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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 where MATH is a primitive MATH root of unity. The element MATH generates a MATH action on MATH. This can be lifted to a map MATH which acts on the spinor bundles MATH and MATH by choosing a square root MATH of MATH and letting MATH where MATH is a spinor in MATH, viewed as a bundle of quaternions. There are two slightly different cases, according to the parity of MATH. If MATH is even, then MATH is even, so MATH has order MATH; if MATH is odd then MATH has order MATH. Thus, letting MATH if MATH is even and MATH if MATH is odd, we see that multiples of MATH generate a free MATH action on MATH together with its spinor bundle. We apply the MATH-index theorem to any MATH, to compute the MATH-eta invariant. First note that the four-ball has non-negative scalar curvature, so the kernel and cokernel of the NAME operator vanish; in particular, the MATH-index vanishes. If MATH, the only fixed point of MATH is the origin, and it is easy to calculate the contribution of that fixed point to be MATH according to whether MATH is even or odd. It then follows from the MATH-index theorem that, for MATH, MATH . From REF , it follows that MATH . Since MATH has symmetric spectrum, the term MATH. The lemma follows. |
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 factor, MATH acts as the differential in the tensor product of infinitely many copies of NAME complexes, while MATH acts trivially; on the second factor, MATH acts trivially, while MATH acts as the differential in the tensor product of infinitely many copies of algebraic NAME complexes of MATH. The proof is completed by taking cohomology in MATH then in MATH. The case of MATH is similar. CASE: Similarly in the case of MATH, we have MATH, where MATH . Again we have MATH. Now MATH . Similar to REF , one sees that cohomology in MATH is MATH. Now any element of MATH is of the form MATH for some MATH. It corresponds to the field MATH . Given two elements MATH and MATH, by NAME 's theorem, we have MATH . Hence on the MATH-cohomology, the product induced from the normally ordered product is isomorphic to the ordinary exterior product on MATH. The case of MATH is similar. |
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 two operators MATH and MATH act such that MATH . In the above we have taken the MATH-cohomology first, then take the MATH-cohomology. We can also do it in a different order. By REF , the MATH-cohomology is MATH or MATH, its MATH-cohomology is the NAME cohomology. This completes the proof. |
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 where MATH and MATH. Hence we have MATH . If MATH, we have MATH. If MATH, we have MATH by the inductive hypothesis. |
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 MATH. |
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. Since MATH by REF , we have MATH. We have MATH, and hence the statement for MATH holds. By the symmetry of MATH it also holds for MATH. CASE: We take the reduced expression MATH such that MATH. We define MATH as in REF. Then MATH is a basis of MATH, and for MATH we have MATH. Hence we have MATH. Since MATH, we have MATH. Moreover, we have MATH for MATH. Hence if MATH for any MATH, then MATH for any MATH. Thus the assertion follows from the non-degeneracy 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 and MATH. We denote by MATH the set of the pairs MATH as above. By REF we have for MATH . On the other hand we have for MATH (see CITE and CITE). Hence we obtain MATH. |
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 from REF . |
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 case where MATH. We can easily show that MATH for MATH. Let us show MATH. For MATH we define MATH by MATH and MATH. We denote the weight of MATH by MATH. Then we have MATH. It is easy to show that MATH for any MATH. In particular MATH for any MATH. On the other hand we have the irreducible decomposition MATH and if MATH, then MATH. Hence MATH, and we have MATH. We obtain MATH for any MATH by the induction. |
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 orthonormal basis MATH, then we have MATH . It follows that MATH . Thus we have proved the lemma. |
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 MATH. Clearly, MATH is MATH-ample. Let MATH. Then MATH and MATH is ample. Therefore MATH can be found in the form MATH for suitable MATH. Take MATH for MATH so that MATH is lc but not klt (and MATH is ample). By CITE (see also CITE) there exists either a MATH, MATH, MATH, MATH, or MATH-complement of MATH which is not klt. |
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 Adjunction we have MATH . This implies MATH, a contradiction. By REF below, MATH, MATH, MATH is smooth along MATH and has only NAME singularities outside. Therefore, MATH and MATH is ample (see REF). Thus MATH is a NAME surface with at worst NAME singularities. Since MATH, we can write MATH, where MATH is an irreducible curve, MATH, MATH and MATH. Further, MATH . Thus MATH. Let MATH be the minimal resolution. By NAME 's formula, MATH. Thus, MATH and MATH. In particular, MATH either is smooth or has exactly one singular point which is of type MATH. By REF, MATH. Similar to REF we have MATH . Hence MATH, so MATH. This means that the curve MATH generates an extremal ray on MATH and MATH. Therefore, MATH is smooth and MATH. In this case, MATH is a rational ruled surface (MATH or MATH). Let MATH be a general fiber of the rulling. Then MATH . Hence MATH, so MATH is a fiber of MATH and MATH. This contradicts REF. |
