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math/0005158 | Let MATH . Then MATH for MATH, so we get REF equations MATH . Eliminating the variable MATH gives MATH and similarly MATH so the vector MATH is orthogonal to the generic NAME hence a rational multiple of the CM cycle. The smallest multiple of the vector such that it is an integral linear combination will be the MATH ti... |
math/0005158 | The idea is to compare the determinants of the intersection matrices of the two bases of the rational NAME. Using the basis coming from MATH and the CM cycle MATH, at a point of discriminant MATH one gets the determinant of the intersection matrix to be MATH . On the other hand, from the computation using the other bas... |
math/0005158 | This is just a consequence of putting together the statements in the previous sections. |
math/0005159 | Since MATH, we clearly have MATH. Notice that if MATH satisfies MATH, then MATH; for otherwise MATH, whence MATH. Therefore MATH, so MATH. Thus, from CITE, we see that MATH. |
math/0005159 | Let MATH be the orthogonal projection onto MATH. First assume that MATH is not hyperatomic. Let MATH be a strictly ascending sequence of projections in MATH such that MATH. Let MATH be a sequence of positive real numbers such that MATH . Let MATH be a sequence of positive integers such that for all MATH, MATH . For eac... |
math/0005159 | CASE: This is proved in CITE. CASE: Obvious. CASE: If MATH is any element of MATH then, since MATH is separable, there is a vector MATH such that MATH is the projection onto MATH. But our hypothesis is that MATH is already closed, so MATH. CASE: Given MATH we may find MATH such that MATH, thus MATH is a closed vector. ... |
math/0005159 | CASE: Let MATH. Since MATH is hyperatomic, REF shows that MATH which, by assumption, is MATH. Thus REF holds. CASE: By assumption, for all MATH, we have MATH. Since MATH is the smallest element of MATH which contains MATH and MATH by hypothesis, we see MATH. Thus, MATH, for all MATH. CASE: Suppose that MATH. We claim t... |
math/0005159 | Observe that MATH is a MATH-algebra which is weakly dense in MATH; thus MATH is an irreducible MATH- -subalgebra of MATH. We first assume that the identity operator MATH is generated by a single atom MATH of MATH. We shall prove that the invariant manifold for MATH generated by a unit vector in MATH is all of MATH. So ... |
math/0005159 | If MATH, then since every invariant manifold for MATH is also an invariant manifold for MATH, the discussion in the introduction shows that MATH decomposes as required. Then every linear manifold invariant for MATH is also invariant for MATH. Conversely, since MATH is a masa in MATH, every invariant manifold MATH for M... |
math/0005161 | Let MATH, MATH and MATH. Then MATH . |
math/0005161 | Indeed MATH and MATH . |
math/0005161 | Indeed the NAME algebra derived from MATH is commutative (see CITE). |
math/0005161 | Assume the converse. In view of REF MATH is trivial and thus MATH. Because of REF the multiplication is trivial (MATH) - which contradicts our assumption that MATH is non-commutative. |
math/0005161 | Indeed, assume that the converse is true. Then there exist non-zero elements MATH and MATH such that MATH, MATH, MATH and MATH is generic. Let us normalize MATH so that MATH. Thus MATH is spanned by MATH, MATH and MATH. Since MATH we must have that MATH. Since MATH we have MATH. But MATH . Thus MATH. Analogously MATH. ... |
math/0005161 | Indeed, for MATH and MATH and MATH different from MATH: MATH . Choosing MATH and MATH . We get MATH, where MATH is also in MATH. A similar argument is used to prove the statement when one of MATH, MATH is infinity. |
math/0005161 | The first part of the statement follows by induction from the fact that MATH. To prove the second part we will make induction on the parameter MATH. The base of induction follows immediately from properties of MATH REF . Assume that the statement is true for all MATH and MATH such that MATH. Let us pick certain MATH an... |
math/0005161 | REF for all MATH we must have MATH . Setting MATH we get MATH, hence MATH. For arbitrary MATH we proceed by induction. Again let us set MATH in REF . We get MATH . But we already know that MATH, thus MATH as well. The case MATH is similar: from the definition we get MATH and from the definition of MATH it follows that ... |
