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math/0004098 | It is given that both MATH and MATH satisfy REF relative to MATH, and further that MATH. Equivalently, by REF , MATH. Using MATH, we get MATH . Now multiplying through from the right with MATH on both sides in REF , the conclusion of the lemma follows. To see this, notice first from REF - REF that MATH . The proof of R... |
math/0004098 | Let MATH. Suppose MATH satisfies MATH, or equivalently MATH. Then by REF , we have MATH . The MATH-indices are in MATH. For the matrices MATH, we have the identities MATH . See REF above. The terms on the left-hand side in REF are MATH again with the convention that the terms are defined to be zero when the subscript i... |
math/0004098 | To show that a subspace MATH is minimal in the sense specified in the lemma, we must check that whenever MATH is a subspace satisfying MATH then MATH cannot be cyclic for the representation MATH of MATH, that is, it generates a cyclic subspace which is a proper subspace of MATH . The cyclic subspace generated by MATH i... |
math/0004098 | The two vectors MATH and MATH, corresponding to the endpoints in MATH, are special in that MATH and when MATH divides MATH, MATH . It follows that, when REF holds, then MATH where MATH the one-dimensional space of the constants, and neither of the two subspaces in this orthogonal sum is zero. Hence REF implies that MAT... |
math/0004098 | Once MATH has been chosen as in REF, the three REF - REF follow from REF . The significance of REF - REF is that they imply that if MATH then the fixed-point set MATH is in fact an algebra. This is a result of NAMECITE. Using REF , we conclude that the projections in MATH are characterized by the condition of REF . Now... |
math/0004098 | We will restrict to the case MATH, although for MATH, we cover arbitrary MATH in CITE. (If MATH, then MATH. Setting MATH, we showed in CITE that the density matrix MATH given by MATH satisfies MATH, where MATH and MATH, and where MATH is the adjoint of MATH with respect to the trace inner product. Defining the state MA... |
math/0004098 | REF : If MATH, then MATH, and therefore MATH that is, MATH. This means that the coefficient matrices in the expansion MATH satisfy the conditions in REF ; and, if we define matrices MATH by REF above, then it follows that MATH where MATH . Hence REF holds. CASE: This is clear from reading REF in reverse. The equivalenc... |
math/0004098 | The details are somewhat technical, and it seems more practical to first do them for the special case when MATH, and then comment at the end on the (relatively minor) modifications needed in the proof for the case when MATH is arbitrary MATH. Using the terminology of REF , we then get MATH where MATH and MATH, MATH, an... |
math/0004098 | Now, the functions MATH in REF are the matrix elements of the loop MATH, and so it follows from REF that each of them is a polynomial of degree at most MATH, say MATH (since MATH itself has degree MATH when MATH). Hence, MATH or equivalently, MATH . Now set MATH where MATH is as in REF . From REF , we then get MATH or ... |
math/0004098 | Since both MATH and MATH are MATH-invariant, the conclusion will follow if we check the inclusion MATH . For the space on the right-hand side in REF , we shall use the terminology-MATH, and similarly, the space spanned by all the spaces MATH will be denoted MATH. In REF , we vary the multi-index MATH over all the MATH ... |
math/0004098 | We will establish the conclusion by proving that if MATH is any subspace of MATH which is both MATH-invariant and cyclic, then MATH. So in particular, MATH does not contain a proper subspace which is both MATH-invariant and cyclic. Now suppose that some space MATH has the stated properties. Since it is cyclic, we must ... |
math/0004098 | The result in the theorem is now immediate from the three lemmas, and we need only comment on the size of the genus MATH. We argued the case MATH; but, for the general case, MATH is spanned by MATH, MATH, and the functions from the list REF , or equivalently REF , will then be MATH . Otherwise, all the arguments from t... |
math/0004098 | The proof comes down to a dimension count. Since MATH is of dimension MATH, we just need to check that the space MATH (MATH), spanned by the MATH functions in REF , is of dimension MATH for a generic set of loops MATH in MATH, and that can be checked by a determinant argument based on the conditions for the matrices MA... |
