paper stringlengths 9 16 | proof stringlengths 0 131k |
|---|---|
math/9904079 | As follows from REF below, MATH. Therefore, it suffices to show that the correspondence MATH is injective map of MATH to MATH. The injectivity is manifest, so we only have to check that the image is MATH-invariant. Clearly, MATH. Therefore, it suffices to verify that MATH for every simple root MATH. This is subject to ... |
math/9904079 | Since the dimensions of both modules are equal, it suffices to show that the natural homomorphism MATH is surjective, that is, the module generated by MATH coinsides with the whole module. The module in the right hand side has a natural filtration induced by filtrations of the modules MATH and MATH. Let the MATH be a b... |
math/9904079 | Let MATH be any positive integer. Consider the map MATH such that MATH. Clearly, MATH is a MATH-module homomorphism because it is induced by projections of MATH and MATH onto their even parts. Take MATH and MATH from REF and consider the map MATH . Clearly, MATH is a MATH-module homomorphism. Let us consider the restri... |
math/9904079 | It is easy to verify that MATH, which immediately implies REF . REF is similarly proved. |
math/9904079 | For NAME tableaux MATH and MATH we have MATH . The dimension of this space is equal to either REF or REF. It is equal to REF only if MATH or both of them contain a MATH rectangle and MATH for MATH and MATH for MATH and any MATH. To prove the theorem, it suffices to show that for the above MATH and MATH the module MATH ... |
math/9904079 | To MATH there corresponds the identity operator MATH; hence, to MATH there corresponds the operator MATH and to MATH there corresponds the operator MATH; finally, to MATH there corresponds the operator MATH. Hence, MATH . The last equality follows from MATH. |
math/9904079 | Since for the representatives of the cosets of MATH we can take a collection of cycles, we may assume in REF that MATH . Hence, MATH. NAME on, MATH if and only if the last row of MATH for some MATH is, up to a permutation, a permutation of MATH. The set of marked tableau MATH is in one-to-one correspondence with the se... |
math/9904079 | Compare REF . |
math/9904079 | Compare REF . |
math/9904079 | Compare REF . |
math/9904079 | On MATH the group MATH and its subgroup MATH naturally act; namely, MATH permutes pairs MATH whereas MATH permutes inside each pair. Clearly, MATH. But, on the other hand, MATH, so MATH. Hence, in the decomposition of MATH only enter MATH for which MATH and their multiplicity is equal to MATH. But MATH so the multiplic... |
math/9904079 | By REF MATH . Hence, the kernel of homomorphism REF is equal to MATH. Let us prove that it is contained in the ideal generated by MATH, where MATH is a MATH rectangle. Let MATH and MATH the corresponding idempotent. Then MATH. Hence, MATH . Thus, MATH and we are done. |
math/9904079 | Set MATH . The map MATH is an isomorphism induced by the invariant bilinear form and MATH is a MATH-invariant. Let MATH be a rectangular MATH tableau as in REF and MATH a MATH-sequence such that after being filled each row MATH is of the form MATH. Denote by MATH the set of such sequences MATH. Then MATH where MATH, , ... |
math/9904079 | Let MATH be a MATH-invariant which is not MATH-invariant. Let MATH depend on MATH even and MATH odd generic vectors MATH, , MATH. Then there exists a MATH such that MATH. Let MATH whereas MATH for MATH. Then MATH and MATH. But, on the other hand, MATH, hence, MATH. This means that MATH-invariants other than scalar prod... |
math/9904079 | Consider the two gradings of MATH: MATH and MATH . It is clear that MATH. If MATH is a MATH-invariant, then MATH (for MATH) and MATH for MATH. Therefore, MATH, where MATH. We similarly establish that MATH, where MATH. Hence, MATH. Taking into account the equality MATH we deduce that MATH and MATH for some MATH and MATH... |
math/9904079 | See REF. |
math/9904079 | From the theory of MATH-rings it follows that MATH, where the sum runs over the MATH of the above described form. One can easily verify that for the tableau as indicated in the formulation of the theorem and a MATH-standard sequence MATH the image MATH in MATH is nonzero. Hence, for a fixed tableau MATH the canonical m... |
math/9904079 | Clearly, MATH where MATH is of the same form as stated in Theorem. Since REF is a MATH-module homomorphism, its kernel is MATH. The fact that this kernel is generated by the elements of the least degreee is proved by the same arguments as for MATH. |
math/9904079 | Clearly, MATH is the product of MATH factors. In each factor, select either MATH or MATH. In the first case, for MATH, set MATH and MATH in the second case set the other way round. We get a matrix with the properties desired. The sign is obtained after reordering of the sequence of the MATH: MATH . This is performed by... |
