paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0009011 | Note that MATH, as there are no further differentials in this part of the spectral sequence. By REF , MATH, which is MATH by REF . |
math/0009011 | We need the fact that an extension MATH is NAME if and only if MATH. This follows from REF . We have shown that MATH is obtained from MATH by adjoining the square roots of elements in MATH, while MATH is obtained by adjoining the square roots of elements in MATH. Therefore we have MATH if and only if MATH. To show that... |
math/0009011 | Since MATH is not a NAME, by CITE there exists an anisotropic binary form MATH over MATH such that the set MATH of values of MATH, regarded as a subgroup of MATH, has at least three elements. (As MATH is an elementary abelian MATH-group, so are any of its subgroups, so ``MATH has at least three elements" immediately im... |
math/0009011 | Recall that we have the exact sequence MATH arising from the long exact sequence in cohomology associated to the short exact sequence of coefficients MATH. Note that since the action of MATH on the coefficient group MATH is trivial, we may identify MATH with MATH. In other words MATH. We shall make this isomorphism exp... |
math/0009011 | We have shown that for MATH-fields, MATH, so that MATH and MATH. So we must show that MATH. From the isomorphism between NAME cohomology and the NAME ring given by the NAME conjecture, it is enough to show that MATH and that MATH, where MATH is the MATH-th power of the fundamental ideal in the NAME ring. Ware CITE has ... |
math/0009011 | Suppose that MATH. Let MATH. Then MATH is an open subgroup of MATH and MATH. Therefore from the above we obtain MATH. However from NAME theory we know that MATH. |
math/0009011 | From local class field theory (see CITE) we know that there is a natural isomorphism MATH for any intermediate extension MATH. If MATH we obtain the first statement. For MATH we observe (see CITE, page REF) that for any proper quadratic extension MATH of MATH, MATH; however MATH. We conclude that MATH cannot be contain... |
math/0009011 | Let MATH denote an extension of MATH such that MATH. Consider MATH; then by restriction MATH will also be a module over this group. Denote the usual module-theoretic norm map by MATH. From the definition of the socle series for MATH it follows that MATH, where the intersection is taken over all extensions as above, of ... |
math/0009011 | From our description of MATH and the triviality of the map induced by the norm, we infer that MATH, whence MATH. Taking a quadratic extension MATH we see that in the socle series for MATH as a MATH-module, MATH. Hence MATH. |
math/0009011 | Let MATH denote a basis for the subgroup MATH. For each MATH one can construct a REF - dimensional real representation of the NAME subgroup of MATH which restricts non-trivially to MATH. This can then be induced to yield a MATH-dimensional representation MATH for MATH. Doing the analogous construction for MATH and pull... |
math/0009011 | Suppose that MATH is as in the statement of our Proposition. In order to show that MATH it is enough to observe that for each monomial MATH that can occur in the expression of the left hand side as an element of MATH, the corresponding coefficient is zero. This can be proved using a case by case analysis, in which divi... |
math/0009015 | We have to show that MATH respects the relations REF , in other words, MATH maps equivalent sums of triples to equivalent ones. It is trivial with REF . To check REF , let us recall REF . Consider a sum of triples MATH belonging to REF , that is MATH, and MATH on the smooth part of MATH. Since the irreducible component... |
math/0009015 | We need to prove this for triples MATH, that is, for forms MATH with normal crossing divisors of poles. The repeated residue at pairwise intersections differs by a sign according to the order in which the residues are taken, see REF. Thus the contributions to the repeated residue from different components cancel out (o... |
math/0009015 | By the NAME duality, MATH is the map MATH and it is sufficient to verify that the pairing REF vanishes if MATH and MATH, or if MATH and MATH. This follows immediately from the NAME - NAME REF : MATH that is MATH. |
math/0009015 | As we have already mentioned, the homomorphism MATH of REF can be conveniently described in terms of the following natural map of polar chains: MATH where MATH is the space of currents of degree MATH (that is, a space of linear functionals on smooth MATH-forms, see CITE). As a matter of fact, this map is described by t... |
math/0009015 | A straightforward verification. |
