paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0006194 | The MATH-invariants of MATH can be calculated using MATH-index theorem, applied to a MATH action on the four-ball MATH endowed with a MATH-invariant metric with non-negative scalar curvature which is product-like near the boundary. To lift the action to the spinor bundle, we follow CITE. Let MATH be the rotation MATH w... |
math/0006201 | Since the vacuum MATH is preserved by the automorphisms, it defines a section which we denote by MATH. Given two sections MATH and MATH, the assignment MATH defines a section denoted by MATH. It is straightforward to check that MATH then defines a structure of a vertex algebra on MATH. |
math/0006201 | CASE: Let MATH be the zero mode of MATH. We have MATH . Since MATH acts as MATH, we actually have MATH, where MATH . It is easy to see that MATH, hence we get a double complex and two spectral sequences with MATH term the MATH-cohomology and the MATH-cohomology respectively (compare CITE). Now MATH . On the first facto... |
math/0006201 | We regard MATH as a bundle of holomorphic vertex algebra, therefore, MATH has a natural structure of a MATH SCVA. By REF , the section space MATH is a MATH SCVA. One can easily verify that MATH is a differential by choosing a local holomorphic frame of MATH. It follows from REF that MATH is a MATH SCVA. Notice that on ... |
math/0006209 | CASE: By CITE we have MATH . On the other hand we have MATH for MATH. Hence we have MATH for MATH. CASE: We show the formula by induction on MATH. By the definition of MATH, it is clear that MATH. Assume that MATH and the statement is proved for any root MATH in MATH satisfying MATH. For some MATH we can write MATH whe... |
math/0006209 | For MATH we have MATH. By REF we have to show MATH. The MATH-analogue MATH is a linear combination of MATH satisfying MATH (see CITE and CITE). By using REF it is easy to show that MATH and MATH. Hence we have MATH . On the other hand there exists MATH such that MATH by REF , hence we have MATH. Therefore we obtain MAT... |
math/0006209 | CASE: It is clear from the definition. CASE: It is sufficient to show that the statement holds for the weight vectors MATH and the canonical generator MATH of MATH. If MATH for MATH, then the assertion is obvious. Let MATH for MATH. By REF we have MATH . On the other hand we have MATH where MATH is the weight of MATH. ... |
math/0006209 | By the definition it is clear that MATH if MATH. In the case where MATH we shall show the statement by the induction on MATH. Since MATH, we obtain MATH. Assume that MATH and the statement holds for any root MATH in MATH satisfying MATH. Then there exists a root MATH in MATH such that MATH where MATH satisfying MATH an... |
math/0006209 | CASE: The uniqueness follows from the non-degeneracy of MATH. If there exist MATH and MATH, then we have MATH. Therefore we have only to show the existence of MATH for any MATH. By REF we have MATH. Let MATH. Then there exists a root MATH such that MATH. By REF we can show that MATH easily. CASE: The assertion follows ... |
math/0006209 | Let MATH, MATH. Since MATH for MATH, we have MATH. Hence we obtain MATH . Similarly we obtain MATH . |
math/0006209 | Let MATH, where MATH or MATH for some MATH. Set MATH. For MATH and MATH we have MATH by the induction on MATH. Note that MATH. Moreover MATH for MATH. Hence we have MATH where MATH and MATH does not depend on MATH. By REF MATH is a linear combination of such MATH satisfying MATH. The assertion is proved. |
math/0006209 | CASE: This is proved easily by the induction on MATH. Note that MATH for MATH. CASE: Since MATH is a central element of MATH, this follows from REF . CASE: Let MATH. Then there exists some MATH such that MATH and MATH for any MATH. Hence we have MATH, and the statement follows. |
math/0006209 | By REF , the statement is proved by the induction on MATH easily. |
math/0006209 | We denote the right handed side of the statement by MATH. It is easy to show that the coefficient of MATH in MATH is equal to that in MATH. Moreover the weight of MATH is equal to that of MATH. Hence it is sufficient to show that MATH is the highest weight vector. Since MATH, we have only to show the statement in the c... |
math/0006210 | Because MATH is the representation of MATH, the relation REF is trivial. So we shall prove that MATH. We take the complexification of REF and may prove MATH. For example, we have MATH . On the other hands, MATH . So we have MATH. Similarly we can prove the other cases. |
