paper stringlengths 9 16 | proof stringlengths 0 131k |
|---|---|
math-ph/9904013 | The proof is similar to the one in REF. Define the two subsets MATH and MATH of the interval MATH, MATH . The intersection of MATH with MATH is by definition empty, and the sets MATH and MATH are open, by continuity of the solution MATH as a function of the initial data MATH. We now show that all MATH sufficiently clos... |
math-ph/9904013 | Define the two subsets MATH and MATH of the interval MATH, MATH . By definition, the intersection of MATH with MATH is empty, and the sets MATH and MATH are open, by continuity of the solution MATH as a function of the initial data MATH. We now show that all MATH sufficiently close to MATH are in MATH and that all MATH... |
math-ph/9904020 | It is a well-known consequence of the NAME Vanishing Theorem (see for example, CITE) that we can find MATH such that if MATH and MATH with MATH for MATH, then there is a section MATH with MATH and MATH for MATH. We write MATH. Suppose on the contrary that MATH, and chose a nonzero vector MATH such that MATH. Then recal... |
math-ph/9904020 | The MATH-point zero correlation MATH is given by REF with MATH. By the NAME formula CITE, the expectation MATH in REF is a homogeneous polynomial (over MATH) of degree MATH in the coefficients of MATH. By REF , the coefficients of MATH are homogeneous polynomials of degree MATH in the coefficients of MATH. The conclusi... |
math-ph/9904020 | We consider the first NAME projector on the reduced NAME group MATH where MATH . (See the remark at the end of REF.) Its kernel can be written in the form MATH where the MATH form a complete orthonormal basis for MATH. (For example, MATH can be taken to be the set of monomials MATH. In fact, MATH is just a ``weighted N... |
math-ph/9904020 | By taking the scaling limit of REF , we obtain MATH . Indeed, since the coefficients of MATH are either of degree REF in the coefficients of MATH or of degree REF in the coefficients of MATH, we see by the proof of REF , using REF - REF , that the leading term of the asymptotic expansion of MATH is MATH times the right... |
math-ph/9904020 | We use REF , which comes from REF - REF as in the proof of REF . To determine the matrices MATH, we let MATH (instead of the right-invariant vector fields we used above). Recalling REF , we have: MATH . By REF , MATH . Recalling REF , we have MATH . We now apply REF ; note that the NAME formula involves terms that are ... |
math-ph/9904026 | If MATH satisfies the pseudoholomorphicity condition then: MATH . That MATH follows by complex conjugation. |
math-ph/9904026 | We perform all computations in a real anholonomic NAME basis. Consider the compatibility equation between the almost complex structure J, symplectic form MATH and metric MATH, viz. MATH. Taking the covariant derivative of both sides and by use of the compatibility equation we find that: MATH . Computing the covariant d... |
math-ph/9904026 | The first three equations are obtained from those of REF by direct substitution from REF . The final equation follows by substitution and by identity REF from REF. |
math-ph/9904028 | Since MATH is diffeomorphic to MATH, we have the NAME cohomology group MATH . The form MATH belongs to its second item which is zero. |
math-ph/9904028 | The proof is based on the relative NAME lemma CITE. |
math-ph/9904028 | Every connection MATH on MATH gives rise to the connection MATH on MATH which is a Hamiltonian connection for the frame Hamiltonian form MATH . Let us consider the decomposition MATH, where MATH is a connection on MATH. The assertion follows from the relation REF . |
math-ph/9904028 | Given locally Hamiltonian forms MATH and MATH, their difference MATH is a REF-form on MATH such that REF-form MATH is closed. By virtue of REF , the form MATH is exact and MATH locally. Put MATH where MATH is a connection on MATH. Then MATH modulo closed forms takes the local form MATH, and coincides with the pull-back... |