math/0004098 | For if not, the greatest common divisor MATH of the family REF would have a root MATH, that is, MATH. By REF , we would then have MATH where MATH is the originally given loop. Recall that, by REF , we may view MATH as an entire analytic matrix function, that is, an entire analytic function, MATH, whose restriction to M... |
math/0004098 | By REF , we have the estimate MATH . If MATH, then by REF , we get MATH . If MATH, then the value MATH is not a common root of the two complex polynomials MATH, MATH, and so by REF the two polynomials MATH and MATH have no common roots at all, by REF . Therefore, when REF are combined, we get MATH. Hence MATH, and the ... |
math/0004098 | CASE: In this example, MATH and MATH . Let MATH be the usual NAME in MATH, viewed as a column vector. By REF , we have MATH where MATH, MATH and MATH are projections, and as before, MATH. Here MATH and MATH, MATH are the MATH, MATH projections specified in REF (see also REF below). We need only check that MATH for some... |
math/0004098 | If MATH is divisible by MATH, then its six coefficients MATH must be of the form MATH, and the associated loop MATH, MATH . The corresponding MATH-conditions then yield: MATH . Substitution of the second into the first yields MATH. Hence, of the three numbers MATH, at most one, and therefore precisely one, can be nonze... |
math/0004100 | Let MATH be a NAME basis of the ideal MATH and MATH be a polynomial set which computed by the algorithm. Apparently, MATH. Consider MATH. If MATH is MATH-autoreduced, then MATH initiated as MATH in line REF does not change in the process of the algorithm, and, hence, MATH. Otherwise, consider the output polynomial sets... |
math/0004100 | Let MATH be a minimal NAME basis. From REF it follows that MATH is MATH-autoreduced. Thus, MATH is MATH-autoreduced. |
math/0004100 | MATH: Suppose MATH which, by REF , is also a MATH-basis of MATH, is not its MATH-basis. Our goal is to prove that a NAME basis of MATH is an infinite polynomial set. From REF it follows that MATH . Among nonmultiplicative prolongations MATH satisfying REF choose one with the lowest MATH with respect to the pure lexicog... |
math/0004100 | This follows from the above proof of REF . |
math/0004100 | MATH has a finite NAME basis CITE. If MATH is homogeneous, then its NAME basis is the reduced NAME basis CITE. |
math/0004103 | Recall the following formula: if MATH is any REF-form on a manifold, and MATH and MATH are vector fields, then MATH . See for example, CITE for an explanation. We choose local bundle coordinates on MATH and apply REF with MATH, MATH and MATH. Since MATH and MATH commute and MATH, we obtain MATH . Note that MATH and eva... |
math/0004103 | Assume to the contrary, that is, assume that for a general point MATH there exists a singular line MATH passing through MATH. Recall that the rational curve MATH can always be dominated by an integral singular plane cubic, that is, by a rational curve with a single node or cusp. We will reach a contradiction by constru... |
math/0004103 | It follows from the definition of the contact structure that MATH. Since MATH, and since vector bundles on MATH always decompose into sums of line bundles we may therefore write MATH where MATH. Thus, the splitting of MATH has exactly MATH positive entries. It follows that the splitting of MATH has at most MATH positiv... |
math/0004103 | The fact that MATH and MATH are immersive follows from CITE and REF . It follows from an argument of NAME that MATH is birational because all lines through MATH are smooth. For this, see CITE. |
math/0004103 | Since MATH, it follows from the estimate REF for the dimension of the parameter space that MATH . By NAME 's bend-and-break argument CITE, for a given point MATH, there are at most finitely many lines containing both MATH and MATH. It follows that MATH . CASE: The subvariety MATH is MATH-integral where it is smooth. CA... |
math/0004103 | We have already seen in REF that MATH is equidimensional of relative dimension MATH. Thus, MATH is a well-defined family of algebraic cycles over MATH in the sense of CITE and the universal property of the NAME yields a map MATH such that MATH is the pull-back of the universal family over MATH. Because MATH, it is clea... |
math/0004103 | Our argument involves an analysis of the deformations of MATH which are obtained by varying the base point MATH. We shall argue by contradiction and assume that the assertion of the proposition is wrong. With this assumption we will construct a family of morphisms MATH which contradicts REF , and we are done. We will n... |
