paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0006131 | Any sublattice MATH of MATH must be interval-connected. Assume by induction that every interval MATH is rank-connected. We will show that MATH is rank-connected. Let MATH be the subposet of MATH that contains all vertices except MATH. Let MATH be the subgraph of the NAME diagram induced by MATH and MATH. Since MATH is ... |
math/0006138 | The proof is similar to that in REF and uses REF together with uniform local finiteness of MATH. More precisely we use the fact that the valency of any vertex in MATH is uniformly bounded, say MATH. A fortiori the same is true (with the same constant MATH) for MATH for all MATH. We now estimate the MATH norm of MATH fo... |
math/0006138 | Observe that MATH is in the kernel of the DML MATH if and only if it is in the kernel of the twisted coboundary operator MATH. That is, MATH for all edges MATH. It follows that MATH for all edges MATH. Since MATH is connected, it follows that MATH is a constant function on MATH. Since MATH is infinite, it follows that ... |
math/0006138 | First observe that if MATH is such that MATH, then MATH . By REF MATH . Therefore we see that MATH . Using REF , we see that there is a positive constant MATH such that MATH . The proof of the lemma is completed by taking the limit as MATH, as the MATH form a regular exhaustion of MATH. |
math/0006138 | See REF. |
math/0006138 | Suppose MATH is not an eigenvalue of any of the MATH, that is MATH. Let MATH be the projection onto the eigenspace for MATH of MATH. Let MATH be the outer MATH-boundary of MATH and let MATH and MATH be the projections onto the functions with support in MATH and MATH respectively. One has MATH . Note that by REF , MATH ... |
math/0006138 | Fix MATH and define for MATH a continuous function MATH by MATH . Then clearly MATH and MATH as MATH for all MATH. For each MATH, choose a polynomial MATH such that MATH holds for all MATH. We can always find such a polynomial by a sufficiently close approximation of MATH. Hence MATH and MATH for all MATH. Recall that ... |
math/0006138 | Let MATH be points of continuity of the spectral density function MATH of the DML MATH with MATH. Then by REF , one has MATH . We also notice that MATH and MATH . This immediately implies the corollary. |
math/0006138 | Notice that the interval MATH is in a gap in the spectrum of the DML MATH if and only if MATH. Then by REF , one has MATH . The converse is proved similarly to the proof above. |
math/0006138 | This corollary follows immediately from REF , together with REF . |
math/0006138 | CASE: In this instance, the operator MATH is invertible, and positivity of the NAME determinant follows immediately. CASE: The argument for the positivity of MATH exactly parallels that of REF, using the lower bound on the modified determinant MATH determined in REF . This case calls for a slightly modified version of ... |
math/0006139 | An element MATH maps the generating monomials MATH to some MATH with MATH. The condition that MATH yields that MATH unless MATH and MATH. On the other hand, if MATH, then MATH, and the value MATH does not matter at all. Hence, we may restrict the knowledge of MATH to MATH. The linearity of MATH translates into the last... |
math/0006139 | CASE: If MATH, then MATH, hence MATH. On the other hand, let MATH with MATH for some MATH. It follows that MATH and MATH. With MATH, this implies MATH; with MATH being a proper subset of MATH, we obtain the MATH statement. The claims in REF are obvious. |
math/0006139 | Denote, just for this proof, the spaces given by REF of the previous proposition by MATH and MATH, respectively. Then we have to ascertain that pulling back via MATH induces isomorphisms MATH and MATH with MATH and MATH being the corresponding spaces from REF . The maps MATH are correctly defined: This is clear for the... |
math/0006139 | If MATH, then we may identify MATH with the union of all MATH in MATH where MATH and MATH. Thus, MATH as subset of MATH is the union of such MATH with MATH. For such a MATH let MATH be the maximal subflag consisting only of faces in MATH. Now we can continuously retract MATH onto MATH and this can be done simultaneousl... |
math/0006139 | We will prove the dual statement in homology using the method of acyclic models (see for example, CITE). If MATH has property U, consider the simplicial complex MATH where the vertices are the elements of MATH and a set of vertices MATH is a face if MATH. If MATH is the chain complex of MATH, then our MATH is the dual ... |
math/0006139 | MATH is the kernel of MATH. Hence, since MATH, the previous theorem implies that an element MATH belongs to MATH if and only if MATH, that is, iff MATH. On the other hand, this means that, for every MATH, the conditions MATH and MATH imply MATH. Since the assumption MATH can be omitted, this translates into MATH. Final... |
