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math/0002018 | Assuming the contrary, there is a nonzero morphism MATH. Composing this with the maps MATH to the left and MATH to the right, we obtain a nontrivial endomorphism of MATH, which contradicts the stability assumption. |
math/0002018 | We have an induced diagram MATH where MATH is the composition of MATH with the inclusion of MATH in MATH. It is clear that MATH preserves MATH if and only if MATH is identically zero. The lemma follows then easily from the definitions. |
math/0002018 | An elementary transformation at MATH points MATH is given by a choice of hyperplanes MATH for each MATH. By the previous lemma, such a transformation preserves MATH if and only if MATH for all MATH. We have a natural diagram MATH where MATH is the restriction to MATH and MATH is the composition with the inclusion of MATH in MATH. For MATH, we have: MATH where the product is taken MATH times. So MATH is irreducible of dimension MATH and this gives that MATH is irreducible of dimension MATH. |
math/0002018 | Fix a point MATH. We can consider an elementary transformation of MATH of length MATH, supported only at MATH: MATH such that MATH. Then as in REF, the only maximal subbundles MATH that are preserved by this transformation are exactly those such that MATH. By REF this implies that only at most a finite number of MATH's can be preserved. If none of the maximal subbundles actually survive in MATH, then any further transformation at one point would do. Otherwise clearly for a generic MATH we have MATH for all the MATH's that are preserved and so we can choose a hyperplane MATH such that MATH for any such MATH. The elementary transformation of MATH at MATH corresponding to this hyperplane satisfies then the required property. |
math/0002018 | Denote by MATH an irreducible component of MATH (recall that we are thinking now of this NAME scheme as parametrizing subsheaves of rank MATH and degree MATH). The first step is to observe that it is enough to prove the statement when MATH is non-special. To see this, note that every nonsaturated subsheaf MATH determines a diagram: MATH where MATH is the saturation of MATH, MATH is a quotient vector bundle and MATH, the torsion subsheaf of MATH, is a nontrivial zero-dimensional subscheme, say of length MATH. We can stratify the set of all such MATH's according to the value of the parameter MATH, which obviously runs over a finite set. If we denote by MATH the subset corresponding to a fixed MATH, this gives then: MATH . The right hand side is clearly less than MATH if we assume that the statement of the theorem holds for MATH. Let us then restrict to the case when MATH is a non-special component. The proof goes by induction on MATH. If MATH, the statement is exactly the content of REF. Assume now that MATH and that the statement holds for all the pairs where this difference is smaller. Recall that MATH denotes the open subset corresponding to vector subbundles and fix MATH. Then by REF , there exists an elementary transformation MATH of length MATH, such that MATH, but MATH for any MATH. Then necessarily MATH (consider the saturation in MATH of a maximal subbundle of MATH) and so MATH. This means that we can apply the inductive hypothesis for any non-special component of the set of subsheaves of rank MATH and degree MATH of MATH. To this end, consider the correspondence: By REF , for any MATH, the fiber MATH is a (quasi-projective irreducible) variety of dimension MATH and so: MATH . On the other hand, for MATH, the inductive hypothesis implies that MATH and since MATH we have: MATH . Combining REF we get: MATH and of course the same holds for MATH. This completes the proof. |
math/0002018 | Recall from REF that the statement of the theorem is equivalent to the following fact: for any stable bundle MATH, there exists a line bundle MATH such that MATH. This is certainly an open condition and it is sufficient to prove that the algebraic set MATH has dimension strictly less than MATH. A nonzero map MATH comes together with a diagram of the form: MATH where MATH is just the image in MATH. Then we have MATH for some effective divisor MATH. Since MATH is stable, the degree of MATH can vary from MATH to MATH and we want to count all these cases separately. So for MATH, consider the following algebraic subsets of MATH: MATH . The claim is that MATH for all such MATH. Then of course MATH and any MATH outside this union satisfies our requirement. To prove the claim, denote by MATH the NAME scheme of coherent quotients of MATH of rank MATH and degree MATH. The set of line bundle quotients MATH of degree MATH is a subset of MATH. On the other hand every MATH can be written as MATH, with MATH as above and MATH effective of degree MATH. This gives the obvious bound: MATH . To bound the dimension of the NAME scheme in question, fix a MATH as before and consider the exact sequence that it determines: MATH . Note that the kernel is isomorphic to MATH since MATH has trivial determinant. Now we use the well known fact from deformation theory that MATH. To estimate MATH, one uses all the information provided by NAME 's theorem. The initial bound that it gives is MATH (note that MATH). If actually MATH, then we immediately get MATH as required. On the other hand if MATH, by the equality case in NAME 's theorem (see for example, CITE III, REF) one of the following must hold: MATH or MATH or MATH is hyperelliptic and MATH. The first case is impossible since MATH. In the second case MATH is a theta characteristic and we are done either by the fact that these are a finite number or by other overlapping cases. The third case can also happen only for a finite number of MATH's and if we're not in any of the other cases then of course MATH. This concludes the proof of the theorem. |
math/0002018 | Let us denote for simplicity MATH. Since the problem depends only on the residues of MATH modulo MATH, there is no loss of generality in looking only at MATH with MATH. The statement of the theorem is implied by the following assertion, as described in REF: MATH . Fix a stable bundle MATH. Note that only in this proof, as opposed to the rest of the paper, MATH in fact denotes the degree of MATH, and not that of MATH. If for some MATH there is a nonzero map MATH, then this comes together with a diagram of the form: MATH where the vector bundle MATH is the image of MATH. The idea is essentially to count all such diagrams assuming that the rank and degree of MATH are fixed and see that the MATH's involved in at least one of them cover only a proper subset of the whole moduli space. Denote as before by MATH the NAME scheme of quotients of MATH of rank MATH and degree MATH and for any MATH and any MATH in the suitable range (given by the stability of MATH and MATH) consider its subset: MATH . The theorem on NAME schemes stated in the introduction then gives us the dimension estimate: MATH where MATH is the minimum possible degree of a quotient bundle of MATH of rank MATH (which is the same as MATH). Define now the following subsets of MATH: MATH . The elements of MATH are all the MATH's that appear in diagrams as above for fixed MATH and MATH. The claim is that MATH which would imply that MATH. Assuming that this is true, and since MATH and MATH run over a finite set, any MATH satisfies the desired property that MATH, which gives the statement of the theorem. It is easy to see, and in fact a particular case of the computation below, that in the case MATH (that is, MATH) MATH has dimension exactly MATH. Let us concentrate then on proving the claim above for MATH. Note that the inclusions MATH appearing in the definition of MATH are valid in general only at the sheaf level. Any such inclusion determines an exact sequence: MATH where MATH, with MATH locally free and MATH a zero dimensional subscheme of length MATH. We stratify MATH by the subsets MATH where MATH runs over the obvious allowable finite set of integers. A simple computation shows that MATH has rank MATH and degree MATH. Denote by MATH the set of all vector bundles MATH that are quotients of some MATH. These can be parametrized by a relative NAME scheme (see for example, REF) over (an étale cover of) MATH and so they form a bounded family. We invoke a general result, proved in CITE and REF, saying that the dimension of such a family is always at most what we get if we assume that the generic member is stable. Thus we get the bound: MATH . Now we only have to compute the dimension of the family of all possible extensions of the form REF when MATH and MATH are allowed to vary over MATH and MATH respectively and MATH varies over the symmetric product MATH. Any such extension induces a diagram MATH . If we denote by MATH the set of isomorphism classes of vector bundles MATH that are (inverse) elementary transformations of length MATH of vector bundles in MATH, then we have the obvious: MATH . On the other hand any MATH is obtained as an extension of a bundle in MATH by a bundle in MATH. Denote by MATH the open subset consisting of pairs MATH such that there exists an extension MATH with MATH stable. Note that by REF for any such pair we have MATH and so by NAME MATH is constant, given by: MATH . In this situation it is a well known result (see for example, REF or REF) that there exists a universal space of extension classes MATH whose dimension is computed by the formula: MATH . There is an obvious forgetful map: MATH whose image is exactly MATH. Thus by putting together all the inequalities above we obtain: MATH where the last inequality is due to the obvious fact that MATH if MATH. Since MATH runs over a finite set, to conclude the proof of the claim it is enough to see that MATH. By the inequality above this is true if MATH or equivalently if MATH for any MATH and MATH. This can be rewritten in the following more manageable form: MATH . The first case to look at is MATH, when we should have MATH and this should hold for every MATH. But clearly MATH, defined above in terms of maximal subbundles, and MATH. Since MATH is stable we then have MATH for all MATH and so MATH as mentioned before. Note though that in general one cannot do better (compare REF ). In any case, this says that the inequality MATH must be satisfied (which would certainly hold if MATH). When MATH, it is convenient to collect together all the terms containing MATH. The last inequality above then reads: MATH . For MATH as before it is then sufficient to have MATH, which again by simple optimization is satisfied for MATH. Concluding, the desired inequality holds as long as MATH. |
math/0002018 | This is clear since MATH. |
math/0002018 | Note that by duality it suffices to prove the claim for one of the moduli spaces, say MATH. In this case MATH and for any MATH, MATH. Following the proof of the theorem we see thus that it suffices to have MATH which is equal to MATH for MATH. For MATH one can slightly improve the last inequality in the proof of the theorem (actually this is true whenever MATH is even) to see that MATH already works. |
math/0002018 | By the projection formula, for every MATH we have: MATH . Also the restriction of MATH to any fiber MATH of the determinant map is isomorphic to MATH and so globally generated for MATH. It is a simple consequence of general machinery, described in REF , that in these conditions the statement holds as soon as MATH is globally generated on MATH, where MATH is a simplified notation for MATH. To study this we make use, as in CITE, of a cohomological criterion for global generation of vector bundles on abelian varieties due to REF . In our particular setting it says that MATH is globally generated if there exists some ample line bundle MATH on MATH such that MATH . We chose MATH to be MATH, where MATH is the theta divisor on MATH associated to MATH. The cohomology vanishing that we need is true if it holds for the pullback of MATH by any finite cover of MATH. But recall from REF that MATH, where MATH is the multiplication by MATH. Since MATH, via pulling back by MATH the required vanishing certainly holds if MATH . |
math/0002021 | NAME 's lemma asserts that if the group MATH acts on a set (the set of maps) and MATH is a function from MATH to a ring containing the rationals which is constant on conjugacy classes, then MATH where the sum on the righthand side is over a complete set of orbit representatives (the unlabeled imbeddings). Taking the function MATH as the cycle-type monomial we observe that the lefthand side is MATH and the righthand side is MATH. |
math/0002021 | From the Decomposition REF we have MATH . There is a unique solution to this system of equations in the unknowns MATH obtained by comparing coefficients of MATH (or MATH) and then, in turn, the coefficients of MATH (or MATH) for each successive divisor MATH of MATH beginning with the largest. Uniqueness of the solution was actually shown to be the case in NAME 's original article CITE. Extracting the coefficient of MATH (or MATH) yields MATH . The latter system of equations (one equation for each divisor MATH of MATH) can be solved explicitly for MATH by NAME inversion. The details of this particular inversion problem can be found in CITE. |
math/0002023 | In the appendix we compute MATH for a disc of radius MATH; the result is MATH . The scattering phase for a disc of radius MATH is then MATH. Thus, for a disc of radius MATH, we also have MATH . Using the first two heat invariants, the area and perimeter are constant on an isophasal class of domains; hence, by REF the inradius and circumradius are uniformly bounded below and above. Thus, we can sandwich any domain in an isophasal class between fixed discs MATH and MATH. By CITE, MATH is monotonic in the domain, so we obtain REF. |
math/0002023 | See appendix. |
math/0002023 | These are all routine. To prove REF, let MATH and differentiate the equations MATH and let MATH to get MATH . It follows that MATH which establishes REF. To prove the next formula, let MATH. Then MATH is continuous at MATH; let MATH. Then MATH is the unique solution MATH of MATH which is continuous at MATH and takes the value MATH there. Thus MATH. This demonstates REF. To prove the final equation, consider the right hand side applied to MATH. This gives us a function MATH such that MATH, with MATH equal to MATH (since MATH). Therefore, MATH, proving REF. |
