paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0006165 | We have to represent an arbitrary vector MATH as MATH for some MATH. It is enough to consider a dense set of vectors MATH (since MATH is always closed). Let MATH and MATH. Clearly, MATH implies MATH in MATH, hence MATH in MATH, therefore MATH in MATH. |
math/0006165 | Let MATH, MATH; we have to prove that MATH weakly. Introduce MATH where MATH is chosen such that for all MATH large enough, MATH in terms of NAME transform it means that MATH such MATH exists due to the asymptotic relation MATH for large MATH (recall REF). The positive definiteness means that MATH for all MATH and then... |
math/0006165 | The construction is straightforward, just MATH equidistant intervals of equal length: MATH . We have to find MATH and MATH such that MATH . Still, the construction is straightforward: MATH the proof of REF is more complicated. NAME transform MATH will be used, MATH. Recall that MATH for all MATH, and MATH for large MAT... |
math/0006165 | Suppose that MATH. REF gives elementary sets MATH such that MATH, and MATH consists of MATH intervals, and the relation MATH is violated. Taking into account that MATH we get MATH by REF . Also, MATH and MATH by REF . So, MATH satisfy REF but violate REF . By REF the product systems are nonisomorphic. |
math/0006165 | Similarly to the proof of REF we have for any MATH . This time, however, the following compactness property holds: MATH which ensures MATH. In order to prove the compactness property we estimate integrals similarly to the proof of REF. We have MATH . For MATH we note that MATH . For MATH, introducing MATH and MATH, we ... |
math/0006165 | Assume the contrary, then there exist MATH such that MATH and MATH whenever MATH. NAME use MATH for MATH only. Introduce a Gaussian type space MATH such that MATH; we have MATH where MATH is the MATH-field generated by the Gaussian subspace MATH. Consider measures MATH; as was explained in REF, MATH is a probability me... |
math/0006168 | First suppose that MATH satisfies REF. Using MATH, we find that MATH . REF follows since MATH. The converse is proved similarly, since, for all MATH, one can always find MATH such that MATH at MATH. |
math/0006168 | It remains to verify the moment map REF , which in this case is MATH . From the definition of MATH we obtain, MATH . These two expressions differ by MATH which vanishes by MATH-invariance of the inner product. |
math/0006168 | Using REF we find that MATH for all MATH. This shows MATH. |
math/0006168 | By REF , MATH takes values in the image of the action map MATH. Equivalently, MATH is tangent to the orbits for the conjugation action. |
math/0006168 | The map MATH is a homomorphism with respect to both the exterior product and the NAME bracket. Moreover, the operator MATH is a generator of the NAME bracket on MATH. Hence, it is sufficient to prove the property MATH on the elements of degree REF. For all MATH, MATH. To compute MATH we consider MATH where MATH denotes... |
math/0006168 | Since the NAME is a derivation of the NAME bracket, for each MATH, MATH. Moreover, MATH. The element MATH defines a cycle in NAME algebra homology. Hence, we obtain MATH and MATH. Choosing MATH with MATH a positive MATH-invariant function on MATH, one obtains the new NAME, MATH. The modular vector field also changes, M... |
math/0006168 | Let MATH be the bi-invariant volume form on MATH defined by the basis MATH of the NAME algebra MATH. Then, MATH, where MATH. Applying the NAME differential yields MATH since MATH and MATH commute, and since MATH is both left-and right-invariant and closed. This implies MATH. |
math/0006168 | The trivector field for the MATH-action is the sum of the trivector fields MATH and MATH for the two MATH-factors. By assumption, MATH is MATH-invariant and MATH. Therefore MATH, and MATH . By REF , this implies that MATH where MATH is the image of MATH under the map extending the infinitesimal diagonal action of MATH ... |
math/0006168 | Since the identity map is a moment map for MATH, REF shows that MATH is a moment map for MATH. Hence it is a quasi-Poisson map by REF . |
math/0006168 | Since MATH is MATH-invariant, MATH. To compute MATH we observe that MATH, since MATH is the vector field generating the left action on MATH, and MATH is equivariant with respect to this action. Thus MATH . Finally, since MATH, we obtain MATH which cancels the term MATH. |
math/0006168 | REF follows by equivariance of the moment map, MATH . To prove REF, write MATH as the composition of two maps, MATH where MATH, and MATH is the action map MATH of the second MATH-factor. The tangent map to MATH is defined by MATH . In particular, MATH and MATH, so that MATH . We now use the fact that MATH and MATH and ... |
