paper stringlengths 9 16 | proof stringlengths 0 131k |
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cs/9912020 | The proof uses induction. First, the case MATH is just REF . If the equality holds for MATH then the first MATH terms are the same as for MATH and only the last term needs further expansion. In this term each summand is a function of MATH and from REF one obtains: MATH . Replacing the last term in the equation for the ... |
cs/9912020 | The statement that MATH is a projection, that is, MATH, is equivalent to showing that MATH is zero for MATH. This follows directly from the fact that any function MATH for which the function values do not depend on the variable MATH one has MATH. As any MATH consists of a sum of functions which depend only on MATH vari... |
cs/9912020 | It is shown the same way as in an earlier theorem that all the terms of the sum defining MATH are orthogonal and so MATH . For simplicity set MATH and application of REF and similar reasoning as for the case MATH gives MATH where MATH . Now one can see that MATH and from this the claimed bound follows. |
cs/9912020 | Because of the equivalence of the measures the error in the original measure is bounded by MATH and REF for the measure MATH then gives MATH. |
cs/9912020 | The lower bound holds by definition. For the upper bound we first use the property that the measures are in the same class to get: MATH . Then we note that MATH is a best approximation with respect to the norm defined by the product distribution by REF . Thus one has for any MATH: MATH and, as the measures are in the s... |
math-ph/9912002 | Choose MATH such that MATH and MATH . Note that the latter condition can be achieved for MATH small enough, since MATH. For this choice of MATH, let MATH be the minimal length scale from REF . We can now use REF to find a sequence MATH and a constant MATH such that for every MATH, CASE: MATH, CASE: MATH, CASE: MATH is ... |
math-ph/9912002 | It is well known that (INDY), (WEYL), (GRI) and (EDI) are satisfied if we take for MATH the operator MATH restricted to MATH with periodic boundary conditions. Due to CITE we have a NAME estimate of the form MATH where MATH denotes the sidelength of the cube MATH. In particular, MATH is satisfied for a neighborhood MAT... |
math-ph/9912002 | We have already checked everything except for the initial length scale estimate MATH, and in particular how large MATH can be taken. By an elementary argument, we can take MATH subject to the condition MATH (see CITE), which gives the claimed result. |
math-ph/9912005 | MATH . |
math-ph/9912005 | NAME has shown that MATH is a MATH CITE. If MATH is not empty, then it contains an entire orbit which is dense by minimality. |
math-ph/9912005 | The assertion follows from MATH which is readily verified. |
math-ph/9912007 | The MATH spectral residue vanishes if and only if MATH . If this is the case, then evidently there exists a monic series solution for every possible choice of MATH. |
math-ph/9912007 | Let MATH, MATH and MATH be as above, and let MATH and MATH be the corresponding monic series solutions. From MATH it follows that MATH . From REF it is evident that the residues of MATH at MATH are all zero. The desired conclusion follows immediately. |
math-ph/9912007 | In the non-commutative setting it suffices to take MATH where MATH, MATH are the coefficient series of the operator MATH. In the commutative setting one can use simpler formulas. Indeed, it suffices to introduce an auxiliary function MATH and then set MATH . |
math-ph/9912007 | Taking MATH and MATH, the desired conclusion follows immediately from REF above. |
math-ph/9912007 | We saw above that the eigenfunctions of the NAME Hamiltonian are the product of a power of MATH and of a hypergeometric series. Hence by the gauge-invariance of spectral residues as asserted in REF above, the spectral residues of the NAME Hamiltonian can just as well be calculated by taking residues of the coefficients... |
math-ph/9912007 | Let MATH be the unique series solution to MATH . By REF , if MATH, MATH, then MATH is MATH times a MATH degree polynomial of MATH whose coefficients are polynomials of MATH over MATH. Hence for all MATH the residue of MATH at MATH must have MATH as a factor. By the gauge invariance of spectral residues, the same must b... |
