paper stringlengths 9 16 | proof stringlengths 0 131k |
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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 times the CM cycle or half the CM cycle depending on the discriminant, at least up to a sign. Define MATH as the multiple which gives it, hence it is either the smallest multiple or twice the smallest multiple. From the description above, and the fact that MATH is even, which we will see presently, it is clear that MATH and MATH. The smallest multiple is determined by MATH, if MATH is even, then it is MATH otherwise it is MATH. The proposition will then follow from the following lemma MATH always. Let MATH be the singular relation corresponding to the elliptic curve MATH. From NAME 's explicit description of the embedding, one can compute MATH and MATH to be MATH . Let MATH . It turns out that MATH satisfies an equation of discriminant MATH. This is even if and only if MATH is even. Hence MATH always, as if MATH is even, then the smallest multiple is MATH hence MATH is twice the smallest multiple. If MATH is odd,then the smallest multiple is MATH and MATH is the smallest multiple. This also says that the discriminant of MATH is MATH up to a square factor. This concludes the proof of the proposition. |
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 basis given by MATH and the fact that the change of basis matrix has determinant MATH, a simple calculation shows that the determinant is MATH where MATH is the quadratic form MATH. Comparing the two shows that MATH where MATH . If MATH and MATH are two elements of the primitive NAME, then the lattice generated by them is the NAME lattice of an abelian surface with multiplication by an NAME order of discriminant MATH. This explains why MATH and MATH are in MATH. Conversely, if MATH and MATH satisfy the conditions that MATH and MATH are in MATH and MATH divides MATH then one can see that the element MATH where MATH satisfies an equation of discriminant MATH, is a primitive integral element of invariant MATH, hence, from CITE, one sees that it is an elliptic curve of degree MATH. |
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 each MATH, let MATH be a vector with MATH. By REF , the sum MATH converges to an element MATH. For every MATH, we have MATH . Hence for all MATH, MATH . It follows from CITE that MATH for any MATH; that is, MATH. Thus MATH, so MATH is not a closed vector. Now suppose that MATH is hyperatomic. Let MATH be a finite set of atoms of MATH such that MATH . By deleting some atoms, if necessary, we may assume that MATH is a minimal set which generates MATH. Thus, if MATH is any proper subset of MATH, then MATH. Let MATH, MATH . Note that MATH, for each MATH. (Otherwise, we have MATH and MATH is in MATH, a contradiction.) Now let MATH be any vector in MATH. Then there exist vectors MATH (not necessarily unique) such that MATH, for each MATH, and MATH. By REF , there is, for each MATH, an element MATH such that MATH and MATH. Let MATH. Clearly, MATH whenever MATH. So MATH. This shows that MATH, and it follows that MATH is a closed vector for MATH. |
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. By REF , MATH is a hyperatomic projection. Since MATH is arbitrary, every projection is hyperatomic and so MATH is hyperatomic. CASE: Obvious. CASE: If MATH is any invariant linear manifold, then MATH is closed by hypothesis, so by the equivalence of REF , we see that there exists a vector MATH such that MATH. |
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 that there is a vector MATH in MATH such that MATH . Write MATH and MATH for the projections onto MATH and MATH. These projections are in MATH and hence in MATH. Let MATH. We will show that MATH. Since MATH, it follows immediately that MATH. Now let MATH. Then there is MATH such that MATH. Since MATH, we have MATH . Since MATH, MATH . This shows that MATH. Combining this with MATH gives MATH. The reverse inequality follows from the fact that MATH and the claim is verified. Now let MATH be an arbitrary invariant linear manifold for MATH. We need to show that MATH is closed. Let MATH be the projection onto the closure of MATH. Clearly MATH and hence is a hyperatomic projection. Let MATH be a family of atoms of MATH such that MATH. Then MATH. Notice that if MATH is a non-zero vector in MATH, then MATH. Inductively applying REF , we see that there exists MATH such that MATH, whence MATH is singly generated. Since MATH is MATH-transitive, MATH is a closed vector, whence MATH is closed. CASE: Obvious. It remains to show that when MATH, then every invariant manifold for MATH is singly generated. Let MATH be an invariant linear manifold for MATH. Then by hypothesis, the orthogonal projection MATH onto MATH belongs to MATH, hence there is a vector MATH such that MATH. Clearly MATH. Now MATH and since MATH is the smallest element of MATH containing MATH, we conclude that MATH. |
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 fix a unit vector MATH and let MATH be any vector. Let MATH be a sequence of projections in MATH such that each MATH is a finite sum of atoms of MATH, MATH, MATH and MATH. Fix MATH. Write MATH as a finite sum of atoms of MATH and let MATH. Since MATH, and MATH is weakly dense in MATH, we may find a norm one operator MATH such that MATH. Hence we may find a unit vector MATH such that MATH. By NAME 's transitivity theorem, there exist unitary operators MATH and MATH such that MATH . Writing MATH we see that MATH . Therefore, if MATH, we find MATH and MATH. Moreover, since MATH, we find that for any MATH, MATH, so MATH . Notice also that MATH by construction. The fact that MATH shows that the series MATH converges uniformly to an element MATH. Clearly, MATH. Thus we have shown that the invariant manifold generated by MATH is all of MATH. Furthermore, notice that our construction shows the following: CASE: MATH; CASE: if MATH is an atom of MATH such that MATH, then MATH; and CASE: MATH, where for each MATH, MATH is a finite sum of elements MATH which satisfy MATH for some atoms MATH and MATH of MATH. Returning to the general case, if MATH is any atom from MATH, we may compress to MATH (that is, replace MATH by MATH acting on MATH) and apply the argument above to obtain the following: if MATH is any non-zero vector in MATH and if MATH, then there is MATH such that MATH, MATH, and MATH implies MATH for all atoms MATH. (There is one delicate point: our hypotheses do not guarantee that MATH, but in the construction of MATH, MATH is a norm limit of elements which are finite sums of elements of the form MATH with MATH and MATH atoms of MATH. Such elements are in MATH, by our hypotheses.) Now let MATH be an invariant linear manifold under MATH. Let MATH be the projection onto MATH. Then MATH is invariant under MATH and, hence, under MATH. So, MATH. Let MATH be independent atoms which generate MATH. So, MATH and MATH whenever MATH. There is a vector MATH such that MATH for all MATH. (Let MATH with MATH and approximate MATH in norm by an element of MATH.) Clearly MATH. We will prove that MATH; this implies that MATH, whence MATH is closed and singly generated. Let MATH be arbitrary. Write MATH, where MATH for each MATH. This can be done since MATH. For each MATH, there is an element MATH such that MATH, MATH, and MATH implies MATH, for all atoms MATH. Let MATH. Then MATH . Thus, MATH and MATH. Finally, since MATH is weakly dense in MATH, for every MATH we have MATH. Since MATH is hyperatomic, we have MATH, so MATH for every MATH. It follows that MATH is MATH-transitive. |
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 MATH decomposes as finite orthogonal sum of invariant manifolds for MATH, whence MATH is closed. |
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. But this contradicts the assumption that our algebra is not commutative (index equals 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 and MATH. Let MATH, MATH, where both MATH. Let MATH be an element corresponding to MATH according to REF and MATH be the element corresponding to MATH. Let MATH be an arbitrary element of MATH. Assume MATH (since MATH and MATH are finite). Then: MATH . Now by assumption of induction we have MATH which concludes the proof. |
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. |
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-standard order of multiplication between vectors and operators). |
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 the following computation: MATH . |
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 operator MATH is given by the matrix MATH . |
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 is defined on MATH - as opposed to MATH as used in REF . |