math/0005161 | We will reduce this problem to the classical theorem about NAME normal form for matrices. Recall the definition of MATH: MATH . The equation MATH can be transformed as follows: MATH . Thus MATH. Hence MATH is nothing more but the space of NAME vectors of rank MATH corresponding to eigenvalue MATH (note also a non-stand... |
math/0005161 | Most of these statements have already been discussed. The fact that MATH follows by induction from MATH - which, in turn, is due to the requirement that characteristic polynomial is non-zero. MATH in view of the symmetry MATH, MATH. Solvability of MATH is due to commutativity of MATH. The last statement follows from th... |
math/0005161 | We will proceed by induction. For MATH the statement is true with MATH. Assume that the statement holds for all MATH. By definition for MATH and MATH we have MATH where MATH. Let us choose a basis MATH in MATH such that first MATH vectors are in MATH. Let MATH be vectors MATH corresponding to vectors MATH. Then the ope... |
math/0005161 | Indeed, consider the multiplication table evaluated in point MATH: MATH (here MATH is a bilinear form defined on MATH) The characteristic polynomial is then given by the following formula: MATH . Thus the only way for MATH to be nonzero is for MATH, MATH and MATH to be invertible. Note, however, that the form MATH here... |
math/0005161 | Assume the converse, that is, there exists MATH such that MATH . If there is more than one candidate let us choose one that has maximum absolute value. By REF MATH should have the smallest possible value (because otherwise there would have existed non-zero MATH and due to symmetry MATH as well). Pick two vectors MATH a... |
math/0005161 | Indeed, MATH acts on MATH for all MATH both from the left and right. The map MATH is eqivariant with respect to left action on the first argument and right action on the second argument. Consequently, the image of MATH is stable under both left and right action of MATH. |
math/0005162 | A projection depends only on the center from which it is done. The effect on the projection of a movement of the center can be achieved by a rigid isotopy defined by a path in the group of projective transformations of MATH. |
math/0005162 | This is obvious. Indeed, the real branch of the projection does not interact with the imaginary branches, it just passes through their intersection point. |
math/0005162 | At the moment of the fourth move take a small ball MATH in the complex projective plane centered in the solitary self-tangency point of the projection of the curve. Its intersection with the projection of the complex point set of the curve consists of two smoothly embedded disks tangent to each other and to the disk MA... |
math/0005162 | Any two nonsingular real curves of the type under consideration can be connected by a path as above. Hence their self-linking numbers coincide. On the other hand, it is easy to construct, for any pair of natural numbers MATH and MATH of the same parity, a pair of nonsingular real algebraic surfaces of degrees MATH and ... |
math/0005164 | This is essentially a special case of REF. We sketch the proof. For any constant MATH-form MATH on MATH define the tangential part of MATH to be MATH . Then MATH decomposes as MATH where MATH is the normal part of MATH defined previously. When MATH, restricting to the unit sphere it follows that MATH . Since MATH and M... |
math/0005164 | Let MATH be a minimal Legendrian submanifold of MATH. It is a standard fact CITE that MATH is minimal if and only if MATH is minimal in the unit sphere. Thus MATH is a minimal Lagrangian cone which from REF must be MATH-special Lagrangian for some MATH. By the previous proposition this implies MATH is MATH-special Lege... |
math/0005164 | CASE: By REF , MATH is a MATH-SLG cone. By rotating MATH by MATH we can assume MATH is MATH-SLG. Thus MATH. Let MATH denote the inclusion of MATH in the sphere, and let MATH be local coordinates on MATH. Then MATH is equivalent to MATH . Let MATH be given by MATH where MATH is some nonconstant smooth complex valued fun... |
math/0005164 | From REF any minimal Legendrian immersion is MATH-special Legendrian for some MATH and hence MATH restricts to zero on the cone. At a point MATH on the cone (where MATH) we have MATH . Since this must hold for all real MATH and MATH, for MATH to restrict to zero we must have MATH as claimed. One can also show necessity... |