math/0002023 | Recalling REF, the zeta function for MATH is MATH . This equation shows why there is an expansion as MATH with local coefficients: the factor of MATH means that the integral from MATH to infinity is exponentially decreasing in MATH, for any MATH, so only the expansion of the regularized heat trace at MATH will contribute to polynomial-order asymptotics in MATH, and this expansion is local. For precisely, for any integer MATH we consider the expansion to MATH terms of the regularized heat trace (see REF), MATH at MATH. Let MATH be the difference between the regularized heat trace of MATH and this finite expansion. Then, MATH is MATH as MATH and, since the heat trace is bounded as MATH, MATH is also MATH at infinity. It is easy to see that if MATH is substituted for the regularized heat trace in REF then both the result, and the derivative in MATH of the result, is MATH as MATH. Thus, to compute the expansion as MATH to this order we need only substitute REF in to REF, namely MATH . Changing variable to MATH, this gives us MATH . Differentiating at MATH gives us an expansion of the form above, with MATH since the two MATH-factors cancel when MATH to give a constant. |
math/0002023 | Constancy of the log determinant of MATH over the isophasal class implies that REF MATH . Expanding this out, we get MATH . Since MATH, we have MATH . Integrating the first term by parts, using sup bounds on MATH and MATH, and using MATH, we get MATH . Since we have REF, and MATH the left hand side is bounded by MATH. The term MATH is bounded similarly. Thus we have MATH . Adding twice REF to this inequality gives us a bound on MATH, and then REF provides a bound on MATH. This gives a bound on the NAME one-half norm of MATH. Next we bound the MATH norm of MATH. Since MATH, it is sufficient to bound the MATH norm of MATH. The derivative of MATH is bounded by MATH. We have uniform sup bounds on MATH. By NAME 's inequality, REF , MATH so MATH which gives us the required bound. |
math/0002023 | The bound on the circumradius MATH, where MATH is the perimeter, is trivial. Consider the inradius. Let MATH be a domain with area MATH and perimeter MATH and MATH a number such that MATH is larger than the inradius but smaller than the circumradius. It is possible to choose of covering of MATH by balls of radius MATH whose centres lie in MATH, such that the balls of radius MATH with the same centres are disjoint. To construct such a covering, start with a finite covering by balls of radius MATH whose centres lie in MATH. Choose any two of the balls. If the balls of radius MATH about their centres intersect, then the ball of radius MATH about any one of the centres contains both balls of radius MATH, and therefore one of the balls can be discarded. By discarding balls successively in this way, we end up with a covering with the required property. Let MATH be the number of balls in the covering. Considering the area of each ball, we have an inequality MATH . Since MATH is assumed larger than the inradius, but smaller than the circumradius, the circles of radius MATH with the same centres as the balls in our covering must all intersect the boundary. Since the circles are all distance at least MATH apart, this gives an inequality MATH . Combining the two inequalities, we find that MATH . Taking the infimum over r, we find that MATH . |
math/0002024 | Let MATH. We want to show that there are isomorphisms MATH such that MATH for every homomorphism MATH in MATH. First of all notice that we can work on an arbitrarily fixed skeleton of MATH: all the notions we are dealing with are invariant under such a passage. Henceforth MATH is the fixed skeleton. By CITE one knows that MATH determines (up to an affine integral isomorphism) the polytope MATH (this is so even in the category of all commutative MATH-algebras). Therefore, we assume that for each object MATH there is a unique polytope MATH such that MATH and different polytopal algebras determine non-isomorphic polytopes. By a suitable choice of the skeleton we can also assume that the objects of MATH are of the type MATH for the unit MATH-simplices MATH, MATH (by convention MATH CITE ). Also, we will use the notation MATH. CASE: We claim that CASE: MATH (NAME dimension), CASE: MATH (MATH-ranks of the degree REF components), CASE: MATH is surjective (the degree REF component MATH is injective) if and only if MATH is surjective (MATH is injective). In fact, it follows from REF below that for any polytopal algebra MATH its NAME dimension is the dimension of a maximal torus of the linear subgroup MATH it is certainly an invariant of MATH - hence REF . REF follow from the observation that both the injectivity and surjectivity conditions can be reformulated in purely categorical terms by using morphisms originating from MATH. CASE: Observe that MATH restricts to an autoequivalence of MATH. This follows from REF and the fact that polynomial algebras are the only polytopal algebras whose NAME dimensions coincide with the MATH-ranks of the degree REF components. Next we correct MATH on MATH in such a way that MATH. We know that MATH. This means that there are elements MATH for all MATH such that MATH . For each MATH put MATH and let MATH be the autoequivalence determined as follows: it is the identity on the objects and MATH for a morphism MATH in MATH. Then MATH represents the neutral element of MATH. Now the functor MATH is isomorphic to MATH and it restricts to the identity functor on MATH. Without loss of generality we can therefore assume MATH. CASE: For any positive dimensional object MATH we fix a bijective mapping MATH from the set of vertices of MATH, MATH to the set of lattice points of MATH. The resulting surjective homomorphism MATH will also be denoted by MATH. Every matrix MATH gives rise to a graded MATH-surjective homomorphism MATH whose degree REF component is given by MATH where MATH is the linear transformation of the MATH-vector space MATH determined by MATH, and MATH is the degree REF component of MATH. By REF MATH for all MATH. Therefore, for a polytope MATH, satisfying the condition MATH, REF implies the following commutative square MATH where MATH denotes the set of surjective homomorphisms. By the definition of MATH the horizontal mappings are MATH-equivariant automorphisms of MATH-varieties. We have the following obvious equalities: MATH . After the appropriate identifications with MATH and MATH respectively we arrive at the commutative square of MATH-varieties MATH whose horizontal arrows are algebraic MATH-equivariant automorphisms for the diagonal MATH-actions. Thus the upper horizontal mapping is a linear non-degenerate transformation of the MATH-vector space MATH, which leaves the subset MATH invariant, i. e. the matrix degeneracy locus MATH is invariant under MATH. Then by REF below there are only two possibilities: CASE: there are MATH such that MATH for all MATH, or CASE: there are MATH such that MATH for all MATH, where MATH is the transposition. Consider the commutative square MATH where MATH is arbitrary matrix and the degree REF component of the upper horizontal mapping is the linear transformation given by MATH. By applying the functor MATH to this square and using the equality MATH REF we arrive at the following equalities in the corresponding cases: MATH . Identifying MATH and MATH along MATH we get the matrix equalities: MATH . First notice that REF is excluded, i. e. there is no matrix MATH for which the following holds: MATH . This follows from running MATH through the set of standard basic matrices (i. e. the matrices with only one entry REFs elsewhere). For REF we have MATH . Then MATH is in the center of MATH (in particular, it is a scalar matrix). So we can write MATH . We arrive at the Claim. For each lattice polytope MATH with MATH, there is MATH, MATH, such that MATH . CASE: Now we show that for every polytope MATH the linear automorphism MATH of the MATH-vector space MATH, determined by the matrix MATH REF , belongs to the closed subgroup MATH provided MATH. The functor MATH induces a MATH-equivariant automorphism of the MATH-variety MATH. On the other hand we have the natural identification MATH where the right hand side denotes the variety of closed points of MATH. Therefore, there exists an automorphism MATH of the MATH-algebra MATH - not a priori graded, such that the mapping MATH is given by MATH, and, moreover, it is MATH-equivariant. It follows that MATH is a MATH-equivariant automorphism of MATH. It is easily seen that a MATH-equivariant automorphism is graded (and conversely). Therefore MATH. (Here we identify elements of MATH with their degree one components, which are linear automorphisms of MATH.) We let MATH denote the homomorphism which sends MATH to MATH. For any toric automorphism MATH (that is, an automorphism for which any element of MATH is an eigenvector, MATH is the group of such automorphisms, see REF) we have the commutative diagram MATH where MATH refers to the diagonal MATH-matrix corresponding to the degree REF component of MATH. In view of what has been said above and of REF , an application of MATH to the last commutative diagram yields the equality MATH . Let MATH be the matrix of the degree REF component of MATH, and put MATH. Then the equality can be reformulated into the condition: CASE: for any MATH the sum of the entries of each row in the matrix MATH is REF, that is MATH where MATH is the linear transformation of MATH determined by MATH, and MATH. Now we derive from REF that MATH, that is MATH. Fix MATH and MATH. We put MATH and MATH . Then MATH . Consider the NAME polynomial MATH . Without loss of generality we assume that MATH and MATH. Observe that the assignment MATH extends uniquely to a toric automorphism of MATH (the torus MATH acts tautologically on its coordinate ring MATH). Conversely, any toric automorphism of MATH can be obtained in this way from some element of MATH. Therefore MATH . In particular, the sum of the coefficients of the NAME polynomial MATH is REF for any MATH, in other words MATH for all MATH. Taking into account the infinity of MATH we conclude MATH, that is MATH. CASE: Let MATH denote the full subcategory whose objects are those polytopal algebras MATH for which MATH. Now we show that there is MATH such that MATH and MATH. As in REF , this means that we have to show the existence of elements MATH such that MATH for any morphism MATH in MATH. Consider the commutative diagram MATH where MATH is any morphism in MATH, MATH, MATH, and MATH is the unique lifting of MATH. By REF we know that MATH transforms this square into the square MATH . Therefore, looking at the degree REF components we conclude MATH . So by REF the system MATH is the desired one. CASE: By the previous step we can assume that MATH. Now we complete the proof by showing that this assumption implies MATH, and therefore MATH. Assume MATH is a lattice polytope. We let MATH denote the unit square and consider all the possible integral affine mappings: MATH . We define the diagram MATH as follows. It consists of CASE: MATH copies of MATH indexed by the elements of MATH, CASE: MATH copies of MATH: MATH, and the following morphisms between them MATH . We claim that MATH whenever the defining ideal of the toric ring MATH is generated by quadratic binomials, where the direct limit is considered in the category of all (commutative) MATH-algebras. In fact, it is clear that the mentioned limit is always a standard graded MATH-algebra, whose degree REF component has MATH-dimension equal to MATH, and that there is a surjective graded MATH-homomorphism MATH . This is so because of the cone over the diagram MATH with vertex MATH where MATH is mapped to MATH and the vertices of MATH from MATH are mapped accordingly to MATH. Therefore, we only need to make sure that MATH in MATH whenever MATH in MATH, where MATH is the image of MATH in MATH, and similarly for MATH, MATH and MATH. But if MATH in MATH then these four points in MATH belong to MATH for some MATH and the desired equality is encoded into the diagram MATH. Having established the equality above for quadratically defined polytopal rings we now show that MATH for every lattice polytope MATH and all MATH. (Here MATH denotes the MATH-th homothetic multiple of MATH.) By REF the defining ideal of the toric ring MATH is generated by quadratic binomials for MATH. Therefore, MATH for such MATH. Now observe that REF imply that either MATH or MATH for a lattice triangle MATH with REF lattice points. Let us show that MATH cannot be a `triangle ring'. Up to isomorphism there are only two lattice triangles with REF lattice points MATH . By REF below the triangle MATH with the property MATH must have the same number of column vectors as MATH (see REF for the definition). But this is not the case because the second triangle has no column vectors whereas the first has REF of them, and the square only REF. (The column vectors are indicated in the figure.) Therefore MATH as well. By the assumption MATH we get the natural homomorphism MATH (MATH as above). Arguing similarly for the functor MATH, we get the natural homomorphism MATH, and applying MATH to the latter homomorphism we arrive at the natural homomorphism MATH. But every natural (that is compatible with MATH) endomorphism MATH must be the identity mapping for reasons of universality. In particular, MATH is a MATH-retract of MATH. By reasons of dimensions (and REF ) we finally get the equality MATH, as required. Now we are ready to show the equality MATH, completing the proof of REF . Let MATH. Consider two coprime natural numbers MATH. We have the commutative square, consisting of embeddings in MATH, MATH where the horizontal (vertical) mappings send lattice points in the corresponding polytopes to their homothetic images (centered at the origin) with factor MATH and MATH respectively. The key observation is that by restricting to the degree one components we get the pull back diagram of MATH-vector spaces MATH . This follows from the fact that the following is a pull back diagram of finite sets MATH . Caution. MATH is in general not a pull back diagram of MATH-algebras. Otherwise, by REF, any polytopal algebra would be normal, which is not the case. That the latter square of finite sets is in fact a pull back diagram becomes transparent after thinking of it as the (isomorphic) diagram consisting of the inclusions: MATH where MATH refers to the lattice polytope MATH with respect to the intermediate lattice MATH. Here again we assume that MATH and MATH. Since MATH and MATH is the identity functor on MATH, an application of MATH to the square MATH yields the square of graded homomorphisms MATH with the same arrows into MATH as MATH. The same arguments as in the proof of REF show that the degree REF component of this square is a pull-back diagram, in other words MATH. Therefore, MATH and MATH generate the same algebras, i. CASE: MATH. |