math/9904079 | Let MATH be a MATH rectangle filled in along columns. Set MATH. Denote by MATH the tensor obtained from MATH by replacing the elements occupying positions MATH, , MATH with numbers MATH, , MATH, respectively. Then MATH where MATH is any of the numbers of the positions occupied by MATH. If MATH, the product being ordere... |
math/9904079 | is similar to that of REF. |
math/9904086 | If MATH, then taking the stalks at MATH, we have an associated cosimplicial abelian group MATH, and a corresponding augmented complex. Clearly MATH unless MATH. Since the partially ordered subset MATH has a minimal element, one sees easily that the stalk complex at MATH is contractible (note that if MATH, and MATH is a... |
math/9904086 | We first claim that if MATH is an irreducible REF locally closed subset such that MATH is locally free, then MATH is a constant sheaf associated to a finitely generated abelian group, which is a quotient of MATH. Indeed, MATH corresponds to a representation of the fundamental group of MATH (with respect to any convenie... |
math/9904086 | We first note that, by an argument with mapping cones and cylinders (rotating the distinguished triangles in REF-diagram), we may assume that MATH without loss of generality. For such a MATH the analogous result for the cohomology diagram arising from a REF-diagram in the category of sheaves has been proved by NAME and... |
math/9904086 | Let MATH. Then by REF , there exists a NAME open cover MATH of MATH such that MATH where MATH . Therefore MATH as shown in the commutative diagram below (where MATH stands for any of the MATH) MATH . This finishes the proof of the lemma. |
math/9904086 | From REF there exists an exact sequence of abelian groups MATH . The natural surjective maps MATH (for MATH) are maps of pure NAME structures of weight one, which are quotients of MATH. Hence MATH is identified with the kernel of a morphism of pure NAME structures of weight REF, and hence itself supports a pure NAME st... |
math/9904086 | Let MATH as above, and let MATH. We note that the natural map MATH is an injection (any section in the kernel must vanish in all stalks). This implies that the map MATH (which is a morphism of mixed NAME structures) is injective in the following commutative diagram (here MATH). MATH . We are done, because all the arrow... |
math/9904086 | It is easy to see, from REF , that MATH . We will prove, using REF , that given any element MATH and any preimage MATH some non-zero (integral) multiple of MATH lies in MATH . This will prove the assertion of the lemma. Since MATH there exists a finite NAME open cover MATH of MATH such that MATH in MATH for all MATH . ... |
math/9904086 | Let MATH be any point and MATH . Let MATH and MATH, where MATH is the maximal ideal. For each MATH we have the restriction maps MATH . We claim that the kernel of MATH for each MATH is a MATH-vector space. To see this consider the short exact sequence of NAME sheaves MATH where MATH denotes the exponential map, which m... |
math/9904086 | Using the short exact sequence of sheaves REF , we get the following commutative diagram, whose right column is exact, MATH . Here, we claim the dotted arrow MATH exists (and is also injective) because the composition MATH is zero. This is obvious as this map can be described in the following way: given the image in MA... |
math/9904086 | Consider the following diagram with exact rows and columns MATH . The above diagram comes from the following MATH-diagram in the category of sheaves (where the bottom row defines MATH and MATH is the inclusion). MATH . Let MATH such that MATH both in MATH and MATH (that is, MATH). By a diagram chase as before we get el... |
math/9904088 | It will suffice to show the pullback vectorial extension is universal. Since MATH, any extension of MATH by a vector group MATH is pulled back from a unique extension of MATH by MATH. This extension of MATH is a pushout from the universal vectorial extension, so the same holds for the pullbacks to MATH. |
math/9904088 | There is a boundary map MATH . Define MATH. Let MATH be the torus with character group MATH. For MATH let MATH be the line bundle on MATH corresponding under the map REF. As a MATH-algebra MATH . The map MATH in REF comes from the above inclusion MATH . For MATH, (as is well known, compare CITE III REF), MATH has trivi... |
math/9904088 | The MATH-algebra MATH is filtered, with MATH . With respect to the exact sequences MATH it suffices to show the boundary map MATH is injective. Composing on the right with the evident map, it suffices to show the maps MATH are injective. But MATH and the map in REF is the map MATH, which is injective. |