math/0005164 | Let MATH be a minimal Legendrian immersion of the form REF, that is, MATH where MATH. By conjugation we may assume MATH where MATH. Let MATH, then MATH and MATH. Moreover, by rescaling MATH and MATH we may assume that MATH. Let MATH . Then the constraints REF together with the constraint that MATH lie on the unit spher... |
math/0005164 | The proof is a straightforward computation using the basic properties of the NAME elliptic functions (for details see CITE). |
math/0005164 | In the case MATH we know explicitly the values of the MATH and hence MATH . REF specializes to MATH . Define MATH by the formula MATH where as previously we set MATH for MATH. Then MATH is a MATH-special Legendrian immersion invariant under MATH. To find the extreme values taken on by the NAME curvature, note that in t... |
math/0005164 | CASE: Differentiating REF with respect to MATH and taking the norm of both sides implies MATH. CASE: If the periods are MATH and MATH then MATH is also a period. CASE: MATH is a period implies MATH, for MATH. So MATH. In particular, MATH. Conversely if MATH then MATH is a period. CASE: If MATH admits two independent pe... |
math/0005164 | Since MATH, the MATH are constant and the second condition of REF of the previous proposition is superfluous. Thus the immersion is doubly periodic if and only if MATH. Let MATH. It is easy to see that MATH and MATH belong to the period lattice of MATH. To find the full period lattice it is sufficient to find all perio... |
math/0005164 | Consider an immersion MATH with MATH and MATH. From REF the minimum and maximum values of the NAME curvature are given by MATH and MATH . Certainly for MATH we have MATH and similarly for MATH. Since MATH gives rise to an embedded minimal Legendrian torus whenever MATH, just choose MATH and the result is proved. |
math/0005166 | For every MATH pick a vector from MATH. In that way we get a function, which will be denoted by the same symbol MATH, from MATH into itself with the property that for every vector MATH, there exists a vector MATH such that MATH for some MATH of modulus REF. Let us do the same with the other transformation MATH. Clearly... |
math/0005166 | Just as in the proof of our theorem above, we can define an "almost" surjective map (that is, which has values in every ray) on the underlying NAME space MATH denoted by the same symbol MATH such that MATH . Set MATH. The proof of our theorem now applies and we find that there is a bounded invertible either linear or c... |
math/0005166 | Since MATH is finite dimensional, it is easy to see that there exists an invertible linear operator MATH on MATH such that MATH. Now, REF applies. |
math/0005167 | Clearly, MATH sends projections to projections. So, MATH and MATH are projections. Since MATH and MATH it is easy to see that MATH can be written in the form MATH where MATH is a *-semigroup endomorphism of MATH having the same continuity property as MATH which sends REF to REF and maps MATH into MATH. Therefore, we ca... |
math/0005168 | First we recall the following fact whose proof requires only trivial calculations. Let MATH be an affine function with MATH where MATH is a linear space. Define MATH . Next let MATH where MATH and MATH are the positive part and the negative part of MATH, respectively. That is, MATH . Finally, set MATH where MATH and MA... |
math/0005168 | First observe that MATH sends projections to projections. Indeed, if MATH is a projection, then we have MATH. Since MATH is a positive operator, by the spectral mapping theorem we obtain that MATH and this proves that MATH is a projection. We next show that MATH preserves the partial ordering MATH among the projections... |
math/0005168 | We first prove that MATH is either MATH or MATH. Since MATH for every MATH and MATH is surjective, it follows that MATH. Therefore, we have MATH . Since this holds for every MATH, by the surjectivity of MATH, it follows that MATH is in the center of MATH and, consequently, MATH is a scalar. This yields that either MATH... |