math/0004103 | By CITE, the canonical bundle MATH is not nef. But then it has already been shown in CITE that MATH is automatically of type MATH if MATH. We may therefore assume without loss of generality that MATH is NAME and that MATH. Let MATH be a general point, and MATH a minimal rational curve through MATH. It follows from the ... |
math/0004104 | Consider the subset MATH which is dense in MATH. To see that MATH is a closed subspace of MATH, it will suffice to show that MATH is a subspace. Let MATH and MATH. Since MATH and MATH, it is clear that MATH and MATH. Moreover, since MATH it is clear that MATH, and hence also MATH, is invariant for MATH. To show that MA... |
math/0004104 | If MATH then MATH so MATH while MATH. This shows MATH. Suppose MATH and MATH. Then MATH. As an operator on MATH, MATH is invertible and its inverse has spectral radius MATH. Therefore MATH and hence MATH, so MATH. This shows that MATH, and therefore that MATH. |
math/0004104 | Let MATH be given by REF and let MATH . Letting MATH denote the algebraic tensor product of vector spaces, we clearly have MATH, so the inequality MATH holds in REF. Given a unit vector MATH let MATH be MATH. Then MATH, so if MATH then MATH for every MATH. Therefore MATH and MATH holds in REF. |
math/0004104 | Let MATH be a NAME space and MATH a normal, faithful MATH - representation. Using CITE, one finds a NAME space MATH such that the representations MATH and MATH are unitarily equivalent, via a unitary MATH. Now applying REF twice, we have MATH . Hence MATH. |
math/0004104 | Write MATH . Let us first fix MATH, MATH and let us denote MATH simply by MATH, MATH by MATH, MATH by MATH and the diagonal entries of MATH by MATH, (MATH). Now each word in MATH and MATH is a sum of words in MATH, MATH, MATH and MATH; hence we will investigate words of the form MATH for arbitrary MATH . Now, using the... |
math/0004104 | If MATH then let MATH. If MATH but MATH then let MATH. If MATH and MATH then let MATH . |
math/0004104 | We may without loss of generality assume that MATH is a transposition of neighbors: MATH . We will use REF to show that there is a unitary random matrix, MATH, such that MATH has the same distribution as MATH, and this will prove the theorem. Recall that MATH is the usual probability space underlying our random matrice... |
math/0004104 | We may introduce a uniformly distributed MATH - valued random variable, MATH, that is independent from MATH. Then MATH has the same distribution as the random matrix, MATH, which at a point MATH takes the value MATH . Now each MATH - moment of MATH is the average over MATH of the corresponding MATH - moments of the mat... |
math/0004104 | We may without loss of generality suppose MATH and MATH. That MATH follows from the last paragraph of CITE and the fact that the limit moments of each MATH are those of a centered semicircle law with second moment MATH. To show MATH we will use CITE and an idea from the proof of CITE. Let MATH be a nontrivial ultrafilt... |
math/0004104 | We will apply REF . Let us first show that the quantity REF remains bounded as MATH. Write MATH for the MATH-th entry of MATH. We have MATH where MATH. Using the generalized NAME inequality, we have MATH . But by REF , there is MATH such that MATH . Now consider MATH . From the nature of the entries MATH, we see that t... |
math/0004104 | We shall use REF and adapt the proof of CITE to our situation. The family MATH of sets of random variables is asymptotically free as MATH. With MATH each of MATH and MATH is in MATH and MATH is an independent family of sets of matrix - valued random variables. Thus, by REF , each MATH and each MATH converges in moments... |
math/0004104 | From MATH - freeness of MATH and MATH we get MATH for every MATH. Let MATH and let MATH be so that MATH for every MATH. For every MATH let MATH be such that MATH. In order for freeness of REF to hold, it will suffice that MATH . But the above equality follows directly from freeness of REF. |
math/0004104 | For brevity we shall prove REF simultaneously; while proving REF , we may from the outset assume that the stronger hypotheses of REF hold, because if we require MATH to be a single element then they will in any case be satisfied. We may write MATH where MATH is a random unitary matrix and MATH. For every MATH let MATH ... |
