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math/0007158 | This lemma is a direct consequence of REF . |
math/0007158 | Let us consider an expression of the following type MATH . We can represent this expression by a graph (see for example CITE) with MATH vertices which are labelled by variables MATH and with vertices MATH and MATH connected by an edge for all MATH. We see that the nonzero summands in REF come from indexes MATH such that to all vertices in the same connected component of the graph is assigned the same value. Therefore REF is equal to MATH, where MATH denotes the number of connected components of the graph. Let us fix an index MATH. We shall apply the above observation to evaluate MATH. Let MATH be the number of indexes MATH such that MATH and let MATH be all such indexes. We consider the following cases. CASE: If MATH then we obtain a graph of type presented in REF . This graph has one connected component and therefore MATH. CASE: If MATH then we obtain a graph of type presented in REF . This graph has two connected components and therefore MATH. CASE: Suppose that MATH. If the lines MATH and MATH do not cross REF then corresponding graphs have three components and and therefore MATH. But if the lines MATH and MATH cross REF then the graph has only one connected component and therefore MATH. CASE: In the general case MATH the graph has at most MATH components and there is a factor MATH, therefore MATH. Therefore if for a given partition MATH sets MATH have a property that the common part of each three of them is empty then MATH . In the general situation the expression MATH is a real number from the interval MATH. |
math/0007158 | The measure MATH on the set of all subsets of MATH such that for any MATH we have MATH is simply a NAME distribution and REF is obviously fulfilled. Therefore if MATH are independent random variables with distribution given by MATH then MATH and REF is fulfilled. Let MATH be independent random variables with distribution given by MATH and let us consider the probability of following event: MATH. The discussion from the previous paragraph shows that this probability is equal (up to an error of order MATH) to the probability of the event: MATH and furthermore MATH for all MATH. We shall now evaluate the latter probability. There are MATH choices of disjoint sets MATH such that MATH. For MATH the probability of the event: MATH and MATH for MATH is equal to MATH . For MATH the probability of the event: MATH for all MATH is equal to MATH . Therefore the probability of the considered event is equal to MATH where all sums and products are taken over MATH. After short calculations one can show that the limit of this expression as MATH tends to infinity is equal to MATH and therefore REF is fulfilled. REF follows from the observation that the distribution of the random variable MATH is binomial and simple computations. |
math/0007159 | The generating function of the cardinalities of strict partition-valued functions is MATH which is the generating function of MATH. |
math/0007159 | CASE: Let MATH be even and let each partition MATH have odd integer parts. Assume on the contrary that MATH and MATH are conjugate in MATH, where MATH is associated to a sequence of tableaux (see REF ). Then for some MATH where we have used the fact that MATH. It follows from REF that MATH, since MATH. Let MATH and MATH be their cycle representations. Then MATH and MATH for some MATH by REF . Since each cycle length MATH is odd, all the cycles mutually commute with each other. Substituting MATH back and rearranging the cycles, we have MATH which is a contradiction. Now suppose that for some MATH there is an even cycle in MATH of the class MATH of type MATH. That is, there is an element MATH such that MATH. Consider the element MATH, where MATH and MATH with MATH for MATH and MATH. We claim that MATH which is shown by two steps. First we consider the MATH-th component of MATH in MATH. It equals MATH when MATH, and it equals MATH for MATH. Secondly we have MATH by using REF and MATH again. Thus MATH is conjugate to MATH. Therefore all partitions MATH must be from MATH if MATH splits. CASE: Let MATH be odd. Assume all partitions MATH are strict partitions. If on the contrary MATH is conjugate to MATH, then using MATH we have as in REF that MATH for the permutation MATH associated to some MATH. Let MATH and MATH be their cycle representations. Then MATH for some MATH. Since MATH cyclically permutes the indices in each cycles of MATH we have MATH. On the other hand, note that each cycle MATH corresponds to one part in MATH for some MATH and any conjugation of MATH still corresponds to a part in the same MATH. When we plug the equations MATH back to MATH we see that MATH is actually the identity since MATH is strict. Therefore MATH. Then MATH, which is a contradiction. Hence MATH splits. Now suppose MATH splits. If there are two identical parts in MATH for some conjugacy class MATH, say MATH for MATH. Then the cycle-products of these two identical parts are conjugate, that is, there exists an element MATH such that MATH . Consider the element MATH such that MATH and MATH for MATH, and MATH . Clearly MATH and MATH by REF . Therefore we have the following equations for MATH which imply that MATH. Note also that MATH and MATH. We see that the conjugation MATH where we used MATH. Observe that by REF MATH . Plugging this into REF we obtain that MATH, and this contradiction says that each partition MATH must be strict. |
math/0007159 | Let MATH be a conjugacy class in MATH. If MATH does not split, then MATH is a conjugacy class in MATH, so its centralizer has the order MATH. Otherwise MATH, and MATH. |
math/0007159 | The first statement in REF is a corollary of REF . To see the second equation in REF , we observe that an irreducible spin representation MATH decomposes as follows when restricting to the subgroup MATH: MATH . Moreover a pair of the associated spin representations, when restricted to MATH, become the same irreducible representation. Applying the counting formulas in REF we obtain REF . |
math/0007159 | Let MATH be the characteristic function on the conjugacy class MATH. Then MATH is spanned by MATH, where MATH ranges over the set of split even classes. Thus MATH. On the other hand we see that the characters of spin supermodules are class functions in MATH since the trace of any odd endomorphism is zero. Let MATH and MATH be the characters of two irreducible spin supermodules of MATH. It follows from REF that the underlying module of MATH or MATH is either irreducible module or the sum of two associated irreducible modules according to their types, which implies immediately the orthogonality relation REF . Therefore the matrix of the inner product is orthogonal on the set of super spin characters. Then by REF MATH . Thus the two inequalities above become equality, and so the irreducible characters of spin MATH-supermodules form a MATH-basis in MATH. The last characterization of irreducible supermodules follows from the semi-simplicity of the superalgebra MATH and the usual orthogonality of ordinary irreducible characters. |
