paper stringlengths 9 16 | proof stringlengths 0 131k |
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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 MA... |
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... |
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 determin... |
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... |
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,... |
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 th... |
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 gl... |
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 fu... |
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... |
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 i... |
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 th... |
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 contribu... |
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. ... |
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 ... |
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 t... |
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... |
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 MAT... |
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 des... |
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 ... |
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 t... |
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, whic... |
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... |
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 s... |
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 functi... |
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. ... |
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 t... |
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 orth... |
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. Ta... |
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 wher... |
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 NA... |
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 mul... |
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 ... |
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 MAT... |
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 ... |
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 ... |
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 de... |
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 ... |
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. Be... |
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 observin... |
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 conc... |
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 ... |
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 MAT... |
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 t... |
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 det... |
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 singula... |
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 M... |
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 ... |
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 MAT... |
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 t... |
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... |
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 ... |
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 s... |
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 dista... |
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 st... |
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 NA... |
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 M... |
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 short... |
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 as... |
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/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... |
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... |
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 li... |
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 ... |
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 ... |
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 ch... |
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 a... |
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 ... |
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... |
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 intro... |
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 r... |
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 M... |
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 ... |
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... |
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 mix... |
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