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 with MATH. It is easy to see that all the assumptions of the lemma holds for this new MATH. Contractions of MATH-curves again preserve the assumptions. Since MATH and MATH are disjoint, whole MATH cannot be contracted. Thus we get MATH or MATH (a rational ruled surface). In both cases simple computations gives us MATH. |
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 (because, in this case, the map MATH must be an isomorphism in codimension one). If MATH, then MATH is not equidimensional, a contradiction. Finally, we consider the case MATH (and MATH is the central fiber of MATH). Let MATH be a general fiber of MATH (a NAME surface with at worst NAME singularities). Consider the exact sequence MATH . By NAME Vanishing CITE, MATH. Hence there is the surjection MATH . Here MATH (because MATH is NAME and ample). Therefore, MATH. Let MATH be any member. Take (positive) MATH so that MATH is lc and not plt. Clearly MATH, so MATH is MATH-nef. By Base Point Free Theorem CITE there is a complement MATH, where MATH for sufficiently big and divisible MATH and this complement is not plt, a contradiction with exceptionality (see CITE). |
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 in CITE one can show that MATH . This group is nontrivial if MATH, a contradiction. |
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 that MATH is a (singular) rational curve. Notation as above. Let MATH be the proper transform of MATH on MATH. If MATH is not a flipping contraction, then MATH is ample over MATH for MATH. In particular, all nontrivial fibers of MATH are contained in MATH. We claim that MATH is not nef over MATH for MATH. Indeed, assume MATH. Take MATH, MATH and let MATH be any compact irreducible curve. Clearly, MATH is an isomorphism along MATH. Let MATH be the proper transform of MATH on MATH. Since MATH, we have MATH. The curve MATH cannot generate an extremal ray (because extremal contractions on MATH are flipping). If MATH is nef over MATH, then taking into account that MATH we obtain MATH, a contradiction. Thus we may assume that MATH is a (prime) divisor. Then the exceptional locus of MATH is compact. If MATH is nef, this implies MATH. Again we have a contradiction because the proper transform of MATH does not coincide with MATH. We prove our lemma by induction on MATH. It is easy to see that MATH is ample over MATH. The NAME cone MATH is generated by two extremal rays. Denote them by MATH and MATH, where MATH (respectively, MATH) corresponds to the contraction MATH (respectively, MATH). Suppose that our assertion holds on MATH, i. CASE: MATH. By our claim above, MATH and after the flip MATH we have MATH. This completes the proof of the lemma. CASE: Let us consider the case when MATH is not a flipping contraction. Let MATH be the proper transform of MATH on MATH. If MATH, then MATH cannot contract MATH. Thus MATH is well defined. Now we need to show only that on each step of REF no flipping curves MATH pass through singular points of MATH. (Then MATH is an isomorphism near singular points of MATH and we are done). By REF all flipping curves MATH are contained in MATH. Therefore we can reduce problem in dimension two. The last claim MATH easily follows by the lemma below. Let MATH be a birational contraction of surfaces and let MATH be a boundary on MATH such that MATH is klt and MATH is MATH-ample. Put MATH. Assume that MATH does not contract components of MATH. Then MATH is smooth on MATH. Assume the converse and let MATH. Let MATH be a component of MATH passing through MATH. Then MATH. There is a MATH-complement MATH of MATH near MATH for MATH (see CITE, CITE, or CITE). By the definition of complements, MATH for all MATH. In particular, MATH whenever MATH, i. CASE: MATH. This means that MATH is lc. Since MATH, MATH is not plt at MATH. Therefore MATH near MATH and MATH is not a component of MATH. We claim that MATH is lc. Indeed, MATH is lc at MATH (because both MATH and MATH are smooth at MATH). Assume that MATH is not lc at MATH. Then MATH is not lc at MATH and MATH for MATH. This contradicts Connectedness Lemma CITE, CITE. Thus MATH is lc and we can apply Adjunction: MATH . Since MATH over MATH and MATH, we have MATH. On the other hand, the coefficient of MATH at MATH is equal to MATH, a contradiction. CASE: Finally let us consider the case when MATH is a flipping contraction. If MATH is ample over MATH for MATH, then we can use arguments of REF. From now on we assume that MATH is nef over MATH for some MATH. Let MATH be an effective divisor on MATH passing through MATH. Take MATH so that MATH is lc but not klt. By Base Point Free Theorem CITE, there is a member MATH for some MATH such that MATH is lc (but not klt). Thus, we may assume MATH. Let MATH be the crepant pull-back. Write MATH, where MATH. Then MATH is nef over MATH and trivial on fibers of MATH. Run MATH-MMP over MATH. Since MATH, this MATH-divisor cannot be nef until MATH is not contracted. Therefore after a number of flips we get a divisorial contraction: MATH . Since MATH the cone MATH has exactly two extremal rays. Hence the sequence REF is contained in REF. MATH is nef over MATH and MATH is ample over MATH for MATH. Clearly, MATH is ample over MATH (because MATH cannot be nef over MATH). After the flip MATH we have that MATH is ample over MATH. Continuing the process we get our claim. Further, MATH has only terminal singularities. Indeed, MATH is MATH-factorial, MATH and MATH is an isomorphism in codimension one. Therefore, one of the following holds: CASE: MATH is ample over MATH, then MATH; CASE: MATH is numerically trivial over MATH, then so is MATH, a contradiction; CASE: MATH is ample over MATH, then MATH is a flip and MATH has only terminal singularities CITE. This shows also that MATH is a plt blowup. Then we can replace MATH with MATH and apply arguments of REF. This finishes the proof of REF . |