math/0002024 | The inclusion MATH follows from REF . Assume MATH is a homomorphism respecting the monomial structures and such that MATH. By a polytope change we can assume MATH for a sufficiently big natural number MATH, where MATH and MATH is taken in the lattice MATH. In this situation there is a bigger lattice polytope MATH and a unique homomorphism MATH for which MATH. By REF MATH is tame. Consider the situation when the ideal MATH is a nonzero prime monomial ideal and there is a face MATH such that MATH and MATH factors through the face projection MATH, that is MATH for MATH and MATH for MATH. In view of the previous case we are done once the tameness of face projections has been established. Any face projection is a composite of facet projections. Therefore we can assume that MATH is a facet of MATH. Let MATH denote the halfspace that is bounded by the affine hull of MATH and contains MATH. There exists a unimodular (with respect to MATH) lattice simplex MATH such that MATH, the affine hull of MATH intersects MATH in one of its facets and MATH for some MATH. But then MATH is a restriction of the corresponding facet projection of MATH, the latter being a homothetic blow-up of the corresponding facet projection of the polynomial ring MATH - obviously a tame homomorphism. |
math/0002024 | We will use the notation MATH. Any lattice point MATH has a unique representation MATH where the MATH are nonnegative integer numbers satisfying the condition MATH. The numbers MATH are the barycentric coordinates of MATH in the MATH. Let MATH be any homomorphism. First consider the case when one of the points from MATH is mapped to MATH. In this situation MATH is a composite of facet projections and a homomorphism from MATH with MATH. As observed in the proof of REF facet projections are tame. Therefore we can assume that none of the MATH is mapped to MATH. By a polytope change we can also assume MATH, MATH. Consider the polynomials MATH, MATH. Then the MATH are subject to the same binomial relations as the MATH. One the other hand the multiplicative semigroup MATH is a free commutative semigroup and, as such, is an inductive limit of free commutative semigroups of finite rank. Therefore, by REF below there exist polynomials MATH, MATH, and scalars MATH, MATH, such that MATH where the MATH are the barycentric coordinates of MATH. Clearly, MATH are subject to the same binomial relations as the MATH. Therefore, after the normalizations MATH REF we get MATH . But the latter equality can be read as follows: MATH is obtained by a polytope change applied to MATH, where CASE: MATH, MATH, MATH, CASE: MATH, MATH, MATH, and MATH is a sufficiently large lattice polytope so that it contains all the relevant lattice polytopes. Now MATH is tame because it can be represented as the composite map MATH (the first map is the MATH-th homothetic blow-up of MATH, MATH for all MATH) and MATH is just a free extension of the identity embedding MATH. CASE: First consider the case of lattice segmental fibrations. Consider the rectangular prism MATH. By a polytope change (assuming MATH is sufficiently large) we can assume that MATH so that MATH is parallel to MATH: The lattice point MATH will be identified with the monomial MATH whenever we view it as a monomial in MATH, where the MATH are the corresponding barycentric coordinates of MATH (see the proof of REF above). (In other words, the monomial MATH is identified with the point MATH.) Assume MATH is a homomorphism of the type MATH, MATH for some MATH satisfying the condition MATH. Consider any homomorphism MATH, MATH that splits the projection MATH, MATH and MATH for MATH, MATH. The description of such homomorphisms is clear - they are exactly the homomorphisms MATH for which CASE: MATH, CASE: MATH, MATH, CASE: MATH for all MATH. Clearly, all such MATH are tame. In case MATH we have the homomorphism MATH which obviously splits the projection MATH defined by MATH . Assume MATH splits MATH. Since the NAME polytope of a product is a NAME sum of the NAME polytopes of the factors, we get: the polynomials MATH and MATH, mentioned in the proof of REF , that correspond to MATH, satisfy the conditions: MATH and MATH. Is is also clear that upon evaluation at MATH we get MATH, MATH. Therefore, after the normalizations MATH, MATH we conclude that MATH is obtained by a polytope change applied to MATH as above with respect to MATH, MATH, MATH. For a lattice fibration MATH of higher codimension similar arguments show that MATH is obtained by a polytope change applied to MATH, where CASE: MATH is a splitting of a projection of the type MATH such that the base polytope MATH is a unit simplex and CASE: MATH is a homomorphism defined by a single polynomial whose NAME polytope is parallel to MATH. We skip the details for splittings of face projections and only remark that similar arguments based on NAME polytopes imply the following. All such splittings are obtained by polytope changes applied to MATH where MATH is a splitting of a face projection onto a polynomial ring and MATH is again defined by a single polynomial. |
math/0002024 | Assume MATH, MATH (the MATH as above). Consider the vector MATH . It suffices to show that all the vectors MATH are MATH-th multiples of integral vectors. But for any index MATH the MATH-th component of either MATH or MATH for some MATH is zero. In the first case there is nothing to prove and in the second case the desired divisibility follows from the fact that MATH is a MATH-th multiple of an integral vector (because MATH is integral affine). |
math/0002026 | See CITE or CITE. |
math/0002026 | Let MATH be an orthonormal basis of MATH, so MATH is a MATH-basis of MATH . Let MATH, so MATH is a closed subspace of MATH-codimension MATH in MATH, MATH, and MATH. Therefore MATH, so MATH. Viewing MATH as functions on MATH, they form a MATH-basis of the MATH-dual space MATH. So we are reduced to an issue about linear algebra over finite fields: for MATH and MATH, do the MATH reduced functions MATH, as constructed in REF , form a basis of MATH? Let MATH be a finite-dimensional MATH-vector space, of dimension (say) MATH. Let MATH be a basis of MATH. Extend the MATH to a set of polynomial functions on MATH by using digit expansions. That is, for MATH write MATH in base MATH and set MATH . So MATH and MATH. By a dimension count, we just need to show the functions MATH are a basis of MATH. It suffices to show the MATH span MATH. Let MATH be the dual basis to the MATH. For MATH, write MATH where MATH. Taking an idea from the proof of the NAME Theorem in CITE, define MATH by MATH . Since MATH is REF when MATH and MATH when MATH, the MATH-span of the MATH is all of MATH. Expanding the product defining MATH shows MATH is in the span of the MATH since the exponents of the MATH in the product never exceed MATH. This concludes the proof. |
math/0002026 | Let MATH, so MATH. For MATH, the maps MATH give well-defined MATH-linear maps from MATH to MATH. By hypothesis they are linearly independent over MATH, and since MATH has dimension MATH as a MATH-vector space indeed, MATH the functions MATH, when viewed in MATH, are a MATH-basis. Therefore the functions MATH separate the points of MATH. (Intuitively, this situation as analogous to a finite-dimensional MATH-vector space MATH and a MATH-basis MATH of MATH. Such a MATH-basis separates any two points of MATH, since a MATH-dual vector MATH does the job and we view MATH to realize MATH as a MATH-linear combination of the MATH; thus one of the MATH separates the two points.) An argument as in the proof of REF then shows that MATH is spanned over MATH by the monomials MATH . This set has size MATH, which is too large (when MATH) to be a MATH-basis of MATH. To cut down the size of this spanning set, note any MATH is MATH-linear, so in MATH we can write MATH as a MATH-linear combination of MATH. Therefore for all MATH, MATH is spanned over MATH by MATH so this set is a MATH-basis. NAME done by REF . |
math/0002026 | For MATH, our hypotheses make the function MATH given by MATH equal to REF for MATH and REF for MATH, so the proof of REF still works. |
math/0002026 | To check that the hyperdifferential operators satisfy these three properties, only the third has some (slight) content. By MATH-linearity it reduces to the case MATH and MATH, in which case the NAME rule becomes the NAME formula MATH . Conversely, properties ii and iii suffice to recover the formula MATH for MATH, which by property i forces MATH to be the MATH-th hyperdifferential operator. |
math/0002026 | It suffices to assume MATH is a finite extension, say MATH where MATH is the root of the separable monic irreducible polynomial MATH. Let MATH be a higher MATH-derivation on MATH. Any extension of MATH to a higher MATH-derivation on MATH must send MATH to an element of MATH which has constant term MATH and is a root to the polynomial MATH, where MATH is obtained by applying MATH to the coefficients of MATH. This polynomial is irreducible over MATH by NAME 's Lemma. Since MATH has MATH as a simple root in the residue field MATH, MATH lifts uniquely to a root MATH by NAME 's Lemma. So MATH extends uniquely to a MATH-algebra map MATH by sending MATH to MATH. |
math/0002026 | We take two cases, depending on whether MATH corresponds to a monic separable irreducible polynomial in MATH or to MATH. If MATH is a place corresponding to a monic separable irreducible MATH in MATH, REF shows that the MATH are all MATH-adically continuous, so they all extend by continuity to the completion MATH and still satisfy MATH-linearity and the NAME rule. The corresponding MATH-algebra homomorphism MATH given by MATH is a higher MATH-derivation on MATH. Since MATH is separable, MATH has a coefficient field, say MATH, which contains MATH. Since MATH is separable, the restriction of MATH to MATH must be the usual inclusion MATH, so MATH is a higher MATH-derivation on MATH and therefore also on the fraction field of MATH. If MATH is the place corresponding to MATH, we set MATH and note that MATH. So the MATH are MATH-adically continuous on MATH. They form a higher MATH-derivation on MATH, since this is a subfield of MATH. The continuous extension of all MATH to the completion MATH (by continuity) and then to MATH (by algebra) is along similar lines to the previous case. |
math/0002026 | By REF , it suffices to show the reductions MATH are an algebraic basis of the space of continuous MATH-linear maps from MATH to MATH. We show MATH form a basis of the MATH-dual space MATH for all MATH. For MATH, MATH since MATH . Indeed, by the definition of MATH, when MATH . So for MATH, MATH is a well-defined function from MATH to MATH. Since MATH vanishes at MATH and MATH, the MATH matrix MATH is triangular with REF's along the main diagonal, so it is invertible. Reducing the matrix entries from MATH into MATH gives an invertible matrix, so MATH forms a basis of MATH for all MATH. |
math/0002026 | Use REF and the digit principle. |
math/0002026 | We may suppose MATH, and have to show MATH, where MATH and MATH are the appropriate NAME objects on the ring MATH. For integers MATH, let MATH, where MATH. In particular, write MATH. Since MATH and MATH, MATH . Since MATH, we're done. |
math/0002026 | Let MATH be the degree of MATH and MATH be any positive integer. By REF , for MATH is a well-defined map from MATH to MATH. For MATH, the MATH matrix MATH is triangular with all diagonal entries equal to REF. Since MATH are a MATH-basis of MATH, it follows that the MATH reduced functions MATH are a MATH-basis of MATH. Therefore MATH is an orthonormal basis of MATH, as we wanted to show. |
math/0002026 | To simplify the notation, we write MATH for MATH. Let MATH be the degree of MATH. Let MATH, so MATH by REF . We want to show MATH, and then we'll be done by REF . By REF , MATH form a MATH-basis of the MATH-linear maps from MATH to MATH. By an argument as in the proof of REF , this implies the functions MATH separate the points of MATH. So any element of MATH which is killed by all MATH for MATH must be in MATH. Therefore MATH. |
math/0002026 | Composing the function MATH with reduction mod MATH, we get a MATH-linear map MATH whose kernel consists of power series with MATH-coefficient REF. The reductions MATH are well-defined elements of the MATH-dual space MATH, and in fact are the dual basis to MATH. We are done by REF . |
math/0002026 | Use REF and the digit principle. |
math/0002026 | First we see why completion at MATH is not being considered. Write MATH. Since MATH, MATH has image in MATH. So these functions are not orthonormal on the completion at MATH. Now we look at the completion MATH. To establish the first claim of the theorem, it suffices by the digit principle to check the MATH are an orthonormal basis of MATH. NAME already checked in REF that they belong to this space. By REF and continuity, the reduced functions MATH, for MATH, annihilate the ideal MATH. We now check the corresponding functions on MATH are a basis of the MATH-dual space, which will end the proof by REF . Consider the effect of these MATH functions on the basis MATH. Since MATH for MATH, the MATH matrix MATH is triangular. Since MATH, REF shows MATH, so the diagonal entries are all nonzero. So this matrix is invertible. |
math/0002026 | For MATH and MATH, MATH so the MATH functions MATH are well-defined maps from MATH to MATH. To prove the theorem, it suffices by REF to show that for each MATH, the sequence MATH determines MATH. Writing MATH with MATH, NAME 's congruence implies MATH, so we're done. |
math/0002026 | Since each MATH has degree MATH and MATH sends MATH to MATH, the transition matrix from MATH to MATH is triangular over MATH with diagonal entries MATH where MATH, MATH. This ratio is a MATH-adic unit, so the reduced functions MATH are a basis of MATH. We are done by REF . |