math/0006210 | REF is trivial. So we shall prove REF. For an orthonormal basis MATH of MATH, we denote the corresponding one of the irreducible component MATH in MATH by MATH. Since MATH is the orthogonal projection from MATH to MATH, the homomorphism MATH is represented as MATH where MATH is inner product on MATH. If we use another ... |
math/0006210 | By direct calculations. |
math/0006210 | We can easily show that the NAME operator is formally self-adjoint (for example, see CITE). In the same way, we can prove REF by using REF ,. |
math/0006210 | We shall prove REF. We fix MATH in MATH and choose an orthonormal frame MATH in a neighborhood of MATH such that MATH for all MATH. Hence, we have MATH for all MATH. Then it holds from REF that MATH . |
math/0006210 | We eliminate the connection Laplacian MATH from REF. |
math/0006210 | We investigate the ellipticity of MATH. The principal symbol of MATH is MATH where MATH is in MATH. There exists MATH in MATH such that MATH . Then we have MATH . It follows that, if MATH is odd, then MATH is not zero for MATH. Hence MATH is elliptic. In the same way, we verify that MATH is elliptic. |
math/0006210 | Since MATH has constant curvature, it holds that, for vector fields MATH, MATH, and MATH, MATH . Then we have MATH. Hence, MATH . Here, we use that MATH is the NAME operator on MATH. |
math/0006215 | Regard MATH as an analytic germ. It was shown in CITE that the general element MATH has a normal NAME singularity at MATH. By Inversion of Adjunction REF, MATH is plt. Consider the crepant pull-back MATH, where MATH is the proper transform of MATH and MATH. Since both MATH and MATH are MATH-Cartier, so are MATH and MAT... |
math/0006215 | Put MATH and let MATH, MATH. Assume that MATH. Let MATH be such as in REF and let MATH be a non-klt MATH, MATH, MATH, MATH, or MATH-complement of MATH. Using that the coefficients of MATH are standard CITE it is easy to see that MATH and MATH CITE. In particular, MATH is lc. Further, MATH. Indeed, otherwise by Adjuncti... |
math/0006215 | By Adjunction, MATH. Hence MATH. This shows that MATH and MATH has no singularities at points of MATH (see CITE). Let MATH be the minimal resolution and let MATH be the proper transform of MATH on MATH. Define MATH as the crepant pull-back: MATH. It is sufficient to show that MATH. Assume the converse. Replace MATH wit... |
math/0006215 | Assume the converse. Then MATH and MATH is also an exceptional contraction as in REF. First, we consider the case when MATH is divisorial. Then MATH is a plt blowup of a terminal point MATH and MATH (see CITE). By REF , MATH has a non-klt complement. This contradicts CITE. Clearly, MATH cannot be a flipping contraction... |
math/0006215 | We need only the second part of the lemma. So we omit the proof of the first part. Let MATH. We use topological arguments. Regard MATH as a analytic germ along MATH. Since MATH is smooth, MATH. On the other hand, for a sufficiently small neighborhood MATH the map MATH is surjective (see CITE). Using NAME 's Theorem as ... |
math/0006215 | The map MATH is a composition of log flips: MATH where every contraction MATH is MATH-negative and every MATH is MATH-negative. NAME Vanishing CITE implies that exceptional loci of these contractions are trees of smooth rational curves CITE. Thus REF is obvious if the curve MATH is nonrational. From now on we assume th... |
math/0006217 | REF follows from the fact that MATH. It follows from MATH that MATH does not admit any nontrivial deformations as an algebra, (see CITE), which proves REF. From the fact that MATH it follows that any deformation of the algebra morphism MATH appears as a conjugation of MATH. In particular, the comultiplication in MATH l... |
math/0006217 | This follows immediately from REF . This follows also from the categorical interpretation of MATH and MATH, CITE, CITE. |
math/0006217 | A direct computation. Another proof is found in CITE. |
math/0006217 | By REF there is a MATH invariant MATH associative multiplication MATH such that MATH with MATH as in REF . Let MATH be the MATH-bracket corresponding to MATH. Then a direct computation shows that the NAME bracket of MATH is as required. |
math/0006217 | Similar to REF . |
math/0006217 | REF, and REF are proven in REF . REF follows from the fact that the weight subspaces of all MATH have dimension one (see NAME, NAMEgèbres NAME, REF ). |
math/0006217 | Follows from REF (see also REF ). |
math/0006217 | Direct computation, see CITE. |
math/0006217 | For proving REF we have to show that for any solution MATH one can choose a system of positive quasiroots in such a way that the denominators appearing in REF by the recursive procedure are not equal to zero. It follows from REF that the set MATH consisting of MATH such that MATH is a linear subset of MATH. Moreover, t... |