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 looks like REF with some MATH such that MATH. It follows from the coassociativity of MATH that MATH satisfies the equation MATH for some invariant element MATH. The element MATH satisfying REF can be obtained by a correction of some MATH only obeying REF , CITE. This procedure also uses a simple cohomological argument, which proves REF. |
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, the function MATH is constant on the cosets of MATH. Let MATH is the set of cosets on which this function has the value MATH. It follows from REF that MATH is a semilinear subset of MATH. Let MATH be a semilinear subset of MATH projecting on MATH. Then it follows from REF that for MATH, which proves REF. Let MATH be the set of simple quasiroots corresponding to MATH. Let MATH. It is clear that starting the recursive procedure with MATH for MATH arbitrary but close enough to MATH, the denominators in REF remain not equal to zero. This proves that any point of MATH is non-singular and MATH has dimension MATH. Let us prove that MATH is connected. Fix a set of positive quasiroots, MATH, and the corresponding set of simple quasiroots, MATH. We say that a MATH-tuple of complex numbers MATH is admissible, if starting with MATH one obtains a solution of REF by the recursive procedure. It is clear that the admissible tuples form a subset, MATH, of MATH complement to an algebraic subset of lesser dimension, therefore MATH is connected. On the other hand, the set of points MATH such that MATH form an admissible MATH-tuple is obviously dense in MATH. This proves the connectness of MATH. |
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 nondegenerate bivector field, and therefore it defines a MATH-linear isomorphism MATH, MATH, which can be extended up to the isomorphism MATH of MATH-forms onto MATH-vector fields for all MATH. Using NAME identity for MATH and invariance of MATH, one can show that MATH gives a MATH invariant isomorphism of these complexes, so their cohomologies are the same. Since MATH is simple, the subcomplex of MATH invariants, MATH, splits off as a subcomplex of MATH. In addition, MATH acts trivially on cohomologies, since for any MATH the map MATH, MATH, is homotopic to the identity map, (MATH is a connected NAME group corresponding to MATH). It follows that cohomologies of complexes MATH and MATH coincide. But MATH gives an isomorphism of complexes MATH and MATH. So, cohomologies of the latter complex coincide with NAME cohomologies, which proves REF for MATH being NAME brackets. Denote by MATH the variety of points MATH satisfying REF . Consider the family of complexes MATH, MATH. It is clear that MATH depends algebraicly on MATH. It follows from the uppersemicontinuity of MATH and the fact that MATH for odd MATH, CITE, that MATH for odd MATH and almost all MATH. Using the uppersemicontinuity again and the fact that the number MATH is the same for all MATH, we conclude that MATH for even MATH and almost all MATH. We have that there is a number MATH and MATH such that REF holds for MATH. Since the cohomologies do not change when MATH replaces by MATH with any number MATH, we conclude that for any MATH there is MATH satisfying REF . Since by REF MATH is connected, it follows that REF holds for almost all MATH. Let us prove REF. Since MATH, MATH coincides with MATH. According to REF , let us choose a system of positive quasiroots MATH such that MATH has the form MATH, and for MATH. By REF , any MATH such that MATH has the form MATH where MATH obey the equation MATH . Starting with arbitrary values MATH for simple quasiroots we find MATH for all MATH recursively using the formula following from REF MATH . Indeed, denominators of REF are not equal to zero by choosing of MATH. Moreover, assume MATH are positive quasiroots such that MATH are also quasiroots. Then the number MATH can be calculated in two ways, using REF for the pair MATH on the right hand side and also for the pair MATH. But it is easy to check that these two ways give the same value of MATH. So we see that MATH the number of simple quasiroots for all MATH. |
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 support of MATH is an algebraic subset of MATH of lesser dimention. Hence MATH. |
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 naturally identified with the space of homogeneous polynomials of two variables of degree MATH. Taking as the variables MATH and MATH, one can consider the space MATH as the space of global sections of a vector bundle, MATH, that is a direct sum of MATH copies of MATH. It is obvious that MATH defines a map of sheaves, MATH. Denote MATH. It follows from REF that MATH is a map of bundles, that is, its image is a direct subsheaf of MATH. Hence, MATH is a sheaf over MATH without torsion. But over a neighborhood of the point of MATH with homogeneous coordinates MATH. This follows from the fact that MATH is a nondegenerate NAME bracket and, therefore, MATH. Since MATH is connected, MATH over MATH. So, MATH. Consider the exact sequence of sheaves over MATH: MATH . To complete the proof of REF one needs to show that any global section of MATH can be lifted to a global section of MATH. But this follows from the fact that MATH. To prove the last fact we observe that MATH is a subbundle of MATH, therefore, is a direct sum of a number of copies of MATH. Since MATH, MATH. This implies that MATH, too. |