math/0002026 | Let MATH be the NAME series attached to MATH. Since MATH, MATH for all MATH. So for MATH, MATH annihilates MATH. Taking MATH, we will show the induced map MATH is a bijection, so we'd done by the digit principle. For MATH, MATH, so MATH . Therefore MATH recovers the MATH-coefficient of any element of the ideal MATH. Take MATH successively in the congruence REF , which is a formal group version of a weak form of NAME 's congruence: MATH for MATH. So we see that REF is a bijection. |
math/0002026 | The functions MATH for MATH obviously separate the points of MATH. Now apply REF . |
math/0002026 | For MATH and MATH an indeterminate, we can write in MATH for some (unique) MATH. So for any monomial MATH in the variables MATH, we can write MATH where all the exponents in MATH are MATH and MATH is in the ideal generated by MATH. Viewing this equation in MATH, note MATH. So for any series MATH, we can write MATH where MATH is MATH-simplified and MATH. By construction, each coefficient of MATH is a (convergent) sum of coefficients of MATH, so MATH. We have proved existence of a MATH-simplified series in each class of MATH. Provided we show uniqueness, we then vary MATH within a congruence class to see that MATH. For uniqueness, it suffices to show the only MATH-simplified series in MATH is REF. Let MATH be MATH-simplified and nonzero. Scaling, we may assume MATH. Then MATH is a nonzero polynomial in MATH, say MATH. Since MATH, MATH vanishes at all points in MATH. Since MATH and all exponents of MATH are at most MATH, we must have MATH, which is a contradiction. |
math/0002026 | Fix a uniformizer MATH of MATH. Using the basis MATH, we can express any MATH uniquely in the form MATH . The map MATH sending MATH to MATH extends by continuity to a MATH-algebra homomorphism MATH. By REF , this map is surjective. Obviously each MATH is in the kernel, so we get an induced surjection MATH. Since the NAME basis of MATH is orthonormal, we focus our attention on MATH-simplified NAME series and see that MATH is an isometry by REF . Therefore MATH is an isometric isomorphism of MATH-Banach algebras. As MATH depends on MATH, so does the isomorphism we've constructed. |
math/0002026 | Let MATH be a closed prime ideal of MATH, MATH the size of the residue field of MATH. Viewing MATH as a prime ideal of MATH which contains MATH, the containment MATH implies MATH for a unique NAME representative MATH of MATH. Therefore MATH contains the closure of MATH, which is the maximal ideal MATH for MATH. |
math/0002027 | If MATH is NAME, then there exist MATH and a finite rank projection MATH on the kernel of MATH such that MATH for all MATH. Replacing MATH by MATH and applying the estimate MATH, it follows that MATH for all MATH. Introducing the isometries MATH and replacing MATH by MATH, we obtain MATH . Because MATH commute with multiplication operators and MATH, we can write MATH . Observe that MATH and MATH strongly as MATH. Hence it follows that MATH and MATH strongly as MATH. Now we take the limit in the above norm estimate. Since MATH weakly and MATH is compact, we have MATH strongly. Moreover, MATH strongly. We obtain that MATH for all MATH. From this it immediately follows that MATH. |
math/0002027 | It suffices to remark that the NAME operator with a continuous generating function is compact. Hence, by making use of REF , it is easy to see that a NAME regularizer for MATH is given by MATH. As to the index formula, we remark that for MATH, MATH, MATH . The last equality is the well known formula for the NAME index of a scalar NAME operator with continuous symbol. For the precise justification of the second last equality see, for example, CITE . |
math/0002027 | Because of the multiplicative relations REF or REF , it follows that REF implies REF , where the inverse of MATH is given by MATH. The implication MATH is obvious. The fact that REF implies REF follows from REF in connection with the inverse closedness of MATH in MATH. |
math/0002027 | As to REF , it follows from REF that MATH. The inverse of MATH equals MATH and the inverse of MATH equals MATH. In regard to REF , we use REF and obtain MATH. The inverse of MATH equals MATH and the inverse of MATH equals MATH. |
math/0002027 | Because of the assumptions on the NAME algebra MATH there exists a factorization MATH with MATH, MATH, MATH and MATH without loss of generality. Taking the inverse and replacing MATH by MATH, it follows that MATH . Because MATH, REF are equal and represent factorizations of the form REF . We can apply REF and write MATH in the form REF with the conditions on parameters MATH, MATH and MATH stated there. We conclude that there exists a matrix function MATH of the form REF such that MATH . Combining REF with REF and introducing MATH, it follows that MATH where MATH is of the form MATH with MATH . Introduce MATH write MATH and observe that MATH and MATH are nilpotent matrix functions. Note also that MATH and MATH. From REF we obtain that MATH, and moreover MATH for each MATH by induction. The matrix functions MATH and MATH are well defined by a series expansion, which is finite due to the nilpotency. Using this series expansion, it follows that MATH . This in connection with REF implies MATH . From REF and the assumption MATH, it follows that MATH. From the representation REF , we further obtain MATH. On account of REF , it now follows that MATH . Hence MATH, and, consequently, MATH. This in connection with REF implies that MATH . From MATH it follows (by putting MATH and MATH) that MATH . Hence MATH for each MATH. Thus we can write MATH with certain MATH and MATH such that MATH. It follows that MATH with certain MATH where MATH. Denoting by MATH the diagonal matrix in REF , it follows in connection with REF that MATH . Because MATH, we may replace the expression MATH by the notation MATH. In this way, we arrive at the desired factorization REF . |
math/0002027 | Because an antisymmetric factorization is automatically also a usual factorization in the NAME algebra MATH (except for the slightly different middle factor, which is irrelevant at this place), it follows that the numbers MATH are uniquely determined up to change of order. Because the order of the characteristic pairs in an antisymmetric factorization can be rearranged in any desired way, we can assume without loss of generality that MATH. Now suppose that we are given two antisymmetric factorizations of MATH, MATH where MATH and MATH are both of the form REF but with the pairs MATH respectively. Introducing the parameters MATH and the integers MATH as in REF , we can write MATH where MATH and MATH are diagonal matrices of size MATH with entries MATH or MATH on the diagonal. Moreover, we can write MATH and MATH, where MATH is of the form REF and MATH. It follows that MATH are two factorizations of the form REF . We apply REF and see that MATH where MATH and MATH are of the form REF . The last equation can be rewritten as MATH. Combined with the first equation, it follows that MATH . Because of the block triangular structure of MATH and MATH with invertible constant matrices MATH on the block diagonal, we obtain that MATH for each MATH. Hence MATH, and, consequently, the numbers of MATH's and MATH's, respectively, on the diagonal of MATH and MATH is the same. From this it follows that the collection of the pairs MATH is the same as the collection of the pairs MATH up to change of order. |
math/0002027 | Putting MATH or MATH in the factorization MATH it follows that MATH and MATH. Now the assertion follows from the facts that MATH and MATH as can easily be seen. |
math/0002027 | We can assume that MATH where MATH and MATH are nonnegative integers with MATH and MATH. With the numbers MATH defined as above in terms of the characteristic pairs appearing in MATH, it follows from REF (with MATH) that MATH . In particular, MATH. Hence there exists a permutation matrix MATH such that MATH where MATH . Moreover, there exists another permutation matrix MATH such that MATH . We define MATH . It can be verified straightforwardly that MATH . Hence, on defining MATH by MATH it follows that MATH. |
math/0002027 | From the definition of MATH it follows that MATH and MATH. By REF there exists an antisymmetric factorizations REF with MATH and MATH of the form REF . From the definition of MATH it follows furthermore that MATH, which in turn implies MATH. Using REF we obtain the existence of a function MATH for which MATH. Now we define MATH which implies immediately the validity of REF . Moreover, MATH . From MATH we obtain MATH . Hence MATH. Because, obviously, MATH, we even have MATH. This settles REF . The proof of REF is similar. By REF (see also the remark made afterwards) and the facts that MATH and MATH, there exists an antisymmetric factorization REF . From MATH we obtain MATH. We apply again REF , but now with MATH replaced by MATH, in order to conclude the existence of a function MATH for which MATH. Finally, we define MATH apply the equation MATH and obtain in this way that MATH. Hence MATH. |
math/0002027 | First observe that MATH. Consequently, MATH . Moreover, because MATH is a diagonal operator it suffices to determine MATH and to take the sum. If MATH, then MATH. Hence the corresponding dimension is zero. If MATH, then the matrix representation of MATH has entries MATH only on the MATH-th diagonal, which connects the entries MATH and MATH, and has zero entries elsewhere. From this it is easy to see that the dimension equals MATH. |
math/0002027 | By considering the scalar case, MATH and distinguishing MATH and MATH, it can be seen straightforwardly that MATH. Moreover, using REF and the fact that MATH it follows that MATH . In order to prove that MATH and MATH, we introduce MATH and MATH. By just using the identity MATH, one can verify that MATH, MATH, MATH. Because MATH and MATH, the desired relations follow immediately. |
math/0002027 | Using REF , it follows that MATH . Hence MATH. By using this, the relations of MATH and MATH, and formula MATH from REF , we obtain MATH . Next, as an auxiliary step, we are going to establish the identities MATH . Indeed, using that MATH and REF , it follows that MATH . Multiplying from the left with MATH and observing that MATH by REF , we obtain the first identity in REF . Similarly, by using MATH, it follows that MATH . Multiplying from the right with MATH by observing that MATH, we arrive at the second identity in REF . Using this and the definition of MATH, it follows that MATH and MATH. Hence we obtain the desired expressions for MATH and MATH. Moreover, using the notation MATH and MATH introduced in the proof of REF , it is easy to see that MATH. Hence MATH . Combining this with REF it follows that MATH . Now we are able to derive the remaining identities: MATH . This completes the proof. |
math/0002027 | Since MATH by REF , it follows that MATH. The relation MATH stated in REF implies that MATH and MATH. Moreover, because MATH, we obtain MATH. Similarly, since MATH, we arrive at MATH. |
math/0002027 | The formulas for MATH and MATH follow immediately from REF in connection with REF . Moreover, by the index formula stated in REF it can be seen that MATH . Because MATH, we have MATH . On the other hand, MATH . Combining all this, it follows that MATH . Since MATH, we obtain from REF , in particular, that MATH. We conclude from the definition of the function MATH that MATH . Using the formulas MATH it is easy to derive the remaining two formulas. |
math/0002027 | Let us first consider REF . By REF we can assume that we are given an asymmetric factorization MATH with the conditions on the factors stated there. In addition, we are given an antisymmetric factorization MATH with MATH of the function MATH. From REF it follows that MATH where both MATH and MATH are invertible. There inverses are equal to MATH and MATH, respectively. Hence the dimension of the kernel and cokernel of MATH is equal to that of MATH, which, in turn, has been given in REF . Next we need to take into account the following fact, which can be proved straightforwardly: if an operator MATH has a pseudoinverse MATH and MATH where MATH and MATH are invertible operators, then a pseudoinverse of MATH is given by MATH. It follows from REF that a pseudoinverse of MATH is given by MATH. Consequently, a pseudoinverse of MATH is given by MATH . Now we use REF in order to conclude that MATH . Remark that MATH (because this is just the equation MATH). Combining these last facts, we arrive at the desired expression for the pseudoinverse. REF can be treated in the same way, but we give the complete proof because the notation differs sometimes here in comparison with previous results. First of all, we may assume that we are given an asymmetric factorization MATH as has been stated in REF . Moreover, we are given an antisymmetric factorization MATH with MATH of the function MATH. From REF it follows that MATH where MATH and MATH are invertible. The inverses are MATH and MATH, respectively. The above relation MATH can be rewritten as MATH. We now have to apply REF with MATH replaced with MATH and MATH replaced with MATH. Correspondingly, the characteristic pairs MATH have to be replaced with MATH. We arrive at the formulas MATH . It remains to note that MATH in order to conclude the desired formulas for the dimension of the kernel and cokernel of MATH. Also in regard to the pseudoinverse we have to apply REF , but with MATH replaced with MATH and MATH replaced with MATH. It follows that the pseudoinverse of MATH is given by MATH. As before, we obtain that a pseudoinverse of MATH is given by MATH . Using REF we derive MATH . The desired pseudoinverse of MATH is now obtained by piecing together these last facts in connection with MATH, which is just the factorization MATH rewritten. |
math/0002027 | The operators are invertible if and only if the sums in REF , or, REF , respectively are zero. Notice that the different terms appearing there are all nonnegative integers. Hence they must be equal to zero. It remains to remark that MATH if and only if MATH or MATH or MATH. |
math/0002027 | The first assertion follows from the fact that the linear operators MATH are NAME space isometries and are both the inverse and the adjoint of each other. In fact, MATH . In order to prove REF it suffices to recall REF , to use the last identity and the relation MATH . One has also to use the definition of MATH and MATH and the fact that MATH. |