math/0006217 | The proof reduces to solving the system of equations defined by REF , see CITE. |
math/0006217 | The proof of REF follows CITE. First, let MATH be a nondegenerate NAME bracket on MATH, in particular, MATH. Then the complex of polyvector fields on MATH, MATH, with the differential MATH is well defined. Denote by MATH the NAME complex on MATH. Since none of the coefficients MATH of MATH are zero, MATH is a nondegene... |
math/0006217 | Since MATH is a NAME manifold, it is enough to prove that MATH. Note that according to REF MATH is a subbundle (direct subsheaf) of MATH. Therefore, MATH is a subbundle of MATH. On the other hand, MATH being a subsheaf of MATH is a torsion free sheaf. It follows that MATH is a torsion free sheaf. According to REF the s... |
math/0006217 | Let us prove REF. If MATH then MATH with MATH. Hence MATH by REF. If MATH then MATH and MATH is a nondegenerate NAME bracket on MATH, therefore, MATH by REF. Let us prove REF. Denote by MATH a linear holomorphic vector bundle of degree MATH over the NAME sphere MATH. The space of global sections of MATH may be naturall... |
math/0006217 | The proof can be proceed in the similar way as for MATH, using that MATH and MATH are NAME manifolds, CITE. |
math/0006217 | The proof is essentially follows to REF , but here we construct the multiplication for all MATH simultaneously using parameterized NAME cohomologies from the previous section. To begin, consider the multiplication MATH. The corresponding obstruction cocycle is given by MATH considered modulo terms of order MATH. No MAT... |
math/0006217 | Indeed, let MATH be a MATH-bracket. The multiplication corresponding to any path MATH with origin in MATH is as required. |
math/0006217 | We have MATH where MATH, MATH, are terms appearing in the expansion of MATH. It is easy to check that MATH is a NAME cocycle, because both MATH and MATH resolve the same obstruction MATH, MATH depending only on MATH, MATH. Hence, one has MATH, where MATH is a NAME REF-chain, that is, a differential operator, and MATH i... |
math/0006217 | The existence of a multiplication which is MATH associative up to and including MATH terms is nearly identical to the proof of REF . So, suppose we have a multiplication defined to order MATH, MATH which is MATH invariant and MATH associative to order MATH. Consider the obstruction cochain, MATH . The same argument as ... |
math/0006217 | The connection MATH induces an equivariant isomorphism of MATH modules between the sheaf of differential operators on MATH and the sheaf MATH, where MATH and MATH denote the sheaves of symmetric tensors over MATH and MATH. It provides an equivariant isomorphism between the space of invariant bidifferential operators on... |
math/0006217 | Indeed, MATH has the form MATH-bidifferential operators vanishing on constants. So, the corollary follows from REF . |
math/0006217 | Let us prove the first relation. The second relations can be proven similarly. Let MATH be a system of positive roots for MATH which projects on MATH. Let MATH be the corresponding parabolic subalgebra of MATH and MATH the radical of MATH, so MATH. Let MATH and MATH denote the corresponding opposite subalgebras, in par... |
math/0006217 | According to REF we put MATH, where the multiplication MATH is from REF . The functions of MATH are restrictions to MATH of holomorphic functions on MATH. It follows that MATH is a well defined multiplication on MATH. Since the coefficients by MATH in MATH are bidifferential operators, this multiplication is defined, a... |
math/0006217 | We put MATH and use the argument as in the proof of REF . |
math/0006217 | We have MATH. Let MATH, where MATH is considered as a right and MATH as a left MATH module (see REF ). It is clear that MATH is the required quantization. |
math/0006217 | Let MATH. Let us define left and right multiplications of elements of MATH by elements of MATH. Let MATH and MATH. Put MATH, where MATH, and MATH, where MATH. Here MATH means, for example, the multiplication when MATH is considered as a left MATH module. It easily follows from the above properties of MATH that the mult... |
math/0006219 | CASE: Should be clear (an easy induction). CASE: Suppose that MATH and MATH are a counterexample with the minimal possible value of MATH. Necessarily MATH is a limit ordinal, MATH, MATH and MATH. Let MATH be the first ordinal such that MATH. By the choice of MATH, the set MATH is finite, but clearly MATH for all MATH. ... |