math/0006139 | First assume that there is a vertex MATH. If MATH, then MATH (otherwise MATH and MATH). Thus, MATH unless MATH. If MATH, then MATH and MATH. Thus, MATH may be contracted to MATH and its reduced cohomology is trivial, so MATH by REF . It is a simple matter to check that the map between the MATH sets is a bijection (with... |
math/0006139 | If MATH is in MATH (respectively MATH), then MATH is in MATH (respectively MATH). Now, if MATH for all MATH, then there is a standard retraction taking MATH to a MATH which fits together to make MATH a strong deformation retract of MATH. (See for example, CITE.) If MATH is a face, then MATH is impossible, since then MA... |
math/0006139 | It is straightforward to check that MATH is well defined and has the necessary properties to induce MATH by REF . Moreover, it is clear that this means exactly dehomogenization with respect to MATH, hence MATH coincides with the localization map. Finally, to prove that MATH is an isomorphism in case of MATH, we use tha... |
math/0006139 | If a graded piece MATH meets the image of MATH, then its pre-image is a unique summand MATH. Indeed, by REF , the only possibility to get degree MATH is MATH, MATH if MATH and MATH, MATH if MATH. This means that images of elements with different multidegree MATH cannot cancel each other, and it remains to consider the ... |
math/0006139 | CASE: We know that MATH if MATH. This would not be the case if MATH. REF is a straightforward calculation. For REF , we must first recall what we did in REF of the proof of REF . Antisymmetric functions MATH on the MATH level had been turned into antisymmetric functions MATH on the MATH level via MATH with certain elem... |
math/0006139 | Via the cup product, we see that each part MATH is responsible for MATH in the quadratic part of the obstruction equations. Moreover, since MATH is concentrated in degree MATH, no higher order obstructions involving only degree MATH deformations can appear, that is, MATH is described by the desired equations. Thus, in ... |
math/0006140 | Assume that there is such a diophantine model MATH, with MATH. Then there is an affine variety MATH over MATH admitting a finite morphism MATH defined over MATH such that MATH. If MATH is discrete (that is, infinite and totally disconnected in the real topology), the traditional proof applies: the real topological clos... |
math/0006140 | In CITE REF NAME proves that, for MATH, projection onto the MATH-coordinate of the MATH-rational points of the space curve MATH given by MATH gives the set MATH . For MATH, NAME CITE proved that the set MATH is the projection onto the MATH-coordinate of MATH . Already the sets MATH for MATH are disjoint, since their MA... |
math/0006140 | The proof is a bit indirect: we show that the polynomial ring has a diophantine model in the positive rational integers, and the latter has a diophantine model in the field of rational functions. More precisely, MATH is a recursive ring (compare CITE), because MATH is recursive (since finite), and hence the same holds ... |
math/0006143 | We have MATH with MATH, MATH and MATH. We omit here the isomorphisms between MATH and MATH. Then MATH . The result follows from the corresponding property in the category MATH. |
math/0006143 | By the induction hypothesis we can apply REF to NAME diagrams of size MATH. The statement MATH follows then from REF applied to MATH and MATH with MATH. The square of MATH is equal to MATH times the skein element represented by the following tangle. MATH . Let us consider the intermediate morphism MATH (the subscript i... |
math/0006143 | We want to show that MATH for any MATH. We write MATH . By the induction hypothesis, we have the result if MATH is in MATH. So it is enough to consider the case where MATH for MATH an element of the matrix units basis described in REF, that is, for MATH, where MATH and MATH are standard tableaus with the same shape, wh... |
math/0006143 | From REF , we can see that the NAME skein relation is respected, whence we have that MATH is well defined. We have that MATH. This implies that MATH. The computation below shows the multiplicativity. MATH . |
math/0006143 | The recursive construction of the preceding section can be done. The only point to check is that at each step the quantum dimensions MATH are not zero. By the induction hypothesis, we have non trivial minimal idempotents MATH for every standard tableau of size MATH. If the shape of MATH is MATH, then we have that MATH ... |
math/0006143 | From REF if MATH, and from REF if MATH, we get MATH . This implies that MATH if MATH. Using REF , we show that, if MATH is the length of MATH, we have MATH . Hence we have that MATH . The independence follows. To show that the family MATH generate MATH, we proceed recursively on MATH. Using the map MATH, we see from th... |
math/0006143 | The coordinates of MATH in the standard basis are obtained by computing MATH. The result is zero unless MATH and MATH have the same shape MATH and MATH; and in the latter case the result is MATH. |