math/0002027 | The proof is based on REF . Because MATH and MATH are unitarily equivalent to MATH and MATH, respectively, the formulas for the dimension of the kernel and cokernel follow immediately. In order to show that the above expression are indeed pseudoinverses, one can apply the C*-algebra isomorphism MATH introduced in REF to these operators. Using the formulas stated there, one obtains MATH . The operators on the right hand side are exactly REF for the pseudoinverses of MATH and MATH. The observation that an operator MATH is a pseudoinverse of an operator MATH if and only if MATH is a pseudoinverse of MATH completes the proof. |
math/0002027 | The dimension the kernel and cokernel of the NAME + NAME operator MATH, which is defined on MATH, coincides with that of the operator MATH which is defined on MATH. Now we write MATH . Because MATH is nilpotent, the first expression on the right hand side (that is, the operator MATH) is invertible. Hence we have to determine the dimension of the kernel and cokernel of MATH which is just the singular integral operator MATH with MATH given as above. Now the result follows from REF . As to the pseudoinverse, we first remark that a pseudoinverse of MATH is given by MATH, where MATH is a pseudoinverse of the above operator MATH. Since MATH, it follows that MATH. Hence MATH because MATH as can easily be seen. From REF we conclude that MATH may be given by MATH . Using the definition of the operators occurring there, we obtain that this is equal to MATH . Hence MATH equals MATH which in turn is equal to the operator REF . |
math/0002027 | As before, the dimension of the kernel and cokernel of MATH coincides with that of the operator MATH. Now we write MATH . Because MATH is nilpotent, the last expression on the right hand side is an invertible operator. Hence we are led to the dimension of the kernel and cokernel of MATH which coincides with the singular integral operator MATH where MATH is given as above. Now the result follows from REF . Again, a pseudoinverse of MATH is given by MATH. Since MATH-Y', it follows that MATH. Hence MATH because MATH as can easily be seen. From REF we conclude that MATH may be given by MATH . We obtain that this is equal to MATH . Hence MATH equals MATH which in turn is equal to the operator REF . |
math/0002028 | Fix a MATH such that MATH, for any MATH and any MATH with MATH. Find a finite MATH-net MATH in MATH and a finite MATH-net MATH in MATH. Any map MATH produces a (nonunique) map MATH defined so that MATH is a point of MATH whose distance to MATH is MATH. Now if MATH and MATH are MATH-equicontinuous maps with MATH, then MATH and MATH are MATH-close, hence homotopic. In particular, MATH fall into at most MATH homotopy classes. |
math/0002028 | For reader's convenience we review the argument in CITE emphasizing its local nature. It is proved in CITE that any domain MATH as above has an atlas of harmonic coordinate charts MATH where MATH is a metric ball at MATH whose radius MATH depends only on the initial data. Further, the metric tensor coefficients in the charts MATH are controlled in MATH topology. An elliptic estimate then shows that the transition functions MATH are controlled in MATH topology. All these results are stated and proved locally. Next, the relative volume comparison implies that one can choose a finite subatlas so that there is a uniform bound on the multiplicities of intersections of the coordinate charts and the balls MATH still cover MATH (this argument involves only small balls and hence is local). The lower injectivity radius bound gives a lower bound for the volume of any small ball that depends only on the radius of the ball CITE. This, together with a upper bound on MATH, implies an upper bound on the number of coordinate charts. Finally, following NAME 's thesis (as outlined in CITE) one can ``glue the charts together" which proves the theorem. Alternatively, one can follow (almost word by word) the argument in CITE where a ``compact domain version" of NAME convergence theorem is proved. |
math/0002028 | Take an arbitrary MATH. To prove precompactness in NAME topology it is enough to show that the number of elements in a maximal MATH- net in MATH is bounded above by some MATH independent of MATH. Fix a maximal MATH-separated nets MATH in MATH so that MATH-balls with centers in MATH are disjoint and MATH-balls cover MATH. The relative volume comparison gives a uniform lower bound for the volume of the MATH-ball centered at any point of MATH; say MATH. Then MATH and MATH converge in the NAME topology to a compact metric space MATH. As we explained above the interior of MATH is necessarily locally interior. |
math/0002028 | By REF MATH subconverges in the NAME topology to a compact metric space MATH whose interior MATH is a locally interior metric space. We are in position to apply NAME 's stability theorem CITE which asserts that MATH is a topological manifold, and moreover, any compact subset of MATH lies in a compact domain MATH such that there are topological embedding MATH. Furthermore, MATH induce NAME approximations which become arbitrary close to the given NAME approximations between MATH and MATH. Choosing MATH large enough, one can ensure that MATH as promised. |
math/0002028 | It is well-known to experts that a vector bundle over a finite cell complex is recovered up to finitely many possibilities by the total NAME class and the NAME class of its orientable (MATH or MATH-fold) cover (see CITE for a proof). We are now going to reduce to this result. In what follows we use the notations of REF. Since MATH is independent of MATH, there is a covering MATH associated with MATH and, for each MATH, a covering MATH associated with MATH. The embedding MATH lifts to an embedding MATH of orientable manifolds. Using that MATH is a finite group, we can partition MATH into finitely many subsequences each having the same first NAME class. It suffices to show that any such subsequence falls into finitely many isomorphism classes, so we can assume that MATH, and hence MATH, is independent of MATH. Let MATH be a covering associated to the subgroup of MATH that corresponds to MATH. This subgroup lies in MATH so MATH is an intermediate covering space between MATH and MATH, that is, we have coverings MATH and MATH. Also let MATH be a covering associated to MATH. The embedding MATH lifts to an embedding MATH. Now the normal bundle MATH is orientable so its NAME class is well-defined (up to sign since there is no canonical choice of orientations). Note that MATH takes the NAME class of MATH to MATH. It is a general fact that finite covers induce injective maps in rational cohomology. (The point is that the transfer map goes the other way, and precomposing the transfer with the homomorphism induced by the covering is multiplication by the order of the covering.) Also, the NAME class of MATH is MATH-image of MATH. Thus, given MATH and MATH, one can uniquely recover the (rational) NAME and NAME class of MATH. As we mentioned above these classes determine MATH up to finitely many possibilities. Therefore, MATH are determined up to finitely many possibilities by MATH and MATH as desired. |
math/0002028 | Since MATH has uniformly bounded geometry, there exists a metric space MATH, and homeomorphisms MATH of MATH onto compact domains MATH such that both MATH and MATH are almost equicontinuous. If REF holds, then MATH is an almost equicontinuous sequence of maps from MATH into MATH. Thus, REF implies that the maps MATH's fall into finitely many homotopy classes. Now we are done by definition of an invariant since MATH's are homeomorphisms. If REF holds, then MATH is an almost equicontinuous sequence of maps from MATH into MATH. Again, by REF there are only finitely many homotopy classes of maps among MATH. It suffices to show that, whenever MATH is homotopic to MATH, the maps MATH and MATH have the same invariants. Let MATH be a homotopy that connects MATH and MATH. The homotopy MATH defined as MATH connects MATH with MATH. Thus, MATH is homotopic to MATH . Since MATH is a homeomorphism, MATH and MATH have the same invariants as desired. |
math/0002028 | For any invariant MATH, MATH. In particular this is true for the rational NAME class and generalized NAME class. The result now follows from REF combined with REF. |
math/0002028 | Since MATH has bounded geometry, there exists MATH and homeomorphisms MATH such that both MATH and MATH are almost equicontinuous. Note that MATH is a smooth submanifold of MATH. If MATH is almost equicontinuous, then so is MATH. Similarly, if MATH is almost equicontinuous, then so is MATH. Thus, REF implies that the set of isomorphism classes of normal bundles MATH is finite. In particular, the set of topological equivalence classes of normal bundles MATH is finite. |
math/0002028 | Since MATH is almost equicontinuous, MATH is uniformly bounded above. The result now follows from REF. |
math/0002028 | Since MATH we can find a compact domain MATH with MATH. By results of REF, any family of such domains MATH has bounded geometry, hence the conclusion follows from REF. |
math/0002028 | Let MATH be a family of nonnegatively curved manifolds satisfying conditions of REF. For any MATH let MATH be a soul of MATH. First,we show that MATH has uniformly bounded geometry. By REF has lower volume bound. Lower sectional curvature bound follows because souls are totally geodesic. Since MATH and there is a distance-nonincreasing retraction of MATH CITE, the diameter of MATH is at most MATH. The map MATH is a homotopy equivalence, in particular, it has nonzero degree, hence it is onto. We conclude that MATH. Thus, MATH has uniformly bounded geometry. Since MATH we can find compact domains MATH with MATH. Again, MATH has uniformly bounded geometry, and the conclusion follows from REF. |
math/0002028 | Start with an arbitrary family of totally geodesic embeddings MATH as above. First, we show that MATH has uniformly bounded geometry where MATH is equipped with the induced Riemannian metric. By a result of CITE MATH implies a lower volume bound on MATH. Since MATH is totally geodesic MATH, and by REF. Thus, NAME 's stability theorem implies that MATH has uniformly bounded geometry (see REF) Note that MATH, hence, REF implies that there is a compact domain MATH such that MATH has uniformly bounded geometry. The result now follows from REF because totally geodesic embeddings MATH are MATH-Lipschitz, in particular, MATH is equicontinuous. |
math/0002028 | First note that, up to topological equivalence, only finitely many of the bundles MATH can have zero NAME class. (Otherwise, there is a sequence of pairwise topologically inequivalent bundles MATH with zero NAME class. Use homeomorphisms MATH to pull the bundles back to MATH. These pullback bundles clearly have zero NAME class as well as zero rational NAME classes since MATH. Thus the bundles belong to finitely many isomorphism classes which implies that MATH belong to finitely many topological equivalence classes.) Now if the NAME class of the normal bundle to MATH is nonzero, then MATH, hence the distance from MATH to MATH is MATH, and the result follows from REF. |
math/0002028 | Let MATH and MATH be chosen to satisfy the assumptions. According to REF we only have to show is that under our assumptions we have a uniform lower bound on MATH. Let MATH. Let MATH and MATH be a round sphere of constant curvature MATH and MATH be any point on this sphere. Consider the exponential map MATH. Denote by MATH the volume of the ball of radius MATH centered at MATH. First of all, notice that by the triangle inequality MATH contains a tubular neighborhood MATH consisting of all points MATH such that MATH and MATH. Here MATH stands for the NAME retraction MATH. For any MATH denote MATH . Since NAME retraction is a MATH-Riemannian submersion CITE we can apply NAME 's Theorem to see that MATH . Here MATH stands for the ball of radius MATH around MATH in MATH. It suffices to show that for each MATH we have MATH. (Indeed, it would imply that MATH. Finally, by volume comparison, MATH is bounded below in terms of MATH and MATH and we are done.) Fix a MATH and consider the normal exponential map MATH. It follows from CITE that this map sends the ball MATH onto MATH. Choose a linear isometry between MATH and MATH and use it to equip MATH with the metric MATH of constant curvature MATH. Let MATH be the induced Riemannian metric on the NAME fiber over MATH. To finish the proof it is enough to establish the following lemma saying that "reverse NAME comparison" holds on MATH. The surjection MATH is a distance nondecreasing diffeomorphism. Let MATH be a unit vector in MATH and MATH be the normal geodesic in direction MATH. We now show that, for any MATH and any MATH with MATH and MATH, we have that MATH. Write MATH as a sum of its horizontal and vertical components. Then MATH . The first term in the right hand side is MATH by assumption and also because MATH . By CITE MATH and therefore MATH . By the symmetry of the curvature tensor the forth term is equal to the third one and hence is also equal to MATH. Thus MATH. Now since MATH and all the two planes along MATH containing MATH have curvature MATH we can apply the NAME comparison theorem to conclude that the differential of MATH does not decrease the lengths of tangent vectors and thus MATH is a local diffeomorphism that does not decrease lengths of curves. It remains to show that this map is MATH. Suppose not. Then the injectivity radius MATH of MATH at MATH is strictly less than MATH. Let MATH be such that MATH and MATH. Denote MATH. Notice that geodesics MATH and MATH connecting MATH and MATH are obviously distance minimizing. By CITE these geodesics must form a geodesic loop at MATH (that is, MATH). This is impossible since according to CITE the soul MATH is totally convex and MATH lies in MATH. |
math/0002028 | The proof of REF gives a uniform lower bound on MATH so the result follows from REF. |