math-ph/9904028 | Given a Hamiltonian form MATH, its exterior differential MATH is a presymplectic form of constant rank MATH since the form MATH is nowhere vanishing. It is also seen that MATH. It follows that the kernel of MATH is a REF-dimensional distribution. Then the desired Hamiltonian connection MATH is a unique vector field MAT... |
math-ph/9904028 | The constrained NAME equations can be written as MATH . They differ from the NAME REF for MATH restricted to MATH which read MATH where MATH is a section of MATH and MATH is an arbitrary vertical vector field on MATH. A solution MATH of REF satisfies obviously the weaker REF . |
math-ph/9904028 | Let MATH be the universal enveloping algebra of the NAME algebra of the symmetry currents MATH, MATH, REF . Then each non-zero element MATH of its center of order MATH can be written as a polynomial in MATH, and defines the desired Hamiltonian form MATH. |
math-ph/9904028 | The relation REF takes the coordinate form MATH . Substituting REF , we obtain the relation REF . |
math-ph/9904028 | Since MATH exists, the structure group MATH of the vector bundle MATH is reducible to the subgroup MATH of general linear transformations of MATH which keep its MATH-dimensional subspace, and to its subgroup MATH. |
math-ph/9904028 | The map REF is a solution of the algebraic equations MATH . By virtue of REF , there exist the bundle slitting MATH and a (non-holonomic) atlas of this bundle such that transition functions of MATH and MATH are independent. Since MATH is a non-degenerate fibre metric in MATH, there exists an atlas of MATH such that MAT... |
math-ph/9904028 | By the very definitions of MATH and MATH, the Hamiltonian map REF satisfies REF . Then MATH is weakly associated with REF in accordance with REF . Let us write the corresponding NAME REF for a section MATH of the NAME bundle MATH. They are MATH . Due to the surjections MATH and REF , the NAME REF break in two parts MAT... |
math-ph/9904028 | Due to the splitting REF , we have the corresponding splitting of the vertical tangent bundle MATH of the bundle MATH. In particular, any vertical vector field MATH on MATH admits the decomposition MATH such that MATH is a vertical vector field on the Lagrangian constraint space MATH. Let us consider the equations MATH... |
math-ph/9904028 | Let MATH be the holonomic bases for MATH with respect to some bundle atlas MATH of MATH with transition functions MATH, that is, MATH. Then BRST functions read MATH where MATH are local functions on MATH, and we omit the symbol of an exterior product of elements MATH. The coordinate transformation law of BRST functions... |
math-ph/9904028 | The proof is based on inspection of the transformation laws of the holonomic coordinates MATH on MATH and MATH on MATH. |
math-ph/9904036 | From the mass-gap assumption we obtain MATH for all MATH, where MATH. We claim that the vectors MATH exist (in the sense of the functional calculus). Indeed, denoting the spectral measure of MATH by MATH, REF implies MATH . Thus, by the functional calculus, MATH where we used that MATH commutes with MATH. |
math-ph/9904036 | The proof proceeds in three simple steps. For simplicity of notation, we will give the proof only for the case MATH, the proof for MATH is obtained by analogous arguments. In view of translation covariance, it suffices also to consider only the case MATH. CASE: We will first show that for all MATH . To this end, let MA... |
math-ph/9904036 | We consider only the case MATH and MATH, the general case is proved analogously. Then we observe that MATH holds for all MATH as can be seen from REF together with the fact that MATH, MATH. REF defines a quadratic form MATH on MATH which is by assumption positive, that is, MATH, MATH. It follows that there is an essent... |
math-ph/9904038 | The structure of MATH allows to identify the NAME algebras over the different fields. Indeed, transitions MATH may be represented as transitions from the real coordinates in MATH to complex coordinates of the form MATH, where MATH is an additional basis element MATH (volume element). Since MATH is odd, then the volume ... |