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 and MATH. We will show that their product is zero. Indeed, MATH forms a MATH-dimensional subalgebra with unity in MATH. Due to classification of MATH-dimensional associative algebras with unity we know that they have trivial spaces MATH for MATH. Thus MATH must be MATH. This proves that the map MATH is trivial. However this contradicts REF which states that characteristic polynomial must be zero in this case (and it is not in view of REF ). |
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 MATH. Under the move each of the disks experiences a diffeotopy. Before and after the move the intersection the curve with MATH is the union of the two disks meeting each other transversally in two points, but before the move the disks do not intersect MATH, while after the move they intersect MATH in their common points. To calculate the writhe at both vanishing solitary double points, let us select the same imaginary branch of the projection of the curve passing through the points. This means that we select one of the disks described above. The sum of the local intersection numbers of this disk (equipped with the complex orientation) and MATH (equipped with some orientation) is zero since under the fourth move the intersection disappears, while in the boundary of MATH no intersection happens. Therefore the local orientations of the projective plane in the vanishing solitary double points defined by this branch define opposite orientations of MATH. (Recall that the local orientations are distinguished by the condition that the local intersection numbers are positive.) On the other hand, under the move the preimages of the vanishing solitary double points come to each other up to coincidence at the moment of the move and their orientations defined by the choice of the same imaginary branch are carried to the same orientation of the preimage of the point of solitary self-tangency. Indeed, the preimages are real lines and points of intersection of their complexifications with the selected imaginary branch of the curve also come to the same position. Therefore the halves of the complexifications containing the points come to coincidence, as well as the orientations defined by the halves on the real lines. It follows that the intersection numbers of MATH with the preimages of the vanishing solitary double points equipped with these orientations are equal. Since the local orientations of the projective plane in the vanishing solitary double points define distinct orientations of MATH, the writhes are opposite to each other. |
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 transversal to each other in three-dimensional projective space such that their intersection has zero self-linking number. |
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 MATH for any simple MATH-vector in MATH, MATH still has comass one (but since MATH is not closed it is not a calibration itself). Moreover, submanifolds MATH of MATH on which MATH restricts to the volume form are exactly those for which MATH is calibrated by MATH. Hence the result follows by taking MATH to be any of the MATH-special Lagrangian calibrations MATH. |
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 Legendrian. The converse is similar. |
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 function. It is straightforward to check that any such MATH gives rise to a Lagrangian immersion to MATH. Let MATH and for MATH denote MATH by MATH. Then for MATH where the first term vanishes because MATH is Legendrian and the second because MATH. For MATH we have MATH and so MATH is Lagrangian as claimed. Now we claim that MATH is MATH-SLG if and only if MATH satisfies MATH for some real constant MATH. To prove this it is enough to show that MATH holds if and only if REF is satisfied. But MATH is equivalent to MATH . Since MATH is MATH-SLG, we have MATH and hence MATH . Thus REF holds if and only if MATH . Hence MATH is MATH-SLG if and only if MATH as claimed. REF are straightforward to verify. |
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 directly from the equations for a minimal Legendrian equation by showing that the constraints REF are not consistent with the equations of motion of the NAME system unless MATH. To see this let us compute the second derivative of the mysterious constraint MATH for a solution of the NAME system at an instant when all the constraints and their first derivatives are satisfied. One finds MATH . Let MATH and MATH. Then MATH may be expressed in terms of the symmetric polynomials in the MATH as MATH . Hence using the constraints we have MATH . A calculation shows MATH and so MATH . Clearly once we have imposed the constraint MATH, MATH if and only if MATH. Moreover, by differentiating REF it is easy to verify that all higher derivatives of the constraint MATH also vanish when MATH. Thus to show existence of minimal Legendrian immersions we need only show there exist initial conditions for the NAME system which satisfy all the constraints together with the first derivative of the mysterious constraint. We will see that this is indeed the case in the proof of REF which we now give. |
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 sphere can be written as MATH and MATH is equivalent to MATH. Let MATH be the cross product of MATH and MATH . The constraints in REF are equivalent to MATH for some constant MATH, while the constraints in REF are equivalent to MATH for some function MATH. The remaining constraint MATH then becomes MATH or in terms of MATH . Since we seek periodic solutions we may assume that MATH, MATH. Then at MATH, REF becomes MATH and thus specifying MATH determines MATH (and vice versa). Let us fix MATH and consider the case where MATH. Given MATH, REF has a unique smallest nonnegative root MATH. Let MATH. Then up to a translation in time any periodic solution of REF (except possibly a solution corresponding to MATH which we shall treat later) arises from such an initial condition. Once the initial conditions for MATH and MATH have been specified, REF fixes MATH and MATH for MATH. Given MATH and MATH, REF fixes MATH. If we define MATH, then MATH is determined by REF for MATH. By a global rotation in MATH (for example, replacing MATH by MATH where MATH) we may rotate MATH so that MATH. We may not assume also that MATH without allowing MATH in which case we will change the value of MATH for which MATH is MATH-special Legendrian. In the case MATH we shall verify later that choosing MATH gives rise to a MATH-special Legendrian immersion (in the case MATH choosing MATH gives rise to a MATH-special Legendrian immersion). Thus for each MATH, and MATH there is a unique solution of the NAME equation (given by specifying initial data in the manner above) which satisfies the constraints REF . Hence by the proof of the previous lemma it gives rise to a minimal Legendrian immersion which we denote MATH. In the case MATH we will explicitly exhibit solutions later in this section and see that as MATH the period of MATH becomes infinite, and that the limiting solution MATH describes a minimal Legendrian sphere (which as previously noted is necessarily totally geodesic). To see which immersions MATH are geometrically distinct consider in greater detail the geometry of these immersions. Since the immersions are all conformal, the metric MATH induced on MATH can be described by a single positive function MATH, where MATH. A calculation shows that MATH and MATH are related by MATH . It follows from REF that MATH satisfies MATH where MATH . The NAME curvature of the immersion satisfies MATH . In the case MATH, the corresponding solution of REF is MATH independent of the choice of MATH and hence MATH has MATH. It follows that MATH must be (a piece of) a generalized NAME torus. Similarly, if MATH then MATH and it follows from REF that MATH. Once again MATH and hence MATH is a (piece of a) generalized NAME torus. All other immersions MATH are geometrically distinct. To begin with, note that the remaining immersions are all invariant under a unique MATH-parameter family of MATH - the subgroup generated by MATH. For MATH these are all inequivalent, hence MATH and MATH are distinct when MATH. Now fix MATH and consider MATH for MATH. We claim that the minimum and maximum values of the NAME curvature MATH are respectively strictly decreasing and increasing functions of MATH on MATH. It follows that MATH and MATH are geometrically distinct when MATH. To proof the previous claim, note that REF shows that for a given immersion MATH the minimum (maximum) value of MATH occurs at the minimum (maximum) value of MATH. From REF it is clear that for fixed MATH, MATH and MATH, the minimum and maximum values attained by y, are strictly decreasing and increasing functions of MATH respectively. Since MATH is a decreasing function of MATH, from REF we see that the minimum and maximum values of MATH are, like MATH, strictly decreasing and increasing functions of MATH respectively as claimed. |