math/9904088 | Take MATH. |
math/9904088 | By REF on p. REF, the cohomology in the middle is a subgroup of the group of extensions MATH. (Note, MATH.) By the classification of commutative algebraic groups in characteristic MATH, this ext group vanishes (compare CITE, pp. REF). |
math/9904088 | Let MATH be an absolute lifting. Replacing MATH with MATH we may assume MATH vanishes. Then MATH vanishes in MATH. Let MATH be as in REF , so MATH. The previous lemma implies there exists MATH with MATH. Then MATH is the desired invariant absolute form. |
math/9904088 | Pullback on invariant relative forms is injective, because MATH is generated by the image of MATH. It follows by dimension count that the first arrow REF above is an isomorphism. For the absolute forms we may consider the diagram MATH . The left and right hand vertical arrows are shown to be isomorphisms in the proof o... |
math/9904088 | This follows because MATH . The second arrow is surjective because we have a diagram MATH . The bottom row is not a priori exact, but MATH (because MATH acts by MATH on MATH.) The middle vertical arrow is onto for example, because the NAME group of the generic fibre of MATH is zero. Indeed, MATH is rationally split, an... |
math/9904088 | One has as in REF MATH . Now MATH . Exactness of the sequence in REF implies that there exists an element MATH with MATH. Take MATH. |
math/9904088 | This is immediate from the lemma. |
math/9904088 | The first step is to compute MATH for MATH. Let MATH be a finite dimensional MATH-vector space, and suppose MATH is a vectorial extension MATH . We know by REF that this sequence pulls back from an extension of MATH by MATH. Let MATH be the exact sequence of functions of filtration degree MATH as in REF , and let MATH ... |
math/9904088 | For example, in the absolute case, the curvature of a line bundle with invariant absolute connection on MATH is a section MATH satisfying MATH. It is easy to see that such a section lies in the subsheaf MATH. The isomorphism REF follows from REF and the fact that pullback to MATH of invariant forms is injective by REF.... |
math/9904088 | Let MATH be a basis for MATH at MATH a point with multiplicity MATH in MATH, and let MATH be a local parameter at MATH on MATH. Write MATH . We must show MATH is regular at MATH. But integrability of MATH implies that MATH, from which the assertion is clear. |
math/9904088 | Our hypotheses imply MATH has dimension MATH. Consider the diagram MATH . We have MATH exterior tensor product on MATH. The NAME formula gives MATH . There is an action of the symmetric group MATH on the pair MATH . The resulting action on MATH is alternating because of the odd degree cohomology, so the invariants are ... |
math/9904088 | Write MATH for the NAME sheaf associated to the presheaf MATH for any NAME sheaf MATH on MATH. For MATH locally free, MATH for MATH by purity. Duality theory gives (here MATH runs through nilpotent thickenings) MATH . We want to show that this map is an isomorphism, compatible with the connection, thus yielding a quasi... |
math/9904088 | Let MATH correspond to a line bundle MATH of degree MATH, we consider the exact sequence MATH . Suppose first MATH. Then MATH, so any trivialization in MATH lifts to MATH, and the space of such liftings is a torseur under MATH, a vector space of dimension MATH. (Note this is an affine torseur, not a projective torseur.... |
math/9904088 | The isomorphism on the right in REF implies we must show MATH. The assumption MATH means we have a MATH action by translation on MATH, and minimality of MATH implies that the connection is nontrivial on the fibres. The fibration is NAME trivial, so the NAME spectral sequence for NAME cohomology reduces us to showing MA... |
math/9904088 | Note MATH. Extending the top sequence in REF one step to the left and using the previous lemma gives the left isomorphism. We have already seen the isomorphism on the right. |
math/9904088 | Let us write MATH, where MATH is the cycle map. Then by definition, the trivialization of MATH associated to MATH depends only on MATH, or equivalently only on MATH. We have MATH where MATH is the pullback of MATH via the MATH-th projection MATH. Suppose for a moment that the divisor of MATH (viewed as a section of MAT... |
math/9904088 | Recall REF the relative invariant forms on MATH are the MATH defined by the expression MATH . Write MATH where the nonvanishing condition comes from the requirement that the form restricted to MATH gives a trivialization along MATH (see REF). If we write REF MATH we get the table MATH . Note if we give MATH all weight ... |