math/0007159 | Using REF twice we observe that the following two embeddings give rise to two conjugate subgroups in MATH (see REF ): MATH . Using this and REF we can easily check the associativity of the product. For a simple supermodule MATH we define MATH . Let MATH, MATH, and MATH be simple supermodules for MATH, MATH, and MATH respectively. It is easy to see that MATH satisfies the cocycle condition MATH . Therefore we can define MATH to be either of the above expressions. Using the cocycle MATH we prove the coassociativity as follows. Let MATH be a MATH-supermodule and suppose that MATH as an irreducible decomposition. Then we have MATH where we used the cocycle condition in the third equation, and the notation MATH stands for the multiplicity-free summation of the irreducible components (compare REF and definition of MATH). Finally we look at the compatibility of multiplication and comultiplication. Fix MATH and MATH, it follows from REF that MATH where MATH runs through the double cosets MATH. Notice that the double cosets are in one-to-one correspondence with the double cosets MATH. Again by the cocycle property of MATH and counting the double cosets we can check that the last summation is exactly MATH. |
math/0007159 | Consider MATH where MATH and MATH is a MATH-cycle, say MATH. Denote by MATH a basis of MATH, and we write MATH, MATH . It follows that MATH . Thus we obtain MATH where we notice that MATH lies in MATH. Given MATH, where MATH and MATH, we clearly have MATH . Thus it follows that for the conjugacy class MATH of type MATH, we have MATH where MATH. |
math/0007159 | The proof is similar to the untwisted case CITE. |
math/0007159 | The character value of MATH is given in REF , and we have MATH . Let MATH be the characters of two representations of MATH. It follows from REF that MATH . Therefore the proposition holds for MATH, and so for any element MATH. |
math/0007159 | By counting dimensions of homogeneous degree subspaces it is easy to see that MATH is an isomorphism of vector spaces. The algebra isomorphism follows simply from the NAME reciprocity. To check the coalgebra isomorphism we use REF to pass from the generators MATH to the character MATH. It is then a simple calculation to verify that MATH is group-like under the comultiplication REF , and this shows that MATH is a NAME algebra isomorphism by using REF . |
math/0007159 | Let MATH and MATH be any two super class functions in MATH. By definition of REF it follows that MATH where we have used the inner product identity REF . |
math/0007159 | Observe that the operator MATH is the adjoint operator of MATH with respect to the bilinear form MATH. Then the theorem follows from REF by invoking the characteristic map. |
math/0007159 | It suffices to compute that MATH . |
math/0007159 | All the commutation relations without binomial coefficients are easy consequences of REF by the usual vertex operator calculus in the twisted picture (see CITE). The corresponding relations with binomial coefficients in MATH are equivalent to MATH . This is again proved by using REF with the same method as in the quantum vertex operators CITE. |
math/0007159 | It follows from the standard vertex operator calculus (compare CITE) by using REF . |
math/0007159 | The generalized NAME algebra structure REF implies that the nonzero elements MATH of distinct indices generate a spanning set in the space MATH. To see that they satisfy the raising operator expansion we compute that MATH . Using the result in CITE it follows that this is exactly the generating function of the raising operator expansion at the case MATH is trivial under the isomorphism MATH. In other words, REF is true when MATH is a characteristic partition-valued function. Next the orthogonality follows from the generalized NAME algebra commutation relations in REF . The orthogonality relations show that the raising operator is not affected by the character MATH, hence the general case follows by multiplying the raising operator formula for each MATH. |
math/0007159 | Observe that the tensor product of two irreducible supermodules of type MATH is a sum of two irreducible supermodule of type MATH (see REF ). Computing its inner product we know that for positive odd integers MATH and MATH the character MATH is twice of some irreducible character. Then by induction we see that MATH is a character in MATH. From REF we see that MATH is a MATH-linear combination of MATH with MATH, hence MATH is a virtual character of MATH. |
math/0007159 | Suppose we know that MATH corresponds to the character of an irreducible MATH-supermodule. By REF and definition of the characteristic map we see immediately that the linear combination REF is given by the vertex operator structure and the matrix coefficient REF give the character table of all irreducible supermodules. First we observe that the number of irreducible spin supermodules of type MATH is equal to the number of even strict partition-valued functions, which are realized by the vectors MATH up to signs. As for the vectors MATH with MATH, it follows from REF that each of such vectors corresponds to a virtual irreducible character in MATH of type MATH, since the case of sum or difference of two irreducible characters of type M is ruled out by the orthogonality. To show that they correspond to actually irreducible characters it is sufficient to show that the value of MATH at the conjugacy class of the identity element of MATH is positive. Let MATH be the class consisting of the identity in MATH. The type of the identity element in MATH is the partition-valued function MATH such that MATH and MATH for MATH. Recall from REF that MATH. By comparing weights and using orthogonality REF we have MATH . By the result of REF we have MATH which implies REF , thus the theorem is proved. |