math/0002028 | Start with an arbitrary family of totally geodesic embeddings MATH as above. First, note that for any MATH the injectivity radius of MATH satisfies MATH. If not, there is a point MATH with MATH. Since MATH and MATH has nonpositive sectional curvatures, the injectivity radius at any point is half the length of the shortest geodesic loop at this point CITE. Take a geodesic loop at MATH of length MATH and project it to MATH by the closest point retraction MATH CITE. The retraction is a distance-nonincreasing homotopy equivalence. (In fact, since MATH, the normal exponential map identifies MATH with the normal bundle to MATH where MATH corresponds to the bundle projection.) Thus, we get a homotopically nontrivial curve of length MATH in MATH. Since MATH is an isometric embedding it preserves lengths of curves. Therefore, we obtain a homotopically nontrivial curve of length MATH in MATH which is impossible because loops of length MATH lift to loops in the universal cover. Find compact domains MATH that lie in the MATH-neighborhood of MATH. Since MATH, the diameter of MATH is MATH. Also MATH and MATH, hence REF implies that MATH has bounded geometry, and the conclusion follows from REF. |
math/0002028 | Since MATH is bounded in absolute value, lower bound on volume implies a lower bound on the injectivity radius. Thus, the result follows from REF exactly as in the proof of REF. |
math/0002029 | Let MATH be a MATH-plane, and let MATH, and suppose, for simplicity, that MATH, so MATH is identified to the element MATH. Then it is easy to see that MATH, evaluated on MATH, is nothing but the evaluation of MATH on MATH, which depends only on the positive (or self-dual) part of the NAME tensor. To prove the second assertion, we remark that MATH, being a quadrilinear symmetric form on MATH, can be identified with a polynomial of degree REF on MATH, which is determined by its values. |
math/0002029 | We have to prove that MATH is included in MATH, defined as follows: MATH . We will prove that MATH, therefore it will follow that the latter is non-empty, and is a linear space of dimension REF. We denote by MATH the parallel displacement, along MATH, of a non-zero vector in MATH, transverse to MATH. Then MATH, because MATH is included in the is totally geodesic surface MATH, thus we can characterize MATH as the set MATH, for any MATH. We then observe that MATH because MATH is in MATH, thus MATH . So the scalar function MATH satisfies to a linear second order equation, hence it it determined by its initial value and derivative. It follows then that it is identically zero, thus MATH everywhere, as claimed. |
math/0002029 | The open set of MATH which is the space of null-geodesics of M can be viewed as the space of integral curves of the geodesic distribution MATH of lines in MATH, the total space of the fibre bundle of isotropic directions in MATH. MATH is defined as the horizontal lift (for the NAME connection on M) of MATH, which is an isotropic line in MATH. This definition is independent of the chosen metric and connection CITE, and, by integrating this distribution (as M is civilized), we get a holomorphic map MATH, where (an open set of) MATH is the space of leaves of this foliation. This map can be used to compute the normal bundle of MATH, MATH, see CITE,CITE,CITE. Indeed, we have lines MATH, such that MATH, which project onto MATH, thus we get the following exact sequence of normal bundles: MATH where we have written the ambient spaces of the normal bundles on the second position. The central bundle is trivial (MATH is trivially embedded in MATH, which is trivially embedded in MATH as a fibre), and it is easy to check that the left hand bundle is isomorphic to the tautological bundle over MATH, MATH. This proves that MATH, in particular the conditions in REF are satisfied. Thus the lines in the integral MATH-cone MATH form an analytic subfamily of the family MATH, that correspond to the sections of the normal bundle of MATH, vanishing at MATH, or, equivalently, to the points MATH of MATH. But, in order to prove the smoothness of MATH, we first remark that the surface MATH, defined as follows, is smooth: MATH where MATH is the MATH-plane containing MATH. MATH is smooth, and MATH. We note now that MATH is everywhere, with the exception of the points of MATH, transverse to the fibers of the submersion MATH. We may conclude that MATH is a smooth analytic submanifold of MATH (not closed). We can use similar methods to prove that MATH is an immersed submanifold of MATH (by using the projection MATH). |
math/0002029 | Let MATH. We will define MATH as follows: Recall that MATH, where MATH (the tangent space in a point to a cone depends only on the line containing the point). We know that MATH corresponds to MATH, the space of NAME fields on MATH, vanishing at MATH, and such that MATH. It will be shown in the proof of the next theorem that MATH consists of classes of NAME vector fields such that MATH, REF . Then, on a representative NAME field MATH, we define MATH to be the class of NAME fields in MATH, represented by the following NAME field MATH on MATH, which is given by MATH. We remark that MATH is what we usually note MATH, when the parameter on MATH is understood. It is straightforward to check that MATH induces an isomorphism MATH for each non-zero MATH. |
math/0002029 | As MATH is a subbundle of the normal bundle MATH, MATH is isomorphic MATH. As in REF , we will construct the inverse isomorphism MATH: Let MATH be a linear application. Let MATH be a representant of MATH (it involves a choice of a complementary space to MATH in MATH). We define MATH as being induced by the following linear application between spaces of NAME fields on MATH: MATH, where the second space corresponds to MATH, that is, it contains NAME fields MATH such that MATH. Consider a parameterization of MATH around MATH, and let MATH. We define MATH by MATH, and it is easy to check that the class of MATH in MATH is independent of the representant MATH, such that MATH is well-defined. It is also obviously invertible. |
math/0002029 | Consider the following analytic map, which parameterizes, locally around MATH, the deformations of the geodesic MATH that correspond to points contained in the integral MATH-cone MATH: MATH where MATH is a neighborhood of the origin in MATH, and MATH is a deformation of the null-geodesic MATH, such that MATH where the parameterization of the geodesic MATH satisfies MATH, and MATH is the MATH-plane in MATH containing MATH. Convention We know that MATH is defined around the origin in MATH, so there exists a polydisc centered in the origin included in MATH, therefore all the relations that we will use are true for values of the variables MATH sufficiently close to REF. For simplicity, we will omit to mention these domains. The geodesics MATH correspond to points in MATH, and the NAME fields MATH on MATH, defined as MATH correspond to vectors in MATH tangent to the above mentioned lines. We suppose that the deformation MATH is effective, that is, MATH and MATH, which is equivalent to MATH. In order to compute the projective curvature of MATH, we need thus to study the (second order) infinitesimal variation of these NAME fields on MATH. As they are determined by their value and first order derivative in MATH, we need to evaluate MATH for the first derivative of MATH at MATH, and MATH for the second. Dots mean, as before, covariant differentiation with respect to the ``speed" vector MATH, thus correspond to the operator MATH. As the covariant derivation MATH has no torsion, we can apply the usual commutativity relations between the operators MATH and use them to differentiate the following equation, which follows directly from the definition of MATH and MATH: MATH . We get then MATH . We recall now that, besides REF , we have MATH, thus MATH is isotropic, which implies that: MATH as MATH. REF prove that MATH which completes the proof of REF . From REF , it equally follows that MATH is isotropic, and, by differentiating REF , we get MATH . From, REF we have that MATH is isotropic, and also MATH which, together with REF , implies that MATH. We have then MATH which proves that the curvature MATH of the MATH-cone takes values in MATH, as it is represented by MATH. In view of REF , it is clear now that the projective curvature MATH is represented by the following application: MATH . From REF , as MATH and MATH, we get MATH . The right hand side actually involves only MATH, as the other components of the Riemannian curvature vanish on this combination of vectors, thus we can replace MATH with MATH inn the above relation. On the other hand, the class of MATH modulo MATH is determined by its scalar product with MATH, which represents a non-zero generator of MATH. The proof of the Theorem is now complete. |
math/0002029 | If MATH admits a projective structure, some of whose geodesics are the lines MATH, then we have, for a fixed MATH, a linear connection around MATH, whose geodesics in the directions of MATH coincide, locally, with MATH. This means that the integral MATH-cone MATH, for MATH a null-geodesic, is part of a complex surface (namely MATH, where MATH is REF-plane corresponding to MATH). Then the integral MATH-cone MATH, the lift to MATH of MATH, is also a complex surface, thus MATH is a subset of the tangent space MATH, thus a flat cone. As this is true for all points of MATH and for all null-geodesics MATH, REF implies that M is flat. On the other hand, it is well-known that the twistor space of a conformally flat manifold admits a flat projective structure, for which the projective lines MATH are geodesics, CITE. |
math/0002029 | The idea is to prove that the integral MATH-cone MATH is a smooth surface. We know that this holds in all its points except for the vertex MATH REF . The fact that all direction in MATH admits a tangent line is a necessary condition for this cone to be a smooth surface, as it needs to be well-defined around MATH. We choose an auxiliary hermitian (real) metric MATH on MATH. Its restrictions MATH to the lines MATH yield NAME metrics on these lines; in fact these metrics are deformations of one another, just like the lines MATH are. This means that the metrics MATH depend continuously on MATH, a parameter in a compact set. We can therefore find a lower bound MATH for the injectivity radius of all MATH at MATH, and a finite upper bound MATH for the norm of all the second fundamental forms MATH . We can also suppose that MATH is smaller than the injectivity radius of MATH at MATH. The first step is to prove that MATH is a submanifold of class MATH. As its tangent space is everywhere a complex subspace of MATH, it will follow that it is a complex analytic submanifold. Consider now the exponential map MATH, defined for the metric MATH; If we restrict it to a ball of radius less than MATH, it is a diffeomorphism into MATH. The image of the complex plane MATH is then a smooth REF-dimensional real submanifold MATH of MATH, and there exists a positive number MATH such that the exponential map in the directions normal to MATH, MATH restricted to the vectors of length less than MATH, is a diffeomorphism. The image of this diffeomorphism is a tubular neighborhood of MATH, and we will denote by MATH such a tubular neighborhood of ``width" MATH, for MATH. The existence of an upper bound MATH for the second fundamental forms of MATH implies the following fact: For any MATH, there is a neighborhood MATH of the origin such that MATH is contained in MATH, and is transverse to the fibers of the orthogonal projection MATH, where MATH. This is standard if MATH is a submanifold; but it is also true in our case, where MATH is a union of submanifolds MATH. Now it is easy to prove that MATH is a MATH submanifold of MATH (the projection MATH yields a local MATH diffeomorphism from a neighborhood of MATH in MATH to a neighborhood of MATH in MATH; it is MATH in MATH because MATH is tangent to MATH at MATH). So MATH is a MATH submanifold of MATH; Its tangent space is complex in each point, thus MATH is a complex-analytic surface immersed in MATH. We have then that MATH, being the lift of MATH, is a smooth analytic surface immersed in MATH, in particular the MATH-cone MATH is a complex plane. REF implies that MATH vanishes on the MATH-plane MATH which contains MATH, for every point MATH. Now, the plane MATH is not the only one admitting projective lines MATH tangent to any of its directions: all planes ``close" to MATH have the same property. Then MATH vanishes on a neighborhood of MATH, hence on the whole connected manifold M. |