math/9904088 | We have seen in the proof of REF that MATH vanishes at a point in MATH. We must show it vanishes at at most one point. Let MATH be a point. Write MATH with respect to the coordinates MATH REF. Staring at REF, the conditions that MATH are seen to be (recall MATH) for MATH . For MATH one gets the same list but with the l... |
math/9904088 | Since MATH is (relatively) closed on MATH, one can write MATH . Lifting to an absolute form forces MATH . Here the MATH are linearly independent in MATH. Using MATH modulo MATH and taking residues along MATH yields MATH. Then computing MATH yields MATH . It follows that MATH, so MATH. Taking MATH again shows MATH is cl... |
math/9904088 | We have seen REF that this point MATH is determined by the condition that MATH. Changing MATH by a closed form pulled back from MATH changes the NAME connection and the connection at MATH in the same way, so we can assume MATH, that is, MATH. Write MATH . Write MATH. Then MATH . We have MATH . Since MATH, we see from R... |
math/9904088 | First we collect some facts about MATH. We have MATH and MATH. It follows from REF of the paper that MATH does not appear in the expression for MATH and MATH only appears with constant coefficient. Restricting to MATH thus has the effect of surpressing the term in MATH and changing the coefficients MATH by a constant f... |
math/9904088 | The strong NAME theorem identifies the determinant connections on MATH and MATH so we need only consider the connection on MATH. As well known, the NAME duality morphism MATH is compatible with the NAME connection, which is trivial on MATH. On the other hand, it is alternating, thus its determinant MATH fulfills MATH w... |
math/9904088 | Since MATH, where MATH is the ideal sheaf of MATH, MATH . Also one has MATH. Thus MATH . |
math/9904088 | The map MATH is surjective, and its kernel is the MATH-module generated by MATH where MATH is the connection with logarithmic poles along MATH with residue -REF. Let MATH be a local coordinate around MATH. Let MATH be a NAME covering of MATH, with MATH, and MATH. Assume MATH is the only zero or pole of MATH on MATH. Le... |
math/9904088 | Note that MATH so the complex in REF is indeed a quotient. From the diagram MATH one deduces that the left hand side of REF is isomorphic to MATH . The right hand side here is identified under the norm with MATH, which proves the second equality. The third one comes from the map MATH and the vanishing of MATH. Note tha... |
math/9904088 | The quasiisomorphism condition is local about each point of MATH, so we may assume our line bundles are MATH for some MATH and MATH. Choose local coordinates MATH near MATH and a NAME covering MATH of MATH such that MATH, MATH, MATH for MATH. Let us denote by MATH the class defined by the local trivialization MATH . Le... |
math/9904088 | Given the main result of CITE, and REF , the theorem is of course equivalent to MATH . Keeping the same notations as in the proof of REF , one has MATH and thus MATH . This lies in MATH and by REF , its trace factors through MATH. But the image of MATH in MATH is the relative NAME class MATH REF, thus the image of MATH... |
math/9904088 | One reduces easily to the case MATH is a single MATH-point. Let MATH be a NAME set containing MATH and MATH. Shrink MATH if necessary so there exists MATH with MATH and MATH. Let MATH so MATH. Shrinking the MATH if necessary, we can assume MATH, so MATH is represented by some cocycle MATH. Then MATH . On the other hand... |
math/9904088 | If all MATH this is just REF . So we assume that MATH in the sequel. Then as in the proof of REF , replacing MATH by MATH changes MATH to MATH and keeps the rest unchanged. Thus the quotient complex MATH is MATH-linear and the map is the multiplication by MATH. In particular, MATH fulfills REF for all MATH and taking M... |
math/9904088 | The assumption that the curvature is vertical implies MATH . Multiplying through by MATH and contracting against MATH we deduce REF. For REF, we must show MATH. From REF, using MATH we reduce to showing MATH. Since MATH has entries MATH, one has MATH . And the residue of an exact form is vanishing. |
math/9904088 | First we show independence of the choice of lifting of MATH. As remarked above, MATH is determined by the local lifting of MATH to MATH, so MATH depends only on that choice. If MATH and MATH are two such local liftings, with MATH, we have MATH for some MATH. It follows immediately that MATH as desired. Next we show ind... |
math/9904088 | The usual exact sequence reduces us to showing the condition is equivalent to MATH . Writing as usual MATH where MATH is the reduced divisor with support equal to the support of MATH, we have a commutative square MATH . The map MATH is a quasi-isomorphism if and only if for all MATH the map MATH on the bottom line of t... |