math/0007159 | Let MATH be the irreducible spin MATH-supermodule corresponding to MATH. It follows from REF that MATH is an irreducible component of MATH. Note that the supermodule MATH is irreducible and is of type MATH (or type MATH) according to MATH even (or odd). Let MATH, MATH, MATH be odd partitions, and let MATH, MATH, MATH be even partitions. Then the parity of MATH equals the parity of MATH, that is, MATH. It follows from REF that MATH decomposes completely into MATH copies of the irreducible supermodule MATH of MATH, and MATH is of type MATH if MATH is even and of type MATH otherwise. The degree of MATH equals to MATH, and we have MATH where we have used the degree REF for the special case of MATH. The exponents of MATH in the first factor and the product sum up to MATH where MATH or MATH according to MATH is even or odd. Thus the degree of MATH is exactly the one given by REF . Hence MATH is the irreducible spin MATH-supermodule MATH corresponding to MATH. |
math/0007161 | The equivalence follows from REF. |
math/0007161 | If MATH and MATH are in MATH and are joined by an edge MATH of MATH, then MATH for some MATH, so that MATH . However, since MATH, we have MATH. |
math/0007161 | Let MATH be a vertex of MATH, corresponding to an edge MATH of MATH. Let MATH be an edge of MATH incident to MATH and let MATH be the two-dimensional subspace of MATH such that MATH. Then MATH is generated by MATH and MATH, therefore is uniquely determined. |
math/0007161 | Assume first that MATH and let MATH be an edge of MATH, given by REF. Denote by MATH the restriction of MATH to MATH. For every MATH we have MATH hence MATH. Conversely, assume that MATH for every edge MATH of MATH. Let MATH. We need to show that MATH where, according to REF, MATH . Let MATH be the set of all linear factors that appear in the denominators of MATH (see REF), for all edges MATH that intersect the MATH-level. For MATH, let MATH be the multiplicative system in MATH generated by elements of MATH that are not multiples of MATH and let MATH be the corresponding localized ring. To show REF it suffices to show that MATH for every MATH. Fix MATH and let MATH be the corresponding subgraph of MATH. Let MATH be a vertex of MATH corresponding to an edge of MATH, and let MATH be an edge of MATH incident to MATH, where MATH is a two-dimensional subspace of MATH. Then MATH is an edge in both MATH and MATH; the density associated to MATH when we integrate over MATH is given by REF and the density associated to MATH when we integrate over MATH is MATH where MATH is the NAME class of MATH in MATH. If MATH is the NAME class of MATH and MATH then the two densities are related by MATH . Then REF differs from MATH by a term in MATH (the first sums on each side are over all edges of MATH). But MATH and, according to REF of the Appendix, there exists a polynomial MATH such that MATH then, since MATH is a morphism of rings, REF becomes MATH which concludes the proof. |
math/0007161 | There is a natural isomorphism of linear spaces MATH, given by MATH . This induces an isomorphism of rings MATH and, together with the identification MATH, an isomorphism MATH . We will show that REF restricts to a bijection MATH . First, let MATH. Let MATH be an edge of MATH, with endpoints MATH and MATH and let MATH be the subgraph of MATH with vertices MATH and MATH. Consider the NAME class of MATH in MATH. This, by definition, is the map MATH that is REF at vertices not in MATH and, at each vertex, MATH, of MATH, is equal to the product of the values of MATH on edges at MATH which are not edges of MATH. Then MATH . Since this is true for all edges of MATH, it follows that MATH. Conversely, let MATH and MATH. The projection on the first factor MATH induces a map MATH. Let MATH then MATH and therefore MATH. A direct computation shows that MATH . REF implies that the right hand side is an element of MATH and hence MATH for all MATH. Therefore MATH which proves the theorem. |
math/0007161 | The proof of this ``localization" theorem is similar to that of the localization theorem in REF. Let MATH be the restriction of MATH to MATH. To show that MATH we need to show that MATH for all MATH, and just as in the proof of REF , we will show that this can be ``localized" to analogous statements about the integrals of MATH over the hyperedges of the hypergraph MATH. Here are the details. Let MATH. For MATH, let MATH be the multiplicative system in MATH generated by elements of MATH which are not collinear with MATH and let MATH be the corresponding localized ring. We will show that MATH for each MATH; this will imply REF. Let MATH. We join two points MATH and MATH of MATH if MATH and MATH are collinear. Let MATH be a non-trivial connected component of this new graph. Then MATH is a complete graph. Let MATH be the two-dimensional subspace of MATH generated by MATH and MATH and let MATH be the connected component of MATH that contains MATH. Then MATH . We now use our hypothesis : MATH admits a generating family. Then the same is true if we change MATH to MATH; let MATH be a generating class at MATH with respect to MATH and define MATH where MATH is the NAME class of MATH in MATH. Then MATH is supported on the flow-down of MATH in MATH and MATH . For every MATH we have MATH since MATH, REF implies that MATH. Let MATH be the polynomial MATH and let MATH be given by MATH . Then MATH, hence, according to REF of the Appendix, there exists MATH such that MATH. In other words, by inserting MATH for MATH in this polynomial of MATH variables we get back our original polynomial in MATH variables. Using this we deduce that MATH where MATH is the polynomial obtained by replacing MATH with MATH, for all MATH. Since MATH, using REF we conclude that the left hand side of REF is an element of MATH. Now REF follows from the fact that the integral of MATH is a sum of terms of this form. |
math/0007161 | Uniqueness: It is clear that MATH should be equal to MATH on MATH. If we restrict REF to each set of the form MATH, we get a linear system whose matrix is a non-singular NAME matrix; therefore, the solution is unique. Hence there exist unique MATH and MATH which satisfy REF. Existence: We need to show that MATH and MATH, where MATH and MATH are the ones obtained above. Using REF of the Appendix, we deduce that MATH where MATH; since MATH for all MATH, we can use REF to conclude that MATH. Let MATH. Again, we use REF to obtain that MATH where MATH. We now use that MATH, MATH for all MATH, and REF to conclude that MATH for all MATH, that is, that MATH. |
math/0007161 | When MATH, this follows immediately from REF. When MATH it follows from REF and MATH and when MATH it follows from REF and MATH . |
math/0007161 | Consider the decomposition in partial fractions MATH where MATH is a polynomial in MATH. We use the expansion MATH to obtain that: MATH . Now REF follows by comparing coefficients of MATH on both sides. |