math-ph/9904038 | Indeed, in accordance with REF MATH. Further, from REF we have MATH, and a square of the element MATH is equal to MATH, therefore MATH. Thus, MATH. For the group MATH a square of MATH is equal to MATH, therefore MATH, MATH is a double unit. As expected, MATH. The isomorphisms for the groups MATH and MATH are analogousl... |
math-ph/9904038 | We start the proof with a more general case of MATH. According to REF the volume element MATH belongs to a center of MATH (MATH), therefore, MATH commutes with all basis elements of this algebra and MATH. Further, recalling that a vector complex space MATH is associated with the algebra MATH, we see that basis vectors ... |
math-ph/9904038 | As known, the transformations MATH at the conditions MATH form an abelian group with the following multiplication table Analogously, for the automorphism group MATH in virtue of the commutativity MATH and the conditions MATH a following multiplication table takes place The identity of the multiplication tables proves t... |
math-ph/9904038 | First of all, since MATH if MATH and MATH if MATH, then in the case of MATH for the matrix of the automorphism MATH we have MATH . Over the field MATH we can always to suppose MATH. Further, let us find now the matrix MATH of the antiautomorphism MATH at any MATH, and elucidate the conditions at which the matrix MATH c... |
math-ph/9904038 | Indeed, over the field MATH in accordance with REF from all the fundamental automorphisms at the homomorphic mappings MATH and MATH only the antiautomorphism MATH is transferred into quotient algebras MATH and MATH. Further, according to REF the antiautomorphism MATH corresponds to time reversal MATH. Therefore, groups... |
math/9904001 | We observe that the composite map MATH verifies MATH. Since MATH is a linear isomorphism REF, we have MATH. Therefore MATH. |
math/9904001 | REF asserts that MATH is not empty and every irreducible component has dimension at least MATH. Suppose that there is an irreducible component MATH of dimension MATH. Then its inverse image MATH has dimension MATH, hence, since MATH is irreducible, MATH and MATH. The last equality can not happen, since otherwise, using... |
math/9904001 | We are going to define MATH as the residual divisor of the restricted divisor MATH, for a given point MATH and then show that it does not depend on the choice of MATH. We first observe that we have an equality of sets MATH which can be seen as follows: for MATH such that MATH the assumption MATH and the formula MATH im... |
math/9904001 | The equality follows from the commutativity of the right-hand square of the diagram MATH . The commutativity of the right-hand square follows from that of the outside square because MATH generates MATH. In other words we need to check the assertion of the lemma only for hyperplanes of the form MATH for MATH. This follo... |
math/9904001 | By the above Lemma we have MATH . If MATH is irreducible, then the support of one of the divisors MATH or MATH, say MATH, is contained in the support of MATH. This is impossible because MATH is the inverse image of a divisor in MATH and MATH is the inverse image of the codimension MATH support of MATH. |
math/9904001 | Suppose that MATH is reducible. Then a local computation shows that the hyperplane MATH is everywhere tangent to the branch locus of MATH. It is immediately seen that the branch locus MATH of MATH is the dual hypersurface of the canonical curve. The components of the singular locus MATH of MATH are of two different typ... |
math/9904001 | For a bi-elliptic curve MATH, the NAME locus MATH has two irreducible components, which are fixed by the reflection in MATH REF . For a smooth plane quintic this NAME locus is irreducible, ruling out REF. For a trigonal curve this NAME locus has two irreducible components, which are interchanged by reflection in MATH, ... |
math/9904001 | CASE: Given a bundle MATH and a line bundle MATH which is anti-invariant under MATH, that is, MATH, we have a natural non-degenerate quadratic form with values in the canonical bundle MATH where MATH is a local section of MATH. Note that we have canonical isomorphisms MATH . Therefore we are in a position to apply the ... |