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 the case MATH we have MATH and MATH. Thus MATH and MATH . From REF we see that MATH as MATH, and MATH as MATH. In these two limits MATH reduces to MATH and MATH respectively. Thus in the limiting case MATH we have MATH . Finally, one can show that in order for any immersion of the form MATH to describe a harmonic sphere, the limit of MATH as MATH must be a fixed point of the action MATH CITE. Moreover, all the conserved quantities of the NAME system must also be zero (since they are zero at a fixed point). For MATH as above, MATH has nonzero fixed points if and only if MATH, in which case any point of the form MATH is fixed. From REF , all three angular momenta MATH are zero if and only if MATH. Thus MATH is the only MATH which could describe a minimal sphere. In this case MATH (in the MATH-special Legendrian case) has the explicit form MATH and we can see directly that the MATH-sphere described is the intersection of the plane MATH with the MATH-sphere (and hence is totally geodesic). |
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 periods then MATH is rational by REF . By REF any period is of the form MATH. Now since MATH is MATH-periodic we have MATH. Then periodicity with respect to MATH is equivalent to MATH. Hence MATH, for MATH. Together with rationality of MATH this implies MATH. Conversely, by REF the rationality of MATH gives us one period MATH. From above MATH is a period if and only if MATH. By REF for some integers MATH and MATH. With MATH and MATH the period condition becomes MATH . For MATH this condition is trivial, while it holds for MATH because MATH . Since MATH and by REF , MATH, we have MATH . So the MATH period condition also holds, and hence MATH is a second period of MATH. |
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 periods in the rectangle MATH formed by MATH, MATH, MATH and MATH. Let MATH be such a period. By REF of the previous proposition MATH must be a integer multiple of MATH, the basic period of MATH. It is easy to see that the smallest period of the form MATH occurs when MATH and so we need only deal with the case MATH. Using the fact that MATH, MATH, MATH we find that MATH is a period if and only iff MATH satisfies MATH . Clearly the third equation is implied by the first two. Moreover, the first equation implies MATH, whereas the second implies MATH. Hence if either MATH or MATH is even (both cannot be even since we assumed MATH) then these two equations are inconsistent. Thus there are no further periods and the period lattice is generated by MATH and MATH. If both MATH and MATH are odd, then one can check that MATH is the unique solution in MATH. Hence MATH is the only new period in the rectangle MATH and in this case the period lattice is generated by MATH (or MATH) and MATH. Let us show embeddedness in the case where one of MATH, MATH is even. The other case is similar, but a little more involved since the period lattice is not rectangular. We need to show that if MATH and MATH and MATH then MATH and MATH. From our explicit formulae for MATH we see that MATH is equivalent to MATH . Certainly this implies MATH, which implies there exists some MATH such that MATH. If MATH and MATH are distinct, there are essentially two different cases, depending on whether MATH or MATH. In the first case the three equations above reduce to MATH where MATH. That is, we have the same equations as occurred in the periodicity part of the proof. Since we assumed one of MATH and MATH was even, the first two equations are inconsistent unless MATH in which case MATH is also forced. In the second case the equations reduce to MATH . Clearly, the first two equations are inconsistent with the third one. |
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, we have MATH . Obviously, for every unit vector MATH we can choose a scalar MATH with MATH such that MATH. By the properties of our original transformation MATH, we can clearly suppose that here in fact we have MATH. We define a function MATH on the set MATH of all finite rank projections (self-adjoint idempotents) on MATH as follows. If MATH, then there are pairwise orthogonal unit vectors MATH such that MATH. We set MATH . Apparently, the operators MATH are pairwise orthogonal rank-one idempotents (two idempotents MATH are said to be orthogonal if MATH). Hence, MATH is a rank-MATH idempotent. We have to check that MATH is well-defined. This follows from the following observation. We have MATH and MATH where MATH denotes generated subspace. Now, suppose that the pairwise orthogonal unit vectors MATH generate the same subspace as MATH do. Let MATH. Then there exist a vector MATH and a scalar MATH of modulus REF such that MATH. We have MATH . This shows that the range of MATH is the same as that of MATH. The same applies for the kernels. Since the idempotents are determined by their ranges and kernels, this proves that MATH is well-defined. It is now clear that MATH is an orthoadditive measure on MATH. We show that MATH is bounded on the set MATH of all rank-one projections which is equivalent to MATH . Suppose, on the contrary, that there is a sequence MATH of unit vectors in MATH for which MATH. Since MATH is bounded, it has a subsequence MATH weakly converging to a vector, say, MATH. We have MATH . Since this holds for every MATH, we deduce that MATH is weakly bounded which implies that it is in fact norm-bounded. The same argument applies in relation to MATH. Hence, we obtain that MATH has a subsequence MATH such that MATH are bounded which is a contradiction. Consequently, MATH is bounded on MATH. By NAME 's theorem MATH can be extended to a NAME homomorphism of MATH. In fact, if MATH is self-adjoint, then there are finite-rank projections MATH (here, we do not require that they are pairwise orthogonal) and scalars MATH such that MATH. Let MATH . Consider a finite dimensional subspace MATH of MATH with dimension at least REF which contains all the subspaces MATH. Since MATH is bounded on MATH, by the variation CITE of NAME 's theorem, for every MATH there is an operator MATH on MATH such that MATH . |
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 conjugate-linear operator MATH on MATH and a scalar function MATH such that MATH. Since MATH it follows that MATH for every MATH. Therefore, the linear operators MATH and MATH are locally linearly dependent which means that MATH and MATH are linearly dependent for every MATH. Since none of the operators MATH and MATH is of rank REF, by CITE we obtain that there is a scalar MATH such that MATH. Let MATH be arbitrary nonzero vectors. Pick MATH such that MATH. From REF we now infer that MATH . This shows that MATH is constant. If MATH denotes this constant, then we have MATH. Let MATH. Then MATH is of modulus REF and we have MATH . Consider the factorization MATH . Since MATH is of modulus REF, the proof is complete. |
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 can assume that our original map MATH satisfies MATH and MATH. The main step of the proof which follows is to prove that MATH is orthoadditive on the set of all projections. This means that MATH for any mutually orthogonal projections MATH. To see this, let MATH be an arbitrary projection and set MATH. Consider the map MATH from MATH into the group of all invertible operators in MATH. This is a continuous one-parameter group and hence there is an operator MATH such that MATH (see, for example, CITE). Since MATH is *-preserving, we obtain that MATH is self-adjoint for all MATH. This yields that MATH is also self-adjoint. The norm and the spectral radius of any self-adjoint operator coincide. So, from the continuity property of MATH we deduce that for every MATH there is a MATH such that MATH if MATH. Therefore, the function MATH is uniformly continuous on the positive half-line for every MATH. This gives us that MATH. Consequently, MATH is a projection. Then we have MATH and thus MATH or, equivalently, MATH for every positive MATH. By the continuity property of MATH we obtain MATH. As a particular case, we have MATH . Indeed, this follows from MATH and REF. Therefore, MATH is positive homogeneous, and referring to REF again, if we divide by MATH and use the continuity property of MATH, then we arrive at MATH. Thus, we obtain MATH for every projection MATH in MATH. If MATH are arbitrary projections with MATH, then we infer from the multiplicativity of MATH and REF that MATH . Consequently, MATH is orthoadditive on the set of all projections. Since MATH, it follows that MATH sends unitaries to unitaries. Consider the map MATH. Clearly, this is a continuous one-parameter unitary group. By NAME 's theorem there is a self-adjoint operator MATH such that MATH . Since MATH, we have MATH. By spectral mapping theorem this yields that the spectrum of MATH consists of integers. So, MATH can be written in the form MATH, where the MATH's are pairwise