math-ph/9912007 | Upon setting MATH, MATH, the above identity follows from the odd, MATH, case of REF . |
math-ph/9912007 | Upon setting MATH, MATH, the above identity follows from the odd, MATH, case of REF . |
math-ph/9912007 | We proceed in the same way as in the proof of REF . By the gauge-invariance of spectral residues, MATH is equal to the MATH spectral residue of the confluent hypergeometric operator. Taking the residue of MATH at MATH yields MATH . Next, using REF we see that in the first case, where MATH, we have MATH . In the second ... |
math-ph/9912007 | Note that MATH, or equivalently that MATH. Furthermore, REF imply that MATH . The desired conclusion now follows immediately. |
math-ph/9912025 | CASE: Since the bound REF ensures MATH for MATH-almost all MATH, it follows that for these MATH's the operator MATH is an operator perturbation of MATH with relative operator bound zero. NAME of MATH on MATH is then guaranteed by the NAME theorem. Since operator boundedness with bound zero implies form boundedness with... |
math-ph/9912025 | Concerning REF we refer to REF. The existence and non-randomness of the integrated density of states is shown in REF for the case MATH by using functional-integral techniques for the NAME transforms of the density-of-states measures. Employing the appropriate NAME formula and so-called magnetic translations, these meth... |
math-ph/9912025 | A covariance function MATH which is continuous at the origin, where it satisfies MATH, is bounded and uniformly continuous on MATH by the NAME theorem. Consequently, REF implies the existence of a separable version of MATH which is jointly measurable with respect to the sigma-algebra MATH and the NAME sigma-algebra on ... |
math-ph/9912025 | Using MATH, we rewrite the left-hand side of REF as MATH . The first term on the right-hand side of REF follows from a double partial integration and the rest from the reality of MATH and the gauge condition MATH. |
math-ph/9912025 | We infer from REF of MATH and REF that MATH . When applied to MATH, where MATH with MATH, REF yields the claim of the lemma, provided the second term on the right-hand side of REF is shown to be appropriately bounded in terms of MATH. To do so we estimate MATH . Since MATH, we can apply REF to the scalar product. Toget... |
math-ph/9912025 | We will estimate from above the probability for the event that MATH is not MATH-frame-regular. Introducing the event MATH one gets from elementary set-theoretic algebra MATH . By assumption one has MATH, MATH and MATH for MATH. Thus, the MATH-independence of MATH implies MATH . With the help of this inequality and REF ... |
math-ph/9912025 | REF is proven by induction on MATH. Pick MATH such that MATH is large enough as required for the applicability of REF . Let MATH stand for either MATH or MATH. Due to REF we can apply REF with MATH and MATH. For given MATH and given MATH this yields MATH . Thanks to REF we thus conclude from REF that for all MATH and a... |
math-ph/9912025 | We show that the theorem is a special case of REF. The inequalities MATH and MATH imply the existence of MATH with MATH. Then estimate REF still holds when MATH is replaced by MATH. Now we choose the quantities MATH of CITE according to MATH, MATH, MATH. The sequences MATH, respectively, MATH, are monotone increasing, ... |
math-ph/9912025 | Let MATH be a complete orthonormal sequence of vectors in MATH and suppose that MATH for all MATH with some constants MATH and MATH as in REF . It suffices to show that the event MATH occurs with probability one. Here, MATH is the scalar product REF on MATH, and MATH denotes the absolutely continuous component arising ... |
math-ph/9912025 | Let us define the function MATH by setting MATH, if a one-parameter decomposition of MATH exists in the sense of REF , the operator MATH is self-adjoint and its spectrum is pure point in MATH. Otherwise we set MATH. The joint measurability of the random potential MATH implies that the random potential MATH is jointly m... |
math-ph/9912025 | We check that the assumptions of REF are fulfilled for MATH. Obviously, REF provides REF. Since REF of a Gaussian random potential implies REF , the requirement REF follows from the NAME estimate in the weakened form REF, viz. MATH for all MATH and all MATH. Thanks to REF there exists a real constant MATH such that MAT... |