math/0006219 | REF, CASE: Straightforward (for REF use REF ). REF, CASE: Easy inductions on MATH using REF above. |
math/0006219 | It follows from REF that if MATH is an open dense set, MATH, then there is a condition MATH such that MATH. Therefore, to win the game MATH, the second player can play so that the conditions MATH that he chooses are MATH - increasing, and thus there are no problems with finding MATH - bounds (remember REF ). Now, to sh... |
math/0006219 | Easy inductions on MATH. |
math/0006219 | Easy inductions on MATH (remember REF ). |
math/0006219 | We prove this by induction on MATH (for all MATH satisfying the assumptions). Step MATH; MATH, MATH. Take the MATH - component MATH of MATH such that MATH. Then, for MATH, MATH, and for each MATH, MATH we have MATH. Also, if MATH and MATH, then MATH. Consequently, MATH, and if MATH then MATH (remember REF ). Moreover, ... |
math/0006219 | CASE: Note that if MATH, MATH then there is a condition MATH such that MATH. Hence MATH. To show that, in MATH, the algebra MATH is of size MATH it is enough to prove the following claim. Let MATH, MATH. Then MATH. Suppose not, and let MATH be a counterexample with the smallest possible MATH. Necessarily, MATH is a suc... |
math/0006219 | By REF we know that MATH, so what we have to show is that there are no increasing sequences of length MATH of elements of MATH. We will show this under an additional assumption that MATH (after the proof is carried out, it will be clear how one modifies it to deal with the case MATH). Due to this additional assumption,... |
math/0006220 | Let MATH be such that MATH. Suppose that MATH is not covered by a finite subcollection of MATH. Choose MATH such that MATH for MATH and let MATH. We have MATH. This set is not covered by a finite subcollection of MATH, for clearly MATH is not covered by MATH and for MATH, MATH is of positive codimension in MATH. With i... |
math/0006220 | Suppose we have another solution MATH with MATH and MATH stable, MATH for all MATH and MATH as MATH. It is enough to prove that the dimension of the stable set MATH is MATH. Since MATH, REF applies and we find that MATH for some MATH. Since every term has dimension MATH, this is also true for MATH. |
math/0006220 | Given MATH, consider the set MATH of MATH with order MATH along MATH. So for MATH we have MATH and MATH. If MATH is the support of MATH, then we have a natural projection MATH. Its composite with the morphism MATH is a restriction of MATH with MATH-component MATH. In other words, MATH . So the transformation REF yields... |
math/0006220 | Since MATH is a MATH-bundle over MATH, the class of the projection MATH is MATH times the class of MATH. |
math/0006220 | Let MATH be the eigenspace decomposition of the MATH-action. Given a subset MATH, denote by MATH the set of vectors in MATH whose MATH-component is nonzero if and only if MATH. We have a natural projection MATH. This has the structure of a torus bundle, the torus in question being a quotient of MATH by a finite subgrou... |
math/0006220 | One verifies that the NAME factorization of the projection MATH has MATH as finite factor with MATH being an algebraic torus bundle of rank MATH. In view of REF the equivariant class of the latter is MATH times the equivariant class of the base. The lemma follows. |
math/0006220 | Start with the identity of REF (with MATH). Omit at both sides the constant terms (on the right this amounts to summing over nonempty MATH only), and restrict the resulting identity to the fiber over MATH. If we take into account the monodromies and use REF , we get the asserted identity, at least if we take our coeffi... |
math/0006220 | The monodromy MATH acts on MATH as a covering transformation of order MATH. So MATH has no fixed point if MATH does not divide MATH and is equal to all of MATH otherwise. It follows from formula for the nearby cycle class that MATH . So MATH . If we reduce modulo MATH only the terms with MATH a singleton remain. REF sh... |
math/0006220 | The preimage of MATH in MATH under MATH decomposes into the following pieces: MATH, MATH and for MATH the preimage MATH of the subset MATH of pairs MATH with MATH for MATH and MATH. We must evaluate MATH on each of these (relative to the diagonal MATH-action). The first piece gives MATH and the second the same expressi... |
math/0006220 | It is easy to see that it suffices to prove this for MATH. The idea of the proof in this case is inspired by a paper of CITE. Fix for the moment MATH. Let MATH be a radius of convergence for the two expansions. Let MATH be such that MATH and choose MATH. Consider the integral MATH . On the circle of integration the exp... |