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 MATH terms appear because MATH is a biderivation and, therefore, a NAME cocycle. The fact that the presence of MATH does not interfere with the cocyle condition and that this equation defines a NAME MATH-cocycle was proven in CITE (see the proof of REF there). By REF the differential NAME cohomology of MATH in dimension MATH is the space of holomorphic families of MATH-polyvector fields on MATH parameterized by MATH. Since MATH is reductive, the subspace of MATH invariants splits off as a subcomplex and has cohomology given by MATH. The complete antisymmetrization of a MATH-tensor projects the space of invariant differential MATH-cocycles onto the subspace MATH representing the cohomology. The equation MATH implies that the obstruction cocycle is a coboundary, and we can find a MATH-cochain MATH, so that MATH satisfies MATH . Assume we have defined the deformation MATH to order MATH such that MATH associativity holds modulo MATH, then we define the MATH-st obstruction cocycle by MATH . In CITE REF it is shown that the usual proof that the obstruction cochain satisfies the cocycle condition carries through to the MATH associative case. The coboundary of MATH appears as the MATH coefficient of the signed sum of the compositions of MATH with MATH. The fact that MATH mod MATH together with the pentagon identity implies that the sum vanishes identically, and thus all coefficients vanish, including the coboundary in question. Let MATH be the projection of MATH on the totally skew symmetric part, which represents the cohomology class of the obstruction cocycle. The coefficient of MATH in the same signed sum, when projected on the skew symmetric part, is MATH which is the coboundary of MATH in the complex MATH. Thus MATH is a MATH cocycle. By REF , this complex has zero REF-cohomology. Now we modify MATH by adding a term MATH with MATH and consider the MATH-st obstruction cocycle for MATH. Since the term we added at degree MATH is a NAME cocyle, we do not introduce a MATH term in the calculation of MATH and the totally skew symmetric projection MATH term has been modified by MATH. By choosing MATH appropriately, we can make the MATH-st obstruction cocycle represent the zero cohomology class, and we are able to continue the recursive construction of the desired deformation. |
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 is a bivector field. Applying MATH to MATH we obtain MATH . Observe now that MATH is a MATH cocycle, that is, MATH. This follows from the fact that MATH is a MATH associative multiplication. Indeed, if MATH is not equal to zero, its contribution to MATH is not a NAME coboundary. So, MATH is a tangent vector to MATH at MATH. Since, by REF, MATH is without singularities, there exists a formal path in MATH of the form MATH. Let MATH be the path that in local coordinates on a neighborhood of MATH in MATH is the sum MATH. It is clear that MATH. Putting in MATH instead MATH does not change the coefficients by MATH, MATH, in MATH and changes the MATH-st coefficient adding MATH to it. So we have MATH as required . |
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 in the proof of REF shows that MATH is a NAME cocycle. This means that all coefficients MATH are NAME cocycles. Hence, MATH for all MATH, where MATH. Therefore, MATH where MATH, MATH. The element MATH is a cocycle from MATH see Subsection REF. By REF there exists MATH such that MATH. This shows that, as in the proof of REF , we can modify MATH adding a multiple of MATH to get a new multiplication to order MATH with MATH-st obstruction cocycle MATH being a NAME coboundary, MATH. So, we are able to continue the recursive construction of the desired two parameter deformation. |
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 and the space MATH. Thus one may regard MATH as a sum of terms of the form MATH where MATH, MATH. Since MATH is invariant, MATH must be of weight zero with respect to the NAME subalgebra MATH. Since MATH is vanishing on constants, either MATH or MATH must belong to a positive symmetric power of MATH, let such be MATH. But the corresponding to MATH differential operator takes the functions of MATH to zero, therefore, the bidifferential operator corresponding to MATH when applies to the pair MATH gives zero, too. |
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 particular, MATH and MATH. Let MATH be the natural projection. Let MATH be an open set in MATH. Sections of MATH over MATH can be identified with functions MATH such that MATH for MATH, MATH. Sections of MATH must, in addition, satisfy the condition MATH, MATH, and the root vector MATH acts on MATH as a complex left-invariant vector field on MATH. In particular, functions of MATH are identified with functions MATH over MATH such that MATH for all MATH. Let us write MATH in the form MATH where each MATH, MATH, belongs to MATH and MATH belongs to MATH (MATH denote the space of symmetric tensors over MATH, and so on). The total degree of each MATH is greater than zero. Let us take MATH, MATH and apply MATH to MATH . Suppose MATH and MATH are not equal to zero. Then there are MATH such that MATH, MATH, where MATH. But since MATH is an invariant element, it must be of degree zero under the NAME subalgebra. It follows that MATH has to be of the form MATH where MATH, MATH and MATH. Hence, MATH. So, we have proven that in REF all terms except for the first when applying to MATH are equal to zero. |
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, actually, for smooth complex valued functions on MATH. The universality of MATH follows from the universality of MATH, see REF . The MATH-commutativity of MATH for holomorphic functions follows directly from the commutativity of MATH for such functions, see REF . |
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 multiplications MATH and MATH make MATH into a two-sided MATH module. The same argument as in the proof of REF shows that those multiplications define, in fact, the structure of a MATH module on MATH. |