math-ph/9912025 | We check the assumptions of REF with MATH. REF is obviously fulfilled for Gaussian random potentials for any MATH. Concerning REF we introduce a one-parameter decomposition of MATH by defining the function MATH for MATH, MATH being a normalization constant, the centred Gaussian random variable MATH with variance MATH a... |
math-ph/9912025 | We deduce the lemma from REF. When adapted to a homogeneous random field and a cube MATH with edges of length MATH, this theorem implies that, if MATH then the realizations of MATH are MATH-almost surely continuous functions on MATH, and the inequality MATH holds for all MATH obeying MATH . First, we verify that REF is... |
math-ph/9912025 | Let MATH and set MATH with some MATH fixed, but arbitrary. Let MATH such that the inequality MATH is satisfied. Then, REF and the NAME REF imply the estimate for all MATH . Here, MATH and we have defined a function MATH on MATH by MATH with suitable constants MATH depending only on MATH. Observe that there exists MATH ... |
math-ph/9912025 | Given MATH and MATH, set MATH and MATH as in REF. As in the proof of REF we infer the existence of a finite length MATH, which is independent of MATH, such that REF holds for all MATH. The proof is completed by requiring MATH, that is MATH with MATH . |
math-ph/9912025 | Let MATH and let MATH, MATH, MATH and MATH as in REF . By REF of the strong-mixing coefficient we have for all integers MATH and all MATH, MATH, with MATH and MATH for MATH that MATH . Iterating this inequality and using MATH, where MATH as required in REF , we obtain MATH . The MATH-sum is bounded from above by MATH f... |
math-ph/9912025 | We denote the NAME (-Plancherel) transform of MATH by MATH. The matrix-valued function MATH with matrix elements MATH is continuous in MATH due to MATH, which follows from MATH and thus MATH. Each MATH is a real-valued and even function in MATH, because MATH. Moreover, since each MATH has a non-negative NAME transform ... |
math-ph/9912025 | We show that REF follows from an application of REF . To do so we have to check the assumptions of this theorem. Let MATH and MATH. Choose the exponents MATH and MATH such that MATH in accordance with REF. The precise value of MATH will be fixed later on. In case that MATH has REF - let us call this simply case (PHM) -... |
math-ph/9912025 | Let MATH. From the NAME representation of the NAME semigroup MATH, MATH, see for example, CITE, and the explicit form of the heat kernel one obtains the inequality MATH . The fact that MATH and MATH are supported in spatially separated regions allows one to bound the right-hand side of REF by MATH . Estimating the scal... |
math/9912001 | Fix MATH, MATH, MATH; we may normalize MATH. Let MATH denote the function MATH, and define the quantity MATH by MATH our task is then to show that MATH . Let MATH be the nearest integer to MATH, where MATH is a small constant to be chosen later. The first step in the argument is to construct group elements MATH such th... |
math/9912003 | Using REF we calculate the discriminant of MATH then, by NAME inequality CITE, MATH is NAME unstable. |
math/9912003 | Assume that not all weights are equal. By REF , MATH is NAME unstable. We will prove that MATH is not parabolic stable by showing that at least one of the subsheaves of the NAME filtration contradicts the parabolic stability of MATH. Since the parabolic degree of MATH is zero, we have to prove that the parabolic degree... |
math/9912003 | Let MATH be the corresponding stable parabolic bundle given by REF . By REF the parabolic structure is trivial (that is, all weights are equal), and then MATH is a multiple of identity. |
math/9912003 | Let MATH be a representation, and MATH the associated parabolic bundle. If MATH is irreducible then, by REF , MATH is a multiple of identity. Then the representation MATH induces a representation MATH. If MATH is irreducible then MATH is also irreducible, and we get a contradiction. |
math/9912003 | Given an irreducible local system on MATH, consider its NAME extension MATH CITE. By construction, the real part of the eigenvalues of the residue of MATH are non-negative and less than REF. Define a parabolic structure on MATH, setting the parabolic weights equal to the real parts of the eigenvalues of the residue of ... |