math/0009015 | This will be similar to the proof of REF and will use the same notations. We first represent the polar cycles MATH and MATH by the currents MATH and MATH respectively, then MATH where MATH on the right is understood as taking the MATH-cohomology class. On the other hand, for MATH introduced in REF , the current represe... |
math/0009023 | This follows from REF . Write the orthogonal projection of MATH onto the span of MATH as the linear combination MATH. Then MATH, which proves REF . Similarly, MATH. This proves REF with MATH. If MATH are linearly independent then the representation MATH is unique. |
math/0009023 | Since MATH are bounded random variables, to prove REF we need only to verify that for arbitrary polynomials MATH, MATH . This is equivalent to MATH see REF . The latter follows from REF , proving REF . To prove REF , we verify that for arbitrary polynomials MATH we have MATH . By REF , this is equivalent to MATH . It s... |
math/0009023 | Using the definition of vacuum expectation state, REF we get MATH. Therefore REF , and REF imply MATH . The latter is zero, except when MATH or MATH. We will consider these two cases separately. If MATH, by orthogonality we have MATH . Clearly, MATH; this can be seen either from REF , or directly from REF . By REF the ... |
math/0009023 | Since the case MATH is well known, we restrict our attention to the case MATH. Our starting point is REF . A computation shows that the conditional variance MATH is as follows. MATH . The right-hand side of this expression must be non-negative over the support of MATH. It is known, see CITE or CITE, that MATH have the ... |
math/0009024 | We follow the proof of REF. Let MATH be the canonical decomposition of the restriction of MATH to MATH into a direct sum of isotypic representations. Since the MATH-module MATH is simple, MATH permutes the MATH transitively. If MATH is some MATH, then the MATH-module MATH is isotypic and REF holds. Assume from now on t... |
math/0009024 | The MATH-module MATH splits into a direct sum of MATH-modules MATH such that the restriction of MATH to MATH is non-degenerate, and either MATH is simple or MATH where MATH is simple. In the latter case choose a MATH-stable MATH-lattice MATH in MATH and apply REF . |
math/0009024 | Let MATH be the non-zero pairing induced by MATH. Its kernel is a MATH-submodule of MATH, so is zero by the MATH-simplicity of MATH; that is, MATH is non-degenerate. By NAME 's Lemma, MATH is perfect. |
math/0009024 | Let MATH choose a section MATH, and let MATH . Note that MATH is independent of the choice of section MATH. Define a homomorphism MATH by MATH for MATH, MATH, MATH. Then the desired conditions are all satisfied. |
math/0009024 | By REF there is an injective homomorphism MATH. Now apply REF to MATH. |
math/0009026 | The proof is by induction on MATH. CASE: MATH. By REF , MATH and MATH are adjacent regions. Let MATH be the common facet of the closures of MATH and MATH and let MATH be the affine span of MATH. Since functions MATH are continuous, MATH for all MATH and therefore for all MATH. We may assume that MATH (the other case is... |
math/0009026 | Let MATH be a central point in MATH. For MATH, let MATH be the unique intersection point of the ray from MATH through MATH with MATH. We define MATH . Clearly, MATH is a piecewise linear function on MATH and MATH. Thus MATH admits representation REF . |
math/0009027 | Choose a regular value MATH such that the symplectic reduction at MATH is defined. Define induced horizontal and vertical spaces at MATH by the intersections with MATH: MATH . By definition, MATH. As MATH is Abelian, MATH and MATH, so that MATH. Using the following set-theoretical identity MATH if MATH, we obtain MATH ... |
math/0009027 | The proof readily follows from the fact that the momentum map factors through the quotient map, so that MATH, and the definition of the map MATH for any MATH given by REF , where MATH is any vector in MATH that satisfies MATH, with MATH and MATH: MATH . |
math/0009027 | For any tangent vector MATH and any NAME algebra element MATH : MATH where MATH for some MATH is a vertical (fiber) component, MATH is a horizontal component, and MATH by definition. By definition of the momentum map MATH, so that MATH where MATH is thought of as a MATH-valued one-form on MATH and we have used the fact... |
math/0009027 | Using REF , MATH and omitting MATH for simplicity we have MATH where for the last equality we used the definition of a symplectic form and considered the one-form MATH as a tangent map MATH acting on vectors in MATH. |
math/0009027 | The proof follows readily from REF of the previous corollary and the MATH-orthogonality of MATH and MATH: MATH where we used that MATH for a subspace MATH. |
math/0009027 | Using REF we obtain MATH: MATH . |