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. CASE: An easy induction on MATH (with fixed MATH). CASE: We show this by induction on MATH. Suppose that MATH, so MATH, and MATH are distinct. If MATH for some MATH, then by the inductive hypothesis we find MATH such that MATH . If MATH, MATH and MATH are distinct, then look at the definition of MATH, MATH - these two values cannot be equal (and both are distinct from MATH). Finally suppose that MATH is limit, so MATH. Take MATH such that MATH and apply the inductive hypothesis to MATH getting MATH such that MATH (and both are not MATH). CASE: Again, it goes by induction on MATH. First consider a limit stage, and suppose that MATH is a limit ordinal, MATH and MATH. Let MATH be such that MATH. By the inductive hypothesis we find MATH such that MATH. Applying clause REF we may conclude that this MATH is as required. Now consider a successor case MATH. Let MATH, and let MATH be MATH if MATH, and be MATH otherwise. Apply the inductive hypothesis to MATH and MATH to get suitable MATH, and note that this MATH works for MATH and MATH too. REF, CASE: Straightforward. |
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 show that MATH is MATH - cc, suppose that MATH is a sequence of distinct conditions from MATH. We may find a set MATH such that CASE: conditions MATH are pairwise isomorphic, CASE: the family MATH forms a MATH - system with heart MATH, CASE: if MATH are from MATH then MATH . Take an increasing sequence MATH of elements of MATH, let MATH, MATH (for MATH), and look at MATH. It is a condition in MATH stronger than all MATH's. |
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 where MATH is the isomorphism from MATH to MATH. Now it should be clear that the mapping MATH is the order preserving isomorphism (remember clause MATH of the definition of MATH), and it has the property described in MATH. REF; MATH, MATH. Let MATH . For MATH, let MATH be the MATH - component of MATH such that MATH. The sets MATH (for MATH) are MATH - closed for every MATH, and clearly MATH. Hence, by the inductive hypothesis, MATH (for each MATH), and the order preserving mappings MATH satisfy the demand in MATH. Let MATH. Then, as MATH and MATH are isomorphic and the isomorphism is the identity on MATH, we have MATH. Hence MATH, and therefore MATH. But since the mappings MATH are order preserving, the last equality implies that MATH, and hence MATH is a function, and it is an order isomorphism from MATH onto MATH satisfying MATH. |
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 successor ordinal, say MATH. So let MATH and suppose that MATH is such that MATH. If MATH then MATH and we immediately get a contradiction (applying the inductive hypothesis to MATH). So let MATH be such that MATH. We know that MATH (remember clause MATH of the definition of MATH), so we may take functions MATH such that MATH, MATH, MATH. Let MATH be such that MATH, MATH for MATH (where MATH is the order isomorphism from MATH to MATH). Now one easily checks that MATH (remember the definition of the term MATH). By our choices, MATH for all MATH, and MATH, and this is a clear contradiction with the choice of MATH and MATH. CASE: Suppose that MATH is a MATH - name for a MATH - sequence of distinct members of MATH and let MATH. Applying standard cleaning procedures we find a set MATH of the order type MATH, an ordinal MATH and MATH and MATH such that MATH, MATH, MATH and MATH where MATH is identified with MATH by the increasing enumeration (so we will think MATH). For MATH let MATH. Since MATH were (forced to be) distinct we know that MATH for distinct MATH. Hence MATH (for each MATH) and therefore we may find functions MATH such that MATH, and MATH, MATH, and if MATH, and MATH is the isomorphism from MATH to MATH, then MATH. Now fix MATH and let MATH . It should be clear that MATH is a function from MATH to MATH, and moreover MATH. Also easily MATH . Hence we may conclude that MATH for MATH (remember the definition of MATH and REF ). Consequently we get MATH, finishing the proof. |
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, and since the forcing notion MATH is MATH - strategically closed (by REF ), it is enough to show that MATH for each MATH. So suppose that MATH is such that MATH. Then we find a Boolean term MATH, an integer MATH and sets MATH (for MATH) such that MATH . For each MATH use REF to choose a finite MATH - closed set MATH containing the set MATH . Look at MATH (see REF ). There are only MATH possibilities for the values of MATH, so we find MATH such that CASE: MATH, MATH, CASE: if MATH is the order isomorphism then MATH is the identity on MATH, CASE: if MATH is the order isomorphism, then MATH. Note that, by REF , MATH and the order isomorphism MATH satisfies MATH and hence MATH is the identity on MATH (remember REF ). For a function MATH let MATH be defined by MATH . For each MATH, MATH. By induction on MATH we show that for each MATH - component MATH of MATH, the restriction MATH is in MATH. If MATH is limit, we may easily use the inductive hypothesis to show that, for any MATH - component MATH of MATH, MATH. Assume MATH and let MATH be a MATH - component of MATH. We will consider four cases. CASE: MATH. Then MATH and MATH for each MATH. Since, by the inductive hypothesis, MATH for each MATH, we may use the definition of MATH and conclude that MATH (remember the definition of the term MATH). CASE: MATH. Let MATH be the increasing enumeration. Then MATH for some MATH and MATH, MATH, MATH. Moreover, if MATH (the increasing enumeration), MATH, then for MATH: MATH . Note that MATH, so if MATH, then we may