math/0006143 | We can decompose MATH by using the minimal central idempotents in MATH. We get MATH . In the above, only those up and down tableaus MATH with MATH contribute. |
math/0006143 | REF follow from the corresponding in the NAME algebra (see CITE). Using the definition of the idempotent MATH, we can bring REF to the form MATH where MATH is depicted below. MATH . By the NAME lemma, MATH with MATH. Taking the quantum trace of this morphism, we get MATH . The first equality is due to the NAME skein re... |
math/0006143 | Let MATH be the element of the algebra MATH defined below. MATH . For MATH the equation in MATH, MATH defines a central element MATH in MATH. We consider the formal power series in MATH, MATH . We have that MATH . We denote by MATH the series given by the action of MATH on the simple component of MATH indexed by MATH. ... |
math/0006143 | Let MATH be as usual obtained by removing the last term in the sequence MATH. We have MATH . We will obtain the diagonal term MATH by computing the quantum trace of MATH. MATH where MATH is represented in the picture below. MATH . Clearly, MATH and we get the required result for MATH; MATH is calculated similarly. |
math/0006143 | We have that MATH . Multiplying on the left by MATH, we can express the above formula in the canonical basis. We denote by MATH the diagonal term of index MATH. From the left hand side, we get MATH . Let us compute the coefficient from the right hand side. MATH . In the the above sum, the term of indices MATH and MATH ... |
math/0006143 | In the following computations, we will drop some MATH. This means that by drawing the figures corresponding to these computation, one may have to add a vertical string on the right in order to get coherent equalities. For example, we write MATH . From the above, using the skein relation, we obtain MATH . This implies t... |
math/0006143 | We will prove REF . We first write the recursive REF in a more convenient form. We denote by MATH the unique cell in the skew diagram MATH. MATH . Here the first big product gives the contribution of the coefficients MATH corresponding to cells in the rows MATH to MATH. Note that some factors cancel if two among these ... |
math/0006143 | Suppose that MATH, and that MATH is obtained from the NAME diagram MATH by removing one cell from the MATH-th row, then we can write the recursive REF as follows. If MATH and MATH, then MATH else MATH . In REF case, MATH, if MATH and MATH, we get MATH else MATH . The announced result follows. In REF case, MATH, if MATH... |
math/0006145 | We argue as in our discussion of the walk on MATH in REF. Assume first that MATH is central, so that MATH has an identity. The lattice associated with MATH is MATH, so REF gives us an eigenvalue MATH as above for each MATH, with multiplicities MATH characterized by MATH for each MATH, where MATH is the number of chambe... |
math/0006145 | Let MATH be an irreducible MATH-module. Then MATH is a submodule of MATH, so it is either MATH or REF. (Here MATH is the set of finite sums MATH with MATH and MATH.) It cannot be MATH, because then we would have MATH for all MATH, contradicting the fact that MATH is nilpotent and MATH. So MATH, and the action of MATH o... |
math/0006145 | Recall from REF that the eigenvalues of MATH are the same as the eigenvalues of MATH acting by left multiplication on the ideal MATH. For each MATH, let MATH be the number of composition factors of MATH given by the character MATH. Then the discussion at the end of REF shows that MATH has eigenvalues MATH with multipli... |
math/0006145 | Consider the NAME decomposition MATH; here the MATH are pairwise orthogonal idempotents summing to MATH, the MATH are nilpotent, and MATH. If MATH is diagonalizable, then each MATH and we have MATH as required. If MATH is not diagonalizable, then for some eigenvalue MATH we have MATH. Since MATH can be computed in each... |
math/0006145 | Suppose MATH is split semisimple with primitive idempotents MATH, and write MATH. Then MATH, MATH, and REF for MATH follows at once. Conversely, suppose MATH is not split semisimple. Assume first that the minimal polynomial MATH of MATH splits into linear factors in MATH, say MATH, where the MATH are distinct. By the N... |
math/0006145 | Let MATH be independent picks from the probability measure MATH. We will get a formula for MATH by computing the distribution of the reduced word MATH; for we have MATH . Given a reduced word MATH with associated chain MATH, we compute the probability in REF as follows. Let MATH, and let MATH. In order to have MATH, th... |
math/0006145 | Fix a reduced word MATH of length MATH, and let MATH be as above. Then MATH . Setting MATH and multiplying by MATH, we obtain MATH; REF now follows from REF. |