math/9912003 | Assume that there is an irreducible representation MATH. Let MATH be the stable parabolic NAME bundle associated by REF . We will use the following fact: MATH . The proof was shown to us by NAME. Consider the sequence MATH where the second map is the residue, and note that the image of the constant function REF under t... |
math/9912003 | Since MATH (and also the associated NAME connection) is singular on MATH, we cannot directly apply the NAME formula. We will modify the metric to make it smooth, and this will produce the second summand. First note that MATH . This can be proved by first showing that both sides are invariant under change of holomorphic... |
math/9912004 | Since MATH is an isolated singular point, there is a simply connected neighborhood MATH which closure does not contain other critical points of the function and which boundary is a smooth closed curve. Let us assume MATH is a restriction of function MATH to the set MATH . Then the levels of both functions coincide in t... |
math/9912004 | Let MATH be a neighborhood with the same properties as in the lemma and MATH be the number of arcs, which an extremity is the point MATH . The critical level devides the neighborhood MATH to domains. Let MATH be one of them and such that MATH for MATH . Then in the points of the intersection MATH with MATH the vector f... |
math/9912004 | Necessity. If a conjugated homeomorphism of the surface MATH is given, it maps critical levels to critical levels. It sets an isomorphism of the diagrams. Sufficiency. Suppose that the diagrams of two functions are isomorphic. Then there is a homeomorphism MATH of a surface MATH that maps critical levels to critical le... |
math/9912004 | Necessity. If the functions are topologically conjugated, there is an isomorphism of their diagrams. The restriction of this isomorphism to the graphs sets isomorphism of the distinguishing graphs. Sufficiency. Let distinguishing graphs are isomorphic. We construct an isomorphism of the diagrams. By the construction, w... |
math/9912004 | The necessity is obvious. Let us prove sufficiency. The conditions of the theorem guarantee that after gluing of cylinders to the graphs we obtain a closed surface. Let us construct the function f. For this purpose we set MATH and for the cylinder MATH with boundary on MATH and MATH: MATH . Thus in vertexes of the grap... |
math/9912005 | Let MATH and let MATH be the multiplication. Let MATH be the open subvariety of MATH of points such that the linear map MATH has maximal rank. Then MATH carries a vector bundle MATH whose fibre above the point MATH is simply the cokernel of MATH. We consider the vector bundle MATH which is an irreducible variety and ha... |
math/9912005 | The injectivity of this linear map implies that the map considered in the proof of the previous lemma has maximal rank. |
math/9912005 | Let MATH be a MATH-standard family over the algebraic variety MATH. Then for each vertex MATH, MATH is a vector bundle of weight MATH for MATH of rank MATH for a suitable dimension vector MATH. Then by the local isomorphism theorem, REF, there exists an open MATH equivariant subvariety MATH of MATH such that the restri... |
math/9912005 | This argument essentially occurs as a special case of the proof of REF to which the reader should refer for greater detail. We deal only with the first case since the second case has the same proof. Let MATH be the associated left MATH-free family on MATH (after replacing MATH by an open dense subvariety). Let MATH be ... |
math/9912005 | If MATH and MATH are representations for which we have short exact sequences MATH and a short exact sequence MATH then it is clear that the induced extension of the subrepresentation MATH over MATH splits; thus MATH has a subrepresentation MATH isomorphic to MATH such that the factor is isomorphic to MATH and MATH must... |
math/9912005 | If MATH is reducible to matrix normal form for the MATH-th NAME quiver we take the family of representations of dimension vector MATH for the MATH-th NAME quiver and apply the functor MATH considered in REF . This gives a family of representations for the quiver with relations MATH that is birationally representation e... |