proceed as in the previous case. Therefore we may assume that MATH. So, for each MATH we may choose MATH such that MATH (remember REF ). Let MATH (if MATH, then let MATH be any element of MATH). Note that then MATH [Why? Remember REF and the clause MATH of the definition of MATH.] By REF , we find a MATH - component MATH of MATH such that MATH and MATH . We claim that then CASE: MATH. Why? Fix MATH. Let MATH be components of MATH such that MATH, MATH, MATH, MATH (so MATH and MATH, and MATH, are isomorphic). The sets MATH and MATH are MATH - closed, and they have the same values of MATH, and therefore MATH and MATH are (order) isomorphic. Also, these two sets are included in MATH and MATH, respectively. So looking back at our MATH, we may successively choose MATH, MATH, and MATH such that CASE: MATH, CASE: MATH, and CASE: MATH. Then we have MATH . To conclude MATH it is enough to show that MATH. If this equality fails, then there is MATH such that MATH. If MATH, then necessarily MATH, and this is impossible (remember MATH for MATH). So MATH. If MATH, then MATH and (by the choice of MATH) MATH. Then MATH and MATH, and also MATH (by the choice of MATH), a contradiction. Thus necessarily MATH (so MATH) and therefore MATH (as the last is not MATH), again a contradiction. Thus the statement in MATH is proven. Now we may finish considering the current case. By the definition of the function MATH (and by the choice of MATH) we have MATH (and MATH is order - preserving). Therefore MATH . By the inductive hypothesis, MATH (for MATH), so as MATH (and hence MATH) we may conclude now that MATH. CASE: MATH . Similar. CASE: MATH . If MATH, then MATH and we are easily done. If one of the intersections is non-empty, then we may follow exactly as in the respective case (REF or REF). Now we may conclude the proof of the theorem. Since MATH we find MATH such that MATH and MATH. It should be clear from the definition of the function MATH (and the choice of MATH) that MATH . But it follows from REF that MATH, a contradiction. |
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 induction we find a sequence MATH so that for all MATH lies over MATH and MATH is not covered by a finite subcollection of MATH. The sequence defines an element MATH. Since MATH, we have MATH and so MATH for some MATH. But if MATH is stable at level MATH, then MATH, which contradicts a defining property of MATH. |
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 . |
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 subgroup. So the class of MATH in MATH is MATH times the class of MATH. Since MATH has also that structure, the classes of MATH and MATH in MATH coincide. Hence the same is true for MATH and MATH. |
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 coefficients in MATH. Inspection of the proof shows that this actually holds in MATH. |
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 shows that this has the same reduction modulo MATH as MATH. |
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 expression with MATH and MATH interchanged. Since MATH, we find that for MATH, the value of MATH on MATH equals MATH (the action of MATH is trivial here). Notice that MATH is embedded in MATH as MATH in MATH. From the above discussion one sees that MATH takes on this set the value MATH. If we substute the defining equation for MATH, the Lemma follows. |
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 expansions converge uniformly and absolutely and so MATH . Since summation and integration may be interchanged, only the terms with MATH remain and hence MATH. If MATH respectively, MATH denotes the set of poles of MATH respectively, MATH, then the integrand has polar set MATH (there is no pole in MATH or MATH) and the poles enclosed by the circle of integration are those in MATH. By the theory of residues, MATH must then be equal to the sum of the residues of the integrand at MATH. This description no longer requires MATH and defines an analytic extension of MATH to the complement of MATH. This extension is easily seen to be meromorphic at MATH. To compute its behavior at MATH, we note that MATH converges for MATH on a neighborhood of MATH absolutely (with all its derivatives) to the constant function MATH. So as MATH, MATH tends to the sum of the residues of MATH at MATH. This sum is opposite to the residue at the remaining pole MATH, hence equal to MATH. In particular, MATH is a rational function with polar set contained in MATH. Assume now that MATH. A pole of an element of MATH in MATH satisfies an equation MATH for certain integers MATH, MATH. A product of such poles satisfies a similar equation, and this implies that a product of MATH and a finite set of polynomials of the form MATH is in MATH. Since the expansion of MATH at MATH has integral coefficients, this product lies in MATH. |
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 righthand side is in MATH and takes the value zero at MATH. Since MATH is in MATH with value zero at MATH it follows from REF that the same is true for MATH. Likewise for MATH. So MATH is in MATH and has value MATH at MATH. |
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 last MATH basis vectors is then just MATH. We investigate which MATH appear as the constant coefficient of a MATH with the property that MATH. We first do this for the complete intersection defined by MATH. This complete intersection contains MATH and the irreducible component that contains the image of MATH lies in MATH. So we want MATH for MATH. By expanding at MATH this amounts to identities of the form MATH with MATH the derivative of MATH at MATH and MATH. Equivalently: MATH . All the terms are regular and the reduction modulo MATH yields the MATH-th unit vector in MATH. NAME 's lemma says that a solution MATH exists if and only if MATH solves this set of equations modulo MATH. This just means that MATH. In particular, we see that for all MATH, MATH is isomorphic to an affine space and hence is irreducible. This implies that all elements of MATH map to the same irreducible component of the common zero locus of MATH. It follows that MATH. The last assertion is easy. |
