by ... |
math/0007190 | Let MATH be an irreducible MATH module, recalling that any such module is unique up to equivalence CITE, with MATH as a complex algebra acting on MATH by the standard representation. (We use MATH to denote the algebra of complex MATH matrices.) Hence, the only MATH-module endomorphisms of MATH are given by complex scal... |
math/0007190 | The isomorphism MATH is given explicitly by CITE MATH if MATH is a local oriented, orthonormal frame for MATH. Using this isomorphism and the fact that MATH when MATH, it is easy to see that the left-hand side of REF is mapped into MATH, the skew-Hermitian endomorphisms of MATH. The elements of MATH and MATH give isomo... |
math/0007190 | The connection MATH is spin by the remarks preceding the statement of the lemma. From REF one can see that MATH for all MATH and MATH, so MATH preserves the subspace MATH, inducing the NAME connection MATH. Recall that sections MATH of MATH can be characterized as sections of MATH having zero commutator with all MATH, ... |
math/0007190 | Plainly, the map is an injective homomorphism, so it remains to show that it is surjective. Suppose MATH. For any MATH, the remarks preceding REF imply that we may write MATH, where MATH and so we have MATH for some MATH. But MATH and as MATH, we must have MATH and thus MATH, as desired. |
math/0007190 | Suppose that MATH is a reducible spin connection on MATH with respect to the splitting MATH. There is a unique complex line bundle MATH over MATH such that MATH. If MATH is a unitary connection on MATH, then both MATH and MATH are spin connections on MATH. But any two spin connections on MATH differ by an element of MA... |
math/0007190 | We closely follow the method described in CITE. Our NAME space of parameters is given by MATH . We define an extended MATH-equivariant map, MATH by setting MATH . The parametrized moduli space MATH is then MATH and MATH. The map MATH has differential at the point MATH given by MATH where MATH. Suppose MATH is MATH orth... |
math/0007190 | If MATH denotes the NAME space of MATH perturbation parameters MATH, then a generic, smooth path in MATH joining MATH to MATH induces a smooth, oriented cobordism in MATH joining MATH to MATH. This proves REF . If the compact space MATH is contained in the open subset MATH, then the same holds for MATH, that is, the la... |
math/0007190 | Any MATH spin connection on MATH can be written as MATH, where MATH. An element MATH of the gauge group MATH acts on this representation of MATH by sending MATH to MATH. The argument in CITE (also see CITE) implies that there is a solution of the equation MATH, which is unique up to a harmonic gauge transformation, so ... |
math/0007190 | The class MATH is the first NAME class of MATH restricted to MATH. REF then follows from the splitting MATH. |
math/0007190 | From the NAME decomposition, MATH where MATH denotes the harmonic, imaginary-valued one-forms on MATH, we see that MATH . Thus, we can write the pre-configuration space MATH as MATH . As usual, the factor MATH of the harmonic gauge group MATH acts trivially on MATH and by complex multiplication on MATH. An element MATH... |
math/0007190 | Since MATH and MATH are free, we can write MATH with respect to the decomposition given by the NAME formula: MATH . For MATH and MATH, the restrictions of MATH to MATH and to MATH are trivial so MATH. We now calculate the first NAME class of MATH restricted to the tori MATH. Let MATH be the path in MATH, starting at ze... |
math/0007190 | Because MATH is a retraction, it suffices to compute the restriction of the universal line bundle MATH to the image of the retraction where MATH is given by MATH . The bundle MATH is the restriction of the framed configuration space, MATH to the image of MATH and thus has first NAME class MATH. Then the restriction of ... |
math/0007190 | Suppose MATH is a fixed point of the MATH action on MATH. Consequently, there is an element MATH such that MATH and hence a gauge transformation MATH such that MATH . Thus, MATH is in the stabilizer of MATH, and hence that of MATH, and MATH. CASE: If MATH has the trivial stabilizer MATH in MATH, then MATH and so MATH b... |