orthogonal projections with MATH and MATH is a suitable positive integer. We compute MATH . Consequently, we have MATH . From REF we infer that MATH . We know that MATH commutes with MATH for every MATH. Thus, for any MATH we have MATH for every MATH of modulus REF. This implies that MATH if MATH. Consequently, MATH can be written in the form MATH or, in another way, MATH . Here, every MATH is a *-semigroup endomorphism of MATH which is uniformly continuous on the commutative MATH-subalgebras. Every orthoadditive projection valued measure on the set of all projections in MATH can be extended to a linear map on MATH. This is a particular case of the solution of the NAME problem obtained in CITE. Since this extension is linear and sends projections to projections, it is a standard argument to verify that this is in fact a NAME *-homomorphism (see, for example, the proof of CITE). A linear map MATH between *-algebras MATH and MATH is called a NAME *-homomorphism if it satisfies MATH . So, we have NAME *-homomorphisms MATH of MATH such that MATH for every projection MATH. Let MATH be pairwise orthogonal projections whose sum is MATH and pick nonzero scalars MATH. Using the orthoadditivity of MATH, for any MATH we compute MATH . Using the continuity property of MATH, the automatic continuity of NAME *-homomorphisms between MATH-algebras and the spectral theorem of normal operators, we deduce that MATH holds for every invertible normal operator MATH (note that MATH and the range of its spectral measure generate a commutative MATH-subalgebra). Every NAME *-homomorphism of MATH is the direct sum of a *-homomorphism and a *-antihomomorphism (see CITE). Let MATH denote the *-homomorphic and let MATH denote the *-antihomomorphic part of MATH. Let MATH be invertible normal operators whose product is also normal. By the multiplicativity of MATH and REF we have MATH . This implies that MATH and MATH . Any *-homomorphism or *-antihomomorphism of MATH is either injective or identically REF which follows from the form of representations of MATH on separable NAME spaces (see CITE). Now, taking REF into account, one can verify that the only values of MATH for which MATH can be nonzero are MATH and MATH. Moreover, because of the same reasons, for MATH we have MATH and for MATH we have MATH. Observe that MATH. Therefore, MATH can be written in the form MATH for every invertible normal operator MATH, where MATH is a *-endomorphism and MATH is a *-antiendomorphism of MATH. By continuity and spectral theorem we clearly have REF for every normal operator in MATH. Define MATH for every MATH. Clearly, MATH is an additive *-semigroup endomorphism of MATH (it is not linear unless MATH is missing). It is easy to see that every rank-one operator is the product of at most three (rank-one) normal operators. This gives us that MATH for every rank-one operator MATH. Now, let MATH be arbitrary. Pick rank-one projections MATH. Since MATH is of rank at most REF, we compute MATH . Since every *-endomorphism of MATH is normal, that is, weakly continuous on the bounded subsets of MATH (see CITE), it follows from REF that for every maximal family MATH of pairwise orthogonal rank-one projections we have MATH. Therefore, we infer from REF that MATH for every MATH. Finally, in order to get the explicite form of MATH we refer once again to the form of linear *-endomorphisms of MATH appearing in CITE and note that one could get the form of linear *-antiendomorphisms of MATH in a similar way. |
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 MATH denote the real part and the imaginary part of MATH, respectively. Then MATH is the unique linear extension of MATH from MATH to MATH. Let MATH be an affine function. We assert that MATH is continuous in the norm topology. To see this, consider the affine function MATH on MATH which sends REF to REF. Since its unique linear extension MATH has the property that MATH for every MATH, we deduce that MATH . Clearly, every element MATH of the unit ball of MATH can be written as MATH for some MATH. It follows that MATH is bounded on the unit ball of MATH. This implies that MATH and hence MATH are norm continuous. Let now MATH be an affine bijection. Then MATH and it inverse are norm-continuous. Moreover, MATH obviously preserves the extreme points of MATH which are exactly the projections in MATH. We claim that MATH is either MATH or MATH. Let MATH be a projection in MATH. If every projection in MATH commutes with MATH, then we obtain that every element of MATH commutes with MATH which, MATH being a factor, would imply that MATH is a scalar multiple of the identity but this is a contradiction. So, we can choose a projection MATH in MATH which does not commute with MATH. Considering the operator MATH we get a unitary element in MATH which does not commute with MATH. So, we have MATH. In any NAME algebra the unitary group is arcwise connected. Therefore, there is an arc within the set of all projections in MATH connecting MATH to MATH. To sum up, every nontrivial projection in MATH can be connected by an arc within the set of all projections to another projection different from the first one. It is trivial that MATH and MATH can be connected only to themselves. Since MATH is a homeomorphism of the set of all projections in MATH, we deduce that MATH sends nontrivial projections to nontrivial projections and hence we have either MATH or MATH. Clearly, we can assume without loss of generality that MATH (otherwise, we consider the transformation MATH). Let MATH be the unique linear extension of MATH onto MATH. We already know that MATH is a bounded linear transformation which sends projections to projections. It is a standard algebraic argument to verify that MATH is a NAME *-homomorphism (see CITE and use the spectral theorem of self-adjoint operators together with the continuity of MATH). We assert that MATH is bijective. If MATH, then we see that MATH. Let MATH denote the positive and negative parts of MATH, respectively, From MATH we infer that MATH. Supposing that MATH, this means that MATH . Using the homogenity of MATH for the scalars in MATH, we conclude that MATH . Since MATH is injective, it follows that MATH which gives us that MATH. Similarly, one can check that MATH is also true, so we have MATH. Therefore, MATH is injective. Since the range of MATH is a linear subspace of MATH which contains MATH (recall that MATH is an extension of MATH), it follows that MATH is surjective. So, MATH is a NAME *-automorphism of MATH. It is well-known that every factor is a prime algebra. This means that for any MATH, the equality MATH implies that MATH or MATH. Now, REF on NAME homomorphisms CITE applies to obtain that MATH is either a *-automorphism or a *-antiautomorphism of MATH. This completes the proof of the theorem. |
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. Let MATH be projections and suppose that MATH. Then we have MATH which yields MATH. This implies that MATH is an idempotent. On the other hand, since MATH and MATH are projections, the norm of their product is not greater than REF. So, MATH is a contractive idempotent. It is well-known that this implies that MATH is a projection and hence, due to the self-adjointness, it follows that MATH and MATH are commuting. Hence, we can compute MATH which yields that MATH. Since MATH has the same properties as MATH, it follows that MATH preserves the ordering MATH in both directions. In particular, we obtain that MATH, MATH and that MATH preserves the nonzero minimal projections, that is, the rank-one projections on MATH. We claim that MATH preserves also the orthocomplementation on the set of projections. To see this, we first show that MATH preserves the mutual orthogonality. Let MATH be projections such that MATH. Then we have MATH which implies that MATH . This gives us that MATH. It follows that MATH is a projection, say MATH. Since MATH and MATH preserves the ordering in both directions, we infer that MATH. This gives us that MATH and, hence, MATH. Therefore, MATH preserves the orthocomplementation on the set of all projections. The form of such transformations, that is, the form of all bijections of the set of all projections on a NAME space with dimension not less than REF which preserve the order in both directions and the orthocomplementation, is well-known (see, for example, CITE). Namely, there is an either unitary or antiunitary operator MATH on MATH such that MATH for all projections MATH on MATH. We next prove that MATH for every MATH and every rank-one projection MATH. In fact, in that case we can compute MATH for some scalar MATH which follows from the fact that MATH is of rank one. We assert that MATH is a multiplicative function. If MATH, then we have MATH which implies that MATH. Choosing MATH, it follows that MATH. We next obtain that MATH. Since this holds for every MATH, we conclude that MATH