math/0007161 | It suffices to show that the lemma is true if MATH is a fundamental symmetric polynomial in MATH and for this we can use an inductive argument. We can also start with the identity MATH and we use REF to deduce that MATH where MATH is the fundamental symmetric polynomial of degree MATH. The lemma follows by comparing coefficients of MATH on both sides. |
math/0007161 | Since MATH the matrix MATH is invertible; the entries of the inverse are MATH where MATH is obtained from MATH by removing the MATH row and MATH column. We start with the identity MATH where MATH is the fundamental symmetric polynomial of degree MATH in variables MATH. Since the left hand side of REF is REF for MATH, MATH, we get: MATH . The left hand side of REF is the last column of MATH; using REF and basic properties of determinants we deduce that MATH . Using REF we obtain that MATH . We now use REF to finish the proof. Since MATH, for MATH, REF becomes REF. |
math/0007161 | The fact that every map of the form REF is in MATH is a direct consequence of REF . To show that every element of MATH can be written as in REF we proceed as follows: We can regard REF as a linear system whose matrix is a NAME matrix; hence there are MATH such that REF is true. Moreover, we can use REF to deduce that MATH . Since MATH, we conclude that MATH. |
math/0007162 | Let MATH be presented by a MATH-plat. We show that this presentation is equivalent to a special one, by using a finite sequence of moves on the plat presentation which changes neither the link type nor the number of plats. The moves are of the four types MATH, MATH, MATH and MATH depicted in REF . First of all, it is straightforward that REF can be satisfied by applying a suitable sequence of moves of type MATH and MATH. Furthermore, REF is equivalent to the following: MATH there exists an orientation of MATH such that, for each MATH, the top arc MATH is oriented from MATH to MATH and the bottom arc MATH is oriented from MATH to MATH. Therefore, choose any orientation on MATH and apply moves of type MATH (respectively, moves of type MATH) to the top arcs (respectively, bottom arcs) which are oriented from MATH to MATH (respectively, from MATH to MATH). |
math/0007162 | Let MATH be presented by a special MATH-plat arising from a braid MATH, and let MATH be the MATH-decomposition described in REF . Now, all arguments of the proofs of REF entirely apply and the condition of REF is satisfied, since the homeomorphism MATH associated to MATH fixes both the sets MATH and MATH. |
math/0007162 | The proof is similar to the one of REF. The definition of weakly MATH-symmetric splitting implies that the canonical projections MATH and MATH, where MATH is the cyclic group of order MATH generated by MATH, are MATH-fold branched cyclic coverings. By iii', there is a map MATH such that MATH. Therefore, the map MATH is a MATH-fold branched cyclic covering. The restriction map MATH turns out to be a MATH-fold cyclic covering of MATH, branched over MATH points MATH, with MATH such that MATH for each MATH. By MATH we get MATH. Since MATH, we have MATH and therefore MATH. Thus, MATH and the statement is achieved. |
math/0007162 | Straightforward, since MATH when MATH is prime. |
math/0007162 | It is well known that MATH lifts if and only if the induced homomorphism MATH on the fundamental group MATH leaves the subgroup MATH of the covering invariant CITE. The subgroup MATH is the kernel of the homomorphism MATH, defined by MATH, for each MATH. Then we have MATH and therefore MATH. |
math/0007162 | Let MATH be a MATH-fold cyclic covering, branched over MATH, and let MATH be presented by a special MATH-plat associated to a braid MATH. If MATH is the MATH-decomposition of MATH described in REF , the map MATH defined by MATH and MATH, for every MATH and MATH, is an homeomorphism which sends MATH onto MATH, for each MATH. Now, orient MATH in such a way that, for each MATH, the arc MATH is oriented from MATH to MATH and the arc MATH is oriented from MATH to MATH (this is possible since the plat is special). Let MATH be the integers associated to the components of MATH, according to the chosen orientations, and defining the covering. For each MATH, the component MATH contains the MATH top arcs MATH and the MATH bottom arcs MATH. Let MATH be the MATH-fold cyclic covering of REF-ball MATH, branched over the set of arcs MATH, with associated integers MATH, where MATH is the component of MATH containing MATH, for MATH. If MATH is another handlebody of genus MATH and MATH is a fixed homeomorphism, then the map MATH is a MATH-fold cyclic covering of MATH, branched over MATH, with associated integer list MATH. Now let MATH be the restrictions of MATH to the surface MATH. The map MATH is a MATH-fold cyclic covering of MATH, branched over the MATH points MATH, for MATH, such that MATH. Because of REF of special plats, the homeomorphism MATH of MATH satisfies the condition of REF . Hence, MATH lifts to a homeomorphism MATH of MATH such that MATH. If we define the homeomorphism MATH, the manifold MATH turns out to be, by construction, a MATH-fold branched cyclic covering of MATH, with the same monodromy as the covering MATH and therefore MATH is homeomorphic to MATH. Moreover, MATH is a MATH-symmetric NAME splitting of genus MATH. Indeed, let MATH be the group of covering transformations of MATH and MATH be a generator of MATH. Then condition i' easily holds. As regards condition ii' we have: MATH. As for condition iii', the map MATH is the lifting, with respect to MATH, of MATH and therefore it is a covering transformation of MATH. This proves point iii'. As far as the genus of the splitting is concerned, we have by MATH: MATH. Therefore MATH. |
math/0007162 | Follows from previous theorem, with MATH. |
math/0007165 | If MATH, MATH, so MATH. Moreover, for all MATH and MATH, MATH so MATH. Finally, MATH so MATH is the orthogonal projection of MATH onto MATH. |