math/0006220 | We start with the convolution REF . It says that MATH . We now assume that MATH and MATH are massless so that MATH and similarly for MATH. We then have MATH . We consider each series on the right separately. By REF , MATH is in the in MATH with value at MATH equal to MATH. We also have MATH . By the same REF the righth... |
math/0006220 | If we apply MATH to the identity MATH we get MATH. Since MATH, it follows that MATH. So MATH pulls back under MATH to MATH. The rest is left to the reader. |
math/0006220 | Assume that MATH is given as a closed subset of MATH as above. There exist MATH and MATH such that the Jacobian matrix MATH has order MATH along MATH, whereas for any other matrix thus formed the order is MATH. By means of a coordinate change we may arrange that MATH so that MATH. The subspace of MATH spanned by the la... |
math/0006220 | Suppose that MATH is of pure relative dimension MATH. Let MATH denote the subset of MATH defined by MATH. It is clear that MATH. It follows from NAME 's theorem CITE that MATH is constructible. Hence MATH is stable by REF . We have MATH. In view of REF it now suffices to see that MATH as MATH. This is not difficult. |
math/0006220 | Let MATH and put MATH, MATH. Suppose MATH is such that MATH and MATH have the same MATH-jet. We first show that MATH and MATH have the same MATH-jet. We do this by constructing a MATH (by successive approximation) with the same MATH-jet as MATH and with MATH. Since MATH, we will have MATH and our injectivity assumption... |
math/0006220 | It is enough to prove this for MATH stable. In that case the theorem follows in a straightforward manner from REF . |
math/0006221 | In CITE this statement is proved for MATH and, as MATH, for MATH. First of all, from the fact that the monomials of the form MATH with the restrictions MATH span the space MATH it follows that the monomials with the restrictions REF span MATH. So to complete the proof it is sufficient to compare the dimension of the sp... |
math/0006221 | First, prove some lemmas. CASE: MATH CASE: MATH . It follows from the identity MATH for MATH and the identity MATH. MATH for odd MATH, Á MATH for even MATH. It follows from the identity MATH . If MATH then MATH. We want to prove that MATH for any MATH and MATH. Let MATH . Then MATH where MATH is the common numerator of... |
math/0006224 | That MATH follows from one of NAME 's key results for MATH-contractions, Stability of Curvature, REF . For our case of a REF-contraction, a proof can alternatively be obtained by a straightforward calculation of MATH, based on its REF . Since MATH it is obvious that MATH is a partial isometry with MATH. Next we show th... |
math/0006224 | The boundedness of the integrand of the curvature justifies application of the NAME Dominated Convergence Theorem to interchange limit and integral in REF of curvature: MATH REF was obtained by interchanging the infinite sum and integration. This is justified because for fixed MATH, the infinite sum converges absolutel... |
math/0006224 | Note that MATH the sums converging in the strong operator topology. Multiply REF by MATH on the left, take the trace of both sides, and suppose we can justify an interchange of sum and trace, obtaining MATH . Then the following simple calculation establishes the lemma: MATH where the last line was obtained from REF wit... |
math/0006224 | Let MATH be an operator on a NAME space MATH satisfying the hypotheses of REF . First we sketch the simple proof that MATH must be NAME. The assumed finiteness of the rank of MATH implies that MATH is finite-dimensional. Since we have already noted that MATH, also MATH is finite-dimensional. That MATH must have closed ... |
math/0006225 | If MATH is a cone with three facets (that is, MATH and MATH) then clearly MATH holds. If MATH is not a cone, then it must have at least two vertices. Thus (by REF ) MATH has at least one edge (which we can tell from the vertex-facet incidences of MATH). This edge is contained in precisely two facets of MATH if MATH; ot... |
math/0006225 | If MATH has a bounded facet, then the maximum length of a chain in MATH is MATH, thus one can compute MATH from MATH in this case. REF proves the second statement of the proposition. |
math/0006225 | One can compute MATH from the vertex-facet incidences of MATH, and thus, one finds the graph of each facet of MATH. If all these graphs of facets are cycles then MATH is bounded and the statement is clear. Otherwise, due to the MATH-connectedness of MATH, there is a unique (up to reorientation) way to arrange the paths... |