math/0006145 | Suppose MATH. Given MATH, let MATH be the face of MATH of type MATH and let MATH; see REF . We know that MATH and MATH are on opposite sides of MATH, so MATH must be strictly on the MATH-side of MATH, hence MATH and MATH. This proves MATH. Conversely, suppose MATH. To show MATH, it suffices to show that MATH and MATH a... |
math/0006145 | Let MATH be the set of simplices of type MATH. There is a REF map MATH, given by MATH, where MATH is the fundamental chamber. It is REF because we can recover MATH from MATH as the face of type MATH. Its image, according to REF , is the set of chambers with descent set contained in MATH. Hence MATH . The proposition no... |
math/0006145 | Let MATH and MATH be the MATH-bases of MATH introduced in REF. We have MATH . As we noted in the proof of REF , the chambers MATH that occur in the sum are those with descent set contained in MATH. Under our bijection between MATH and MATH, a given chamber MATH corresponds to the element MATH such that MATH. So we can ... |
math/0006145 | A minimal gallery from MATH to MATH crosses only the hyperplanes that separate MATH from MATH. This shows that REF implies REF . For the converse, it suffices to show that if REF holds and MATH is a chamber not in MATH, then there is a hyperplane MATH separating MATH from MATH. Choose a gallery MATH of minimal length, ... |
math/0006145 | For the ``if" part see the beginning of REF. To prove the converse, we may assume that MATH is an idempotent semigroup for which the relation is antisymmetric, and we must prove REF. Note that MATH, so MATH. On the other hand, MATH, so MATH. Thus antisymmetry implies that MATH, as required. |
math/0006145 | The construction of MATH is forced on us by REF: Define a relation MATH on MATH by MATH. This is transitive and reflexive, but not necessarily antisymmetric. We therefore obtain a poset MATH by identifying MATH and MATH if MATH and MATH. If we denote by MATH the quotient map, then REF holds by definition. To see that M... |
math/0006145 | We have MATH for all MATH. In view of REF, this holds if and only if MATH, so REF are equivalent. If REF holds then MATH is maximal, because MATH. For the converse, note that MATH for all MATH; so if MATH is maximal then REF holds. |
math/0006145 | The inequality is immediate by induction on the size of MATH. We have already observed that equality holds if MATH has exactly one atom. If MATH has no atoms, then MATH and MATH. If MATH has more than one atom, then consideration of the term MATH in REF shows that MATH. |
math/0006145 | Let us temporarily take the statement of the proposition as a new definition of MATH. It then suffices to show that, with this definition, REF holds. We may assume that MATH and that REF holds for smaller posets. Then MATH . So we are done if we can show MATH, that is, MATH . The sum on the right counts all chains MATH... |
math/0006145 | Let MATH denote the right-hand side of REF if MATH, and let MATH. It suffices to show that MATH satisfies the recurrence REF, that is, MATH where MATH is the set of elements of MATH of rank MATH. We may assume MATH. Group the terms on the right-hand side of REF in pairs, where MATH is paired with MATH. This leaves one ... |
math/0006145 | Take MATH to be the subspace lattice of MATH, so that MATH. It is known CITE that MATH . The corollary now follows at once from the proposition. |
math/0006147 | All the preceding computations amount to show that MATH . Then MATH represents a class since the double complex MATH computes the hypercohomology. |
math/0006147 | It is sufficient to show that changing the definition of the various logarithm branches in MATH amounts to change it by a coboundary. First, we change these branches, MATH where MATH. The effect of these changes on the representatives of the NAME classes of MATH and MATH is MATH . While the term MATH is obviously invar... |
math/0006147 | The procedure is to compute the variation of the various components of MATH by applying MATH to the descent equations. Start with MATH, that can be computed in two different ways: from REF, and from the variational relation MATH. Since MATH, we have MATH and we deduce, using NAME Lemma, that MATH for MATH. An explicit ... |
math/0006147 | The MATH-form MATH in the relation MATH is a source form CITE, hence it is uniquely determined by the NAME class of MATH. Moreover, given a specific MATH, the form MATH is also determined (so MATH is determined up to an exact form). Since MATH, we must have MATH as both MATH and MATH are source forms for the same Lagra... |
math/0006147 | Indeed, the push-forward of MATH is MATH, where MATH. It is a projective connection on MATH because of the transformation law MATH . The NAME equation is equivalent to the equation MATH, which is precisely the condition that projective connection MATH is holomorphic on MATH. |
math/0006147 | Let MATH be a local section of MATH. As a result of a direct calculation we have (omitting the coordinate index MATH) MATH where MATH is the NAME derivative operator on MATH. Thus the ``if" part is clear. For the ``only if" part, assume the Right-hand side of REF is zero for all MATH. Therefore, if we consider MATH for... |