math/9904001 | It will be enough to show that the canonical divisors MATH and MATH are equal for a general element MATH. In both cases the divisor coincide with the divisor MATH, where MATH is the unique effective divisor in the linear system MATH. The computations are straight-forward and left to the reader. |
math/9904001 | First we need to show that for a general semi-stable bundle MATH with MATH the divisor MATH is smooth at a general point MATH. For this decompose a general NAME divisor into two effective divisors of degree MATH, that is, MATH. Put MATH. Then MATH. If MATH, then MATH, MATH and MATH. At a general point MATH, we see imme... |
math/9904001 | Consider the invertible sheaf MATH and the corresponding embedding MATH . The curve MATH is the curve MATH translated by MATH. A straight-forward computation shows that MATH and by a result of NAME (see CITE page REF) the induced linear map on global sections MATH is surjective. We observe that MATH and that the projec... |
math/9904005 | It's clear that one may suppose that MATH is base point free. Taking the NAME of MATH, we can get MATH where MATH has connected fibers. A generic irreducible element of MATH is a smooth projective curve. If MATH is not composed of a pencil of curves, then it's sufficient to verify the birationality of MATH by virtue of... |
math/9904005 | We may suppose that MATH is basepoint free. Denote by MATH a generic irreducible element of MATH. Then the vanishing theorem gives the exact sequence MATH where MATH is a divisor of positive degree. It is obvious that MATH since MATH is a curve of genus MATH. The proof is completed. |
math/9904005 | First we take a birational modification MATH, according to NAME, such that REF MATH is smooth; REF the movable part of MATH defines a morphism; REF the fractional part of MATH has supports with only normal crossings. Denote by MATH a generic irreducible element of the movable part of MATH. Then MATH is a smooth project... |
math/9904005 | This is obvious according to REF . |
math/9904005 | Taking the same modification MATH as in the proof of REF , we still denote by MATH the general member of the movable part of MATH. Note that both MATH and MATH have the same movable part. For a given integer MATH, we have MATH . It is sufficient to prove the birationality of the rational map defined by MATH . Because M... |
math/9904005 | First we take the same modification MATH as in the proof of REF . We also suppose that MATH is the movable part of MATH. For a given integer MATH, we obviously have MATH . Thus it is sufficient to verify the birationality of the rational map defined by MATH . By REF , MATH is effective. Thus we only have to prove the b... |
math/9904005 | For all MATH, denote by MATH the movable part of MATH. By NAME 's method (CITE or see CITE), we have MATH. By the vanishing theorem, one has MATH . By REF , we see that MATH . Repeatedly performing this process, one has MATH for all integer MATH. This means that we can write MATH where MATH is an effective MATH-divisor... |
math/9904005 | Denote by MATH a generic irreducible element of the movable part of MATH. Then it's well-known that MATH is a smooth curve of genus REF. According to Claim in REF, we have MATH. For a positive integer MATH, we have MATH . Since MATH, taking MATH and applying REF , one has MATH. Taking MATH and applying REF once more, o... |
math/9904005 | We still denote by MATH a generic irreducible element of the movable part of MATH. It's well-known that MATH defines a generically finite map and MATH is a smooth curve of genus REF. By a parallel argument as in the proof of REF , we have MATH for any positive integer MATH. This means MATH where MATH is an effective MA... |
math/9904005 | Denote by MATH a generic irreducible element of the movable part of MATH. Recall that MATH is the contraction onto the minimal model. MATH is the movable part of MATH. It's easy to see that MATH has two types: CASE: MATH, where MATH and MATH is smooth curve of genus REF. CASE: MATH , where MATH is a smooth curve of gen... |