math/0007165 | : Let MATH be primitive vectors such that for every MATH there exists a unique MATH such that MATH is a multiple of MATH. If MATH are all the occurrences of multiples of MATH among all the weights, let MATH . Similarly we define MATH. Then MATH with MATH. We will show that MATH divides MATH in MATH. The vertices of MATH can be divided into two categories: CASE: The first subset, MATH, contains the vertices MATH for which none of the MATH's with MATH, is a multiple of MATH CASE: The second subset, MATH, contains the vertices MATH for which there exists an edge MATH such that MATH and MATH is a multiple of MATH. (Notice that there will be exactly one such edge.) The part of REF corresponding to vertices in the first category will then be of the form MATH with MATH . If MATH then there exists an edge MATH issuing from MATH such that MATH with MATH; let MATH. Since MATH it follows that MATH as well and thus the vertices in MATH can be paired as above. Let MATH and MATH be the edges issuing from MATH and MATH respectively, with MATH, MATH. Then, by REF, the MATH's can be ordered so that MATH which implies that MATH . Similarly, from MATH we deduce that MATH . The part of REF corresponding to MATH and MATH, MATH can be expressed as MATH . From the congruences REF we conclude that MATH divides the numerator of REF, so we deduce that MATH with MATH. Therefore MATH with MATH. Adding REF we obtain MATH with MATH, hence MATH divides MATH. The same argument can be used to show that each MATH divides MATH. The proof of the theorem now follows from: If MATH and MATH, MATH are linearly independent weights such that MATH divides MATH, then MATH divides MATH. If MATH and MATH are pairwise linearly independent weights such that MATH for all MATH then MATH . |
math/0007165 | : Let MATH be a weight that is not in the convex hull of MATH. Then there exists MATH and MATH such that MATH and MATH for all MATH. If MATH and MATH then MATH and using this we deduce that MATH where MATH in the exponent is the sum MATH . From REF we deduce that MATH . Suppose MATH is a weight of MATH; then there exists MATH and non-negative integers MATH such that MATH which implies that MATH . But when we evaluate REF at MATH, the right hand side is non-negative, while the left hand side is strictly negative ! This contradiction proves that MATH is not a weight of MATH. |
math/0007165 | : Let MATH be an extremal weight, that is, a vertex of MATH. Then there exists MATH such that MATH for all MATH. In this REF implies MATH . Since each term on the right hand side is non-negative, REF is only true if CASE: MATH (which also implies that MATH for all MATH with MATH, that is, that there are no terms in the first sum), and CASE: MATH for all MATH. This proves that the multiplicity with which MATH occurs in MATH is REF. |
math/0007165 | : From REF, the multiplicity with which a weight MATH appears in the term corresponding to the vertex MATH is equal to MATH times the number of distinct ways in which MATH can be written as a sum MATH with MATH's non-negative integers; and this number is MATH. Counting the contributions given by all the vertices we obtain REF. |
math/0007165 | Let MATH be real numbers, let MATH be integers and let MATH. As above we will order the MATH's so that MATH for MATH and MATH for MATH. Let MATH be the function REF, that is, MATH . Suppose that, for MATH, MATH and REF are linearly independent over the rationals. Then MATH has simple poles on the unit circle. Let MATH . Then these poles are at the points MATH so if MATH and REF are linearly independent over the rationals these poles are distinct. Let us compute the residue of MATH at the pole REF. The quotient MATH evaluated at MATH is equal, by l'Hopital's rule, to: MATH evaluated at MATH, and since MATH, this quotient is just MATH . Thus the residue at MATH of the function MATH is just MATH which, if we set MATH and MATH can be written MATH . We will now show that if we give MATH and MATH the values REF - REF the sum of these residues is identical with the right hand side of REF. If MATH is equal to REF and MATH is equal to REF, then by REF MATH and MATH where MATH and MATH . Let's now give a more ``intrinsic" definition of MATH and MATH: Let MATH be the NAME algebra of the group, MATH, and MATH the NAME algebra of MATH. Since MATH is by definition the kernel of the homomorphism, MATH, MATH is the annihilator of MATH; so, by REF, MATH is the unique element of MATH which is annihilated by MATH and has the same restriction to MATH as MATH. Similarly, MATH is the unique element of MATH which is annihilated by MATH and has the same restriction to MATH as MATH. Note, by the way, that since MATH and MATH are annihilated by MATH, they are in the dual vector space to MATH; or, in other words, in the dual of the NAME algebra of MATH. Consider the kernel of the map MATH. This consists of the elements MATH and by REF MATH and MATH . Thus the sum MATH of the residues of MATH over the poles MATH, MATH is by REF identical to the expression MATH if MATH (in which case MATH) and is equal to MATH when MATH (in which case MATH). |
math/0007165 | If MATH and MATH, then MATH, and if MATH and MATH, then MATH. Moreover, in both cases, MATH by REF. Thus, if MATH, MATH are the elements of MATH with MATH, and MATH, MATH, are the elements of MATH with MATH, the difference between MATH and MATH is, by REF, equal to MATH or, also by REF, to MATH which, by REF, is identical with MATH . |
math/0007166 | REF is an obvious consequence of REF ; and it is easy to see that if the hypotheses of REF hold, MATH is GKM. For the other implications see CITE. |
math/0007166 | If MATH divides MATH the map MATH is in MATH. |
math/0007166 | It suffices to check this for MATH, that is, it suffices to check that MATH . However, by REF, this sum is equal to MATH which is equal to MATH by REF, with MATH. |
math/0007166 | This is equivalent to asserting that MATH however, REF is, by definition, the common value of MATH, MATH, at the vertices, MATH, at which MATH intersects MATH. CASE: In particular let MATH be an arbitrary vertex of MATH; and choose MATH and MATH such that there are no critical values of MATH on the interval, MATH and such that MATH, MATH. Order the edges MATH in MATH so that MATH are descending and MATH are ascending. For MATH let MATH be the vertex at which MATH intersects MATH and let MATH be the map defined by MATH . |
math/0007166 | By REF, MATH. (This proof assumes that there exists a NAME class, MATH, having the properties listed in REF . Alternatively, REF can be proved directly using a more sophisticated definition of MATH than that which we gave in REF. For more details see CITE.) |
math/0007166 | Let MATH be the totally geodesic subgraph of MATH consisting of the single edge, MATH, and vertices MATH and MATH. The NAME class, MATH, of MATH is defined by MATH and MATH . It is easily checked that MATH. A cohomology class, MATH is supported on MATH iff MATH, MATH. Suppose now that MATH satisfies the hypotheses of REF . Then MATH is supported on MATH; so, by the lemma, MATH . In particular MATH so MATH. |
math/0007166 | A direct computation shows that MATH hence, REF follows from REF. |