math/0007190 | That MATH is closed under the MATH action follows directly from the definitions. It is clear that the function MATH is smooth on each stratum of MATH: we claim it is continuous on MATH. Let MATH be a sequence of points in MATH which converge to MATH. We may assume, by an appropriate choice of a sequence of MATH gauge t... |
math/0007190 | With respect to the decomposition MATH, any element MATH takes the form MATH where MATH, for MATH. Thus, MATH, after identifying MATH. If MATH, then MATH and MATH with MATH and MATH, so any element of MATH takes the shape MATH and the desired isomorphism MATH is given by MATH. We recall from CITE that the induced fiber... |
math/0007190 | The map MATH is clearly a MATH embedding. Furthermore, MATH . Since MATH, we see that MATH acts on MATH as MATH (see also REF) and so MATH. Thus, MATH, as desired. Next, we characterize the image of the map MATH. Suppose MATH is fixed by the MATH action REF on MATH: this action descends to the action REF on MATH, which... |
math/0007190 | Suppose that MATH solves the MATH monopole REF . With respect to the splitting MATH, REF implies that MATH, where MATH is the product connection on MATH and MATH is a unitary connection on the complex line bundle MATH. Let MATH and observe that MATH is a reducible unitary connection on MATH which is a lift of MATH on M... |
math/0007190 | Given REF , we need only characterize the image of MATH. If MATH represents a point in MATH and is reducible with respect to the splitting MATH, with MATH, then REF implies that MATH, for some pair MATH. Then REF implies that MATH satisfies the NAME REF since MATH satisfies the MATH monopole REF . |
math/0007190 | The isomorphisms identifying the cohomology of the tangential deformation complex with that of the NAME complex follow immediately from a comparison of these two complexes. If MATH then MATH and REF implies MATH and so MATH. Moreover, if MATH, then MATH. |
math/0007190 | Let MATH be a point in MATH: by the slice theorem for MATH, the restriction of the projection map MATH to a small enough open neighborhood of MATH in the slice MATH gives a smooth parameterization of an open neighborhood of MATH in MATH. Similarly, by the slice theorem CITE for MATH, the restriction of the projection m... |
math/0007190 | If MATH is a vector bundle over MATH with fiber a NAME space MATH, then REF yields a MATH trivialization MATH, that is, a MATH section MATH of the MATH . NAME bundle MATH over MATH which gives a linear isomorphism on each fiber. Now suppose that MATH is a MATH section of the bundle MATH. If MATH is chosen so that MATH ... |
math/0007190 | By REF , there is a smooth trivialization of MATH given by a smooth isomorphism of complex NAME bundles, MATH as the space MATH (of MATH gauge transformations) is a subgroup of the unitary group of the NAME space MATH. We have an equality of MATH-orthogonal complements, MATH for every MATH, by REF and the fact that MAT... |
math/0007190 | The space MATH is the zero locus of the section MATH of a vector subbundle MATH of MATH constructed in REF , for some open neighborhood MATH of MATH in MATH. For any point MATH, we have MATH . (The differential of MATH does not appear here since MATH.) According to REF , the MATH-orthogonal projection MATH gives a surj... |
math/0007190 | From the final statement of REF , we see that MATH vanishes transversely on MATH and so MATH cuts out the zero locus, MATH, as a regular submanifold of MATH. Then for generic values of MATH, the zero locus will intersect MATH transversely. |
math/0007190 | As an element of MATH, the index bundle MATH depends only on the homotopy class of the leading symbol CITE. Thus, the index bundle of the family MATH is equivalent to that of the family MATH, where MATH. |
math/0007190 | The associated line bundles MATH are given by the quotients of MATH by the relation MATH, for MATH and MATH; under the same relation, MATH. A tensor product of a complex line bundle MATH with a complex vector bundle MATH is given by the quotient of the fiber product MATH by the relation MATH, for MATH, and MATH. Hence,... |
math/0007190 | There are isomorphisms, MATH where, from REF , the action of MATH on MATH is given by MATH. Let MATH be the unit sphere bundle of MATH. If MATH acts on MATH by MATH - where MATH, MATH, MATH and MATH - then we obtain an isomorphism of complex line bundles MATH . REF then implies that MATH and REF gives MATH . The desire... |