is multiplicative. We now claim that MATH does not depend on the rank-one projection MATH. If MATH are rank-one projections which are not mutually orthogonal, then MATH and we have MATH . This gives us that MATH. If MATH are mutually orthogonal, then there is a rank-one projection MATH such that MATH and MATH. Thus we have MATH. So, there is a multiplicative function MATH such that MATH for every MATH and rank-one projection MATH on MATH. We show that MATH is also additive on MATH. To see this, for any unit vector MATH denote by MATH the rank-one projection onto the linear subspace of MATH spanned by MATH. Let MATH be mutually orthogonal unit vectors and MATH such that MATH. Then MATH is a unit vector. We compute MATH in two different ways. On the one hand, since MATH, we have MATH. On the other hand, we compute MATH where we have used the fact that MATH is orthoadditive on the set of all projections (this follows from the form of MATH on that set). Therefore, we have MATH. By multiplicativity, we obtain the additivity of MATH. We claim that MATH is in fact the identity on MATH. Since MATH maps into MATH, one can easily check that MATH is monotone increasing. Moreover, as MATH, the additivity of MATH implies that MATH for every rational number MATH in MATH. If MATH is arbitrary, then approximating MATH by rationals MATH from below and above, respectively, by the monotonity we can infer that MATH. We already know the form of MATH on the set of all projections. It is easy to see that without loss of generality we can assume that MATH holds for every projection MATH and we then have to prove that MATH is the identity on the whole interval MATH. But this is now easy. Indeed, let MATH. Pick an arbitary rank-one projection MATH, where MATH is a unit vector. Then we compute MATH . Since MATH was arbitrary, we obtain MATH for every MATH. This completes the proof of the theorem. |
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 or MATH. Clearly, the function MATH is a bijective mapping of MATH satisfying the equation appearing in the statement. So, without loss of generality we can assume that MATH. We prove that MATH sends projections to projections. If MATH is a projection, then MATH is self-adjoint and we have MATH which shows that MATH is an idempotent. Now, we can follow the argument in the proof of REF . One can verify that MATH preserves the partial ordering MATH in both directions and the orthocomplementation on the set of all projections. So, we have an either unitary or antiunitary operator MATH on MATH such that MATH for every projection MATH on MATH. One can check that for every rank-one projection MATH there exists a function MATH such that MATH. We next obtain that MATH and choosing MATH, this gives us that MATH. In particular, by the injectivity of MATH, from MATH we deduce that MATH. Since MATH, we get that MATH is multiplicative. One can next show that MATH does not depend on the rank-one projection MATH. So, there is a multiplicative function MATH such that MATH for every real number MATH and rank-one projection MATH on MATH. As for the additivity of MATH, just as in the proof of our previous theorem we get that MATH for every real MATH, where MATH. To show that MATH is additive, it is enough to verify that MATH for every real MATH. If MATH, then we already know this. If MATH, say MATH, then we can refer to the equality MATH what is known to be valid since the numbers MATH belong to MATH and their sum is REF. So, we have MATH . Consequently, we obtain that MATH is an injective multiplicative and additive function. This means that MATH is a nontrivial ring endomorphism of MATH. It is well-known that MATH is necessarily the identity (anyway, this can be proved quite similarly to the corresponding part of the proof of REF ). Finally, one can complete the proof of the statement just as in the case of our previous theorem. |
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 follows that MATH and this implies that MATH. So, we have the rank preserving property of MATH for MATH. It follows from the first part of the proof of CITE that every tripotent on a NAME space is the difference of two mutually orthogonal idempotents (to be honest, the theorem in question is about NAME spaces, but the part of the proof that we need here applies also for NAME spaces). Suppose now that MATH is a rank-MATH tripotent if and only if MATH is a rank-MATH tripotent holds true for MATH. Let MATH be a rank-MATH tripotent. Then the rank of MATH is at least MATH. Let MATH be a rank-MATH tripotent such that MATH and MATH. The existence of such a tripotent follows from the representation of tripotents as differences of mutually orthogonal idempotents mentioned above. Let MATH. We have MATH . Clearly, the rank of MATH is at least MATH. On the other hand, the first equality in REF gives us that the range of MATH is included in the range of MATH, so the rank of MATH is exactly MATH. The tripotent MATH is the difference of two mutually orthogonal idempotents. These idempotents induce a splitting of MATH into the direct sum of three closed subspaces. With respect to this splitting every operator has a matrix representation. In particular, we can write MATH . Let MATH be the representation of MATH. It follows from the first equality in REF that the only possibly nonzero entries in the matrix of MATH are MATH and MATH. The second equality in REF now implies that MATH and MATH are idempotents. By the equality of the ranks of MATH and MATH we conclude that MATH. Therefore, we have MATH and this gives us that the rank of MATH is MATH. By symmetry, we obtain that if MATH has rank MATH, then the same must be true for MATH. The key step of the proof now follows. We intend to prove that if MATH are mutually orthogonal rank-REF idempotents, then MATH. In order to verify this, let MATH be a rank-REF idempotent. Then it follows that MATH is a rank-REF tripotent. Let MATH, where MATH are idempotents with MATH. If MATH is any operator satisfying MATH, then we have MATH. We compute MATH which implies that MATH. We similarly have MATH. It follows that MATH. Conversely, if MATH, then we infer MATH . The algebra of all operators MATH for which MATH holds is isomorphic to MATH. Let MATH be the rank of MATH and let MATH be the rank of MATH. Clearly, we have MATH. The algebra of all operators MATH for which MATH is isomorphic to MATH. Therefore, MATH induces a bijective transformation MATH which satisfies REF. We assert that either MATH or MATH. Suppose on the contrary that, for example, MATH and MATH. One can see that there are five rank-REF tripotents MATH on the NAME space MATH such that MATH. Indeed, choose an orthonormal basis MATH in MATH and consider the operators MATH . They fulfil the requirements. It follows that there are five rank-REF tripotents in MATH with similar properties. This readily implies that there are four rank-REF tripotents MATH in MATH for which MATH. But this cannot happen. In fact, applying a similarity transformation or the negative of a similarity transformation we can suppose that MATH for some unit vector MATH in MATH. Choose a unit vector MATH which is orthogonal to MATH. Since MATH, it follows that in the `vector-tensor-vector' representation of any of MATH either the first or the second component is a scalar multiple of MATH. Clearly, at least in two of MATH, the vector MATH appears in the same component. For example, suppose that MATH and MATH. Since MATH, we obtain that either MATH or MATH. But this implies that either MATH or MATH which contradicts the fact that MATH are nonzero tripotents. We have proved that the induced transformation MATH is a bijection from MATH onto itself which satisfies REF. Moreover, observe that we have also obtained that either MATH (this is the case if MATH) or MATH (this is the case if MATH). Without loss of generality we can suppose that MATH. It follows from REF that MATH which gives us that MATH preserves the idempotents in both directions. If MATH are idempotents in MATH such that MATH (that is, if MATH), then we obtain MATH . Since MATH are idempotents, multiplying the second equality in REF by MATH from the left and from the right respectively, we find that MATH. So, we obtain that MATH preserves the partial ordering MATH between the idempotents in MATH in both directions. We now apply a nice result of CITE describing the automorphisms of the poset of all idempotents on a NAME space of dimension at least REF. It is a trivial corollary of his result that our transformation MATH is orthoadditive on the set of all idempotents in MATH, that is, if MATH are mutually orthogonal idempotents in MATH, then we have MATH. Turning back to our original transformation MATH we see that if MATH are mutually orthogonal rank-REF idempotents, then MATH. If MATH is a rank-REF tripotent and MATH is a scalar, then we have MATH for some scalar MATH. This