math/0007166 | We first note that MATH . (This is a consequence of the compatibility conditions MATH from which one concludes that MATH.) Let MATH be the path joining MATH to MATH and MATH the path joining MATH to MATH. By REF MATH and by REF MATH . Hence MATH . However, by REF, MATH; hence we can rewrite this as MATH so MATH is equal to the expression : MATH . However, MATH so the term in parentheses can be rewritten MATH . |
math/0007168 | The proof requires MATH steps, and is performed inductively. First, let MATH, and MATH, where MATH is the approximation to an ideal signal MATH (``ideal" in the sense that if we had MATH we would have a globally asymptotically stable closed loop without need for the stabilizing term MATH), and MATH will be given below. Let MATH be a constant such that MATH, and MATH . Since the ideal control MATH is smooth, it may be approximated with arbitrary accuracy for MATH and MATH within the compact sets MATH and MATH, respectively, as long as the size of the approximator can be made arbitrarily large. For approximators of finite size let MATH where the parameter vectors MATH, MATH, are optimum in the sense that they minimize the representation error MATH over the set MATH and suitable compact parameter spaces MATH, and MATH are defined via the choice of the approximator structure (see CITE for an example of a choice for MATH). The parameter sets MATH are simply mathematical artifacts. As a result of the stability proof the approximator parameters are bounded using the adaptation laws in REF , so MATH does not need to be defined explicitly, and no parameter projection (or any other ``artificial" means of keeping the parameters bounded) is required. The representation error MATH arises because the sizes MATH are finite, but it may be made arbitrarily small within MATH by increasing MATH (that is, we assume the chosen approximator structures possess the ``universal approximation property"). In this way, there exists a constant bound MATH such that MATH. To make the proof logically consistent, however, we need to assume that some knowledge about this bound and a bound on MATH are available (since in this case it becomes possible to guarantee a priori that MATH is large enough). However, in practice some amount of redesign may be required, since these bounds are typically guessed by the designer Let MATH denote the parameter error, and approximate MATH with MATH . Hence, we have a linear in the parameters approximator with parameter vectors MATH. Note that the structural dependence on time of system REF is reflected in the controller, because MATH can be viewed as using the functions MATH to interpolate between ``local" controllers of the form MATH, respectively. Notice that since the functions MATH are assumed continuous and MATH bounded, the signal MATH is well defined for all MATH. Consider the dynamics of the transformed state, MATH . Let MATH, and examine its derivative, MATH . Using the expression for MATH, MATH . Choose the adaptation law MATH with design constants MATH, MATH, MATH (we think of MATH as a ``leakage term"). Also, note that for any constant MATH, MATH . We pick MATH . Notice also that, completing squares, MATH . Finally, observe that MATH with MATH. Then, we obtain MATH . This completes the first step of the proof. We may continue in this manner up to the MATH step, where we have MATH, with MATH and MATH defined as in REF . Consider the ideal signal MATH with MATH. Notice that, even though the terms MATH appear in MATH through the partial derivatives in MATH, MATH does not need to be an input to MATH, since the resulting product of the partial derivatives and MATH can be expressed in terms of MATH, MATH and MATH. To simplify the notation, however, we will omit the dependencies on inputs other than MATH and MATH, but bearing in mind that, when implementing this method, more inputs may be required to satisfy the proof. Also, note that by REF , MATH for bounded arguments. Therefore, we may represent MATH with MATH for MATH and MATH. The parameter vector MATH, MATH is an optimum within a compact parameter set MATH, in a sense similar to MATH, so that for MATH, MATH for some bound MATH. Let MATH, and consider the approximation MATH as given in REF . The control law MATH yields MATH . Choose the NAME function candidate MATH and examine its derivative, MATH. One can show inductively that MATH with constants MATH, MATH. The choice of adaptation laws for MATH and of MATH in REF , together with the observations that MATH, MATH, with MATH and MATH imply MATH where MATH contains the combined effects of representation errors and ideal parameter sizes, and is given by MATH . Note that if MATH or MATH, then we have MATH. Furthermore, letting MATH, MATH, and defining MATH, MATH and MATH we have MATH and MATH . Then, letting MATH, we have that if MATH with MATH, then MATH and all signals in the closed loop are bounded. Furthermore, we have MATH, which implies that MATH so that both the transformed states and the parameter error vectors converge to a bounded set. Finally, we conclude from the upper bound on MATH that the state vector MATH converges to the residual set REF. |
math/0007171 | Let MATH denote the topological euler number of a MATH-complex MATH. Then MATH is equal with the sum of topological euler numbers of singular fibers of MATH. Every singular fiber has a positive topological euler number. We have defined MATH in such a way that, if MATH, then MATH holds, and if MATH, then the type of the fiber MATH is either MATH or MATH. Hence MATH does not exceed the sum of the topological euler numbers of reducible singular fibers, and if MATH, then there is an irreducible singular fiber or a singular fiber of type MATH or MATH. |
math/0007171 | By the surjectivity of the period map of the moduli of MATH surfaces (compare CITE), there exist a MATH surface MATH and an isomorphism MATH of lattices such that MATH. By CITE, the MATH surface MATH has an elliptic fibration MATH with a section such that MATH, where MATH is the cohomology class of a fiber of MATH, and MATH is the orthogonal complement of MATH in the NAME lattice MATH. Because MATH is equal with MATH, and because MATH coincides with MATH, we see that MATH is isomorphic to MATH. |
math/0007171 | Suppose that there exists an extremal elliptic MATH surface MATH with data equal with MATH. It is obvious that MATH and MATH satisfies the condition MATH. Via the isomorphism MATH, the overlattice MATH of MATH corresponds to an overlattice MATH of MATH, which satisfies the conditions MATH by REF . Conversely, suppose that MATH satisfies the conditions MATH and MATH. By REF , the condition MATH and MATH imply that there exists an even unimodular overlattice of MATH into which MATH and MATH are primitively embedded. By REF (see, for example, CITE) on the classification of even unimodular lattices, any even unimodular lattice of signature MATH is isomorphic to the MATH lattice MATH. Then REF implies that there exists an elliptic MATH surface MATH satisfying MATH and MATH. The condition MATH implies MATH. Combining this with REF , we see that MATH. Then REF implies that MATH. Using REF and the condition MATH, we see that MATH. Thus the data of MATH coincides with MATH. |