math/9904005 | We consider the natural map MATH where MATH is the image of MATH. Since MATH, MATH. CASE: MATH. In this case, MATH defines the bicanonical map of MATH. By CITE, MATH is base point free. Thus we see that MATH. We can write MATH, where MATH is an effective MATH-divisor. Denote by MATH a general member of MATH. Now we hav... |
math/9904005 | By CITE, MATH is basepoint free for all MATH. Denote by MATH the movable part of MATH. Then MATH. It's sufficient to prove the birationality of MATH. By NAME 's method, MATH. We only need to verify the birationality of MATH. The vanishing theorem gives MATH . Applying REF , we get MATH. Repeatedly proceeding the above ... |
math/9904005 | We keep the same notations as in the proof of REF . The proof is almost the same except that we have here MATH. Thus we can prove that MATH is birational by the same argument. This, in turn, proves the birationality of MATH. We conclude the claim. |
math/9904005 | We can suppose MATH is a free pencil. Otherwise, we can blow-up MATH at base points of MATH. Denote MATH. Then MATH. Suppose MATH. Then MATH. Because MATH, we can see that MATH. From MATH, we get MATH, that is, MATH. Thus MATH and MATH. This means MATH by virtue of CITE, which is impossible because MATH. So MATH. |
math/9904006 | For each pair MATH the MATH-vector space MATH is graded. Let MATH be the homogeneous component of degree MATH, and MATH . This is a connected algebraic variety. Since MATH is generated as MATH-algebra by the idempotents and the arrows, and the only relations in MATH are the monomial relations arising from incomposabili... |
math/9904006 | CASE: Since MATH is basic we have MATH. By NAME theory we have MATH. Any auto-equivalence of the category MATH extends to an auto-equivalence of MATH (using projective resolutions), and this induces an isomorphism of groups MATH. The class of auto-equivalences MATH is actually a group here. In fact MATH can be identifi... |
math/9904006 | Let MATH be a MATH-linear triangle auto-equivalence of MATH. By REF there exits a two-sided tilting complex MATH with MATH in MATH. Replacing MATH with MATH we may assume that MATH. Hence MATH, and MATH is an equivalence. NAME theory says that MATH for some invertible bimodule MATH. So replacing MATH by MATH we can ass... |
math/9904006 | The group homomorphism MATH is injective, say by REF , and it is surjective by the theorem. |
math/9904006 | CASE: This is implicit in REF and I. REF. In particular REF shows that for any indecomposable object MATH the ring MATH is local. CASE: See REF . According to REF , for each MATH there exists such an NAME triangle. By REF these are all the NAME triangles, up to isomorphism. CASE: Since source and sink morphism depend o... |
math/9904006 | This is essentially REF . |
math/9904006 | Consider a sink morphism in MATH ending in MATH, MATH. By REF , it is of the form MATH with MATH (compare Notation REF). From the definition of a sink morphism we see that this is also a sink morphism in the category MATH. According to REF (dual form), both MATH-modules MATH and MATH have the morphisms MATH as basis. A... |
math/9904006 | Let MATH be the full subquiver with vertex set MATH. Given a vertex MATH in MATH, denote by MATH the number of its predecessors, that is, the number of vertices MATH such that there is a path MATH in MATH. For any MATH let MATH be the full subquiver with vertex set MATH. MATH is a translation quiver with polarization, ... |
math/9904006 | The fact that MATH is in the center of MATH is trivial. As for MATH, this follows immediately from REF (or by REF , since MATH is the NAME functor of MATH). |
math/9904006 | Given an auto-equivalence MATH of MATH, the formula MATH iff MATH defines a permutation MATH of MATH that preserves arrow-multiplicities. Hence it restricts to a permutation of MATH. By REF , MATH commutes with MATH and MATH. |
math/9904006 | Let MATH. By REF we know that MATH. Hence by REF , MATH iff MATH acts trivially on the set MATH. In particular we see that MATH. Now use REF . |
math/9904006 | According to REF, the group MATH is abelian in all cases except MATH. But a direct calculation in this case (compare REF ) gives MATH. |