follows from the fact that MATH has rank REF. We have MATH which gives us that MATH for every MATH. Choosing MATH, we see that MATH. From REF we now obtain that MATH is a multiplicative function. We next assert that MATH does not depend on MATH. Let MATH be a rank-REF tripotent with the property that MATH. We compute MATH . On the other hand, we also have MATH . This yields that MATH and then we have MATH. If MATH, then we can choose a rank-REF tripotent MATH such that MATH and MATH. Hence, we can infer that MATH. This means that MATH really does not depend on MATH. In what follows MATH denotes this common scalar function. Let MATH be arbitrary. Then we have MATH . Since this holds for every rank-REF tripotent MATH on MATH and MATH runs through the whole set of rank-REF tripotents on MATH, we obtain that MATH for every MATH which yields that MATH . We prove that MATH is additive. Let MATH be linearly independent vectors, and choose linear functionals MATH such that MATH and MATH. Let MATH be arbitrary and let MATH, MATH, MATH. By the orthoadditivity property of MATH we can compute MATH and this proves that MATH is additive. We now verify that MATH is additive. Let MATH be arbitrary and pick any rank-REF tripotent MATH on MATH. We have MATH such that MATH. We compute MATH . Since this holds true for every rank-REF tripotent MATH on MATH, we obtain that MATH. Consequently, MATH is an additive bijection satisfying REF. Since every standard operator algebra MATH on a NAME space is prime (this means that for every MATH, the equality MATH implies MATH or MATH), we can apply a result of CITE (also see CITE) to obtain that MATH is necessarily a homomorphism, or an antihomomorphism, or the negative of a homomorphism, or the negative of an antihomomorphism. In the homomorphic cases the satement follows from CITE, while in the antihomomorphic cases one can apply analogous ideas (compare CITE). |
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 MATH and MATH, MATH being the subspace corresponding to the class of the MATH divisor. Now consider the subspace MATH of MATH spanned by all possible MATH's as above. Since the unitary group preserves the inner product, it follows that MATH is a subrepresentation of MATH. Clearly MATH is not equal to MATH, hence it must contain MATH. If there were no MATH's with a non-zero pairing with MATH then the pairing would be zero on all of MATH, contradicting the assumption of real regulator indecomposable. (Note that our assumptions imply that the space spanned by the rational NAME classes is MATH). |
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 and MATH transversally in a single NAME point. By choosing a path in a suitable parameter space, we can identify MATH as a symplectic vector space with MATH, where MATH, MATH. Let MATH and MATH be generic hyperelliptic curves of genus MATH, with MATH specializing to MATH, MATH. Let MATH and MATH, the two components of each curve intersecting transversally in a single point. Again, by choosing paths we may identify MATH and MATH with MATH in such a way that MATH is identified with MATH. Now consider the family of NAME. Since MATH is generic, MATH, the NAME group of MATH, contains the NAME groups MATH and MATH of MATH and MATH respectively. Using induction and the above identifications, we see that MATH contains both MATH and MATH. One easily checks, by an explicit computation using NAME algebras, that the smallest subgroup of MATH containing both these two subgroups is MATH. The NAME group is always contained in MATH so MATH must equal MATH. From the representation theory of symplectic groups it follows that MATH is an irreducible representation of the NAME group, MATH. Since the sub-Hodge structures of MATH are precisely the subrepresentations of the NAME group, it follows that the MATH's are also irreducible as NAME structures. |
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, we choose a standard parameter on the domain of MATH so that MATH maps to MATH, and similarly for MATH and for MATH. In this manner MATH is a well defined rational function on MATH, and we denote by MATH its transform under the involution of MATH associated with MATH. Letting MATH, we see that MATH, MATH a constant. One can take MATH to be the form MATH on MATH and thus MATH . It then follows that MATH for the general bielliptic curve C because MATH . |
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 which coincide with points of ramification for MATH. We have MATH. By induction, REF holds for MATH. It then also holds for MATH by using the arguments given for REF , where we now take MATH to be the form MATH, the lift to MATH of MATH. |
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 that MATH is an element of MATH. If we use MATH to embed MATH in MATH, then the image of MATH in MATH is equal to MATH. The involution MATH of MATH preserves MATH, hence MATH must be the pullback of a cycle from MATH. MATH is always decomposable for any curve MATH, hence MATH must also be decomposable. Thus MATH in MATH. This implies that all the components of MATH except the one in MATH must be zero. |
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 MATH is the one in MATH. The first part of the theorem then implies that the image of MATH in MATH is uncountable. We then use that MATH, MATH, and descending induction on MATH starting from MATH to show that the image of MATH in MATH is uncountable for MATH. The statement for MATH follows by using the NAME transform and the motivic hard NAME theorem: Letting MATH, we see by REF that MATH is uncountable for MATH. By REF, intersection with MATH induces an isomorphism from MATH to MATH, hence intersection with MATH must induce an injection from MATH to MATH, MATH. Since MATH is uncountable, it follows that MATH is also uncountable. |
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 inverse image of MATH. Let MATH be the element of MATH determined by MATH, and thus MATH is such that its restriction to the smooth part is given by functions MATH on the components MATH of MATH, each MATH being the pull-back of a rational function with the same name on MATH. Using devissage we restrict MATH to MATH, where MATH is a point in the inverse image of MATH. Then the claim is that for any such point MATH this restriction is equal to MATH. Here the MATH's may depend on MATH and they are such that the product doesn't have a pole or zero at MATH. Note that contravariant functoriality for local complete intersections yields here the equality MATH, where we set MATH if MATH. This argument shows that at the points of intersection of curves like MATH with MATH the values of the functions MATH lie in a countable set, and therefore MATH must be constant on the irreducible component MATH. To prove the claim, we let MATH be the normalization of an irreducible component of the fibre product of all the MATH's over MATH. MATH is flat over MATH, so we can replace MATH by MATH and MATH by the fibre product of MATH and MATH over MATH. The advantage is that here we are reduced to the case that there is only one point in the inverse image of (the new) MATH because we now have sections, and so the claim follows by functoriality from the simple case, which we describe next. The simple case is when each MATH is an isomorphic copy of MATH and moreover the scheme MATH, is constructed by gluing at one point MATH (= MATH on MATH). Let the projection map be MATH, then we write MATH, and MATH, and set MATH. The map MATH is the identity on each component MATH. Give rational functions MATH on MATH, (MATH is not necessarily complete, and we assume that the only zero or pole of MATH is at MATH) and assume that MATH. Then MATH defines an element MATH say in MATH, which comes from MATH, because it has boundary REF in the exact sequence MATH. Consider now MATH as a NAME divisor on MATH and let MATH be the scheme, which is the pull back of MATH on MATH. The map MATH is well defined because MATH is a local complete intersection in MATH. By devissage MATH; the question is to understand the value of the image of MATH in MATH. Each MATH is a function MATH on MATH. The answer is consider the product MATH, then MATH is a rational function which is in fact regular at MATH, because of our assumption, and then we have that the image value is MATH. The reason is once more the commutativity from REF (it is so below at MATH and then it must have been so at MATH). |
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 intersections with MATH can be recovered by using MATH-symmetry. Note that if MATH is not hyperelliptic then MATH and MATH. |