math/0005170 | First note that MATH preserves the tripotents in MATH and MATH in both directions, that is, an operator MATH is a tripotent if and only if so is MATH. We show that for every MATH and tripotent MATH we have MATH if and only if MATH. First observe that MATH. Indeed, since MATH, there exists a MATH such that MATH. It foll... |
math/0005171 | The inner product on MATH allows us to view it as a representation MATH of the unitary group MATH. MATH is then also a representation of MATH, and is isomorphic to a twist of MATH by a REF-dimensional representation. Hence by NAME 's formula we see that it decomposes as a direct sum of two irreducible representations M... |
math/0005171 | By iteration the proof is the same for all MATH. Say MATH, then we have MATH and MATH, thus MATH . Here MATH is a form from C and MATH is from MATH, and it is either MATH or else MATH, and therefore it is of volume MATH, because of the orthogonality assumption. We have then : MATH, because MATH. |
math/0005171 | The previous lemma along with REF implies that MATH gives a non-zero element of MATH if MATH is orthogonal to MATH. The assumptions on the primitive cohomology imply that MATH. Since MATH, it follows that the condition is indeed satisfied. |
math/0005171 | We shall use induction on MATH, the result for MATH and MATH being well known. Assume that MATH and the result is known for smaller MATH. We degenerate MATH to a stable curve MATH with MATH smooth irreducible components MATH and MATH, with MATH and MATH of genus MATH, and MATH of genus MATH intersecting each of MATH an... |
math/0005171 | Multiplication of MATH by MATH shows that MATH and MATH are isomorphic models of the same curve MATH. On MATH the volume forms coincide, while MATH. Thus MATH hence MATH cannot be constant. |
math/0005171 | We prove it for a bielliptic curve MATH which is a double cover of MATH and of MATH. Consider the diagram MATH . Here MATH is ramified over MATH, MATH is the double cover ramified over MATH and MATH, and MATH is the hyperelliptic cover ramified at MATH. On the range of MATH we have already fixed a standard parameter, w... |
math/0005171 | The proof for arbitrary genus is obtained by induction. Starting from a hyperelliptic curve MATH and a double cover MATH we construct the commutative diagram MATH . Here MATH is the normalization of the Cartesian product, hence MATH is branched at MATH points, where MATH is the number of ramification points of MATH whi... |
math/0005171 | We use REF that is, MATH. If MATH, MATH is itself zero. The action of MATH on MATH is trivial hence MATH is also zero. We conclude the proof by observing that for any MATH-dimensional variety MATH, MATH for MATH and MATH . |
math/0005171 | Consider the following copies of MATH embedded in MATH: MATH, MATH, MATH, MATH, MATH, and MATH. Here MATH and MATH are two distinct NAME points, MATH is the diagonal, and MATH is the image of MATH via the embedding MATH, with MATH the hyperelliptic involution. If MATH is a NAME function with MATH then one easily checks... |
math/0005171 | The first part of the theorem follows directly by combining REF . Let MATH be the zero cycles of degree REF (with MATH coefficients). Then for MATH, MATH and if MATH, then MATH. Using REF and the fact that the subscripts are additive under NAME products, we see that if MATH then the only nonzero component of MATH in MA... |
math/0005171 | The only problem is when MATH. Consider the symbol MATH on MATH, where MATH is a rational function with simple REF at MATH and therefore MATH. The boundary of MATH is MATH, and thus MATH is equivalent to a chain of the stated kind. |
math/0005171 | By taking general linear sections of MATH one may construct curves MATH such that MATH is finite. Our program is to prove that at the points of intersection of MATH with MATH the value of the relevant function MATH is constant, independent of the chosen section MATH. We may assume that MATH is smooth outside the invers... |
math/0005171 | The only possible difficulty is to see where the curves intersect. MATH intersects MATH in two points; on both curves the points come from MATH and MATH under the isomorphism with MATH, but the point which comes from MATH in MATH comes from MATH in MATH and conversely. A similar statement holds for MATH, and then inter... |