math/9904006 | Since MATH we get a permutation MATH. Let's prove that MATH commutes with MATH in MATH. Consider a vertex MATH. In the Notation REF, there are vertices MATH and irreducible morphisms MATH and MATH that form bases of MATH and MATH respectively. Since we have MATH this must be a multiple of a mesh relation. Hence MATH. F... |
math/9904006 | Choose an equivalence MATH as in REF . If MATH has infinite representation type then the isomorphism MATH we have chosen (as in REF ) tells us how to extend MATH to an equivalence MATH that commutes with MATH (compare REF ). Let MATH be a triangle auto-equivalence of MATH. Then MATH induces a permutation MATH of the se... |
math/9904006 | The proof has three parts. CASE: We show that the homomorphism MATH of REF is injective. Let MATH be a two-sided tilting complex such that MATH. Then the permutation MATH fixes the vertices of MATH. Using the fact that MATH we see that MATH in MATH. Replacing MATH with MATH we may assume MATH is a single bimodule. Acco... |
math/9904006 | CASE: By REF the homomorphism MATH is surjective. REF identifies MATH. CASE: If MATH has finite representation type then MATH is a tree, so MATH by REF . By REF we get MATH . CASE: If MATH has infinite representation type then MATH by REF . We know that MATH is in the center of MATH. |
math/9904006 | The isomorphisms are by REF . The data in the third column of REF was calculated in REF, except for the shift MATH which did not appear in that paper. So we have to do a few calculations involving MATH. Below are the calculations for types MATH and MATH; the rest are similar and are left to the reader as an exercise. T... |
math/9904006 | The isomorphisms follow from REF . The structure of MATH is quite easy to check in all cases. In type MATH, MATH odd, the automorphism MATH is MATH . |
math/9904006 | As in the proof of REF , the group of auto-equivalences of the path category is MATH. Hence MATH. Given MATH let MATH be its matrix with respect to to the basis MATH, and let MATH. Define an auto-equivalence MATH with MATH and MATH, MATH. Then MATH preserves all mesh relations, and by a linear algebra argument we see t... |
math/9904006 | CASE: This is because MATH. CASE: Here the group of auto-equivalences of MATH is, in the notation of the proof of REF , MATH, and the group of isomorphisms is MATH. Therefore MATH is isomorphic to MATH as varieties, and as matrix group MATH. The auto-equivalence associated to MATH is MATH and MATH. The quiver MATH has ... |
math/9904006 | There is an NAME sequence MATH in MATH. Applying the functor MATH to this sequence, and using REF , we get a triangle MATH in MATH. Hence MATH . On the other hand for MATH we have MATH. This proves that MATH; but MATH. |
math/9904006 | For an orientation MATH let MATH be the quiver of REF . As usual MATH denotes the set of vertices of MATH. Let MATH be the groupoid with object set MATH, and morphism sets MATH for MATH. The groupoid MATH acts faithfully on the family of sets MATH. According to REF there is an injective map of groupoids MATH. Let us fi... |
math/9904006 | We will only treat the NAME case; the general case is proved similarly with modifications like in the previous proof. Let MATH. From the proof above we see that MATH for some MATH. A quiver map MATH with MATH must have MATH for all MATH, since MATH is a tree. Therefore MATH. |
math/9904008 | See REF or REF. |
math/9904008 | See REF . |
math/9904008 | For MATH, let MATH be the set of MATH such that MATH and MATH. By definition, MATH . As MATH is smooth of pure dimension MATH over MATH, NAME 's lemma implies that MATH . Consequently, MATH . As MATH, we have MATH . Now, MATH therefore MATH . Finally, MATH, hence MATH . |
math/9904008 | By definition, MATH . We compute these sums separately. The integral over MATH is equal to MATH. Then MATH . Concerning the integrals over MATH, we have MATH . Finally, MATH . Adding all these terms gives MATH . |
math/9904008 | Indeed, we have MATH . The integral of a non-trivial character over a compact group is MATH, hence this integral equals MATH if MATH, equals MATH if MATH and equals MATH if MATH. This proves the lemma. |