math/0006147 | Consider the cone of MATH: MATH . Its cohomology sheaf complex equals MATH, thus by standard homological algebra arguments (see, for example, CITE) one has MATH and from the canonical sequence MATH one gets REF. On the other hand, the Right-hand side of REF is the first cohomology group of the complex MATH which is equ... |
math/0006147 | We just repeat the computation of the vertical variation with additional term MATH, where MATH. Since the main term, given by MATH vanishes "on shell", this proves the result. |
math/0006147 | For MATH we have MATH where MATH is the function MATH . Using the infinitesimal action, MATH where MATH is the NAME derivative on MATH, and the NAME bracket in MATH is the usual vector field NAME bracket: MATH. Using the identity MATH we are left with MATH because of the commutativity condition and the skew-symmetry of... |
math/0006153 | A simple proof of this Lemma can be constructed using induction on MATH. |
math/0006153 | MATH . The double sum over permutations, MATH and MATH is equivalent to summing each MATH independently over MATH (terms for which two or more components of MATH are equal make zero contribution) and the first result follows. The second result follows in the same way by interchanging the roles of MATH and MATH. |
math/0006153 | (of REF MATH) REF We first obtain the cyclic REF as follows. MATH . The critical step, and the whole reason for introducing the NAME, is to enable one to go from the restricted sums of REF to the unrestricted sums in REF. This is justified for two reasons, CASE: since MATH is a determinant, if any of the MATH's are equ... |
math/0006153 | If the conditions of REF hold then we have that REF are valid. Using REF it follows that MATH . Also, using REF it follows that MATH . Substituting these into REF and using REF gives the result immediately. |
math/0006153 | Using REF (which follows from REF by REF ) MATH which is an expansion of the required determinant. |
math/0006155 | If MATH is right-orderable (bi-orderable), just take MATH, the set of positive elements. Conversely, if there exists MATH verifying the required hypothesis, then define the order MATH by: MATH if and only if MATH. |
math/0006155 | Notice that, under the action of such a MATH, the degree and the lexicographical order on the monomials are preserved. Hence, MATH preserves the order we defined on the monomials, thus the order MATH on MATH. Therefore, the NAME order on MATH is also preserved. |
math/0006155 | We argue by induction on MATH. If MATH, then MATH is a free group (of infinite rank), so it is bi-orderable. Suppose then that MATH, and that MATH is bi-orderable. By REF , we have an exact sequence MATH where MATH. By definition of bi-order, conjugation by an element of MATH is an automorphism of MATH which preserves ... |
math/0006155 | Let MATH. For all MATH, we write MATH, where MATH is a path on MATH. We denote by MATH the permutation induced by MATH on MATH. Consider the braid MATH. For all MATH, one has MATH . Now consider the isotopy MATH given by MATH . Then MATH and MATH. |
math/0006156 | Consider the following cartesian diagram where MATH is a stable map to MATH, and MATH is the forgetting map, that is, MATH is a prestable curve: MATH . Remember that if we have two morphisms of schemes (or stacks) MATH we get a distinguished triangle of cotangent complexes: MATH . Moreover MATH naturally maps to MATH, ... |
math/0006156 | First we will construct a two - term resolution of MATH that is MATH - equivariant if such an action exists on MATH. We will use similar arguments as NAME does for MATH (compare CITE). Let MATH be an ample invertible sheaf on MATH and let MATH. Then by CITE, for MATH sufficiently large and MATH a vector bundle on MATH ... |
math/0006156 | The equality of the two virtual fundamental classes can easily be seen by looking at the complex MATH dual to MATH, MATH . Here we have used that by CITE there exists a morphism of functors MATH and that this morphism MATH is an isomorphism. For convenience we also use the notation MATH . Therefore, the MATH's fit into... |
math/0006156 | We will prove the lemma for the obstruction complex MATH, the arguments for the dual complex MATH are similar. There is a natural morphism MATH and we have to show that this morphism is a isomorphism in the derived category, that is, a quasi - isomorphism between complexes. Let MATH, indexed at MATH and MATH. We then h... |
math/0006156 | For a stable map MATH to be a fixed point of the MATH - action on MATH means that for any element MATH in the torus MATH, the stable map MATH is isomorphic to the original curve MATH, that is, that there exists a morphism MATH such that the following diagram is commutative (compare REF ): MATH . Now, it is obvious that... |