math/9912005 | We shall proceed by induction on MATH. The associated left MATH-free family, MATH, has a NAME reduction of type MATH to the dimension vector MATH for the MATH-th NAME quiver MATH for some integer MATH. Now suppose that the dimension vector MATH is reducible to matrix normal form. Then by REF , MATH, is reducible to mat... |
math/9912005 | Let MATH be the canonical family on the left general component MATH of representations of dimension vector MATH. Let MATH, MATH and MATH. Then MATH has a two-step NAME reduction of type MATH and so by REF , MATH is reducible to matrix normal form. |
math/9912005 | Once stated, this result is also clear. In the derived category MATH, the object MATH is equivalent to the complex MATH which after our identification is the complex of sheaves MATH and this itself is a sheaf if and only if it has homology only at the penultimate term as required. |
math/9912005 | Consider the complex of sheaves on MATH . Let MATH be the support of the homology of this complex except at the penultimate term. Then MATH is closed and so is its image in MATH; therefore the complement of the image of MATH in MATH is open and it is the set of points MATH where MATH is a sheaf. In the second case, the... |
math/9912005 | After tensoring by a suitable line bundle we may assume for a general MATH in the moduli space that MATH is a representation but MATH is not; that is one of MATH and MATH is non-zero. Since MATH has natural cohomology, it follows that MATH; it also follows that MATH since by NAME duality this is dual to MATH and since ... |
math/9912005 | This follows from the known results on matrices up to simultaneous conjugacy. A good summary of the known results may be found in CITE. |
math/9912006 | We proceed by induction on MATH, the number of components of MATH. The theorem is true for knots in MATH since, by the cyclic surgery theorem in CITE, MATH and MATH surgery on a nontrivial knot cannot both produce a manifold with trivial fundamental group. Now, assuming the theorem is true for all links of MATH or fewe... |
math/9912006 | When MATH has one component, this is the well-known result of NAME and NAME. Suppose MATH has MATH components. Because MATH is homologically trivial, its MATH-th meridian has order MATH in MATH. Thus MATH for all MATH. The MATH-th component of MATH is unknotted because MATH is NAME, and so MATH is a link in MATH. The l... |
math/9912006 | Let MATH be a homeomorphism. (By a homeomorphism of link complements, we always mean an orientation preserving homeomorphism, as is required for link equivalence.) Since MATH and MATH are homologically trivial, the longitudes of the components of each link are null-homologous in the link complements. These are the only... |
math/9912006 | Let MATH be a homeomorphism. Since MATH is homologically trivial, the image of each longitude under MATH must be an essential simple closed curve on the boundary of MATH that is null-homologous in the link complement. Therefore MATH is also homologically trivial, and MATH maps longitudes to longitudes. As in the proof ... |
math/9912006 | We proceed by induction on the number of components in MATH. For knots, the corollary is true by CITE. Suppose that MATH has more than one component. It must be homologically trivial by the argument in the proof of REF . Performing MATH surgery on any one component results in a trivial link by induction, so MATH is NAM... |
math/9912006 | REF imply that MATH is an HTB link. If REF is true, then MATH so L is trivial by REF . |
math/9912014 | If MATH, then there is some term order MATH such that one of MATH and MATH is a minimal generator of MATH, and the other is standard for MATH. Since MATH is also MATH-graded, it suffices to prove the lemma for monomial MATH-graded ideals, where MATH for all MATH. Suppose there exist a MATH such that MATH is not a Grave... |
math/9912014 | It suffices to prove that MATH is weakly MATH-graded, where MATH is any term order, since MATH has the same NAME series as MATH. If MATH is a Graver binomial, then since there is some MATH with MATH, one of MATH and MATH lies in MATH. Let MATH and MATH be two monomials of degree MATH, and let MATH be a Graver binomial ... |