math/0005171 | Let MATH. Recall that we have a finite etale map MATH, where MATH is thought of as MATH and MATH is the discriminant locus. Let MATH be an element of MATH and let MATH be distinct from the MATH's. We may choose loops MATH based at MATH and going around MATH in such a way so that MATH generate MATH with the single relat... |
math/0005171 | Let MATH be small topological disc in MATH containing MATH and such that the only point where the map from MATH to MATH is ramified is MATH. By construction of MATH, if MATH is small enough then the ramification locus of the map MATH intersected with MATH consists of MATH disjoint punctured discs, each mapping isomorph... |
math/0005171 | Let MATH be the family constructed in the discussion preceding REF . We blow down the genus MATH curve MATH in the fibre over MATH and call the resulting family of curves MATH. By replacing MATH by a NAME open subset we may assume that all fibres are smooth and then by replacing MATH by a finite cover we may also assum... |
math/0005171 | The statement for MATH follows from the theorem. For MATH, the generic curve has a MATH-parameter family of MATH-configurations. Consider a family of genus REF curves with special fibre a hyperelliptic curve as in the theorem. By varying REF-configuration in the Jacobian of the generic fibre in the MATH-parameter famil... |
math/0005171 | By a partition of a positive integer MATH we shall mean a tuple of non-increasing positive integers MATH such that MATH and by a partition of zero we mean the empty tuple MATH. If MATH is as above and MATH is a positive integer, we let MATH be the partition of MATH obtained by reordering MATH. The orbits of the action ... |
math/0005171 | The lemma can be proved by considering a suitable NAME scheme as in REF. We do not know the number of components if MATH, but a suitable choice of the monodromy representation allows us to single out a component which gives rise to the desired specializations (compare the discussion before REF ). For example, if MATH w... |
math/0005171 | Consider the cycle MATH. The straight embedding MATH maps it to MATH in MATH. The twisted embedding MATH gives instead MATH, with section MATH, and therefore MATH is the section at MATH of MATH. We use the same type of notations as we did in REF , in particular the holomorphic form MATH comes from MATH. We need to cons... |
math/0005172 | Let MATH. For MATH, by induction we construct a distinguished triangle MATH as follows. If MATH, then we set MATH. Otherwise, we take a direct sum MATH of copies of MATH and a morphism MATH such that MATH is an epimorphism, and let MATH. Then, by easy calculation, we have the following: MATH . Let MATH be a homotopy co... |
math/0005172 | CASE: For any object MATH, we take an object MATH and a morphism MATH satisfying the conditions of REF . Then for any MATH, by REF , we have MATH . By REF , MATH implies that MATH for all MATH. Since MATH is a generating set, we have MATH, and hence MATH. It is easy to see that MATH and MATH. For any object MATH, we ta... |
math/0005172 | According to REF , MATH contains direct sums. Since MATH is a bounded complex of small projective objects of MATH, MATH is a compact object in MATH. By REF MATH has a MATH-structure MATH, and MATH is an equivalence. CASE: By the construction of MATH in REF , MATH also has a MATH-structure MATH and hence by REF we have ... |
math/0005172 | For MATH, applying MATH to a distinguished triangle MATH we have a short exact sequence MATH . Also, by REF we get MATH . Since the MATH are small objects, the above short exact sequence commutes with direct sums. |
math/0005172 | CASE: For any MATH, by REF , MATH for all MATH and hence MATH. CASE: Let MATH with MATH for all MATH. Then by REF , MATH. |
math/0005172 | By REF . |
math/0005172 | CASE: By REF . CASE: Let MATH. Since MATH, it follows that MATH. Next, apply MATH to the canonical exact sequence MATH. It then follows by REF that MATH is an isomorphism. Thus MATH and hence MATH. CASE: Obvious. CASE: By REF . |
math/0005172 | We have exact sequences in MATH . Also, MATH, MATH and MATH in MATH. Thus we get a desired distinguished triangle in MATH. |
math/0005172 | By REF . |
math/0005172 | REF According to REF can be applied. CASE: Note first that by REF we have MATH and MATH. Also, it is trivial that MATH. Let MATH. Then by REF we have a distinguished triangle in MATH of the form MATH . It follows that the sequence in MATH is exact. Thus by REF MATH is a torsion theory for MATH. |