math/9904088 | First MATH . Next computing mod the ideal of MATH and so ignoring the tilde, MATH . By our commutation REF MATH . Also terms with no poles (that is, terms not involving MATH) can be ignored. We get MATH which is what we want. Note the right hand equality holds because we are computing mod forms regular along MATH, and ... |
math/9904088 | Another representative of MATH in MATH is of the shape MATH, thus its derivative in MATH is of the shape MATH since MATH. Then one applies the commutativity of the diagramm MATH . |
math/9904088 | Since, under the integrabiltiy assumption, one has in particular MATH, one can apply REF . One has to compute MATH . We omit the indices since we compute only with one index. Note in the calculations which follow MATH is regular and MATH has poles along MATH. We write MATH to indicate that the polar parts of MATH and M... |
math/9904088 | We first check injectivity. Let MATH be a local defining equation for MATH. Suppose for some MATH with MATH we had MATH . Multiplying by MATH and taking residue along MATH, it would follow that MATH, that is, MATH. To show surjectivity, write a local section of MATH (here MATH) in the form MATH where MATH does not invo... |
math/9904088 | Using the lemma, we get a diagram with exact rows MATH . We view this as a diagram of complexes written vertically. Using the standard hypercohomological interpretation of line bundles with connection, this yields an exact sequence MATH . We claim the map MATH above is zero. In the derived category, MATH factors MATH .... |
math/9904091 | We proceed by induction on MATH. For MATH, there is nothing to prove. Let us suppose now MATH and let MATH where MATH and MATH. Let MATH then MATH where MATH. If there exist MATH such that MATH, and MATH then, since MATH, it easily follows MATH: thus the lemma is true by induction. Otherwise, if such MATH do not exist,... |
math/9904091 | Let MATH and let MATH be a basis of MATH: if MATH, then MATH is a basis of MATH. The weights of MATH are MATH (compare CITE, pag. REF) and since the weights of MATH are given by the sums of couples of different weights of MATH, it easily follows: MATH . Indeed if MATH, then MATH. Let us prove now that MATH does not con... |
math/9904091 | Let MATH. By the NAME criterion CITE, it suffices to show that MATH for any MATH. By the sequence: MATH obtained raising the sequence REF to the MATH-th symmetric power, we see that: MATH for any MATH and MATH. On the other hand by REF , we have the long exact sequence: MATH . This sequence immediately implies that MAT... |
math/9904091 | The proof of this lemma is very similar to the proof of REF. |
math/9904091 | By joining together the sequences REF , we get: MATH . By REF and by the fact that MATH, the last sequence is the minimal resolution of MATH: hence MATH is directly defined by this resolution. |
math/9904091 | By the sequence MATH and since MATH we get MATH hence the lemma is proven. |
math/9904091 | It follows from the definition of MATH, (compare CITE). |
math/9904091 | For brevity's sake, we will write MATH instead of MATH and MATH for MATH. Let also MATH be such that MATH. Let MATH be the sub-variety of the irreducible component of MATH composed by all the quotients of the maps MATH for some weighted bundle MATH and containing the point MATH corresponding to MATH: the morphisms MATH... |
math/9904095 | Assume first that MATH. The class MATH is represented by the disjoint union MATH. It is clear that the NAME scheme of a disjoint union satisfies MATH . Hence we get MATH. By induction, we get MATH whenever there is a relation MATH for positive integers MATH, MATH and MATH. The corollary follows formally from this. |
math/9904095 | Both sides of this equality are multiplicative in MATH, so it suffices to check it for MATH and MATH. Now the NAME schemes of both MATH and MATH have a MATH-action with isolated fix-points. For such varieties the MATH-genus is given by MATH where the first sum is over all fix-points and MATH is the subspace of the tang... |
math/9904095 | For a line bundle MATH on MATH let MATH, where MATH and MATH are the two natural maps and where MATH is the projection of MATH onto the MATH-th factor. We will show later that MATH, compare REF . Assume that MATH is a surface such that MATH for some line bundle MATH. Then we claim that the same is true for the NAME sch... |
math/9904095 | Let MATH. Then MATH . |
math/9904095 | As MATH is a birational proper morphism of normal varieties, it follows that MATH. MATH has rational singularities, as the quotient of a smooth variety by a finite group (see REF ). Therefore its resolution MATH satifies MATH. |
math/9904095 | Let MATH be a locally free sheaf on MATH. Apply the functor MATH to the exact sequence REF and observe that MATH that is, MATH and similarly MATH. Using MATH we get MATH that is, MATH. |