math/0007190 | For convenience in the proof, we write MATH and MATH, as in REF . The NAME identification REF of MATH, REF , and the homomorphism property of the NAME character (see, for example, CITE) imply that MATH . Recall that MATH denotes the index bundle of the family of operators obtained, as in REF , by twisting the NAME oper... |
math/0007190 | For convenience, we write MATH. The NAME character determines the NAME polynomial as an element of rational cohomology (see the formula in CITE or CITE), so MATH determines MATH where the MATH are the NAME roots of MATH. REF implies that, with the constraint on MATH, MATH . Suppose MATH and MATH. If MATH, then MATH for... |
math/0007190 | We may assume without loss that MATH is primitive because a sublattice MATH is orthogonal to MATH if and only if it is orthogonal to MATH, where MATH and MATH is primitive. From the classification of indefinite, integral, unimodular forms (for example, see REF) we have MATH . By hypothesis, MATH, so if MATH then MATH i... |
math/0007190 | We consider the intersection form MATH to be a form on MATH. By NAME 's theorem (see REF), we know that MATH and so the classification REF of forms yields MATH with MATH. A result of CITE then implies that MATH, where MATH and MATH. If MATH and MATH, then MATH and so MATH; if MATH, then MATH and NAME 's theorem implies... |
math/0007190 | We may assume without loss that MATH is primitive for, if not, write MATH where MATH and MATH is primitive. Then MATH for all MATH since MATH is characteristic. If MATH were even we would have MATH for all MATH, contradicting our hypothesis that MATH is an odd form. Hence, MATH is odd and MATH for all MATH and thus MAT... |
math/0007190 | Let MATH be a positive integer. The set of squares modulo MATH is MATH. Hence, if MATH is the sum of four odd squares, then we must have MATH. Conversely, suppose that MATH. By NAME 's theorem we can write MATH . Since MATH is even, we must have one of the following possibilities (up to rearranging the terms on the rig... |
math/0007190 | Interchanging the role of MATH and MATH in the proof of REF takes care of REF . Thus we need only consider REF . Continuing the notation of the proof of REF , we choose MATH . Plainly, MATH is characteristic, primitive, and is orthogonal to the hyperbolic sublattice MATH of MATH, while (as MATH) MATH . Since MATH is od... |
math/0007190 | Let MATH be a compact, complex algebraic, simply connected surface with MATH. Suppose MATH is minimal. The NAME classification then implies that MATH is one of the following (see CITE): CASE: A MATH surface, CASE: An elliptic surface, CASE: A surface of general type. The cases where MATH is diffeomorphic to MATH, MATH,... |
math/0007191 | This follows from CITE. |
math/0007191 | By CITE the ring of invariants MATH is generated by the MATH. Now MATH is a closed subvariety of MATH, so the restriction map on functions MATH is surjective. Since MATH is reductive and the base field MATH has characteristic zero, there is a NAME operator, and so it remains surjective on taking invariants. |
math/0007191 | Since MATH the vertex MATH must be loopfree. Now some composition factor must have dimension MATH with MATH. Then MATH by CITE. Since there is no loop at vertex MATH, the relevant composition factor is isomorphic to MATH. Now because MATH, the choice of a decomposition MATH induces an embedding MATH and hence a map MAT... |
math/0007191 | Clearly any real root MATH in MATH must be indecomposable since MATH. Conversely, by CITE any indecomposable element is in MATH. If MATH is not MATH then MATH is a root with some positive component, hence a positive root. But MATH, contradicting indecomposability. |
math/0007191 | Say MATH are elements of MATH with MATH. By induction it suffices to find a subset MATH of MATH with MATH. We prove this by another induction: if MATH is a subset for which the sum is a root MATH, we show how to enlarge MATH so that the sum is a root MATH. Now MATH and MATH, so MATH. Thus MATH for some MATH. Clearly MA... |
math/0007191 | Since MATH, the trace function MATH for a path which starts and ends at MATH involves the trace of a MATH matrix, which is just the unique entry of the matrix. The assertion thus follows from CITE. |