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 then implies MATH. The difference MATH defines a MATH-derivation MATH over MATH and hence a MATH. Since MATH, this element annihilates the torsion of MATH. This is then also true for its reduction modulo MATH and it follows from the fact that MATH that this reduction is of the form MATH for some MATH. Regard MATH as a MATH-derivation MATH and let MATH be such that MATH represents MATH. Then MATH and MATH. Replace MATH by MATH and continue with induction on MATH. So MATH defines a MATH-derivation MATH and hence a MATH-derivation MATH (because MATH). The latter is zero if and only if MATH. This proves that the fiber of MATH through MATH is an affine space over the kernel of MATH, MATH, which has length MATH. The last assertion is easy. |
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 space MATH and the number of elements of the presumed monomial basis. By MATH denote the set of monomials with the restrictions REF . Also let MATH and let MATH be the number of elements in MATH. We have CASE: MATH. CASE: MATH. CASE: Let MATH. Then either MATH and MATH, or MATH and MATH. And conversely, for any element of MATH or MATH we can find the corresponding element in MATH. CASE: Introduce the map MATH defined by MATH. It is easy to see that the image of the ideal MATH under this map is zero, so MATH induces the map MATH. In particular we obtain MATH. It is also clear that the image of the ideal MATH under the natural projection MATH is zero, so MATH. As MATH, we have the statement of the lemma. From the equality MATH for MATH and also for MATH it follows that the inequality in REF is in fact an equality. So we have the equality MATH for all MATH and the statement of the theorem. |
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 these fractions that doesn't depend on MATH. It easy to prove by induction that for MATH we have MATH . Using the second formula for MATH we obtain the statement of the lemma. Now prove the theorem. For MATH the proof is straightforward. Here we already have two cases depending on MATH and MATH. Suppose that all the statements of the theorem are proved for MATH and prove them for MATH. Let MATH be odd. By REF we have MATH. As MATH, the induction hypothesis implies the equality MATH. Induction on MATH based on REF shows that MATH if MATH, which is REF follows from REF by applying the transformation of the space MATH mapping MATH to MATH. Now let MATH be even. The proof of REF is similar to the proof of REF . For MATH REF follows from REF . Then REF for an arbitrary MATH can be proved by similar induction on MATH. |
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 that MATH. We have MATH . Since the rank of an operator matrix is at least as large as the rank of any entry, if MATH is infinite, so is MATH. Thus we may assume that MATH is finite. Let MATH, denote the MATH'th column of the matrix for MATH, considered in the obvious way as operators, for example, MATH. Then MATH . Since the domain of MATH is MATH, this implies that MATH, and hence MATH. It is well known (for example, REF) that MATH, so MATH . Since MATH is an isometry, the map MATH, defines an isometric bijection between MATH and MATH. Hence MATH. |
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 absolutely with sum of absolute values bounded above by MATH. REF is justified as follows. For fixed MATH, consider the decreasing sequence MATH and set MATH. Then for any positive integer MATH, MATH . For sufficiently large MATH, the right side is arbitrarily close to MATH, showing that MATH thus proving REF . This proves the asserted formula for MATH. To prove the alternative formula for MATH, define MATH to be the partial isometry of REF with MATH. Recall from REF that MATH and check that MATH . The formula for MATH follows immediately upon combining these observations, REF , the formula just proved for MATH, and the cyclic property of the trace: MATH . |
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 with MATH and MATH interchanged and the summation index MATH replaced by MATH. The interchange of sum and trace required to justify the above calculation is not immediate because the trace is not continuous in the strong operator topology. However, the trace is well-known to be a normal linear functional, which implies that for any increasing sequence of trace class positive operators MATH converging in the strong operator topology to a trace class operator MATH, we have MATH . This property is a slight specialization of the definition of normality. It follows routinely from the definition MATH, with MATH an orthonormal basis. Noting that MATH and applying REF with MATH proves REF . I thank NAME for suggesting the above proof to replace the unattractive direct calculation of an earlier draft. |
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 range under these circumstances can be easily seen by noting that closed range is equivalent to a gap above REF in the spectrum of MATH. If there were not such a gap, then MATH would not have finite rank. To show that REF equals MATH, let MATH be the operator on MATH defined by the operator matrix: MATH . Let MATH be the partial isometry MATH of REF . Since MATH is a compact perturbation of MATH, MATH . Also, since MATH is a partial isometry, MATH and MATH. Hence by REF , MATH . |