math/9904095 | We have the following isomorphisms MATH . Here MATH are the higher derived functors of the composite functor MATH. For the first isomorphism see for example, CITE. The last equality is a consequence from the spectral sequence MATH and the observation that the sheaves MATH are supported on the universal family MATH, whi... |
math/9904095 | We have MATH . Now use REF to replace MATH by MATH, where MATH denotes the diagonal. This yields MATH . The two last summands can be simplified as follows: MATH and similarly MATH, since MATH. Finally, MATH, which follows for example, by REF . |
math/9904095 | The morphism MATH is generically finite of degree MATH, so that MATH . Because of an index shift resulting from the insertion of the additional factor MATH between MATH and MATH we have MATH . Using REF we get MATH . And finally by REF MATH . It follows that there are polynomials MATH depending only on MATH, in the NAM... |
math/9904095 | Suppose we are given a polynomial MATH in the NAME classes of MATH. Applying the proposition repeatedly, we may write MATH for some polynomial MATH, which depends only on MATH, in the NAME classes of sheaves on MATH of the form MATH and MATH. Any such expression MATH can be universally reduced to a polynomial expressio... |
math/9904095 | The proof goes along similar lines as that of REF . The result immediately follows from a modified version of REF : We now allow MATH to be a polynomial in the MATH and the NAME classes of the MATH on MATH, and get MATH to be a polynomial in the MATH and the NAME classes of the analogously defined bundles on MATH. To p... |
math/9904095 | Let MATH, and let MATH be the map MATH. The images of the five elements MATH, MATH, MATH, MATH, and MATH under MATH are the linearly independent vectors MATH, MATH, MATH, MATH, and MATH. Now, if MATH we may decompose MATH as MATH, where MATH, and get MATH. Moreover, there is a decomposition of the class of the tautolog... |
math/9904095 | Consider the cartesian diagram MATH where MATH is the open subscheme of zero cycles of length MATH whose support consists of at least MATH points and MATH and MATH are the preimages of MATH under MATH and MATH. It is easy to see that MATH is the blow-up of MATH along the (disjoint) diagonals MATH . Let MATH denote the ... |
math/9904095 | Using new formal variables MATH and MATH that are related by MATH, MATH, and MATH, so that MATH, and, in the last line of the computation, the NAME formula, we have MATH . This gives MATH and MATH. Now MATH and hence MATH . Differentiating MATH we find MATH . As both MATH and MATH are power series in MATH with constant... |
math/9904095 | Let MATH. Then MATH, MATH, MATH. Therefore, by REF , MATH for suitable power series MATH. For the second equality we have used the identities MATH and MATH. It is well-known that MATH. We get by NAME duality MATH . Using MATH and MATH, this gives MATH, MATH, MATH and MATH. To determine MATH and MATH explicitly, let MAT... |
math/9904095 | REF follows from REF by putting MATH. In order to prove REF consider the cartesian diagram for MATH where MATH is the NAME morphism and MATH is the image of the universal family MATH in MATH. We denote by MATH, MATH the restriction of the projection. It is easy to see that there is an isomorphism MATH which identifies ... |
math/9904104 | In order to make sense of this statement, we will first review both the definition of a holomorphic vertex algebra, and the definition of an algebra in a category. A holomorphic vertex algebra is a vertex algebra without singularities. In other words, the vertex operator of a holomorphic vertex algebra is a map: MATH .... |
math/9904104 | This is clear from the action defined in REF and the coassociativity of MATH. |
math/9904104 | The MATH-invariance of the singular multilinear maps allows the action of MATH on the domain, MATH, to carry over to an action on the domain of MATH as described in REF. Hence the action of MATH on MATH passes through to an action on MATH exactly when MATH is MATH-invariant. |
math/9904104 | The proof follows immediately from the fact that the evaluation of MATH on MATH gives the same result when carried out by either first evaluating MATH, or by first evaluating MATH or by evaluating both together. |
math/9904104 | We shall only prove that the composition works for the classical vertex group. For the general theory, see CITE. Since composition is pointwise, the composite REF factors through both REF, and so we can focus on those compositions. In order to prove that the composites map into the pullback, we shall first prove that t... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.