math/9912014 | Let MATH be a Graver binomial, with MATH. By REF it suffices to show that either MATH, or MATH. Since MATH, there is some (possibly identical) Graver binomial MATH with MATH a minimal generator of MATH, and MATH, and MATH. If MATH, then MATH, and so MATH. If MATH, then MATH, so MATH. |
math/9912014 | Let MATH be the initial ideal of MATH with respect to MATH. We first show the containment MATH. Let MATH be a minimal generator of MATH. If MATH, or MATH is a minimal generator of MATH other than MATH, then MATH. So we need only consider the case that MATH, where MATH and MATH, as this is the only other form minimal ge... |
math/9912014 | Since MATH is MATH-homogeneous, MATH is the initial ideal of MATH if and only if MATH is a MATH-graded ideal. But by REF MATH is an initial ideal of MATH, so is MATH-graded exactly when MATH is. |
math/9912014 | Suppose MATH is a flippable binomial for a monomial MATH-graded ideal MATH such that MATH and MATH. If MATH then MATH and if MATH then MATH both of which are contradictions. To see that the converse is false, consider MATH which has REF monomial MATH-graded ideals all of which are coherent. The binomial MATH, but neith... |
math/9912014 | The dense torus in MATH is MATH, and the action of this torus on MATH is by coordinate-wise multiplication. It thus follows that if MATH and MATH are in the same torus orbit, they have the same support. Suppose MATH have the same support. If this support is the entire set MATH, then define MATH. Then if MATH is a binom... |
math/9912014 | It is straightforward to see that MATH lies in every irreducible component of MATH in which MATH does (this follows from the fact that MATH is a flat MATH module). All points in the MATH-torus orbit of MATH have the same support, and thus lie on the same ambient torus orbit by REF . |
math/9912014 | Let MATH be the ideal corresponding to a vertex MATH of MATH. The orbit of MATH under the ambient torus corresponding to MATH is just the ideal MATH. By REF the MATH-torus orbit of MATH is contained in any ambient torus orbit, so MATH is MATH-torus fixed as well, and thus is a monomial ideal. For the other implication,... |
math/9912014 | Let MATH and MATH be the two initial ideals of MATH, and let MATH be the universal NAME basis of MATH. The set MATH contains a reduced NAME basis for MATH with respect to a term order for which MATH is the initial ideal, and so there exist binomials MATH with MATH for which MATH is a minimal generator of MATH, MATH. Su... |
math/9912014 | Suppose MATH and MATH are connected by an edge MATH. Let MATH be the ideal corresponding to a point MATH in the relative interior of MATH. By REF the MATH-torus closure of MATH is contained in MATH. Thus MATH has at most two initial ideals. If MATH had only one initial ideal, it would be a monomial ideal and thus corre... |
math/9912014 | It suffices to show that the reduced scheme MATH is connected if and only if the flip graph of MATH is connected. Since passing to an initial ideal is a flat deformation, each irreducible component contains a monomial MATH-graded ideal. It thus suffices to show that all monomial MATH-graded ideals lie in the same conne... |
math/9912014 | By computing the MATH-degree of both terms in each binomial of MATH, it can be seen that MATH is a subset of MATH. It can also be checked that for each binomial in MATH, the positive term is the leading term with respect to MATH. Hence MATH is contained in the initial ideal of MATH with respect to MATH and no generator... |
math/9912014 | We will show that the binomials in MATH are flippable for MATH while those in MATH are not. In order to show that a binomial MATH is flippable for MATH we need to show that every MATH-polynomial (monomial in our case) formed from the binomial MATH (with MATH as leading term) and a minimal generator MATH of MATH differe... |
math/9912014 | The same proof as in REF shows that the generators of MATH form a NAME basis with respect to MATH with initial ideal MATH, for every choice of scalars MATH from the underlying field MATH. REF proved this claim for the case MATH, for a MATH and MATH for all MATH. Since MATH is MATH-graded, the MATH-homogeneous ideal MAT... |
math/9912014 | The implication MATH is immediate in the first statement so we need only show that MATH implies MATH. Suppose MATH. Let MATH be a minimal generator of MATH. Then either MATH is a minimal generator of MATH, MATH, or MATH for some MATH. In each case MATH, so MATH. If the containment is proper, REF gives a proper containm... |