math/0007171 | This is checked by listing up all MATH satisfying the conditions MATH and MATH using computer. |
math/0007171 | This can be proved easily by the NAME theorem. |
math/0007171 | By CITE, the primitive embedding of MATH into the MATH lattice MATH is unique up to MATH. Hence the assertion follows from NAME 's connectedness theorem CITE. |
math/0007174 | We start by recalling the content of the hypotheses stated in the theorem. The algebra MATH as a vector space is the tensor product of MATH and MATH, whereas its product law is obtained combining the product laws of these two tensor factors with the cross-product law, MATH for any MATH and MATH. MATH being an algebra homomorphism means that for any MATH . For MATH, MATH this implies MATH . Hereby we have also used REF . After these preliminaries, note that under REF , for any MATH and MATH which proves REF in this case. Moreover MATH proving that MATH is a homomorphism. To prove injectivity we show that MATH can be inverted on MATH, and the inverse is given by MATH where MATH is the invertible element defined by MATH (MATH is invertible because MATH is). In fact, MATH . In fact, if MATH can be extended to an algebra homomorphism MATH a little calculation with the help of REF shows that MATH, where MATH is the invertible central element defined by REF . We know that MATH. To prove that MATH note first that by REF any element in MATH can be written as a sum of products MATH, with MATH and MATH. So we need to show that MATH for some MATH, MATH (at the right-hand side a sum of many terms is implicitly understood). Now this can be proved as follows: MATH which is of the form REF . The proof for MATH under the corresponding assumptions is completely analogous. |
math/0007174 | Under the first assumptions, for any MATH, MATH . Similarly one proves REF for MATH. Under the second assumptions, for any MATH, MATH . By similar arguments one proves the claim for MATH. |
math/0007176 | If the algebra is MATH-abelian, it is simply an algebra whose derived algebra is abelian CITE. If it is MATH-abelian, then there exist MATH such that MATH. From the above equations it is immediate to derive the possibilities: CASE: MATH CASE: MATH and MATH such that MATH CASE: MATH and MATH such that MATH . |
math/0007176 | Suppose MATH for MATH. We can take MATH and MATH for MATH, as well as MATH. The NAME conditions imply MATH . Let MATH for MATH, so that we can choose MATH for MATH. A change of basis allows to take MATH. From the conditions above we deduce MATH. Consider tha change MATH with MATH. Then we have MATH . There are two cases : CASE: If MATH we suppose MATH and MATH through a linear change-MATH . Reordering the forms MATH we can suppose MATH for MATH for the remaining. We obtain a unique class of nonsplit NAME algebras in even dimension and isomorphic to MATH . CASE: If MATH CASE: If MATH with MATH we can suppose MATH and MATH for MATH . Reordering the MATH we obtain one algebra in even and one algebra in odd dimension, which are respectively isomorphic to MATH and MATH . CASE: If MATH for MATH we obtain in an analogous way two even dimensional algebras, respectively isomorphic to MATH and MATH . |
math/0007176 | MATH for all MATH . The characteristic sequence implies MATH for all MATH . Moreover, a change of basis of the type MATH allows to suppose MATH . CASE: If MATH we suppose MATH A change of basis allows MATH . There are two possibilities: an even dimensional algebra isomorphic to MATH and an odd dimensional one isomorphic to MATH . CASE: MATH . CASE: If MATH we put MATH and MATH with a linear change of basis. We obtain a unique algebra in odd dimension isomorphic to MATH . CASE: If MATH there are two possibilities, depending on MATH : we obtain two odd dimensional algebras which are respectively isomorphic to MATH and MATH . |
math/0007176 | The starting assumptions are MATH for all MATH and MATH for some MATH . We can suppose MATH, and from the NAME conditions we obtain MATH . From the characteristic sequence we deduce the nullity of MATH for all MATH so that MATH as well. CASE: MATH : a combination of the linear changes of basis allow to take MATH . CASE: MATH :. CASE: If MATH, MATH we reorder the MATH such that MATH-for MATH and the remaining brackets zero. The decisive structure constant is MATH . If it is zero we obtain an odd dimensional NAME algebra isomorphic to MATH . If not, MATH is an essential parameter. So we obtain an infinite family of odd dimensional NAME algebras isomorphic to the family MATH . CASE: MATH, MATH . Without loss of generality we can choose MATH and the remaining zero for MATH . It is easy to deduce MATH . Reordering MATH in the previous manner we obtain an even dimensional NAME algebra and an infinite family of even dimensional algebras, which are respectively isomorphic to MATH and MATH . CASE: Now take MATH CASE: If there is an index MATH such that MATH we can suppose MATH and the remaining zero. Reordering the MATH as before we obtain two even dimensional NAME algebras, respectively isomorphic to MATH and MATH . CASE: MATH . A similar reordering of the MATH gives two NAME algebras in odd dimension isomorphic to MATH and MATH . CASE: Suppose now MATH . CASE: MATH A linear change allows to annihilate MATH . CASE: MATH , and the remaining zero. There are two possible cases, depending on MATH if it is nonzero we obtain an odd dimensional algebra isomorphic to MATH and if it is zero an algebra in even dimension isomorphic to MATH . CASE: MATH for any MATH : we obtain an odd dimensional NAME algebra isomorphic to MATH . CASE: MATH CASE: MATH and MATH . With a linear change we can suppose MATH . A-REF MATH and MATH for MATH A linear change allows to suppose MATH for all MATH . If MATH for all MATH we obtain the algebras MATH and MATH. If not, reorder MATH such that MATH . We obtain the algebras MATH . A-REF MATH . If MATH we obtain the NAME algebras MATH and if MATH we obtain the algebras MATH and MATH . CASE: MATH . B-REF MATH . We obtain two NAME algebras in even dimension isomorphic to MATH and MATH . B-REF MATH for MATH : there is only one algebra in odd dimension which is isomorphic to MATH . |