math/0004104 | Since any noncountable standard NAME space is NAME isomorphic to the unit interval, and since MATH is NAME isomorphic to the one - point compactification of MATH, it is no loss of generality to assume that MATH is a separable compact NAME space and MATH is the NAME MATH - algebra associated to this topology. For any co... |
math/0004104 | It will be more convenient to consider the measure MATH whose density with respect to NAME measure on MATH is MATH for some constant MATH; thus MATH is the push forward measure of MATH under the transformation MATH. We will find the limit as MATH of the marginal distributions, MATH, of MATH corresponding to the variabl... |
math/0004104 | Clearly for every non - random MATH unitary matrix MATH, the distribution of MATH is equal to the distribution of MATH. Let MATH be the symmetrized joint distribution of the eigenvalues of MATH and let MATH be the symmetrized joint distribution of the eigenvalues of MATH. For MATH let MATH, respectively MATH, be the ma... |
math/0004104 | Let MATH be the manifold of matrices in MATH having MATH distinct eigenvalues. Then MATH has full NAME measure in MATH. Let MATH be the NAME group of MATH unitary matrices, and let MATH be the manifold of all upper triangular MATH complex matrices, no two of whose diagonal elements are the same. Let MATH be given by MA... |
math/0004104 | This is quite similar to the proof of REF . Let MATH. Consider first the case MATH. Let MATH be the polynomials obtained via NAME - NAME orthonormalization of the sequence MATH in MATH. Then MATH . Using the NAME determinant we have MATH for some constant MATH. Therefore MATH . Writing MATH and noting that as MATH rang... |
math/0004104 | Using the NAME determinant we find MATH . Averaging REF over the action of MATH gives MATH . From this one easily sees that-MATH and that MATH is obtained from MATH by averaging. In order to show that MATH is obtained from MATH by averaging, it suffices to note that for any measure MATH on MATH, the average over the ac... |
math/0004104 | For every MATH let MATH be a MATH random matrix whose distribution has density with respect to NAME measure MATH . Then by REF , MATH converges in MATH - moments as MATH to a circular free NAME element of parameter MATH. By REF , each MATH has the same MATH - moments as MATH, where MATH, MATH is a diagonal MATH random ... |
math/0004104 | We may without loss of generality assume that MATH, which implies MATH and MATH is a trace. Let MATH and let MATH be the circular free NAME element of parameter MATH as in REF , where we take MATH to be a MATH - noncommutative probability space. Thus MATH is a circular free NAME element of parameter MATH, each MATH is ... |
math/0004104 | The positive part MATH of MATH has the same moments as the measure MATH on MATH whose density with respect to NAME measure is MATH with MATH and MATH. Since MATH has the polar decomposition MATH where MATH is a NAME unitary and where MATH and MATH are MATH - free, the MATH - moments of MATH can be expressed as certain ... |
math/0004104 | We know from general principles that MATH if MATH, and from REF we have that MATH; as MATH is faithful we conclude REF . Let us now prove REF . Arguing as in the proof of REF , we have MATH whenever MATH. We may take the MATH-noncommutative probability space MATH so that MATH is a faithful trace, in which case, since M... |
math/0004104 | Note that REF implies MATH. Let MATH be normally and faithfully represented on a NAME space MATH. For MATH we have MATH where the last inequality is because the spectral radius of MATH is MATH. Hence MATH. In order to prove the reverse inequality, it will suffice to show MATH for all MATH, because MATH is strong-MATH -... |
math/0004105 | CASE: Let MATH be a smooth point. It is clear. CASE: Let MATH be a NAME terminal singular point. Since we have MATH, then MATH or MATH. Hence, MATH . MATH . CASE: Let MATH be a terminal quotient singular point of type MATH with MATH . Let MATH for MATH. For MATH, we have MATH . Then by REF , MATH . |
math/0004105 | (compare [KMMT REF ]) By [MM REF ], there is a covering family of rational curves MATH such that MATH. If MATH has a fixed point MATH, then REF , we have MATH. We have MATH. By CITE, we have MATH. Hence MATH in this case. If MATH has a fixed point MATH, the proof is the same as the one of [KMMREFa, Theorem.]. By [KMMT,... |
math/0004105 | The proof is the same as the one of [KMMT REF ] except that we can use REF instead of [KMMT REF ]. |