math/9912014 | MATH is the only minimal non-face of MATH, so to show that MATH is a subcomplex of MATH, we need to show that MATH is the only minimal generator of MATH with support in MATH. Suppose MATH is a minimal generator of MATH, with MATH. Then there is some MATH such that MATH. Write MATH, where MATH, and MATH. If MATH, then M... |
math/9912014 | If MATH then MATH by REF , and so MATH. Suppose MATH. Then REF implies that MATH is a circuit of MATH. By REF MATH is a subcomplex of MATH with each maximal simplex of MATH having the same link in MATH. It remains to show that MATH is the result of performing a bistellar flip on MATH. Let MATH be the result of performi... |
math/9912014 | For the matrix MATH, the monomial ideal MATH is MATH-graded. The flippable binomials of MATH are MATH and MATH. In this example, there are REF monomial MATH-graded ideals in total and the flip graph of MATH is connected. Consider the monomial ideal MATH and a degree MATH. All the monomials in MATH of MATH-degree MATH t... |
math/9912015 | One has to show MATH . NAME if the verification is a bit lengthy, we insert it here, for sake of completeness. We have a complex MATH, sum of two complexes MATH, with maps MATH. We denote by MATH. Then we define MATH, with differential MATH, and MATH on MATH given by MATH, that is concretely MATH. Now take local sectio... |
math/9912015 | The NAME sequence can be written MATH . Here MATH is interpreted as infinitesimal symmetries of the bundle MATH. A connection MATH relative to MATH gives rise to an action of MATH, that is, a lifting MATH of the inclusion MATH. The corresponding sequence for infinitesimal symmetries of the projective bundle MATH looks ... |
math/9912015 | From the exact sequence MATH the splitting principle on MATH and MATH, and the compatiblity of the products, one just has to see that MATH for MATH. To this end, we will define a quasi-isomorphism MATH . CASE: We write the term in degree MATH in MATH as MATH. CASE: The exact sequence REF defines a decreasing filtration... |
math/9912015 | Let MATH, and let MATH. Let MATH and MATH be the natural maps. Define MATH by restricting the cone of complexes of sheaves used to define MATH to MATH. One has a pullback map MATH and MATH. Finally, in degree MATH we have MATH . Compatibility of the connections on MATH and MATH leads to a commutative diagram of sheaves... |
math/9912015 | A connection defines an infinitesimal action of vector fields over MATH on MATH, that is, a splitting of the NAME sequence REF. Thus, a relative connection leads to a diagram MATH . The descent comes by defining MATH . |
math/9912015 | We start with the resolution MATH which we filter MATH . This filtration is compatible with the differential, and (using MATH), we find MATH . The complex MATH is obtained by tensoring the NAME resolution MATH with MATH and is thus exact. It follows that MATH is a resolution of MATH. Taking MATH, we obtain the desired ... |
math/9912015 | Briefly, one has diagrams of complexes of sheaves on MATH and on MATH . Using NAME resolutions and Cousin complexes, one defines a trace map MATH which yields a commutative diagram MATH . |
math/9912021 | It is enough to consider the case when MATH. Suppose first that MATH where each MATH is an integer. Then MATH takes either MATH or MATH. Conversely, suppose that all the MATH with MATH are real. Then MATH equals MATH and MATH for all MATH. This implies that , for each MATH, MATH. Since MATH is semisimple and the MATH f... |
math/9912021 | It was shown in REF that the elements MATH with MATH such that MATH is real for all MATH are those for which MATH has the form MATH with MATH as in Notation REF, and MATH are integers. Moreover as in REF we also have that MATH is real for any MATH exactly when MATH for some integers MATH . Therefore all the root spaces... |
math/9912021 | Let MATH denote the representation of MATH on MATH. Then we have MATH. If MATH is odd, MATH is isomorphic to MATH. Denote MATH with MATH as in REF . we set MATH when MATH is even and MATH when MATH is odd. Let MATH. We have that for each MATH, MATH for MATH. The MATH generate the group MATH with MATH elements where MAT... |
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