math/0006168 | Suppose that the action is locally free at the point MATH. To prove that MATH has maximal rank at MATH, we need to show that the equation MATH does not admit a non-trivial solution MATH. Let MATH be any solution of this equation. By equivariance of the moment map, MATH for all MATH. Hence, MATH, and the moment map REF ... |
math/0006168 | Let MATH . We want to compute MATH . The first term vanishes since MATH is a NAME structure. For any function MATH, MATH. Using the moment map REF of MATH and the invariance property MATH of MATH, we obtain MATH . To calculate MATH we use the relation MATH . Taking symmetries into account, we obtain MATH . Together wit... |
math/0006168 | At MATH, the fiber MATH is the space of covectors MATH with MATH. Let MATH. By the moment map REF , it follows that, MATH for all MATH. |
math/0006168 | Replacing MATH by MATH, we may assume that MATH. It is clear that the moment map condition for MATH is equivalent to that for MATH. We have to show that MATH if and only if MATH . Let MATH denote the extension of MATH to a MATH-invariant bivector field on MATH. Also, let MATH be the MATH-invariant extension of MATH. By... |
math/0006168 | We show that MATH is integrable near any given point MATH. Let MATH, and MATH be a MATH-invariant slice through MATH, and MATH the corresponding cross-section. Since MATH, it suffices to show that the distribution MATH induced by the quasi-Poisson structure MATH is integrable. However, as explained in the previous sect... |
math/0006168 | Given MATH, let MATH be a slice through MATH, and let MATH be the corresponding cross-section. We first evaluate both sides of REF on elements of MATH, and then on orbit directions. REF-form MATH vanishes on MATH, and the bivector MATH vanishes on MATH. Hence MATH, which agrees with the right-hand side of REF since MAT... |
math/0006168 | Given MATH, let MATH be a cross-section at MATH, as in the proof of REF , and let MATH. Thus MATH is a non-degenerate Hamiltonian quasi-Poisson MATH-manifold that corresponds to a quasi-Hamiltonian MATH-space, MATH. Let MATH be the unique REF-form on MATH such that MATH is a quasi-Hamiltonian MATH-space, and MATH is th... |
math/0006168 | Choose MATH and a slice MATH containing MATH and let MATH be the cross-section. We observe that in both the quasi-Poisson and the quasi-Hamiltonian settings, the reduction of MATH at MATH is canonically isomorphic to the reduction of MATH at the group unit of MATH. The cross-section MATH carries the NAME bivector MATH ... |
math/0006168 | We have to compute MATH . We compute the four terms in the expansion of the right-hand side. By REF , the first term is MATH . Next, using MATH we find that MATH . From MATH we obtain MATH . Finally, MATH . Putting everything together, we get MATH as required. |
math/0006171 | This is equivalent to MATH which is obvious. |
math/0006171 | The sufficiency of the conditions in the lemma is clear. To prove necessity, define the map MATH by MATH where MATH is arbitrary. This map MATH is well defined because MATH for any closed path MATH. The required properties of MATH follow easily from the definitions. |
math/0006171 | The NAME imaginary transformation (see for example, REF) yields MATH . We have MATH . It is obvious that this series, together with all derivatives, is dominated in the MATH limit by only two terms, namely the terms with MATH and MATH which combine into a multiple of MATH. All other terms differ by a factor of at least... |
math/0006171 | Observe that all MATH-functions appear in MATH in the following combinations: either they appear in pairs of the form MATH where MATH is a subset of MATH, or else they appear as the nullwerts MATH . It is clear from REF that the asymptotics in the either case is a polynomial in MATH of degree MATH with coefficients inv... |
math/0006171 | Follows, as explained in REF , from the corresponding facts for MATH, see REF or REF. |
math/0006171 | Follows from the difference equation satisfied by MATH, see REF or REF. |
math/0006171 | Observe that the binomial theorem and the definition of the function MATH imply that MATH where the sum is over all subsets MATH of MATH and hats mean that the corresponding terms should be omitted. Interchanging the order of summation in the partition MATH and in the subset MATH one obtains MATH where MATH is a partit... |
math/0006171 | By induction on MATH. The case MATH is clear, see REF . Suppose MATH. Denote by MATH the right-hand side of REF. We know that MATH satisfies the same difference equation as MATH does. We claim that it also has the same singularities as MATH does. Indeed, MATH is regular at the divisors MATH, MATH, because MATH is regul... |