math/0004105 | We have MATH and MATH. The rest of the proof is the same as the one of [K, REF .]. |
math/0004106 | In the sequel the admissibility of every block basis of MATH will always be considered with respect to MATH. Given MATH, we set MATH . In the above, the admissibility of MATH is measured with respect to MATH, the sequence of functionals biorthogonal to MATH. Because MATH, we have that MATH. Indeed, suppose that MATH an... |
math/0004106 | Suppose the assertion is false. A standard perturbation argument yields a normalized block basis MATH of MATH equivalent to a block basis MATH of MATH. Let MATH be an isomorphism from MATH onto MATH such that MATH, for all MATH. We can choose MATH such that MATH, for every MATH. Our assumptions allow us to choose MATH ... |
math/0004106 | We first choose MATH such that every MATH is MATH-good. To see that such a MATH exists, set MATH . We can easily verify that MATH is closed in the topology of pointwise convergence in MATH, and therefore it is a NAME set. Because MATH, for every MATH, the infinite NAME theorem yields MATH such that MATH, as claimed. It... |
math/0004106 | Let MATH be an infinite block basis of MATH, and let MATH. Set MATH, where MATH. We can find MATH such that MATH, for every MATH. Let MATH. Choose MATH in MATH, according to the hypothesis. There exists MATH such that MATH. Put MATH, MATH, and note that MATH is non-increasing. We now have that MATH . On the other hand,... |
math/0004106 | By induction on MATH. If MATH the assertion of the lemma is trivial. Assuming the assertion true when MATH, MATH, let MATH with MATH. If MATH there is nothing to prove. So assume MATH. Let MATH be the root of MATH and let MATH for some MATH. We denote by MATH the set of immediate successors of MATH in MATH. Given MATH ... |
math/0004106 | Let MATH denote the set of all branches of MATH (a branch is a maximal well ordered subset of MATH). If MATH the assertion is trivial. So assume that MATH for some MATH. Given MATH set MATH . Note that MATH is well defined and that MATH since MATH (MATH being the root of MATH). Let us say that MATH is of type MATH if M... |
math/0004106 | Observe that MATH and hence MATH, for all MATH. Let MATH, where MATH. Set MATH and suppose that MATH, for some MATH. Let MATH denote the collection of the ranges of the MATH's, and let MATH denote the collection of those MATH's whose support intersects at least one member of MATH. Put MATH, MATH. Because MATH is MATH-a... |
math/0004106 | Let MATH, where MATH for MATH. We can assume without loss of generality that MATH for every MATH where MATH. We are going to show that there exists a normalized block basis of MATH admitting a generic MATH average of norm at least MATH. Suppose instead that this were false. Then it is easy to construct for every MATH, ... |
math/0004106 | Note first that REF guarantees the existence of the block basis MATH. Let MATH, where MATH. By passing to a subsequence of MATH, if necessary, we can assume that the union of any MATH subsets of MATH belongs to MATH. Choose MATH, MATH, such that MATH and MATH, for every MATH (if MATH, then MATH). Let MATH. Suppose firs... |
math/0004106 | Let MATH be a block subspace of MATH spanned by the normalized block basis MATH of MATH. Let MATH and choose a block basis MATH of MATH satisfying the conclusion of REF . Applying REF (compare also REF), we obtain that for every subsequence of MATH which is a MATH spreading model, it must be the case that MATH and thus... |
math/0004106 | We choose MATH and MATH as we did in the proof of REF . Suppose first that MATH. Because MATH, the argument in the proof of REF shows that MATH. When MATH, we apply the decomposition REF to find a MATH admissible subset MATH of MATH and scalars MATH satisfying the conclusion of REF . Note that if MATH and MATH, then MA... |
math/0004106 | Set MATH. Let MATH, MATH and MATH. There exists MATH so that MATH, where MATH and MATH. Note that MATH. Applying a splitting argument similar to that of REF and taking in account REF , we obtain that MATH is MATH-admissible. The assertion follows from REF and the fact that MATH. |
math/0004106 | It follows from REF and our preceding remarks that MATH satisfies the MATH distortion property. We show that MATH is H.I. This is accomplished through REF . Let MATH be a normalized block basis of MATH and let MATH. Set MATH, where MATH. We can assume that the union of any MATH subsets of MATH belongs to MATH. Successi... |
math/0004109 | Apply REF to MATH. |