math/0002029 | We first remark that the main difficulty is the definition of MATH, the space of null-geodesics , and of MATH, the twistor space of MATH, as M is not necessarily civilized. This is only possible on small open sets, but, in general, we can not expect to have any global construction of this kind. Thus, things that were almost obvious in the twistorial framework (like the existence of compact deformations of the null-geodesic MATH), seem much more difficult to prove directly. The idea is to prove that all null-geodesics close to MATH are diffeomorphic to MATH. Then, we show that, conversely, every projective line which is a deformation of MATH as a compact curve is a null-geodesic. In particular, sections in the normal bundle MATH are induced by (local) NAME fields. We obtain then directly that MATH. Let MATH be an immersed null-geodesic, diffeomorphic to a projective line MATH. Then any local NAME field MATH with MATH induces a global normal field MATH on MATH. If MATH is just a compact geodesic, it may have points of self-intersection, but it is always an immersed curve. It is more convenient then to think of MATH as a projective line immersed in M rather than the image of this immersion. The tangent, normal bundles, etc. are also to be thought as bundles over this projective line, still denoted by MATH. Tubular neighborhoods of MATH are then neighborhoods of the zero section in the normal bundle MATH, small enough to be immersed (non-injectively) in M as a neighborhood of the image of MATH. We first notice that MATH may be decomposed in the union of a two open sets MATH, both biholomorphic to the unit disk in MATH, and such that MATH is connected. Then, for any local metric in MATH, we have a NAME equation around a point MATH, and a NAME field MATH corresponding to prescribed MATH. It is easy to prove that the (local) normal field induced by MATH is independent of the chosen metric. Moreover, this normal field is the unique solution, for the prescribed REF-jet in MATH induced by MATH, of a second order differential equation on MATH: The NAME equations for null-geodesic induce a second order linear differential operator MATH on MATH, depending only on the conformal structure MATH of M. For a NAME connection MATH of a local metric on M, we locally define the following differential operator on MATH: MATH by MATH. It obviously induces a (local) differential operator on MATH, and all we need to show is that, for a different connection MATH, the corresponding operator MATH induces the same one on MATH. First we write MATH then we recall that another NAME connection MATH is related to MATH by the formula CITE: MATH so we directly obtain: MATH thus they induce the same operator on MATH. This one is, therefore, globally defined (the topology of MATH is not important). Now, for any MATH, we have an unique solution MATH of MATH, for a prescribed REF-jet MATH (which consists in the values in MATH of MATH and of his first-order derivative), globally defined on every contractible open set MATH in MATH. This is because on any such contractible set the equation MATH becomes a second order ordinary linear equation on a disk in MATH, which admits global holomorphic solutions (unique if we fix the initial conditions). Take now MATH. Then, the two solutions MATH and MATH, defined on MATH, respectively, MATH, coincide on the connected MATH, so they yield a global solution MATH with the prescribed initial conditions in MATH. In particular, this solution is a global section of the normal bundle MATH. After the infinitesimal result, the local one: Small deformations of MATH are also compact immersed projective lines. Remark. The tubular neighborhoods considered below are always seen as images, by a local diffeomorphism, of subsets - which are, generally, fiber bundles over MATH - of the normal bundle of MATH, respectively, MATH, the lift of MATH to MATH. We need this because of the possible self-intersections of MATH; MATH is always embedded. Consider an auxiliary hermitian metric MATH on M. Then MATH induces the same topology like a round metric MATH on the sphere MATH. We can define a tubular neighborhood MATH of MATH, as the open set of points MATH with MATH. We choose MATH small enough for MATH to be a fiber bundle over MATH (the fiber MATH, for MATH, being the image of the real REF-plane of MATH, MATH-orthogonal to MATH, by the exponential of MATH). Take now a finite number of contractible open sets whose union covers MATH, and we choose holomorphic metrics in MATH on each of this open sets. Then, on these sets we have connections, and it is well-known that any point in such a set has a basis of geodesically connected neighborhoods CITE, CITE. We choose a finite number of such geodesically convex open sets, that cover MATH, and such that they are all included in MATH. We have then MATH . (Of course, they are geodesically convex only with respect to some particular metric in MATH, but we are interested only in the implications involving the null-geodesics, which are independent of the metric.) It is immediate, CITE, that a geodesically convex set MATH has the following property: all maximal geodesics are closed submanifolds of MATH and are contractible as topological spaces. This is particularly true for null-geodesics included in MATH. We refine now the covering by another one, MATH, such that MATH, where MATH, such that MATH. In fact, we ask for MATH to be restrictions to MATH of a tubular neighborhood MATH, for MATH sufficiently small (MATH are ``cylindrical" neighborhoods). We can easily imagine how to find such a refinement of the initial covering. The proof of the proposition now follows two ideas: first, we consider a very special covering by disks (for the round metric MATH) of MATH; second, we define a neighborhood of MATH in MATH of isotropic directions for which, using the special covering of MATH, we can extend the associated null-geodesics and eventually get compact ones. We call MATH the canonical lift of a null-geodesic in MATH; it is an integral curve of the geodesic distribution of lines in MATH. The first idea has nothing to do with complex analysis; it is just a matter of metric topology on the round sphere MATH. Let MATH be a covering of MATH by open sets. Then there exists a positive number MATH and a finite set of points MATH, for MATH, such that: CASE: All disks MATH are contained in at least one of the open sets MATH; CASE: The disks MATH cover MATH; CASE: MATH, where MATH is the diameter of MATH. An important property of this covering is that all sets, as well as the intersections of a finite number of them, are convex for the round metric, thus contractible. We intend to extend null-geodesics which can be projected diffeomorphically onto MATH by means of the fiber projection MATH. We will do that step-by-step, extending it over disks MATH of increasing radius. But before that, we need to restrict ourselves to some particularly ``close to MATH" null directions. We need two things: the extensions need to remain within MATH, and they should also be transverse to the fibers of MATH, otherwise the projection into MATH would not be an invertible diffeomorphism. First, we consider the following compact subset of MATH: MATH where we say that a complex line MATH is tangent to a real manifold MATH if it contains a non-zero (real) vector tangent to this real submanifold. The hermitian metric MATH on M induces a metric on MATH, and also one on MATH. We can, then, evaluate the distance between MATH and MATH: MATH as they are disjoint compact sets. Following CITE, we can define the complex REF-manifolds MATH as the spaces of null-geodesics of MATH, equivalently the space of integral curves of the geodesic distribution in MATH. The projections MATH, which send an isotropic direction to the null-geodesic tangent to it, is a submersion and the (closed) fibers are precisely the lifts of the null-geodesics of MATH. This construction is possible because MATH are geodesically convex (for a particular local holomorphic metric in MATH), see CITE for details. We first consider a tubular neighborhood MATH of MATH in MATH which projects, by MATH inside MATH, and such that MATH. This second condition ensures that all directions in MATH are transverse to the fibers of MATH. Consider then the following neighborhoods of MATH: MATH is an open application, so we define MATH to be MATH, where MATH is an open neighborhood of MATH contained in MATH. It is important to note that MATH have the following property: for any point MATH, the null-geodesic MATH tangent to MATH, contained in MATH, lies into MATH and is always transverse to the fibers of MATH. Hence, its restriction to the points of MATH projects diffeomorphically onto MATH. Moreover, all the points of MATH that lie over MATH are in MATH. Obviously, the crucial property of these open sets is that every null-geodesic starting there is totally contained in MATH, at least the part that ``lies over" (in the sense of the projection MATH) MATH. After constructing these sets MATH, we define MATH small enough for the tubular neighborhood MATH, restricted to MATH, to be contained into MATH, for all MATH. For each MATH, this means that MATH has to be less than the minimum of the following continuous functions defined on the compact set MATH: MATH . This neighborhood MATH of MATH has the following property: For each MATH, and for each MATH such that MATH, we have MATH, and thus the whole null-geodesic MATH, contained in MATH, is lifted to MATH. The disadvantage of MATH is that it does not necessarily contain MATH. But we know that the latter is contained in MATH, which contains the union of all MATH. We recall now that the idea of proof is to extend a null-geodesic MATH close to MATH over the disks MATH. Every extension over a disk brings MATH from MATH to the larger set MATH. As we have a finite, well-determined, number of disks MATH, all we need now to apply our extending idea is a sequence of open sets MATH such that MATH . To do that, we construct MATH as we have done for MATH, and then MATH by repeating the same procedure. We stop after MATH (the number of disks covering MATH, see REF ) steps and claim: MATH, the null-geodesic MATH extends to a compact curve which projects (via MATH) diffeomorphically onto MATH. Fix MATH. We can define MATH inside MATH, where MATH is an open set containing MATH. In particular, MATH is well-defined over MATH, where MATH, see REF . Of course, this is because MATH is transverse to the fibers of MATH. We intend to extend it over disks centered in MATH. Consider the domains MATH which are ``quadrilaterals" contained in MATH and containing MATH, as in the following picture (the ``vertical" parts of the border of MATH are segments of circles centered in MATH): We change, if necessary, the order of the indices MATH of MATH and MATH such that it coincides with the ordering of increasing distances MATH. We define than the open sets MATH . Remark. The closed disk MATH is included in MATH as soon as MATH, where MATH is the point of MATH opposed to MATH. Then, because of the specific geometry of the domains MATH (see REF ), we easily conclude that the domains MATH are contractible (along the geodesics of the sphere passing through MATH) and so are the intersections MATH, too. We prove, then, by induction, that MATH can be extended over MATH, and that all the corresponding points of MATH are contained in MATH. This is obvious for small values of MATH. When we add MATH to MATH, we consider a point MATH in the connected (see above) intersection MATH. It is contained in the MATH that contains MATH and, as MATH is contained in MATH, the connected piece MATH is contained in MATH, and thus the whole extension MATH in MATH is contained in MATH, hence in MATH. The connectedness of the considered piece implies that MATH coincides with MATH. Thus we have obtained an extension of MATH to MATH, such that the corresponding lift lies in MATH, as claimed. For large values of MATH, the MATH are all identical to MATH, which contains MATH. We have thus proven that there is an extension of MATH over this disk, such that the corresponding points of MATH are in MATH. Consider then the disk MATH; it is contained in some MATH: The intersection MATH of this disk with MATH is a connected open subset of MATH, and we know that MATH is contained in MATH. This implies that we can extend in a unique way MATH to MATH, in particular to MATH, and the corresponding points in MATH are in MATH. We have proven that MATH extends over MATH, that is, there is a maximal extension (obviously unique) of the null-geodesic tangent to MATH, such that it projects (via MATH) diffeomorphically onto MATH. The projection is MATH, but the extended null-geodesic is clearly an analytic submanifold of M. The proof of REF is now complete. Remark. Generic deformations of MATH are embedded projective lines. Indeed, let MATH be the nodes (self-intersection points) of MATH, and consider the manifold MATH, obtained by blowing-up the points MATH. Then, generically, any null-geodesic MATH close to MATH avoids these points, hence is diffeomorphic to its lift to MATH, which is a deformation of the lift of MATH, thus it is an immersed projective line. But the lift of MATH is embedded, and so must be its deformations, hence MATH is embedded. The next step is to prove that all deformations of MATH as a compact curve are null-geodesics , by a dimension-counting argument; we need to compute the normal bundle of MATH. We ask now if the family of projective lines in M defined as the deformations of MATH is locally complete in the sense of CITE. For this, we need to prove that the dimension of the space of global sections in MATH is equal to REF, that is, to the dimension of MATH. The extensions of the null-geodesics close to MATH yield local diffeomorphisms between neighborhoods of MATH in MATH, respectively, MATH. In fact, we have a projection MATH onto the space of integral curves of the geodesic distribution in MATH. MATH is the space of complex null-geodesic close to MATH in M. But essential for us is that MATH is a submersion, fact that has important consequences for the normal bundle of MATH in M. The normal bundle of MATH in M is isomorphic to MATH. It is well-known that all holomorphic bundles over MATH are direct sums of line bundles, all of which are isomorphic to MATH. We have the subbundle MATH of MATH, represented by vectors orthogonal to MATH. We have the following exact sequence: MATH . The right hand term of this sequence is a line bundle, and it admits global non-zero sections (extensions of NAME fields MATH such that MATH is a non-zero constant, see REF ). MATH is then isomorphic to MATH, with MATH. We denote by MATH the subbundle of the normal bundle represented by vectors in MATH. It admits global sections, namely the extensions of NAME fields contained in MATH, see REF . It is a line bundle, thus isomorphic to MATH, as it contains global sections with prescribed REF-jet in a point. Remark. In general, MATH is the line bundle over MATH admitting global sections for any prescribed MATH-jet in a point MATH. This section is unique, and it gives a unique value of the MATH-jet in MATH. We have the following exact sequence: MATH . It is easy to check that the right hand term admits local sections represented by NAME fields for any prescribed REF-jet in a point MATH. Hence, MATH. All we can obtain now is that MATH, with MATH. Actually MATH, otherwise all sections of MATH that vanish somewhere would be contained in a line subbundle, which would contradict REF . We have then MATH. We want to prove that there is equality in all these three inequalities. We know, from REF , that there is a tubular neighborhood MATH of MATH in MATH such that the null-geodesic distribution yields a foliation with compact leaves, and such that the projection onto the space MATH of these compact curves is a submersion. (Of course, this space is nothing but the space of complex null-geodesics close to MATH.) It is obvious then that the normal bundle of MATH in MATH is trivial, as MATH is e fiber of a submersion. We have now the following exact sequence of bundles, related to the projection MATH: MATH where MATH is the normal subbundle of MATH represented by vectors tangent to the fibers of MATH. In a point MATH, the fiber of MATH is equal to MATH, so the tangent space to it is isomorphic to MATH, for the projective variety MATH. Thus MATH . The central bundle in the exact sequence REF is trivial. The equation above then implies that the NAME number of MATH is subject to the following constraint: MATH thus, as MATH and MATH, we have MATH and MATH. As observed above, generic, compact, simply-connected null-geodesics are embedded. From now on, we suppose MATH is one of them. Then MATH, so we can apply the theory of NAME to deform MATH, CITE, so the dimension of the space of global sections of MATH is MATH, the same as the space of complex null-geodesic close to MATH, hence The deformations of MATH as a compact curve are null-geodesics in MATH. This means that any global section in MATH can be represented, locally, by the NAME fields that yield the same element in MATH, the space of jets of order REF in MATH. Recall now the exact sequence REF ; we conclude that MATH . We have the canonical isomorphism MATH, coming from the restriction of the projection MATH (we denote by MATH, respectively, MATH, the subbundle of the normal bundle of MATH, such that its fiber at MATH is MATH, respectively, MATH). We have thus MATH, and all REF-jets of MATH yield global sections, thus local NAME fields. But the existence of a NAME field in MATH implies, by the NAME equation MATH, that MATH, the self dual NAME tensor vanishes on MATH, the MATH-plane generated by MATH. We recall now that, for a fixed point MATH, MATH is a polynomial of order REF, and it is zero on MATH for the compact null-geodesic MATH. The same is true for the compact deformations of MATH, which implies that MATH. This holds for all the points of MATH, and also for all points covered by the deformations of MATH. As these deformations cover at least an open set around MATH, we conclude that MATH vanishes on a non-empty open set, thus, being holomorphic, MATH on the whole (connected) manifold M. This completes the proof of REF . |