math/0009027 | Consider any vectors MATH and MATH. By definition of the induced metric MATH where MATH are horizontal components of the pre-images: MATH, and MATH. From MATH it follows that MATH. But MATH, so that by REF MATH . For the vector MATH it holds MATH and, hence, by the commutativity of the diagram in REF , MATH for any of ... |
math/0009027 | By the definition of the momentum map, MATH is a Hamiltonian for the vector field MATH of the infinitesimal transformations, that is, for any vector MATH . The one-form MATH can be thought of as the tangent map MATH acting on vectors in MATH and paired with MATH. Take MATH to be MATH for some infinitesimal generator MA... |
math/0009027 | The proof is quite straightforward and relies on the constructions discussed in this section. Using the definition of the transverse derivative, REF and MATH-invariance of the abstract locked inertia tensor and the almost complex structure we obtain MATH where the last equality follows from REF . As it was pointed out ... |
math/0009027 | We start with the definition of the two-form MATH above and shall demonstrate that the following three special cases hold for any MATH : CASE: MATH, here MATH is the characteristic distribution and MATH, CASE: MATH, CASE: MATH for any MATH and MATH, which all together prove the statement of the theorem, according to th... |
math/0009034 | REF is trivial. If MATH is MATH-absorbing, there exists MATH such that MATH. Then, of course MATH. Thus MATH . Since MATH is convex we obtain MATH . But this is true for every MATH. Hence we can take limits: MATH which gives MATH and by REF , MATH is norming. That REF is trivial. To see that REF does not imply REF does... |
math/0009034 | If MATH is norming for MATH, MATH for some MATH. Thus MATH and MATH is invertible. If MATH and MATH are isomorphic by MATH then, for some MATH, MATH . But, by the construction of MATH, MATH for all MATH. We use that MATH inductively for a constant MATH: MATH . Continuing this way gives after MATH steps MATH . This is t... |
math/0009034 | REF clearly implies REF . Suppose MATH is thin. Since MATH is thin we can pick a countable, increasing covering, MATH of MATH, consisting of sets which are non-norming for MATH, and a sequence MATH such that MATH but MATH. Let MATH be an arbitrary element of MATH. Then there is a natural number MATH such that MATH. Thu... |
math/0009034 | Suppose MATH is covered by an increasing family MATH. Since MATH is of second category some MATH, contains a ball. Then MATH contains a ball centered at the origin, and hence MATH is norming. Since MATH was arbitrary, MATH must be thick. |
math/0009034 | REF implies of course REF . The proof that REF implies REF is completely analog to the corresponding part of the proof of REF . The proof that REF implies REF is also very similar to the corresponding part of the proof of REF . Just put MATH . |
math/0009034 | Of course REF implies REF implies REF . To show that REF implies REF suppose REF is not true, that is, MATH is thin. We will construct a NAME injection which is onto MATH but not onto all of MATH. Let MATH be an increasing family of subsets of MATH such that MATH. Since MATH, we may assume each MATH to be contained in ... |
math/0009034 | Only the implication MATH needs proof since the remaining implications are trivial. Suppose MATH is thin. Write MATH an increasing union of sets which are non-norming for MATH. Then MATH, an increasing union of sets. Since MATH is thick, there exists a number MATH and a MATH such that MATH. Thus MATH . Thus MATH . This... |
math/0009034 | That REF implies REF is trivial. To show that REF implies REF we make necessary adjustments in the corresponding proof of REF . First substitute MATH's with MATH's. Then define MATH by MATH-closure. Note that MATH is now a non-MATH-norming set. Define MATH to be the MATH-closure of MATH. Two cases must be considered: C... |
math/0009034 | We will use REF . Let MATH be a sequence in MATH such that MATH for all MATH and all MATH. Since MATH is MATH- thick we conclude that MATH for all MATH. Thus MATH for all MATH and every MATH. Since MATH is MATH- thick, MATH for all MATH and the result follows since MATH is thick. |
math/0009034 | CASE: Note that the restriction to a subspace MATH of a NAME boundary is a NAME boundary. CASE: Put MATH. By NAME 's generalization of the NAME lemma, there is a sequence MATH on MATH which converges weakly to MATH. By the NAME selection principle MATH can be assumed to be a basic sequence. Let MATH. We look for MATH i... |
math/0009034 | Since MATH and MATH does not contain a copy of MATH the sets MATH and MATH are both MATH-thick. Hence, by REF , MATH is MATH-thick. |