math/0004109 | By REF , MATH contains a primitive set. Let MATH be the primitive curve class corresponding to this primitive set. Write MATH. Now MATH, so we are done by induction on the degree of MATH (with respect to a fixed projective embedding of MATH). |
math/0004109 | Let MATH be a primitive curve class, and write MATH with MATH and MATH very effective. Then MATH and REF follows since MATH is a non zero divisor in MATH. If MATH is NAME, then a presentation for MATH is obtained by starting with a presentation for MATH in terms of generators and relations, and replacing each relation ... |
math/0004109 | For REF , we induct on the dimension MATH. The case MATH is trivial, and the base case MATH is easily verified. For the inductive step, let us suppose MATH satisfies REF , but that REF fails to hold. Then there is a primitive set MATH whose associated primitive relation REF satisfies MATH. Let MATH be a maximal cone co... |
math/0004109 | Induct on MATH, and apply REF . |
math/0004109 | We induct on MATH. Suppose MATH. Then MATH must contain a primitive set. The set MATH itself cannot be a primitive set, since MATH is not a ray generator in MATH. So, we may suppose MATH is primitive, with MATH. Then we have MATH for some ray generator MATH, and now MATH, with MATH, MATH, MATH, MATH linearly independen... |
math/0004109 | We need to show that for all MATH, MATH, and every cone MATH with MATH, MATH . Suppose REF fails for MATH. We may suppose MATH, and in fact, MATH is a primitive set, with MATH. Hence MATH for some MATH. Now MATH, MATH, MATH, MATH are linearly independent and MATH. So, we have a contradiction to REF . Suppose REF fails ... |
math/0004109 | Suppose not: MATH, say. In the case MATH, then we find MATH, a contradiction. In the case MATH, then by REF , the fact that MATH implies that MATH and MATH are two sets of cone generators. Now MATH and we have a contradiction. |
math/0004109 | Since a NAME toric variety is determined uniquely by the set of ray generators we have REF , and REF is clear. We obtain REF from the characterization of how primitive relations behave under blow-down. By CITE, if MATH is the blow-down corresponding to the primitive relation MATH, then the primitive sets of MATH are pr... |
math/0004109 | For a maximal cone MATH, let MATH denote the affine span of the generators of MATH, and let MATH denote (signed) integer distance to MATH in MATH. Then the quantity MATH appearing in the statement is MATH. We prove the statement by induction on the degree MATH of a tree of MATH's. The induction hypothesis is, first, th... |
math/0004109 | Let MATH, and let MATH be a maximal cone containing MATH, MATH, MATH, with MATH. For each ray generator MATH, let MATH be a tree of MATH's joining MATH to a point of MATH, with MATH. For each MATH, let MATH; we have MATH for all MATH. Now the sum over all MATH of MATH copies of MATH has homology class MATH. |
math/0004109 | Suppose not. Since MATH and MATH, it follows that MATH is special for MATH. By REF , then, if MATH denotes the unique element of MATH not in MATH, then MATH for every MATH. So MATH, and hence some MATH has intersection number MATH with MATH. Without loss of generality, then, MATH. Then MATH is special exceptional, say ... |
math/0004109 | We know MATH is the number of negative entries in the coordinate expression for MATH, in the coordinate system dictated by MATH. Let us suppose MATH is generated by the second through MATH standard basis elements plus one additional vector. Based on REF there are two possibilities. First, the additional generator can b... |
math/0004109 | Let MATH be a torus-invariant genus MATH stable MATH-pointed map, which stabilizes (upon forgetting the map to MATH) to MATH distinct points on a single irreducible component MATH, such that the MATH marked point maps into MATH for MATH and such that the image of the MATH point is MATH. By REF , MATH, and in fact (exer... |
math/0004113 | Let us start with a combinatorial description for the objects involved in REF: By the NAME - NAME interpretation of NAME functions as generating functions of nonintersecting lattice paths, we may view the left - hand side of the equation as the weight of all pairs MATH, where MATH and MATH are MATH-tuples of noninterse... |
math/0004113 | Without loss of generality we may assume that all even points are white and all odd points are black in MATH. By recolouring changing trails, all the points MATH are matched with points of opposite colour and parity. So if MATH is odd and black, then the recolouring trail starting at MATH connects it which some other p... |