math/0006171 | The NAME coefficients of MATH are certain contour integrals and hence by REF they represent the asymptotics of the corresponding coefficients of MATH. |
math/0006171 | Recall that, by definition, MATH . We have MATH where, we recall REF , MATH means that MATH is a subset of a block of MATH. The first factor in REF is a common factor for all MATH. The NAME series expansion of the second factor in REF can be interpreted as summation over certain graphs MATH with multiple edges MATH whe... |
math/0006171 | By definition, we have MATH . Substituting REF in this sum yields MATH . Interchanging the order of summation we obtain MATH . Using REF we conclude that MATH where dots stand for lower order terms. We know from REF that the exponent MATH takes its minimal value MATH precisely when MATH. This concludes the proof. |
math/0006171 | We can assume that the number of variables MATH is finite and equal to MATH and switch to the variables MATH. It is known, see REF , that MATH where MATH. Expand all fractions in geometric series assuming that MATH . We have MATH . Now let MATH is a partition of weight MATH and let us compute the coefficient of MATH in... |
math/0006171 | By definition of the cumulants and of the elementary cumulants, we have MATH and therefore MATH . By REF , for any MATH such that MATH, we have MATH with the equality if and only if MATH. |
math/0006171 | Follows by extracting terms of degree MATH in MATH from the formula MATH . |
math/0006171 | REF specializes to MATH where the summation is over MATH satisfying the conditions described above. In particular, MATH. Recall that the MATH term in REF is to be understood as MATH. This is in agreement with the coefficient of MATH in REF. Therefore, in what follows we can assume that MATH. We know from REF that all M... |
math/0006172 | The nest algebra MATH and the MATH- algebra it generates have a MATH block structure induced by the MATH atoms in the nest for MATH. In this block structure, the diagonal blocks are square matrices of varying size and the off-diagonal blocks are rectangular. Elements of MATH and MATH will be written as MATH matrices, M... |
math/0006172 | We first prove the theorem for the special case in which every atom of MATH has rank one (that is, MATH is some MATH). Select a matrix unit system for MATH compatible with MATH; for each pair of (rank one) atoms MATH and MATH, let MATH denote the matrix unit with initial projection MATH and final projection MATH. Note ... |
math/0006172 | Assume that MATH is not MATH-degenerate. Let MATH have atomic interval projections MATH and let MATH have atomic interval projections MATH. The hypothesis for MATH implies that there exists a matrix unit system MATH, MATH, MATH, for a copy of MATH in MATH such that MATH is supported in MATH the MATH block subspace of M... |
math/0006172 | Fix matrix units for MATH and MATH so that MATH maps matrix units of MATH to sums of matrix units in MATH, and then choose matrix units for MATH such that MATH maps matrix units in MATH to sums of matrix units. Let MATH be the bimodule over MATH generated by MATH. Since MATH is, in particular, a bimodule over the diago... |
math/0006172 | The proof that REF implies REF is immediate. For the converse, let MATH be a star extendible isomorphism. Since each algebra in the systems MATH and MATH is finitely generated, there is a commuting diagram isomorphism of the systems with crossover maps MATH, MATH: MATH . Since MATH, the range of MATH is contained in MA... |
math/0006172 | The proof of this theorem follows easily from REF below. If MATH and MATH are star extendibly isomorphic, then REF gives a regular system isomorphism in which any adjacent pair of crossover maps have a composition which is order preserving or order conserving, as appropriate. From the lemma, any consecutive triple of c... |
math/0006172 | The proof of the Lemma in the order preserving context differs considerably from the proof in the order conserving context, so we present the two arguments separately. We start with the the order preserving context, where the argument is simpler. First, suppose that MATH is a rank one partial isometry in MATH such that... |
math/0006172 | It will be enough to show that the assertion of the lemma follows if it is assumed that the lemma is true in the case of matrices of sizes MATH, MATH and MATH, where MATH, MATH, MATH, MATH, and at least one of these inequalities is strict. We may assume that all the entries MATH, MATH and MATH, MATH are non-zero. The r... |
math/0006172 | The usual scheme of proof that REF. implies REF. can be completed as follows. With the aid of REF , obtain a lifting of a suitable restriction MATH to a locally order conserving star extendible homomorphism MATH. In the same way, the restriction of MATH to MATH may be lifted to a locally order conserving embedding MATH... |