math/0005172 | Let MATH. According to REF , it is easy to see that if MATH belongs to MATH (respectively, MATH), then MATH for MATH (respectively, MATH and MATH). Thus we have MATH so that REF can be applied. |
math/0005172 | We have MATH . |
math/0005172 | By REF . |
math/0005172 | We have MATH and for any MATH we have MATH . |
math/0005172 | This is due essentially to CITE. We have an exact sequence in MATH with the MATH finitely generated projective, and an exact sequence in MATH with the MATH finitely generated projective. CASE: Let MATH. For any MATH, we have a functorial homomorphism MATH which is an isomorphism if MATH is finitely generated projective... |
math/0005172 | CASE: By REF . CASE: Since MATH vanishes on MATH, MATH. Thus MATH contains the modules generated by MATH. Conversely, let MATH. Then, since REF implies REF , MATH and hence MATH. Thus MATH, which is generated by MATH. Next, since by REF MATH, MATH contains the modules cogenerated by MATH. Conversely, let MATH. Take a s... |
math/0005172 | By REF . |
math/0005172 | CASE: It follows by REF that MATH is a tilting complex such that MATH (MATH). Let MATH and for MATH denote by MATH the composite of canonical homomorphisms MATH . Then for MATH we have an equivalence MATH which sends MATH to MATH, so that the MATH are tilting complexes for MATH, that is, projective generators for MATH.... |
math/0005172 | By REF . |
math/0005172 | See REF. |
math/0005172 | By REF , there exists MATH satisfying MATH. Since MATH we have MATH in MATH. Also, since MATH we have MATH in MATH and MATH in MATH. Thus, we can assume MATH for MATH and MATH. |
math/0005172 | For any MATH, we have MATH . Thus MATH for MATH and MATH. Also, MATH . |
math/0005172 | CASE: According to REF , we can apply REF for a tilting complex MATH to conclude that MATH is a torsion theory for MATH. CASE: For any MATH, by REF we have MATH . Also, since by REF MATH, MATH for all MATH. The last assertion follows by REF . CASE: For any MATH, by REF we have MATH . Also, since MATH, for any MATH we h... |
math/0005172 | For any MATH and MATH, we have MATH . |
math/0005172 | By the same arguments as in the proof of REF . |
math/0005172 | By the same arguments as in the proof of REF . |
math/0005172 | For any MATH, we have functorial isomorphisms MATH . Thus MATH and hence MATH . |
math/0005172 | CASE: Let MATH. Then by REF MATH is generated by MATH and hence MATH is generated by MATH. CASE: Since MATH, by REF we have MATH. CASE: By the dual arguments. |
math/0005172 | By REF . |
math/0005172 | By REF . |
math/0005172 | According to REF , we have only to show that MATH and MATH. It follows by REF that MATH and MATH. Since MATH is a direct summand of MATH and MATH is a direct summand of MATH, it follows that MATH generates MATH and MATH cogenerates MATH. It now follows by REF that MATH and MATH. |
math/0005176 | The first step is to prove the inequality MATH where MATH is the total volume of MATH. Any self-dual REF-form MATH on any oriented REF-manifold satisfies the NAME formula CITE MATH . It follows that MATH . However, MATH simply because MATH is trace-free. Thus MATH and hence MATH . On the other hand, the particular self... |
math/0005176 | REF implies MATH . On the other hand, REF asserts that MATH . Now multiply REF by MATH, multiply REF by MATH, and add. The result is MATH . Applying the same NAME inequalities as before, we now obtain MATH . Passage from this MATH estimate to the desired MATH estimate is then accomplished by the same means as before: e... |
math/0005176 | Equality in REF implies equality in REF . However, MATH times REF plus MATH times REF reads MATH . Equality in REF therefore implies that MATH forcing the MATH-form MATH to be parallel. If MATH, we conclude that the metric is NAME, and the constancy of MATH then follows from the NAME portion of the argument. On the oth... |
math/0005176 | We begin begin with REF MATH and elect to interpret the left-hand side as the dot product MATH in MATH. Applying NAME, we thus have MATH . Thus MATH and hence MATH as claimed. In the equality case, MATH would be a closed self-dual form of constant norm, so MATH would be almost-Kähler unless MATH. |
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