math/0004113 | In the notation of REF, let MATH and MATH; and choose horizontal offset MATH. That is, interpret MATH as the generating function of two - coloured graph objects consisting of two MATH-tuples of nonintersecting lattice paths, coloured green and blue, respectively, where green path MATH starts at MATH and ends at MATH, a... |
math/0004114 | Let's prove the statement by induction on MATH. When MATH, by CITE MATH is a group algebra and if MATH then MATH and MATH. Now assume the statement is true for MATH. Consider MATH of dimension MATH. MATH and thus, by CITE, there exists a central grouplike of order MATH in MATH and therefore MATH contains a normal subHo... |
math/0004114 | In REF we have described all possible NAME ring structures of non-commutative semisimple NAME algebras of dimension MATH and there are exactly MATH of them. Only one of these MATH-rings is not commutative, namely MATH, which corresponds to nonabelian MATH. Therefore, by CITE all NAME algebras with non-commutative MATH-... |
math/0004117 | First of all, choose an open covering MATH of MATH such that MATH and MATH have transition functions MATH and MATH respectively, both relative to the same open cover MATH. Then, as we have already seen, choices for MATH and MATH are given by MATH and MATH where the MATH are defined as above. If one performs the calcula... |
math/0004117 | Consider first the case of a map MATH which is onto. Here MATH and MATH only have to be sets and MATH is a function. Denote as usual by MATH the MATH-fold fibre product over MATH. Then define MATH by MATH where MATH. In the case that MATH is a point there is a standard proof that this complex has no cohomology, see for... |
math/0004117 | Choose an open cover MATH of MATH such that there exist local sections MATH of MATH. Let MATH. Define a MATH bundle MATH on MATH by MATH, where MATH sends MATH to MATH, where MATH. The section MATH of MATH induces a section MATH of MATH and hence an isomorphism MATH. The coherency condition on MATH ensures that the MAT... |
math/0004117 | We first define a map MATH. If MATH then choose MATH and define MATH . Here MATH is the bundle gerbe multiplication in MATH. Clearly MATH is independent of the choice of MATH. We need to show that MATH is a descent isomorphism, that is, that the following diagram commutes MATH . So let MATH, MATH, and MATH. Then MATH .... |
math/0004117 | The bundle gerbe product in MATH defines an isomorphism MATH. We need to check that this isomorphism is compatible with the descent isomorphism defined in REF . That is, we need to check that the following diagram over MATH commutes: MATH where MATH is the descent isomorphism constructed in REF . Let MATH, MATH, and MA... |
math/0004117 | First of all define a map MATH by mapping a point MATH of MATH to the point MATH of MATH. We then have MATH. Note that MATH. We therefore have the following series of isomorphisms of MATH bundles MATH where MATH denotes the projection MATH and so on. The MATH bundle map MATH pulls back to define an isomorphism MATH . W... |
math/0004117 | To show this we need to use the group multiplication in MATH using the non-homogenous coordinates on MATH as reviewed in CITE and see also REF. We get MATH where we have used REF-cocycle REF to write MATH equal to MATH and so on. |
math/0004117 | We will calculate the NAME class of MATH relative to the same open cover MATH of MATH used above. We choose sections MATH of MATH above each MATH and form the pullback MATH bundles MATH over MATH. Since the sections MATH and MATH are related by the transition cocycles MATH, we see that a classifying map for MATH is giv... |
math/0004117 | First of all note that a singular REF-chain in MATH is a linear combination of REF MATH with integer coefficients. Since MATH is just a point, it follows that we may identify a singular REF-chain of MATH with a formal linear combination MATH. A singular REF-cochain is a linear map MATH. There is a canonical choice of a... |
math/0004117 | The problem is to show that MATH is a representative in singular cohomology of the NAME class of MATH. A singular representative for the NAME class of MATH can be constructed by observing as above that MATH gives rise to a MATH bundle gerbe MATH on MATH - the lifting MATH bundle gerbe associated to the short exact sequ... |
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