math/0006172 | In view of REF , it will be sufficient to show that if MATH are star extendibly isomorphic then not only is the induced isomorphism of MATH regular, but the commuting diagram isomorphism may be implemented by order conserving embeddings. However this follows immediately from REF . |
math/0006174 | This is just the statement that MATH. |
math/0006174 | With MATH defined as above, MATH is in the kernel of all simple roots MATH distinct from MATH, and thus, since MATH is primitive, MATH is an embedding of MATH into the center of MATH. Also, since MATH, if MATH, then MATH if and only if MATH, if and only if MATH. Thus MATH is the cyclic subgroup of order MATH in MATH, a... |
math/0006174 | We have MATH . Then MATH is the subgroup of MATH which is in the kernel of the character MATH. The restriction of MATH to MATH is non-trivial, and hence surjective, and for an element of MATH, written as MATH, we clearly have MATH. Thus, for each MATH there is an element MATH of the form MATH. This element MATH is in t... |
math/0006174 | Let MATH be the fundamental weight corresponding to the simple root MATH. The unique primitive dominant character of MATH is MATH, viewed as a character on MATH, and MATH. |
math/0006174 | Since MATH is primitive MATH for some integer MATH. Since the character MATH of MATH is given by MATH, the character MATH of MATH is given by MATH. The action of MATH is diagonal with respect to the decomposition of MATH as a sum of root spaces, and the character on the one-dimensional subspace spanned by a root MATH i... |
math/0006174 | By definition, MATH as additive characters on MATH. By the previous lemma, MATH. On the other hand, MATH, where MATH lies in the MATH-span of the simple coroots MATH, MATH, and hence in the NAME algebra of the derived group MATH. Thus MATH and so MATH. |
math/0006174 | As before, MATH for some integer MATH. To compute MATH, note that, if MATH, then MATH acts on MATH via raising to the power MATH. Since MATH, we must have MATH . Hence, MATH. |
math/0006174 | Let MATH be the subset of MATH consisting of roots MATH such that MATH divides MATH. Clearly MATH is again a root system. Let MATH be the real span of MATH and let MATH be the subspace of MATH spanned by MATH. Then clearly MATH is the set of all roots which are linear combinations of elements of MATH. Thus MATH is a ro... |
math/0006174 | Since MATH, MATH. Suppose inductively we have shown that MATH for all MATH. If MATH, then MATH is not a root. If MATH is not a root, then MATH is the highest root and MATH. Otherwise, MATH. |
math/0006174 | Clearly MATH has the property that MATH is a set of simple roots for MATH. By the proof of REF , MATH, proving REF . To see REF , note that MATH, and thus MATH. To see REF , write MATH, where MATH. Then MATH . On the other hand, MATH, and plugging this back in gives the first part of REF . The second part is proved in ... |
math/0006174 | The first equality follows since, by REF , MATH . The second follows since MATH. The third is an easy consequence of the fact that MATH, using MATH. |
math/0006174 | By REF , it suffices to prove that MATH. By REF , since MATH, MATH. Since MATH exchanges positive and negative roots in MATH, it follows that MATH is a permutation of MATH. Clearly, given MATH, MATH . Thus, since MATH permutes MATH, we have MATH . Next we claim: MATH . Suppose that MATH, and consider the MATH-string de... |
math/0006174 | In the notation of REF , set MATH and let MATH, with simple roots MATH. Although MATH need not be irreducible, we can still define the integer MATH with respect to the root system MATH as in REF . By REF , which holds even if MATH is reducible, MATH, where MATH is the sum of the fundamental weights of the root system M... |
math/0006174 | For MATH, we have MATH and MATH, and the corollary is clear. |
math/0006174 | If MATH, it is easy to check that the above conditions follow from REF and the fact that MATH. The remaining cases are: MATH with MATH the root such that MATH, MATH with MATH the root such that MATH, or MATH with MATH a root such that MATH. These cases may be checked by hand. |
math/0006174 | By CITE, with MATH and MATH as above, every pair MATH such that MATH is conjugate to such a pair with MATH and MATH. This proves that the map is surjective. Clearly, it is finite-to-one. If MATH is in the interior of MATH, then it is regular, and the only further possible conjugation is via an element MATH, which acts ... |
math/0006174 | The dominant character MATH lifts to the character MATH on MATH. By CITE, MATH is the unique point MATH in the center of MATH such that MATH. Thus MATH, showing REF . The congruence condition MATH follows from CITE. REF follows from CITE. The remaining statements are clear. |