math/0006225 | Let MATH be a vertex of a simple MATH-polyhedron MATH and let MATH be the facets of MATH that contain MATH. Then the edges and extremal rays containing MATH are precisely MATH . Since we can compute the edges of MATH from a vertex-facet incidence matrix, we can thus also deduce (combinatorially) the extremal rays of MA... |
math/0006225 | By REF , MATH is homeomorphic to a sphere. The induced subcomplexes MATH and MATH cover all vertices (that is, one-element chains of MATH) of MATH. Using barycentric coordinates, it is seen that MATH retracts onto MATH. Thus, MATH has the same homotopy-type as MATH, where the latter is a simplicial sphere minus an indu... |
math/0006225 | Setting MATH defines an order preserving map from MATH into itself such that each face MATH is comparable with its image MATH. From the Order Homotopy Theorem CITE, we infer that MATH is homotopy-equivalent to the image MATH. In fact, MATH, and hence MATH is a strong deformation retract of MATH. This proves the lemma, ... |
math/0006225 | The reduced NAME characteristic of a MATH-sphere equals MATH, while the reduced NAME characteristic of a contractible space vanishes. Thus the claim follows from REF , and REF . |
math/0006225 | Suppose that there are two facets MATH and MATH of MATH REF with MATH. Since MATH, and since MATH is connected, there must be a vertex MATH that is a neighbor of some vertex MATH. Hence, we have MATH. Because of MATH and MATH this implies MATH or MATH, which in both cases yields a contradiction to MATH. |
math/0006225 | If MATH, then the claim obviously is correct. Therefore, assume MATH. Since MATH is connected, the vertices in MATH REF can be ordered such that MATH is adjacent to some vertex of MATH for each MATH (additionally, define MATH). Clearly MATH, since vertex MATH has MATH facets in common with some vertex in MATH. Define M... |
math/0006225 | Since MATH is a connected subgraph of the connected graph MATH (which has MATH vertices), there is a chain MATH with MATH, such that MATH and MATH is connected for all MATH. For every MATH, the vertex MATH with MATH is connected to some vertex MATH. From MATH we infer MATH, and thus, MATH. Together with MATH (since MAT... |
math/0006225 | Let MATH be a cycle of size MATH in MATH. In the following, all indices are taken modulo k. For MATH define the set MATH of size MATH. Clearly, MATH is connected, and, by REF , there exists exactly one facet MATH with MATH. Due to MATH, the facets MATH are pairwise distinct. This means that MATH (since MATH is simple) ... |
math/0006225 | Let MATH be a cycle in MATH of size MATH. For each MATH define MATH. Taking all indices modulo MATH, we have MATH for each MATH, and hence, there are facets MATH and MATH with MATH . It follows that MATH . If MATH is not complete, then MATH holds, and we infer from REF that MATH, which implies MATH (with MATH). Due to ... |
math/0006225 | Assume MATH is a tree. Let MATH be a leaf of MATH with MATH being the unique vertex of MATH adjacent to MATH. Due to MATH, there is one facet that induces a subgraph of MATH in which MATH is isolated. This, however, is a contradiction to REF . |
math/0006225 | For the ``if"-direction of the proof, let MATH be a polyhedron with a vertex-facet incidence matrix MATH (MATH). The cases MATH (implying MATH) as well as MATH (implying MATH) are trivial. Therefore, let MATH. Obviously, it suffices to show MATH. To each row MATH of MATH there corresponds a facet MATH of MATH. For MATH... |
math/0006225 | If MATH, then MATH is a vertex-facet incidence matrix of a MATH-simplex. Hence, by REF , MATH cannot be unbounded, and thus it must be a MATH-simplex as well. Therefore, in the following we will assume MATH. Let us first treat the case MATH. Consider the facets MATH and MATH corresponding to rows MATH and MATH, respect... |
math/0006227 | For MATH, MATH, MATH, we put MATH . With this notation the modules MATH, MATH, MATH, MATH, MATH and MATH are mutually isomorphic, as well as the modules MATH, MATH and all obtained from them by cyclic permutation of colors. For example, the map MATH and its inverse are depicted below. MATH . Identifying these modules a... |
math/0006227 | If MATH is transparent, then this morphism is equal to MATH, which is nonzero. Conversely, if this morphism is nonzero, it is equal to MATH for some MATH. Then, for any MATH, we have MATH . The second equality holds by the sliding lemma. |
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