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; otherwise, it is contained in more than two facets. |
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 that are the graphs of the unbounded facets of MATH as a cycle. From this cycle, it is easy to determine the incidences of extremal rays and facets of MATH, which then allow to reconstruct the entire combinatorial structure of MATH. |
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 MATH and the information which ray is contained in which facets. From that, we can deduce the entire face poset of MATH. |
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 induced ball. Hence, MATH is contractible. |
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, since MATH is a poset isomorphism from MATH onto MATH. |
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 MATH to be the last MATH, such that MATH, that is, MATH is the last index, where we encounter a new facet (MATH is well-defined due to MATH). Since this facet must contain MATH vertices from MATH, we have MATH, which yields MATH. Furthermore, we have MATH, since MATH intersects all facets. It follows MATH. |
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 MATH is simple) and MATH (by REF ), this implies MATH for all MATH. |
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) and MATH (since MATH is simplicial). Hence, every vertex that is adjacent to one of the nodes MATH must be contained in at least one (more precisely, in MATH) of the facets MATH, and thus it lies in MATH. Since MATH is connected, this means that MATH. For MATH this immediately yields that MATH is a complete graph on MATH nodes, while for MATH one finds that MATH is the cycle MATH (since, in this case, MATH if and only if MATH). |
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, REF implies MATH contradicting REF . |
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 define MATH. Clearly, MATH holds for MATH. Due to MATH it follows MATH and therefore MATH. Now MATH is a (decreasing) chain of length MATH in the face poset of MATH. Hence we have MATH. Since each vertex must be contained in at least MATH facets it follows that MATH (because each vertex of MATH is contained in precisely MATH facets). Conversely, let MATH be a simple and simplicial MATH-polyhedron REF with MATH vertices. The case MATH is checked easily. Thus, assume MATH. By REF , MATH either is a complete graph on MATH nodes or it is a cycle. In the first case, every vertex-facet incidence matrix of MATH is the complement of a permutation matrix, which can be transformed to MATH by a suitable permutation of its rows. In the second case, consider any vertex-facet incidence matrix MATH of MATH, where the columns are assumed to be ordered according to the cycle MATH. Call two positions MATH and MATH in MATH REF mates if MATH and MATH. Walking around the cycle MATH, we find that the total number of mates in MATH is precisely MATH (because every edge is contained in precisely MATH facets). But then, since every row of MATH has only MATH ones (because MATH is simplicial), it follows that in each row the ones must appear consecutively (modulo MATH). Denote by MATH the starting position of the block of ones in row MATH. Because there are no equal rows in MATH (by REF ) we deduce that MATH defines a permutation of the rows of MATH which tells us how to transform MATH to 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, respectively. If we identify the vertices of MATH with the column indices MATH of MATH, then the vertex set of the face MATH is MATH, where MATH (due to MATH). By REF , MATH is a MATH-cycle (due to MATH). Since neither vertex MATH nor vertex MATH, which are the only neighbors of MATH in MATH, are contained in MATH, we conclude that the subgraph of MATH induced by MATH is disconnected, which is a contradiction to REF . Hence, we can assume MATH. This implies MATH (where, again, all indices are to be taken modulo n), that is, MATH consists of all (cyclic) intervals of MATH with at least one and at most MATH elements. We will compute the NAME function MATH (see REF) on the lattice MATH (which arises by adding artificial top and bottom elements MATH and MATH to MATH). For each MATH let MATH. Obviously, for every MATH with MATH we have MATH. In particular, one readily deduces MATH and MATH. For MATH we then infer (by induction) MATH. Thus, we finally calculate MATH which by REF implies that MATH is bounded (and, hence, a MATH-gon). (Alternatively, one could derive from the Nerve Lemma CITE that MATH is homotopy-equivalent to a circle for MATH, and thus, MATH must be a polygon.) |
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 along the isomorphisms we get a symmetrized multiplicity module MATH; here only the cyclic order of colors is important. We will represent the elements of MATH by a circle with one incoming line (colored with MATH) and two outgoing ones (colored with MATH and MATH), the cyclic order of lines is MATH. The module MATH is dual to MATH. The natural pairing is non-degenerate, since we have no negligible morphisms. We denote by MATH, MATH, a basis of MATH, and by MATH the dual basis with respect to this pairing. Applying the domination axiom we get that the natural map MATH is an isomorphism. By writing the identity of MATH in the basis corresponding to MATH, we get the following decomposition formula (fusion formula) MATH . The calculations below establish the sliding property. MATH . In the first and third equalities we use the fusion formula, the second equality holds by isotopy. |
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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