math/0002029 | We need first to define the anti-self-dual NAME tensor as an irreducible component of the NAME tensor of M. Convention We note MATH the NAME tensor of MATH; we will not use this letter for the isotropic cone in this Section, nor in the following one. The NAME tensor is not conformally invariant; its definition depends on a (local) metric MATH in the conformal structure, which is supposed to be fixed CITE: MATH where MATH is the normalized NAME tensor of M, MATH being the trace-free NAME tensor, respectively, the scalar curvature of the metric MATH, and MATH. In our case, MATH, but the formula applies in all dimensions greater than REF. Remark. The NAME tensor MATH of M is a REF-form with values in MATH, thus it has two components MATH, and MATH. MATH satisfies a first NAME identity, as MATH is a symmetric tensor, and also a contracted (second) NAME identity, coming from the second NAME identity in Riemannian geometry, CITE : MATH . That means that MATH, and is orthogonal on MATH and on MATH, which is identified with the image in MATH by the metric adjoint of the contraction REF . Now, the NAME operator MATH induces a symmetric endomorphism of MATH, which maps the two above spaces isomorphically into each other. This implies that MATH and MATH satisfy REF (note that these two relations are equivalent in their case). The NAME tensor is related to the NAME tensor of M by the formula CITE: MATH where MATH is induced by the codifferential on the second factor, and by the NAME connection MATH. Then, MATH has to be the component of MATH in MATH, and we know that the restriction of MATH to MATH is identically zero. This means that MATH . Hence, as M is self-dual, MATH vanishes identically. We can prove now that the NAME tensor of a MATH-surface MATH is identically zero: first we prove MATH . We recall that the suspension MATH, viewed as an endomorphism of MATH, is defined by CITE: MATH where MATH is identified with a symmetric endomorphism of MATH. We have then the following decomposition of the Riemannian curvature CITE: MATH . Of course, if M is self-dual, MATH and MATH if MATH (in fact, the elements in MATH, for any MATH-plane MATH, correspond to the isotropic vectors in MATH), because MATH. Then, if we choose the basis MATH in MATH, we get MATH which proves REF . The NAME tensor of the projective structure of MATH has the following expression (see REF ): MATH and, as MATH vanishes on the anti-self-dual REF-form MATH, we conclude MATH . |
math/0002029 | We cannot use directly REF , as the ``ambient" self-dual manifold M can only be defined for a civilized (for example, geodesically connected) CASE: Hence, we follow the steps in REF and prove NAME close to a compact, simply-connected one are also compact and simply-connected, and they are embedded. Still using the same arguments as in REF, we get an embedded null-geodesic MATH diffeomorphic to MATH, and we have: The deformations of MATH as a compact curve coincide with the null-geodesics close to MATH. We cover MATH with geodesically convex open sets MATH, such that: MATH where MATH is still geodesically convex (with respect to a particular NAME connection). This is possible by choosing MATH, small enough. Then we choose a relatively compact tubular neighborhood MATH of MATH, such that its closure is covered by the MATH's. We consider then the twistor spaces MATH, the spaces of null-geodesics of MATH. The compact, simply-connected, null-geodesics close to MATH identify (diffeomorphically) the neighborhoods of MATH with the space MATH of the deformations of MATH as a compact curve. We can see then MATH as an open set common to all the MATH's: Following NAME, we define the self-dual manifolds MATH as the space of projective lines in MATH, with normal bundle MATH. Then MATH is an umbilic hypersurface in MATH. The local twistor spaces MATH admit contact structures, which coincide on MATH, and contain projective lines MATH corresponding to points MATH. If we denote by MATH the twistor space of MATH, then MATH and MATH are identified to open sets in MATH, in particular the lines MATH and MATH are identified, thus their intersections with the common set MATH coincide. We denote by MATH this (non-compact) curve in MATH, and by MATH the canonical contact structure of MATH (restricted from the ones of MATH). Remark. We already have obtained that the MATH-cone corresponding to MATH is a part of a smooth surface: the union of the lines MATH, MATH, thus, from REF , the NAME tensor MATH of the self-dual manifold MATH vanishes on the MATH-planes generated by MATH. But this is nothing new: we know, from REF , that MATH vanishes on MATH. We intend to apply REF to prove that MATH vanishes on points close to MATH, but in MATH. We do that by showing that the integral MATH-cones corresponding to planes MATH are parts of smooth surfaces, then we conclude using REF . First we choose hermitian metrics MATH on MATH, such that they coincide (with MATH) on MATH. We have a diffeomorphism between MATH and MATH, so we choose relatively compact open sets in MATH, covering MATH, with the following properties: As the metrics MATH induce metrics on MATH, we first choose a small enough distance MATH such that CASE: MATH, there is a sub-covering MATH of MATH such that the ``tubular neighborhoods" MATH are compact (MATH is the ``orthogonal projection" - for the hermitian metric - from MATH to MATH; it is well defined because of the condition below); CASE: MATH is less than the bijectivity radius of the (hermitian) exponentials for the points of MATH in MATH, and for the points of MATH in MATH (if MATH . ). We have then For any MATH, MATH such that the curves MATH, MATH are tangent to the same direction in MATH, the respective curves MATH coincide. We first note that the projection MATH from MATH is equivalent to the MATH- orthogonal projection of the direction of MATH to a direction in MATH, so MATH; thus MATH belongs to both MATH and MATH, and we use again the twistor space MATH to conclude that MATH and MATH are ``restrictions" to MATH of the same projective line (as they both have the same tangent space at MATH) MATH, for a point MATH. Now we have a tubular neighborhood MATH of MATH, of radius MATH, such that, for any REF-plane MATH, the conditions in REF are satisfied (considering any of the local twistor spaces MATH). We conclude that the NAME tensor MATH of MATH vanishes along all null-geodesics of MATH, close (in MATH) to MATH and included in the MATH-surface MATH, determined by MATH. This means that MATH vanishes everywhere on MATH. By deforming MATH, we obtain that MATH vanishes on a neighborhood of MATH in MATH, hence MATH, as well as MATH, are conformally flat (by REF ). It follows from REF that MATH is conformally flat. |
math/0002030 | Let MATH be a point of the classifying space MATH and define MATH . Then, as discussed in CITE, the subalgebra MATH is a vector space complement to MATH in MATH, and hence the map MATH restricts to a biholomorphism from some neighborhood of zero in MATH onto some neighborhood of MATH in MATH. Consequently, we may introduce a hermitian metric on MATH by first identifying MATH with MATH via the differential MATH and then applying the rule: MATH . To see that MATH act by isometries with respect to MATH, one simply computes using MATH. |
math/0002030 | The construction of REF defines a map MATH by virtue of MATH and the standard commutator relations of MATH. To obtain an map MATH such that MATH one simply considers how the NAME blocks of MATH interact with the grading MATH of MATH. |
math/0002030 | By virtue of the definition of the relative weight filtration and the mutual compatibility of MATH with MATH, it follows that the induced map MATH grades the monodromy weight filtration MATH. Moreover, by REF . Thus, application of REF defines a collection of representations MATH which we may then lift to the desired representation MATH via the grading MATH. To see that MATH and MATH, observe that if MATH is the decomposition of an endomorphism MATH according to the eigenvalues of MATH then the lift of MATH with respect to the induced isomorphism MATH is exactly MATH. Thus MATH. Likewise, MATH since the mutual compatibility of the gradings MATH and MATH implies that MATH and MATH may be simultaneously diagonalized. |
math/0002030 | One simply expands out the flatness condition MATH and taking note of the additional requirements imposed by REF. |
math/0002030 | If MATH admits a decomposition into a sum of NAME bundles MATH then the desired isomorphism MATH may be obtained by setting MATH on MATH. Conversely, given a NAME bundle MATH which is a fixed point of the MATH action MATH one may obtained the desired decomposition of MATH into a system of NAME bundles as follows: Let MATH be an isomorphism from MATH to MATH for some element MATH which is not a root of unity. Then, because MATH is holomorphic and MATH is compact, the characteristic polynomial of MATH is constant. Upon decomposing MATH into a sum of generalized eigenspaces, one then obtains the desired system of NAME bundles. |
math/0002030 | Since the weight filtration MATH of MATH is by definition flat, it will suffice to verify condition MATH of REF by simply computing the action of the NAME - NAME connection MATH of MATH upon a smooth section MATH of MATH at an arbitrary point MATH. Accordingly, we recall from REF that MATH may be represented near MATH by a holomorphic, horizontal map MATH from the polydisk MATH into a suitable classifying space MATH upon selection a system of holomorphic local coordinates MATH on MATH which vanish at MATH. In particular, if MATH and MATH denotes the vector space complement to MATH in MATH constructed in REF [compare Theorem MATH], it then follows that there exists a neighborhood of zero in MATH over which we may write MATH relative to a unique MATH-valued holomorphic function MATH vanishing at the origin. To continue our calculations, we recall CITE that for sufficiently small values of MATH, one may write MATH relative to a MATH function MATH which has a first order NAME series expansion given by the the formula MATH where MATH denotes projection with respect to the the decomposition MATH and MATH denotes the linearization of MATH about MATH. To determine how MATH acts upon a smooth section MATH of MATH observe that by virtue of the preceding remarks, we may write MATH relative to a unique function MATH taking values in the fixed vector space MATH. Consequently, MATH . In particular, since MATH on account of the horizontality of MATH, MATH . |
math/0002030 | By definition, a complex variation of graded-polarized mixed NAME structure consists of a complex variation of mixed NAME structure together with a collection of parallel hermitian forms on MATH which polarize the induced complex variations. Accordingly, the associated decomposition MATH given by REF has the additional property that MATH . On the other hand, by REF, if MATH denotes the decomposition of MATH defined by REF, then MATH and hence the decomposition MATH obtained by setting MATH and MATH coincides with usual decomposition of MATH defined by REF via the system of NAME bundles MATH. In particular, MATH is an operator of NAME - type. |
math/0002030 | According to REF, the mixed NAME metric of a complex variation of graded-polarized mixed NAME structure is filtered harmonic. Conversely, given a filtered harmonic metric for which the above two conditions hold one defines MATH . To prove the bigrading MATH does indeed define a complex variation of graded-polarized mixed NAME structure for which the associated mixed NAME metric equals the given hermitian metric MATH, we note that for each index MATH, the decomposition MATH defines a system of NAME bundles with respect to MATH. Indeed, this is automatic given conditions MATH and MATH since MATH is just the component of MATH which preserves MATH. On the other hand, by definition, MATH coincides with the NAME bundle structure obtained by applying Construction MATH to the pair MATH. Consequently, by REF, there exists a unique complex variation of polarized NAME structure on MATH for which MATH is the NAME metric, MATH is the flat connection and equation MATH represents the decomposition of MATH into a system of NAME bundles. Therefore, the decomposition MATH given by REF has the additional property that MATH . Returning to REF, it then follows that if MATH denotes the decomposition of MATH defined by REF then MATH since MATH and MATH on account of the filtered harmonicity of MATH. Consequently, MATH . |