math/9904008 | The inequality is trivially true for MATH. We prove it for any MATH by induction: to lift a point in MATH to a point in MATH, one needs to solve two equations in MATH: MATH . A point in MATH which reduces to a point in MATH modulo MATH has MATH lifts in MATH. On the other hand, a point reducing to a point in MATH has M... |
math/9904008 | The set MATH is defined by the two equations MATH. Fix the coordinates MATH so that MATH is the first vector. Up to a constant, one may write MATH for some homogeneous polynomials MATH of degree MATH. Then, denoting MATH, MATH is defined by the equations MATH . On MATH, MATH and on MATH, MATH. As MATH is smooth, MATH d... |
math/9904008 | We prove this by induction on MATH. If MATH, the result is clear. Then, one can assume that MATH is reduced, irreducible and not contained in any hyperplane. For any hyperplane MATH which is rational over MATH, MATH is a closed subscheme of MATH of dimension MATH and of degree MATH. By induction, we have MATH . Finally... |
math/9904008 | Indeed, we see from REF that for MATH, MATH the MATH being uniform in MATH. Consequently, MATH converges to a holomorphic bounded function on MATH. As the finite number of remaining factors converge uniformly in MATH, the existence of MATH is proven. The growth of MATH in vertical strips follows from NAME 's estimates ... |
math/9904008 | Recall REF : MATH the right hand side of which we have to estimate all terms. The first one is MATH. Then, as MATH is bounded, the second one is MATH . For the last term MATH, we use REF so that, denoting MATH, MATH . Moreover, MATH so that MATH . The lemma is proved. |
math/9904008 | Write MATH . The convergence of the first infinite product to a bounded holomorphic function follows from the preceding lemma. As in REF, there exists a constant MATH such that MATH . Using the rapidly decreasing behaviour of MATH as a function of MATH established in REF, the proposition is proved. |
math/9904014 | We use an algebraized version of arguments in CITE. CASE: Consider the double filtration for MATH. Since MATH and MATH, the sum in the definition is actually direct. Obviously, MATH. It follows from the definition of double filtration that MATH. Therefore MATH . Now, an easy summation proves that MATH. Thus, MATH . Sin... |
math/9904014 | It is already proved that the inclusion ``MATH" holds. Since MATH is NAME and the pair MATH is not principal, it follows from CITE that MATH. Then the assertion follows for dimension reason. |
math/9904014 | By symmetry, it suffices to prove the first half of each item. The proof applies to both pn- and almost pn-pairs. CASE: This part is essentially the same as in CITE. Consider the MATH-limit: MATH, which lies in MATH. Since different summands have different weights relative to MATH, the sum is direct and therefore MATH.... |
math/9904014 | CASE: For CASE : Suppose that MATH, that is, all the eigenvalues of MATH in MATH are integral. We need to prove here that the case MATH is impossible. Assume not and MATH, MATH. A standard calculation with the Killing form on MATH shows that MATH if and only if MATH. By definition, put MATH. For each MATH, consider the... |
math/9904014 | Define MATH by MATH. It is an inner automorphism of MATH. Then MATH, MATH, and MATH. As MATH contains no semisimple elements, MATH is semisimple. |
math/9904014 | CASE: Let MATH be the minimal element in MATH with respect to the lexicographic ordering. Then MATH, while MATH (MATH). It is not hard to prove that MATH, but we do not need this. CASE: Now the eigenvalues of MATH are integral and the bi-weights of MATH lie in the upper half-plane. The same argument as in CITE shows th... |
math/9904014 | In view of REF can be restated as: MATH and MATH. Assume now that MATH is special, that is, MATH for some MATH. Then MATH and MATH. A contradiction! |
math/9904014 | CASE: By REF , any almost pn-pair of non-MATH-type yields an inner involution MATH such that MATH is semisimple. But MATH has no such involutions. CASE: Since all nilpotent orbits in MATH are NAME and hence special, there are no almost pn-pairs of MATH-type as well. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.