math/0006156 | First we will describe the family of curves MATH. For each edge MATH, let us number its vertices: MATH. For each such edge MATH we also fix a map of degree one MATH, such that MATH and MATH. We set MATH to obtain a map MATH of degree MATH. Note that up to parametrization such a map is a unique. Also set MATH. Let MATH ... |
math/0006156 | The MATH - dimensional cone MATH gives a local chart MATH of our toric variety MATH, and the coordinates on MATH are given by (compare REF ): MATH . The MATH - dimensional submanifold corresponding to MATH is given by the equations MATH. In the coordinates of MATH, these equations are equivalent to MATH . Hence we have... |
math/0006156 | The action of MATH on MATH is just the pull back by MATH of the action on MATH. Since MATH has multiplicity MATH, the formula follows immediately from REF . |
math/0006156 | Let MATH be the complex MATH indexed at MATH and MATH. It fits into the following short exact sequence: MATH . The corresponding long exact sequence of higher direct image sheaves corresponding to MATH is then REF where exactness on the left, that is, the injectivity of the map MATH induced by the natural map MATH is e... |
math/0006156 | The bundle MATH parameterizes infinitesimal automorphisms of the pointed domain. The induced MATH - action on MATH is obviously trivial on all automorphism MATH of MATH that restrict to the identity MATH on all irreducible components MATH corresponding to edges MATH in the graph MATH. Thus, the moving part of MATH spli... |
math/0006156 | Just apply NAME duality: MATH where MATH is the dualizing bundle of MATH. |
math/0006156 | Just assemble the long sequence REF , and REF . To obtain the exponent of MATH just observe that there MATH flags at the vertex MATH. |
math/0006156 | First we observe that formally MATH . Since MATH is the product of NAME - NAME spaces of stable curves MATH corresponding each to some vertex MATH of the graph MATH, the integral above is the product of integrals over the NAME - NAME spaces MATH of the classes corresponding to their associated vertex. So let MATH be a ... |
math/0006156 | Remember that by REF we have MATH . Hence, by the binomial formula, we obtain MATH . |
math/0006156 | By REF , the formula is obviously true for the parts coming from vertices with at least three flags (that is, special points). Thus it remains to show that MATH . The first term on the right hand side of the equation obviously coincides with the first term on the left hand side. The third factor on the right hand side ... |
math/0006156 | Given a graph type MATH, a graph MATH in this graph type is given by the positions MATH of the MATH marked points (where the classes MATH are attached), and vice versa any such tuple MATH of vertices of MATH yields a graph MATH. Hence, up to automorphisms, the first part of the lemma is obvious. For the automorphism gr... |
math/0006156 | Let MATH be a graph type of MATH. Remember, that the weights MATH in the sum of the right hand side of REF are given by MATH where MATH are the flags at MATH corresponding to its MATH edges. Hence we can write MATH . Note that as the notation suggests, MATH only depends on the class MATH and the vertex MATH of the poly... |
math/0006158 | This can be seen by pulling back the extension MATH along MATH and then applying REF . |
math/0006158 | The first statement follows from the previous proposition as the mapping MATH is MATH-equivariant. The second follows from the first by taking MATH to be the adjoint representation. The third statement is clear. Set MATH . The inclusion of this into MATH induces a surjection on MATH as MATH has all negative weights. Si... |
math/0006158 | Since MATH and since the action MATH is compatible with the gradings, each term MATH of the weight filtration is a MATH-module. Since MATH, the image of the action MATH is contained in MATH, from which the result follows. |
math/0006158 | If MATH is reductive, this follows directly from the classical NAME decomposition REF . When MATH connected, the existence of the splitting follows from the existence of a splitting of MATH as MATH is simply connected. The existence of such a NAME algebra splitting follows by first taking a lifting MATH of MATH, and th... |
math/0006158 | Suppose that MATH is another lift of MATH. Then, by REF , there is MATH such that MATH. Write MATH where MATH is the subspace of MATH where MATH acts with weight MATH via MATH. Then MATH. Since each term of the weight filtrations associated to MATH and MATH are MATH-modules, it follows that, for each MATH, MATH . The r... |
math/0006158 | We use the notation of Paragraph REF. Since MATH and MATH are finite dimensional, we may assume that MATH and MATH are algebraic. To prove that the weight filtration is preserved by the morphism MATH, choose a lift MATH, and define MATH to be MATH. Then both MATH and MATH become MATH-modules, and MATH is MATH-equivaria... |
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