math/0007191 | By REF we know that MATH is surjective. Thus it suffices to prove that it is a closed embedding, that is, that the map on functions MATH is surjective. By REF the ring MATH is generated by the trace functions MATH for MATH a path in MATH starting and ending at REF. Since the ring is finitely generated, a finite number ... |
math/0007191 | By CITE the ring of invariants is generated by polarizations of the elementary symmetric polynomials, so by elements of the form MATH where the sum is over all distinct MATH in the range MATH to MATH. Now the elementary symmetric polynomials can be expressed as polynomials in the power sums by NAME 's formulae, and on ... |
math/0007191 | We consider the pairs MATH which can be obtained from MATH by a sequence of such admissible reflections. Always MATH is positive, since it is in MATH by CITE. Thus we can choose a pair MATH with MATH minimal. Clearly we have MATH. For a contradiction, suppose that MATH. Since MATH is unchanged by these reflections, we ... |
math/0007191 | Some sequence of admissible reflections at vertices MATH sends MATH to MATH. If MATH and MATH then by CITE the reflections send it to a positive root MATH, still with MATH. Thus MATH, and so MATH which is one of the inequalities. The other one is obtained by replacing MATH with MATH. |
math/0007191 | Apply CITE to MATH, and then consider the sequence of reflections as reflections for MATH. Of course non-admissible reflections can be omitted, for if MATH and MATH then MATH. |
math/0007191 | Any indecomposable element of MATH which vanishes at MATH is MATH, so it suffices to prove that if MATH is a vector with MATH and MATH, then either MATH for all MATH with MATH, or MATH for all MATH with MATH. Since MATH is an extending vertex for MATH we have MATH, and so by replacing MATH by MATH if necessary, we may ... |
math/0007191 | First suppose that MATH. If MATH then the expression for MATH as a sum of coordinate vectors is a non-trivial decomposition into elements of MATH. Since MATH by REF , this contradicts the fact that MATH. Thus we may suppose that MATH. Replacing MATH by the pair MATH of REF , we may assume that the restriction of MATH t... |
math/0007191 | We cannot have MATH for all MATH, for otherwise MATH, so MATH since MATH is NAME. Embed MATH in an extended NAME quiver of the same type by adding an extending vertex MATH, and consider MATH as a dimension vector for this quiver. Let MATH be the minimal positive imaginary root. Since MATH is a root for MATH we have MAT... |
math/0007191 | We can identify MATH with a MATH-stable closed subvariety of MATH (defined by the vanishing of all arrows with one end in MATH and the other end in MATH). The inclusion thus induces a closed embedding MATH and by the assumption on composition factors this is a bijection. |
math/0007191 | The statement does not depend on the orientation of the arrows in MATH, so we may suppose that the arrow connecting MATH and MATH is MATH. By CITE the condition that MATH is that there is a MATH-module of dimension MATH. Similarly for the other two conditions. Now if the module is given by an element MATH, then for any... |
math/0007191 | Because of the existence of a module of dimension MATH we have MATH, hence also MATH. For a contradiction, suppose there is a composition factor whose dimension MATH does not have support in MATH or MATH. Then MATH. Since the dimension vector MATH of any other composition factor must have support in MATH or MATH, and h... |
math/0007191 | Because of the existence of a module of dimension MATH, we have MATH. Since the field MATH has characteristic zero, we deduce that MATH. For a contradiction, suppose there is a composition factor whose dimension MATH does not have support in MATH or MATH. Then MATH. Since the dimension vector MATH of any other composit... |
math/0007191 | We prove this for all MATH, MATH and MATH by induction on the maximum possible number of terms in an expression for MATH as a sum of elements of MATH. If MATH then the assertions are vacuous, so assume that MATH. By CITE and REF we can always apply a sequence of admissible reflections to the pair MATH. Let MATH be the ... |