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 relation as MATH. Now degree MATH covers of MATH branched over MATH correspond to transitive representations of MATH in the symmetric group MATH. For the covers to correspond to points of MATH, the images of MATH of the MATH's must be transpositions and the images of the other two must be products of MATH disjoint transpositions. Two such representations give isomorphic covers if they differ by an inner automorphism of MATH. Thus MATH can be identified with classes of MATH-tuples MATH of elements of MATH , all but two of the MATH's being transpositions, the remaining two being products of two disjoint transpositions, and MATH. Two such MATH-tuples are identified if they differ by coordinatewise conjugation by an element of MATH. The action of the monodromy on MATH contains elements MATH, MATH, which act on MATH-tuples as above by: MATH . To show that MATH has two components we use the MATH's to prove that the monodromy action on MATH has two orbits. The proof is a case by case analysis; we shall describe the main steps and leave some simple verifications to the reader. Let MATH, MATH, denote the transposition which switches MATH and MATH and let MATH, MATH and MATH be the permutations MATH, MATH and MATH respectively. Let MATH. Using the action of the MATH's one sees that each orbit contains a MATH-tuple MATH such that MATH and MATH are in MATH. NAME conjugation, we may assume that MATH and MATH or MATH. Suppose MATH, hence MATH. Let MATH be the subgroup of MATH generated by MATH. Using the fact that MATH and MATH is generated by transpositions we analyze the two possibilities for the action of MATH on REF transitive and REF intransitive. In both cases ( REF requires some computations with the MATH's) we conclude that we may assume that at least one of MATH or MATH occurs among the MATH's (without changing MATH and MATH). Again using the action, we may assume that MATH. Then replacing MATH, by MATH we may also assume that MATH. As above, let MATH be the subgroup of MATH generated by MATH and note that now MATH. We now consider four possibilities for the action of MATH on MATH. CASE: The action is transitive. CASE: The action has a unique fixed point. CASE: The action has two fixed points. CASE: The action has no fixed points but fixes two disjoint subsets of two elements each. CASE: Since MATH, it follows by a result of NAME (see CITE) that we may assume MATH . Note that a priori we may also have to use an inner automorphism of MATH in order to achieve this, but it is easy to check using the action of the MATH's that we may choose an automorphism preserving MATH. Hence we may assume that MATH . CASE: Without loss of generality, we may assume that the fixed point is MATH. Again, by NAME result we may assume that MATH . Moreover, it is easy to check that if we replace MATH by MATH or MATH, we stay in the same orbit, hence all elements of type REF lie in one orbit. Further, note that using the MATH's , we can switch pairs of adjacent transpositions, that is, MATH so we may replace MATH by MATH. One checks that MATH . Since we have assumed that MATH, it follows that the MATH corresponding to this element acts transitively on MATH. Hence elements of types REF lie in the same orbit. CASE: In this case we must have MATH where MATH is one of MATH, MATH, MATH, MATH, since the representation is assumed to be transitive. It is clear that all choices are conjugate by an element of MATH which fixes MATH, hence all elements of type REF are in the same orbit. Assume, without loss of generality that MATH. Then MATH . Since MATH and MATH commute, MATH, for MATH, just switches MATH and MATH. Thus we may repeat the above procedure and deduce that MATH is in the same orbit as elements of the form MATH where each MATH for MATH, is either MATH or MATH, there being an even number of both kinds. CASE: Elements of type REF are precisely those considered in the previous paragraph, so all such elements lie in the same orbit as elements of type REF . We thus see that MATH has at most two components. Now observe that covers of type REF have an automorphism of order MATH commuting with the covering map, corresponding to the representation of MATH in MATH induced from the original representation in MATH by identifying MATH with MATH and MATH with MATH. Covers of type REF have no automorphisms commuting with the covering map, hence MATH has precisely two components. |
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 isomorphically to MATH. The closures of all these discs remain disjoint in MATH except for two of them which meet at MATH. By construction, the product of the local monodromies around these two discs is trivial, hence the map MATH is unramified over the complement of the closures of all the above discs. To complete the proof we examine the induced monodromy representation of MATH in MATH. The local monodromy around MATH is equal to the product of the local monodromies around MATH and MATH of the original representation used in the construction of MATH and the local monodromies around the other points are the same as those in the original representation. Then we see that the representation is no longer transitive but breaks up into two representations, each of degree MATH. The conclusions of the lemma follow immediately on inspection of these two representations. |
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 assume that branch locus of the map MATH is a union of sections. Finally, by replacing MATH by a further open subset (containing MATH) we may assume that there exist a rational function MATH on MATH such that MATH, where MATH restricts to the divisor used to define the MATH-configuration on each fibre. MATH allows us to construct an element MATH of MATH which restricts to the MATH-configuration on each fibre. By REF , we see that upto labelling the restrictions of MATH, MATH, MATH, MATH to MATH must be MATH, MATH, MATH, MATH, where MATH and MATH are distinct NAME points on MATH and MATH is a point lying over the point MATH. It is then clear that MATH has divisor MATH and is therefore a NAME function. One easily checks that the element of MATH obtained by translating MATH by MATH is equal to MATH, where MATH is the basic hyperelliptic cycle and the subscript denotes translation. Since MATH, MATH, , MATH and MATH are arbitrary points in MATH, it follows that we may assume that MATH is a generic hyperelliptic curve of genus MATH and MATH a generic point on MATH. By REF it follows that MATH is indecomposable and then by REF it follows that MATH restricted to a generic fibre is also indecomposable. The second statement follows by considering the NAME product of the MATH-configuration with the zero cycles MATH, MATH, where MATH is a generic point of MATH. Taking MATH, we see from the proof of REF that the components of elements of MATH give uncountably many elements of MATH (note that specialization preserves the decompositions). We then deduce that MATH is uncountable for MATH, as well as the statement for MATH, in the same way as in REF . |
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 family, we obtain as specializations cycles of the form MATH where now MATH varies in a MATH-parameter family. By REF , it follows that the set of specializations is an uncountable subset of MATH. Hence by REF the MATH-configurations also form an uncountable subset of MATH. |
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 generated by all the above groups on MATH are in MATH correspondence with the set MATH . Here the elements of MATH (and MATH below) are viewed as unordered pairs. The boundary of an element MATH is supported on points of MATH, each of whose coordinates is one of the MATH's or MATH's. The orbits of the set of all such points under the group action are in MATH correspondence with the set MATH . MATH is not included because the number of MATH's and MATH's is MATH. Let MATH be the set MATH . Consider the projection MATH and observe that MATH for all MATH except for those of the form REF MATH, MATH, with MATH a divisor of MATH and REF MATH with MATH odd and MATH a divisor of MATH. For both these cases MATH. Using the fact that for any integers MATH, if MATH then MATH cannot divide MATH, one checks that if MATH, then MATH. On the other hand, MATH for all MATH except for those with MATH or MATH, MATH, and MATH. For both these cases MATH. If MATH denotes the number of partitions of MATH, one sees that the number of such elements is MATH which is greater than MATH for all MATH. Elements of MATH give us sufficient relations among the MATH's for the boundary of MATH to be zero, but if MATH is odd then elements of the form MATH, MATH, give trivial relations because of symmetry. Using this observation along with explicit computations for MATH, one sees that the number of relations is always strictly less than the number of variables that is, MATH, hence the space of invariant cycles is positive dimensional. |