math/0009034 | Since MATH is a separable dual it has the RNP and thus doesn't contain a copy of MATH. By the main result from CITE the sets MATH and MATH are both MATH-thick. Hence, by REF MATH is MATH-thick. But by CITE MATH is exactly the set of exposed points of MATH. |
math/0009034 | Use that MATH. |
math/0009048 | If a simply degenerate honeycomb has a loop, we can breathe it in and out; one direction will increase the value of any (generic) superharmonic functional. A largest lift is by assumption already at the maximum value of the functional, so there can be no loops. |
math/0009051 | It is straightforward to see that the intersection of any number of the MATH is smooth. This implies that MATH is smooth, and it is clearly MATH-equivariant. It will now suffice to prove that MATH. We claim that every MATH for MATH is a proper preimage of some MATH for MATH that is not contained in MATH and vice versa.... |
math/0009051 | First, it is easy to see that the MATH stratification is conical. Moreover, this can be said about any stratification induced by some connected components of fixed point sets of some subgroups of MATH. Indeed, for any MATH there exists a MATH-equivariant isomorphism of a neighborhood MATH and a neighborhood of the orig... |
math/0009051 | The statement is clearly local in MATH, so it could be assumed that MATH is a neighborhood of the origin in a MATH-vector space MATH. Denote by MATH and MATH the sets of strata in the MATH and stabilizer stratifications respectively. According to CITE the maximal blowups MATH and MATH can be described as closures of th... |
math/0009051 | Combine REF . |
math/0009051 | Suppose MATH has points with non-solvable stabilizers. Consider the point MATH with a non-solvable stabilizer MATH that lies in the minimum number of exceptional divisors of MATH. If this minimum number is zero, then MATH has a point with non-solvable stabilizer that lies outside the image of the exceptional divisors. ... |
math/0009051 | In view of REF , one needs to show that every point MATH with non-solvable stabilizer is contained in an irreducible stratum of codimension bigger than one. Since all MATH have codimension at least two, it remains to observe that each point MATH with a non-solvable stabilizers is contained in MATH. |
math/0009055 | The ``only if" direction is clear. For the ``if" direction choose disjoint neighborhoods MATH, each with a Riemannian metric coming from REF for every critical point of MATH. Now choose finitely many contractible open sets MATH with MATH-critical points-MATH that together with the MATH's cover MATH. Using REF., it is e... |
math/0009055 | It is to be shown that MATH vanishes on boundaries. So let MATH satisfy MATH, then MATH since MATH and MATH is a homomorphism. |
math/0009055 | CASE: We need to show that MATH is a well defined matrix over MATH. Note that MATH. We will look at terms of the form MATH and get an estimate for MATH. The idea is to write MATH as a word MATH where the length of the word MATH is smaller than MATH and the words MATH are of the form MATH. So assume that MATH where MATH... |
math/0009055 | As before denote the image of MATH in MATH by MATH. Then MATH . It is sufficient to show that MATH . Both sides are clearly MATH for MATH with MATH. Call the left side of REF MATH and let MATH with MATH. Then MATH . Let MATH act on MATH by the cycle MATH and on MATH by rotation. For MATH denote by MATH the orbit of MAT... |
math/0009055 | Let MATH be a triangulation of MATH which has MATH as a subcomplex MATH. This induces a triangulation MATH of MATH which has two copies of MATH as subcomplexes. Denote the one corresponding to MATH by MATH and the one corresponding to MATH by MATH. Assume MATH has the following properties: CASE: MATH is adjusted to MAT... |
math/0009055 | We have chosen a basis of MATH by choosing MATH-cells in MATH, call these cells MATH. If MATH supp MATH, then there exist negative flowlines from MATH to MATH by the construction of MATH. We need to show that there exists a MATH with the property that given MATH and indices MATH and MATH supp MATH, MATH supp MATH supp ... |
math/0009055 | Since MATH we can form the MATH complex MATH from the proof of REF which is simple homotopy equivalent to MATH. The matrices MATH are MATH-regular by REF , so the projection of MATH coker MATH is a chain homotopy equivalence. We have already seen that the boundary homomorphisms of MATH and MATH are the same when viewed... |
math/0009055 | There exists a filtration MATH of MATH such that MATH is a compact cobordism containing all critical points of index MATH and such that MATH and the boundary homomorphism comes from the long exact sequence of the triple MATH, see CITE. Also, MATH, where MATH denotes the MATH-skeleton of the triangulation. A simplex MAT... |