math/0007191 | CASE: If MATH is a real root in MATH, then MATH is a point by CITE. CASE: If MATH is an isotropic imaginary root in MATH, then it is indivisible, for if MATH then MATH is a root, it has MATH since the base field MATH has characteristic zero, and the decomposition MATH has MATH, contrary to the definition of MATH. By CI... |
math/0007192 | The first cohomology group of the product of two abelian varieties is the direct sum of the first cohomology groups of the individual abelian varieties. Moreover, the NAME cycles on the individual varieties pull-back to give NAME cycles on the product. Thus it follows that the NAME group of the product contains the pro... |
math/0007192 | Let MATH be the standard representation of MATH. Let MATH be its decomposition into isotypical components as a representation of MATH. Let MATH be the decomposition of MATH as a representation of MATH. Then each MATH is either MATH or MATH or MATH. The result follows by dimension counting. The lemma also follows from t... |
math/0007192 | The first cohomology group of MATH decomposes as a direct sum of two (polarised) sub-Hodge structures. It follows that MATH is the product of two abelian subvarieties. Hence we have the result. |
math/0007192 | We begin with the case where the base curve has genus zero. In this case the NAME varieties are the NAME of the corresponding hyperelliptic double cover. The result is classical for elliptic curves which can be considered as the NAME varieties associated with double covers of smooth rational curves branched at REF poin... |
math/0007193 | Write MATH . By induction on MATH we have that MATH for MATH, from which it is clear that MATH for MATH. We write MATH, or MATH . Then MATH so MATH which has positive entries for MATH. |
math/0007193 | We present an outline of the more detailed proof in CITE. Suppose that MATH is a hyperbolic fixed point of MATH. Let MATH be a generator of the cyclic group of matrices fixing MATH. MATH is determined up to inverses and MATH . We show in CITE that if MATH is a principal ideal domain, there is a uniquely determined posi... |
math/0007193 | Suppose, by way of contradiction, that MATH but MATH. By the first relation REF we have MATH, then by the second relation REF we have MATH for some MATH, MATH. Repeating this, we produce a sequence of poles MATH, none of which are real, with MATH and MATH for each MATH. REF and a geometric argument show that MATH for e... |
math/0007193 | Suppose that MATH. As in the proof of REF , MATH for some MATH, MATH. By REF we may take each entry of MATH to be non-negative, which implies that MATH. Repeating this process gives a sequence of poles MATH, with MATH and MATH for each MATH. Since MATH is finite we must have that MATH for some MATH. Thus we have a fini... |
math/0007193 | A routine exercise showing containment in both directions establishes the Lemma. |
math/0007193 | Fix MATH and put MATH and MATH. For REF we apply MATH to the second relation REF, use the first relation REF and rearrange to get MATH . We claim that the right hand side of this expression approaches MATH as MATH. To prove our claim it suffices to show that for MATH, MATH . Write MATH, so MATH and MATH . Now MATH as M... |
math/0007193 | Fix MATH and put MATH. Note that if MATH has a pole at MATH and MATH is a linear fractional transformation, then MATH has a pole at MATH. In fact, MATH . By REF with MATH, we have MATH . In a similar way, we use REF with MATH, MATH an integer, to get MATH . Suppose that MATH. Then by the proof of REF , MATH is fixed by... |
math/0007193 | Functions of the given form exist, since MATH has a pole of order MATH at MATH and can be normalized so that the coefficient of MATH is MATH. For uniqueness, suppose that MATH and MATH are functions satisfying the hypotheses of the Lemma, with MATH. Then there exist RPFs MATH and MATH and nonzero constants MATH and MAT... |
math/0007193 | Write MATH. Let MATH be an element of MATH which fixes both MATH and MATH. We calculate MATH . We have used the fact that MATH. From this it follows that MATH satisfies REF, as does MATH. By an argument similar to the proof of REF , MATH is a nonzero multiple of MATH, which is in turn a multiple of MATH. Thus MATH is a... |
math/0007193 | By REF we have that any RPF MATH of weight MATH on MATH has the form MATH where each MATH is a hyperbolic MATH-equivalence class of MATH-BQFs and MATH is given by REF. From REF we see that the coefficients MATH alternate in sign as MATH goes through any cycle MATH. Thus for each irreducible system of poles MATH there i... |