math/0006174 | On the level of NAME algebras, there is a direct sum decomposition MATH, where MATH, and the proof follows. |
math/0006174 | The first two statements are immediate from NAME on MATH. The final one follows from REF . |
math/0006174 | We compute the degree of the line bundle MATH. Let MATH be the dominant character for MATH. The line bundle MATH is associated to MATH by the character MATH. By REF , MATH lifts to the character MATH of MATH. Since MATH lifts to MATH on MATH and since the line bundle associated to MATH by the character MATH has degree ... |
math/0006174 | By REF applied to MATH, the character of MATH defined by the determinant on MATH lifts to the character MATH on MATH. The degree of MATH is thus MATH. It follows that the slope of MATH is MATH. |
math/0006174 | It follows from the definition of a special root that MATH. By REF , there is an isomorphism MATH where the image of MATH is mapped to MATH and to MATH. From this, we must have MATH. The map from MATH to MATH which is the natural inclusion MATH and which maps MATH to MATH then factors to give an induced homomorphism MA... |
math/0006174 | If the center of MATH is trivial, then MATH is a primitive element of MATH, and hence MATH. This handles the cases MATH. Next suppose that MATH is simply laced and not of type MATH, so that the NAME diagram of MATH is a MATH diagram, with MATH or MATH with MATH. Let MATH, so that MATH, MATH in the respective cases abov... |
math/0006174 | Recall that, for every MATH, there is a unique stable vector bundle MATH of rank MATH over MATH such that MATH. Given the structure of MATH as in REF , it is clear that there is a unique principal MATH-bundle, up to isomorphism, satisfying REF above, with MATH for every MATH. Since the vector bundles MATH in REF are si... |
math/0006174 | The cases MATH and MATH follow easily from the explicit descriptions of the maximal parabolic subgroups and are left to the reader. In case MATH, the corresponding maximal parabolic of MATH is the set of MATH preserving an isotropic subspace of MATH of dimension MATH. The corresponding NAME factor MATH is the subgroup ... |
math/0006174 | First, by REF , MATH since MATH. Next we show that MATH has the following minimality property: If MATH is special, then MATH. If MATH is not special, then MATH. To see that MATH, it suffices by REF to show that MATH. First, by CITE, MATH is always a coroot. Clearly, MATH for all MATH, MATH if MATH is an end of the NAME... |
math/0006174 | The group MATH acts on MATH with weight MATH. By REF , MATH . Thus, it suffices to show that MATH. If we define MATH then it clearly suffices to show that, for all MATH, MATH, in the notation of REF . By REF , the integers MATH have the circular symmetry property with respect to MATH and MATH, since MATH is a long root... |
math/0006174 | The minimally unstable points MATH are listed in REF. From this list, it is easy to check that MATH and MATH. To prove the remaining statements, we make a case-by-case analysis. MATH: We choose an identification of MATH with MATH in such a way that MATH with MATH with MATH being the standard unit vector in the MATH-coo... |
math/0006174 | If MATH is cyclic and MATH is a generator, then MATH. The remaining cases are MATH and MATH, and these can be checked directly. |
math/0006174 | The weight for the action of MATH on MATH is MATH. By REF , the dimension of MATH is MATH. By REF , the integers MATH have the circular symmetry property with respect to MATH and MATH. By REF , MATH. If we define MATH then the integers MATH satisfy: MATH, and the MATH have the circular symmetry property with respect to... |
math/0006174 | Let MATH be the closed subgroup of MATH whose NAME algebra is MATH. Then the filtration MATH is a decreasing filtration of MATH by normal, MATH-invariant subgroups such that MATH is in the center of MATH for every MATH, and MATH. By REF of the appendix, it suffices to check that MATH. The second statement is clear sinc... |
math/0006174 | Let MATH be represented by the MATH-cocycle MATH, where MATH is an open cover of MATH and MATH is a morphism. Then MATH acts on the cocycle MATH. Define morphisms MATH as follows: MATH . There is a MATH-equivariant morphism from the unipotent subgroup MATH to the affine space MATH (see for example CITE). Using this MAT... |
math/0006174 | Let MATH denote the dual numbers. The space MATH is the set of maps from MATH to MATH such that the closed point is mapped to the origin. By the functorial interpretation of MATH, such a morphism corresponds to a MATH-bundle MATH over MATH, which is the pullback of MATH, together with an isomorphism from MATH to MATH, ... |