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 we would consider the representation in MATH corresponding to the tuple MATH . We let MATH be the closure in MATH of points of the form MATH, MATH, MATH, where MATH is a cover of MATH corresponding to a point on the chosen component of the NAME scheme, and MATH, MATH are the inverse images of the two points of MATH over which the cover is not simply ramified. Since the dimension of each component of the NAME scheme is MATH and MATH, it follows that MATH. To prove the statement about the MATH, we consider a degeneration of the cover such that three of the simply ramified points come together (generically) at a point. For instance, in the above example we would let MATH, MATH and MATH come together. If MATH, the special fibre of the stable model of the degenerating family of genus MATH curves then consists of two smooth components, one of them a curve of genus MATH which is a cover of MATH of the same type, and the other a generic elliptic curve. If MATH, we obtain a nonconstant family of elliptic curves and so the statement is true in this case. The statement for MATH follows, since we may then assume that both components are generic elliptic curves. If MATH, we may assume using induction that the point of intersection of the two components is a generic point on the curve of genus MATH (also generic). Thus MATH, and so MATH must be dominant. Finally, if MATH we see that MATH. Since MATH and MATH, it follows that we must have equality for all MATH. |
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 consider also the forms MATH on MATH and MATH on MATH. The procedure of REF gives here again: CASE: MATH. The NAME space of divisors with rational coefficients on MATH is isomorphic to the same space on the product of the two associated elliptic curves. On the gneric bielliptic Jacobian MATH is orthogonal to the NAME group, because it is orthogonal to the elliptic curves. A look at the proof of the main theorem of CITE shows that this property of MATH and REF imply that the generic section MATH is indeed weakly indecomposable. |
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 colimit MATH and MATH a structural morphism MATH. According to REF , the conditions (MATH), REF imply that MATH belongs to MATH and satisfies REF . For an object MATH, we have an exact sequence MATH . Since MATH for MATH, MATH is an isomorphism for any MATH and MATH. Then, we have an epimorphism MATH for any MATH. Therefore, we have an exact sequence MATH . Hence we have MATH . |
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 take an object MATH and a morphism MATH satisfying the conditions of REF , and embed MATH in a distinguished triangle MATH . Applying MATH to the above triangle, by REF , we have MATH. Since MATH is a generating set, it is easy to see that MATH is non-degenerate. CASE: Since MATH is the core of the MATH-structure MATH, the assertion follows by REF . CASE: According to REF , the short exact sequences in MATH are just the distinguished triangles MATH with MATH and MATH belonging to MATH. It follows that MATH is exact. Let MATH and take a free presentation MATH. We take MATH and MATH satisfying the conditions of REF . Then there exist sets MATH and a collection of morphisms MATH such that MATH is commutative, where the vertical arrows are isomorphisms. We take an exact sequence in MATH . Since MATH is compact, by the exactness of MATH, we have MATH. CASE: We show that MATH reflects isomorphisms. Let MATH be a distinguished triangle in MATH with MATH and with MATH an isomorphism. Then, by applying MATH, we get MATH for all MATH, and hence MATH. It follows that MATH is an isomorphism. Next, we show that MATH is faithful. Let MATH be a morphism in MATH with MATH. By the exactness of MATH, MATH. Since MATH, we have MATH for all MATH, and hence MATH and MATH. Let MATH be the full subcategory of MATH consisting of objects MATH such that there exists an exact sequence MATH in MATH, where MATH are direct sums of copies of MATH. Since MATH is faithful, by the same technique as in REF , it is not hard to see that MATH is full dense, and hence an equivalence. It remains to show that MATH. For an object MATH, we have a commutative diagram MATH with the top row being exact and with the vertical arrows being isomorphisms. And we have a commutative diagram in MATH with MATH. By REF , there exists MATH such that MATH. Since MATH, we have MATH. Then there exists MATH such that MATH, where MATH is a canonical morphism. Then MATH is an isomorphism, and hence MATH is an isomorphism and MATH. |
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. According to REF , we have a fully faithful MATH-functor MATH. Also, since MATH, MATH sends MATH-modules to objects in MATH. Then we have a fully faithful MATH-functor MATH which sends MATH-modules to objects in MATH. For any MATH, there exist integers MATH such that MATH. Let MATH. If MATH, then there exist obviously a MATH-module MATH and an integer MATH such that MATH. If MATH, then we have a distinguished triangle MATH with MATH and MATH. Since MATH is full, by induction on MATH, there exists MATH such that MATH. CASE: By the assumption, MATH also has a MATH-structure MATH. Thus MATH, and hence MATH. By REF, we have a MATH-functor MATH, and then we have a MATH-functor MATH which sends MATH-modules to objects in MATH. Let MATH with MATH and MATH. Take a distinguished triangle in MATH such that MATH is a direct sum of copies of MATH and MATH is an epimorphism. By easy calculation, MATH, and hence we get an exact sequence in MATH . Since MATH, we have MATH, that is, MATH effaces MATH. Thus the epimorphic version of effacibility in REF can be applied. Finally, it is easy to see that MATH implies MATH. |
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. Since the MATH are reflexive, we have MATH and MATH. We have a commutative diagram MATH with the top row exact. Since the MATH are isomorphisms, MATH is embedded in MATH. The assertion follows by REF . CASE: Let MATH. For any MATH, we have a functorial homomorphism MATH which is an isomorphism if MATH is finitely generated projective. We have a commutative diagram MATH with the bottom row exact. Since the MATH are isomorphisms, MATH is a homomorphic image of MATH. The assertion follows by REF . |
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 set of generators MATH for a MATH-module MATH and set MATH . It is obvious that MATH is surjective. Also, by REF we have MATH. Applying MATH to the canonical exact sequence MATH we get MATH. Thus MATH and hence MATH. CASE: By REF . |
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. It follows by NAME Theory that MATH in MATH. Thus MATH in MATH. CASE: It is obviously deduced that MATH and 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 have MATH . The last assertion follows by REF . |
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-dual REF-form MATH satisfies MATH . Setting MATH, we thus have MATH . But REF tells us that MATH so we obtain MATH . By the NAME inequality, we thus have MATH . Since the NAME inequality also tells us that MATH we thus have MATH where the last inequality is exactly REF . This completes the first part of the proof. Next, we observe that any smooth conformal MATH class on any oriented MATH-manifold contains a MATH metric such that MATH is constant. Indeed, as observed by CITE, this readily follows from the standard proof of the NAME problem. The main point is that the curvature expression MATH transforms under conformal changes MATH by the rule MATH just like the ordinary scalar curvature MATH. We will actually use this only in the negative case, where the proof is technically the simplest, and simply repeats the arguments of CITE. The conformal class MATH of a given metric MATH thus always contains a metric MATH for which MATH is constant. But since the existence of solutions of the NAME equations precludes the possibility that we might have MATH, this constant is necessarily non-positive. We thus have MATH so that MATH . Thus we at least have the desired MATH estimate for a specific metric MATH which is conformally related to the given metric MATH. Let us now compare the left-hand side with analogous expression for the given metric MATH. To do so, we express MATH in the form MATH, where MATH is a positive MATH function, and observe that MATH . Applying NAME, we thus have MATH and hence MATH exactly as claimed. |
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: every conformal class contains a metric for which MATH is constant, and this metric minimizes MATH among metrics in its conformal class. |
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 other hand, since MATH and MATH is a monopole class, MATH does not admit any metrics of positive scalar curvature. If MATH and REF is saturated, one can therefore show that MATH is MATH or MATH with a NAME metric. The details are left as an exercise for the interested reader. |
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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