math/0007193 | Suppose that MATH is an RPF of weight MATH on MATH with NAME MATH-symmetric irreducible systems of poles. Then MATH for each MATH-equivalence class of MATH-BQFs associated with poles of MATH. Thus the expression for MATH in REF simplifies to MATH where MATH is given by REF and MATH and MATH are all constants. A calcula... |
math/0007194 | Immediately, by definitions and induction on MATH. |
math/0007194 | MATH . Let MATH; choose MATH such that MATH. Since MATH avoids MATH, we see that MATH for all MATH, and since MATH avoids MATH, we get that MATH, where MATH, and so on. MATH . Let MATH, and let MATH; choose MATH such that MATH. Similarly to MATH, MATH, therefore MATH which means MATH for all MATH, where MATH. Hence MAT... |
math/0007194 | Let MATH; put MATH. Since MATH avoids MATH, we see that MATH contains MATH, since MATH avoids MATH, we get that MATH, and since MATH avoids MATH, we have two cases: either MATH for MATH, or MATH such that MATH. Now let us consider the two cases: CASE: Let MATH, where MATH, MATH. Similarly to the above, we have two case... |
math/0007194 | MATH . Immediately, by REF , MATH where MATH, MATH. MATH . Again, by REF , for all MATH hence, this theorem holds. |
math/0007194 | MATH . Let MATH, and let MATH. Since MATH avoids MATH, we see that MATH contains MATH, and since MATH avoids MATH, we get that MATH, where MATH, and so on. MATH . Let MATH, and let MATH; similarly to MATH, MATH. Therefore, for MATH we have MATH if and only if MATH, and for MATH we have MATH if and only if MATH. Hence M... |
math/0007194 | By the proof of REF , MATH which means that MATH . Besides, MATH, and MATH for MATH (see REF). Hence, the corollary is true. |
math/0007194 | MATH . Let MATH; if MATH, then MATH contains either MATH or MATH, which means MATH contains either MATH or MATH, hence MATH or MATH, and so on. MATH . Let MATH; similarly to MATH, MATH or MATH. Let MATH, hence in the above two cases (MATH or MATH) we obtain MATH for all MATH. Besides, MATH, MATH, and MATH for all MATH ... |
math/0007194 | Let MATH; since MATH avoids MATH and MATH we have either MATH or MATH. If MATH, then, since MATH avoids MATH, we see that MATH. Now we consider the two cases: CASE: Let MATH, MATH. Similarly to the above, either MATH, or MATH, so evidently MATH . CASE: Let MATH, MATH. Similarly to the above, either MATH, or MATH, so ev... |
math/0007194 | Let MATH; by REF , there are two cases: CASE: MATH. So MATH, MATH, and for all MATH . Besides, MATH for all MATH (see REF), where MATH is the MATH-th NAME number. Hence MATH . CASE: MATH. So MATH, MATH, and for all MATH, MATH . Besides, MATH for all MATH (see REF), where MATH is the MATH-th NAME number. Hence MATH . He... |
math/0007194 | MATH . Let MATH; put MATH. Since MATH avoids MATH, we see that MATH contains MATH, since MATH avoids MATH, we see that MATH, and since MATH avoids MATH, we get that MATH. MATH . Let MATH; similarly to MATH, MATH for MATH, hence MATH . |
math/0007194 | MATH . Let MATH and choose MATH such that MATH. Since MATH avoids MATH, we get that MATH for all MATH, since MATH avoids, MATH we see that MATH contains MATH, and since MATH avoids MATH, we get that MATH. MATH . Let MATH; similarly to MATH, MATH for MATH, hence MATH. |
math/0007194 | MATH . Let MATH; put MATH. Since MATH avoids MATH, we get that MATH contains MATH, and since MATH avoids MATH, we see that MATH, and since MATH avoids MATH, we get that MATH for MATH. MATH . Let MATH and MATH; similarly to MATH, MATH for MATH. Hence MATH. |
math/0007194 | MATH . Let MATH; put MATH. Since MATH avoids MATH, we get that MATH contains MATH, since MATH avoids MATH, we see that MATH, and since MATH avoids MATH, we get that MATH for MATH. MATH . Let MATH; similarly to REF , MATH for MATH. Hence MATH. |
math/0007195 | In any NAME loop, each MATH is a pseudo-automorphism with companion MATH, and each MATH is a pseudo-automorphism with companion the commutator MATH REF . In general, if MATH is a companion of the pseudo-automorphism MATH, then MATH is in the nucleus iff MATH is an automorphism. Thus all cubes and commutators are in the... |
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