math/0009055 | Let MATH be a homotopy between MATH and MATH. As above we can change MATH to a homotopy MATH between MATH and MATH such that MATH intersects MATH transversely for all critical points MATH with MATH, where MATH is a MATH-simplex. Describe the NAME complex as in the proof of REF . Then we define MATH by MATH. Here MATH i... |
math/0009055 | Let MATH be a subdivision of MATH. If MATH is adjusted to MATH, so is MATH. Moreover, the diagram MATH commutes, where sd is subdivision, a simple homotopy equivalence. By CITE it is good enough to show the theorem for a special smooth triangulation. As in the proof of REF we have the filtration MATH where MATH is a co... |
math/0009055 | For MATH choose triangulations MATH of MATH adjusted to MATH and let MATH be the flow of MATH. Then there is a MATH such that REF. is satisfied for MATH. Extend MATH to a triangulation MATH of MATH. Choose a diffeomorphism MATH so close to the identity that MATH is adjusted to MATH and REF. and REF. still hold. Modify ... |
math/0009061 | The second statement follows immediately from the first, since MATH is a domain. To prove that MATH, we first show that MATH is a binomial ideal. We can factor the map MATH as follows. Let MATH and MATH be defined like MATH, except that if MATH, then it appears as an exponent of the variable MATH instead of MATH. Then ... |
math/0009061 | The proof is similar to that of CITE. It was shown in the proof of the previous theorem that any binomial in MATH is of the form MATH for some MATH. Hence the NAME basis MATH of MATH is of the form MATH . We first show that the set MATH is a generating set for MATH. Suppose not, so there exists MATH which is not a MATH... |
math/0009061 | According to CITE the dimension of MATH is equal to the number of linearly independent column vectors in the matrix MATH . The theorem now follows. |
math/0009061 | Obviously, every solution of REF is also a solution of REF . Conversely, let MATH be a solution of REF . Then MATH and MATH for MATH . Taking into account that MATH is a linear operator, we conclude that MATH on the right-hand side of REF is non-negative. |
math/0009061 | It is enough to show that the above REF form a NAME basis of the ideal MATH. To compute MATH, one can use the Symmetry Component Algorithm, with any computer algebra system. Here we are using the specialized system CITE which performs NAME basis calculations substantially faster than most general purpose symbolic calcu... |
math/0009061 | Similar to the proof of REF . To simplify notation we rename the variables MATH and MATH: MATH . REF is the NAME session used to find MATH. |
math/0009061 | Similar to the proof of REF . |
math/0009061 | It is sufficient to show that the ideal MATH is prime in MATH. Consider the ring homomorphism MATH defined by MATH. It is sufficient to prove that MATH . It is clear that MATH. We will show the other inclusion by induction. Let us suppose that MATH and MATH is linear in MATH, that is, MATH . Then MATH where MATH. There... |
math/0009061 | It is easy to check (using, for example, the radical membership test, see for example, CITE) that MATH . Thus we only have to show that MATH for all MATH and MATH . Indeed, the systems corresponding to the points of MATH are Hamiltonian. For the variety MATH in the case MATH one can easily find an invariant conic MATH ... |
math/0009061 | Computing the intersection of the ideals MATH we find MATH . Hence MATH is radical because due to REF MATH are prime, and according to REF MATH is prime as well. Similarly, for quadratic system we easily check that, for MATH, the ideals MATH are prime and MATH . This yields that the ideal of focus quantities of quadrat... |
math/0009061 | CASE: MATH REF and the ideal MATH REF is a radical ideal. Therefore MATH . |
math/0009066 | If MATH has all MATH non-negative and some MATH, then MATH meets the conditions for the Descent Axiom to apply, so we have MATH . But by REF , we have MATH . |
math/0009066 | This follows from repeated application of the Descent Axiom and REF . |
math/0009066 | The convexity axiom of a MATH-spin virtual class (compare CITE) gives that MATH where MATH and MATH is the MATH-th root sheaf of the universal MATH-spin structure. The MATH-th root sheaf MATH of the universal MATH-spin structure of type MATH is isomorphic to MATH under the identification MATH . Since MATH is a subsheaf... |
math/0009066 | Consider the exact sequence of sheaves on MATH . Since MATH and MATH, the corresponding long exact sequence for the functor MATH gives a short exact sequence of bundles on MATH . Now the statement of the lemma follows from multiplicativity of the total NAME class. |