math/0006174 | Let MATH be the affine coordinate ring of MATH, where MATH is defined by MATH, and let MATH be a basis for the linear functions on MATH. The finite-dimensional subspace of MATH spanned by the MATH is contained in a finite-dimensional MATH-invariant subspace MATH of MATH, by the NAME lemma CITE (or by using the grading ... |
math/0006174 | The MATH-weights are of the form MATH for MATH, and so the lemma is clear. |
math/0006174 | Let MATH be the set of points of MATH where MATH acts freely and effectively. We have seen that there is a lifted action of MATH on MATH, which in fact is free. The action of the isotropy group MATH of a point in the base on the fiber is via multiplication by elements of the center of MATH, and thus there is an induced... |
math/0006174 | By REF , if MATH is unstable and MATH, then MATH. As we have noted above, the same holds for a semistable MATH-bundle MATH which is not regular. Thus, a MATH-bundle MATH is semistable and regular if and only if MATH. To prove REF , we shall show that, for all MATH, MATH. Let MATH. We have the inclusion of the NAME alge... |
math/0006174 | As in the proof of REF , the tangent space MATH can be identified with bundles MATH over MATH which restrict to MATH over the closed fiber and reduce mod MATH to MATH. If MATH is given by the MATH-cocycle MATH, where the MATH take values in MATH, and MATH by the MATH-cocycle MATH, where the MATH take values in MATH, th... |
math/0006174 | Since the MATH-action is via strictly positive weights, the differential of the action is injective at every nonzero point MATH, so the hypothesis of the previous theorem is satisfied. This proves the first statement. To see the second statement, the induced map from MATH modulo the tangent space to MATH to MATH is sur... |
math/0006174 | It is an elementary exercise to check that, if MATH acts linearly and with positive weights on two affine spaces MATH and MATH and if MATH is a MATH-equivariant morphism whose differential at the origin is injective, then there exist coordinates on MATH for which MATH acts linearly and such that MATH is a linear embedd... |
math/0006174 | REF follows easily from the explicit description of the special root. To see REF , it follows from REF that, if MATH is a nonzero element of MATH, then the image of MATH in MATH is also nonzero. In particular, if MATH is the corresponding MATH-bundle, then MATH is semistable. It is easy to check that, in this case, the... |
math/0006174 | We must show that the MATH-linearization on MATH is given by the character which is raising to the power MATH. To compute the MATH-linearization, it suffices to compute the action of MATH on the fiber of MATH over the origin, which is a fixed point for the MATH-action on MATH. The fiber over MATH is canonically MATH . ... |
math/0006174 | The morphism MATH defined by MATH is MATH-equivariant, where MATH acts with all weights equal to MATH on the domain and with weights MATH on the range. Thus there is an induced cover MATH, and it is easy to check that the degree of this cover is MATH. There is always a natural inclusion MATH, and one checks that this i... |
math/0006174 | This is immediate from REF , with MATH, MATH, and MATH. |
math/0006174 | There is a universal MATH-bundle MATH over MATH, which in fact arises from a universal MATH-bundle, which we shall also denote by MATH. One can describe MATH as follows. A MATH-bundle over MATH is the same thing as an element of MATH. The inclusion MATH induces an inclusion MATH and we take the image of the element MAT... |
math/0006174 | By REF , it clearly suffices to show that, for every MATH, we have an equality (under the obvious identifications) MATH . As such, this equality is a general fact about lattices MATH: suppose that MATH is a lattice and MATH is a homomorphism. There is an associated morphism MATH which we shall denote by MATH. We can de... |
math/0006174 | The proof is similar to that in the simply connected case for REF , and we shall be a little sketchy. Suppose that MATH is a flat MATH-bundle corresponding to the MATH-pair MATH. Let MATH be a fixed, general line bundle of degree zero on MATH. We compute when MATH is in the support of MATH. As we have seen in REF, MATH... |
math/0006174 | Let MATH be the MATH-form corresponding to MATH. First suppose that MATH is diagonalizable with respect to some MATH-basis for MATH, corresponding to a given isomorphism MATH. Then MATH is of the form MATH where MATH, say, MATH is projection onto the MATH factor, and MATH is the positive generator for MATH. In this cas... |
math/0006174 | Let MATH be the degree of the covering MATH. We see by REF that it suffices to prove that MATH which we can rewrite as MATH . This is exactly the statement of REF . |
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