math/0009079 | Let MATH. For a condition MATH, let MATH be the set of all accumulation points of MATH. Clearly, MATH . Let MATH be a generic filter. Then MATH is generic in MATH. If MATH then there is some MATH so that MATH. So MATH and MATH. Thus MATH is closed under finite intersections. Clearly, MATH is upwards closed. Thus MATH i... |
math/0009079 | Since MATH is a complete subalgebra of MATH, the universe MATH is contained in MATH. Therefore, there is an onto function MATH in MATH. Since MATH is MATH-complete, MATH is preserved in MATH. Thus, the cardinality of MATH in MATH is MATH. |
math/0009079 | By REF there exists a dense subset of MATH which is isomorphic to the dense subset MATH of MATH, namely, there are conditions MATH so that MATH. Let MATH. Define MATH. It should be clear that MATH is a maximal antichain in MATH. Let MATH be arbitrary. By density of MATH there exists some MATH with domain MATH so that M... |
math/0009083 | Recall from CITE that the map MATH defines a group morphism MATH . It follows that MATH if and only if MATH, that is, if MATH is a MATH-th root of unity. |
math/0009083 | As a first step, we will prove that MATH. In order to see this, recall from deformation theory that - after shrinking MATH, if necessary - MATH is of the form MATH where MATH is a function MATH. In particular, if MATH is the non-normal locus and MATH its preimage in the normalization, then MATH is a unit disc and MATH ... |
math/0009083 | Let MATH be a line bundle such that MATH; the existence of MATH is guaranteed by REF . Note that MATH is locally free of rank one. Thus, after shrinking MATH if necessary, a section MATH exists whose restriction to any fiber of MATH is not identically zero. But since the relative degree of MATH is one, the restriction ... |
math/0009083 | Let MATH be any point and MATH a small unit disk centered about MATH. The preimage MATH will then be isomorphic to MATH. Let MATH be the MATH-osculating section whose existence is guaranteed by REF and find an isomorphism MATH such that MATH. Apply REF to see that MATH is then given as MATH . Hence the claim. |
math/0009083 | Choose a unit disk MATH centered about MATH and equip MATH with a coordinate MATH. Then MATH is a cuspidal curve. After shrinking MATH we may assume that MATH is the only point in MATH whose preimage is cuspidal. By REF , we can find an index MATH such that MATH. Therefore we can choose a bundle coordinate on MATH so t... |
math/0009083 | It follows immediately from the construction of the elementary transformation that the intersection number between the strict transforms of MATH and MATH drops exactly by one with each transformation in the sequence MATH. In particular, the strict transforms MATH, MATH of MATH and MATH are disjoint. Likewise, it follow... |
math/0009085 | MATH for MATH by the naturality property of the first obstruction. |
math/0009085 | Since MATH is a proper oriented submanifold MATH has a NAME dual MATH in MATH. Therefore MATH. On the other hand MATH (if follows from the analogous statement for non equivariant cohomology see for example, in CITE). |
math/0009085 | Let MATH denote a tubular neighborhood of MATH in MATH. Replacing MATH with its maximal compact subgroup doesn't change the equivariant cohomology groups, so we can assume that MATH is a MATH-equivariant subset of MATH. Now looking at the MATH-equivariant NAME sequence of MATH: MATH we can see that the Lemma is equival... |
math/0009085 | Let MATH. By REF is equivalent with the statement that MATH is equal to MATH where MATH . So it is enough to show that the inclusion MATH induces an injection in degree MATH. From the relative cohomology exact sequence MATH we can see that it is enough to show that MATH. By excision MATH where MATH and by REF (and by l... |
math/0009085 | Easy computation shows that the maximal compact symmetry group of MATH is MATH, with cohomology ring MATH (MATH are the NAME classes of rank MATH and MATH are the NAME classes of rank MATH). In other words this ring is MATH, MATH with the ``substitutions" MATH . Let us call these substitutions MATH. With this notation ... |
math/0009085 | The orbits of MATH are described by the NAME normal forms. A generic MATH has MATH different eigenvalues and the stabilizer group MATH of the orbit MATH is isomorphic to MATH. The inclusion MATH induces a map MATH. This map is injective. (This fact is usually called the splitting lemma. For semisimple NAME groups the c... |
math/0009085 | Since MATH the restriction map MATH factors through MATH so the homogeneous equations contain no extra information. |
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