math/0007176 | The starting assumptions for this case are MATH . In particular the NAME condition forces MATH . CASE: Suppose MATH and MATH . We observe that if there exists a MATH then a linear change of basis allows to suppose MATH . So we have the conditions MATH CASE: MATH with MATH . We can suppose MATH and MATH CASE: If MATH for all MATH we obtain two even dimensional algebras isomorphic to MATH and MATH . CASE: If MATH for an index MATH we can suppose MATH . We obtain two odd dimensional algebras isomorphic respectively to MATH and MATH . CASE: MATH CASE: If MATH then MATH by MATH . Reordering MATH we obtain an algebra isomorphic to MATH . CASE: If MATH and MATH we obtain two algebras isomorphic to MATH and MATH . CASE: If MATH we obtain two algebras in odd dimension isomorphic to MATH and MATH. CASE: Suppose MATH and MATH . Additionally we can suppose MATH . CASE: If MATH for an index MATH let MATH and MATH and MATH by MATH . We obtain two algebras isomorphic MATH and MATH . CASE: If MATH we obtain two even dimensional algebras respectively isomorphic to MATH and MATH . |
math/0007176 | If MATH, the assertion follows immediately from the linear system MATH associated to the algebra, as this system admits nontrivial solutions. If MATH, the only case for which the system could have zero solution is MATH, and the distinct values of MATH. For any of these starting conditions it is routine to prove the existence of a nonzero semisimple derivation. |
math/0007176 | Consider the endomorphism defined by MATH and zero over the undefined images, where MATH is the dual basis of MATH. Clearly MATH is a nonzero semisimple derivation of MATH. |
math/0007176 | As the sum of characteristically nilpotent algebras is characteristically nilpotent REF , the assertion follows from the previous proposition. |
math/0007176 | Let MATH be the semidirect product of MATH by the torus MATH defined by its weights : MATH over the basis MATH dual to MATH. Then the law is given by MATH . Thus the only nonzero brackets not involving the vector MATH are MATH . Now NAME implies MATH, and by a change of basis MATH. Thus the law is rigid, and MATH is a maximal torus of derivations of MATH. |
math/0007176 | As known, for the ideals MATH of the descending central sequence we have MATH . Thus, if the derived subalgebra is not abelian, it suffices to show the existence of two roots MATH such that MATH and that for any two roots MATH we have MATH . Moreover, let MATH CASE: MATH : take MATH CASE: MATH : MATH CASE: MATH : MATH CASE: MATH : MATH CASE: MATH : MATH CASE: MATH : MATH CASE: MATH : MATH CASE: MATH : MATH CASE: MATH : MATH CASE: MATH : MATH CASE: MATH : MATH . Let MATH be generators of the weight spaces MATH and MATH : then the preceding relations show that MATH proving that the derived subalgebra is not abelian. Finally, it is trivial to see that for any subset MATH listed above we have MATH . |
math/0007179 | Thanks to CITE MATH is a projective MATH-module of rank MATH. By a result of Marlin, CITE, the NAME group of projective modules over MATH is trivial, in other words MATH . In particular, if MATH is a finitely generated projective MATH-module then MATH is stably free. Hence if the rank of MATH is greater than the NAME dimension of MATH, then MATH is necessarily free, CITE. Since MATH we have MATH, so the theorem follows. |
math/0007183 | Let MATH be an infinite dimensional NAME space and MATH is a bounded selfadjoint operator in MATH such that MATH, MATH, and the spectrum of MATH accumulates to MATH, equivalently, its range is not closed. Consider the NAME space MATH as well as the bounded selfadjoint operator MATH . Let MATH be the NAME space MATH with the indefinite inner prodcut MATH defined by the symmetry MATH . Consider the operator MATH . It is a straightforward calculation to see that MATH and, by performing a NAME factorization, it follows that MATH has dense range. Thus, MATH is a NAME\ĭn space induced by MATH and we show that it does not have the lifting property. Let MATH be an operator in MATH such that, with respect to its MATH block-matrix representation, all its entries MATH, MATH, commute with MATH. Define the operator MATH and note that MATH. Let us assume that there exists a bounded operator MATH such that MATH. Then, there exists the constant MATH such that MATH or, equivalently, that MATH where MATH . Taking into account that MATH commutes with all the other operator entries involved in REF , it follows that REF is equivalent to MATH where we denoted MATH . Note that, by continuous functional calculus, MATH is an operator in MATH such that MATH and its spectrum accumulates to MATH. The use of the NAME factorization MATH suggests to take MATH and this choice is consistent with our assumption that all its entries commute with MATH. Since MATH is bounded invertible, from REF we get MATH . Looking at the lower right corners of the matrices in the previous inequality we get MATH which yields a contradiction since the spectrum of the operator MATH accumulates to MATH. |
math/0007183 | CASE: Let MATH be the NAME space obtained by the completion of the quotient space MATH with respect to MATH, where MATH is the isotropic subspace of the inner product space MATH. Let MATH be the NAME operator of MATH with respect to MATH and let MATH be the NAME decomposition of the kernel MATH described in the previous section. Since MATH is MATH-bounded, it follows that each MATH extends to a bounded operator MATH on MATH. We notice that MATH . From the definition of MATH we have that for MATH, MATH . By our REF , MATH, so that MATH . Since MATH, we deduce that MATH . This relation can be extended by linearity to MATH for all MATH. We deduce that MATH which implies that MATH . REF implies that there exists a unique operator MATH such that MATH . Moreover, for MATH, MATH . We also notice that MATH . We deduce that MATH and this relation implies that MATH . Since the set MATH is dense in MATH, we deduce that MATH is a projective representation of MATH on MATH. For MATH we have MATH and MATH . Since MATH, we deduce that MATH so that REF holds. Finally, the relation REF implies that MATH for all MATH, which implies that MATH. We now notice that the relation REF implies that MATH, which concludes the proof of the relation MATH for all MATH. CASE: Let MATH be a fundamental symmetry on MATH. Then MATH is a selfadjoint operator with respect to the positive definite inner product MATH. Let MATH be the NAME decomposition of MATH and define the hermitian kernels MATH . From MATH and MATH we get MATH and MATH. To prove that MATH and MATH are positive definite kernels let MATH. Then MATH . It remains to show that MATH is MATH-bounded. If MATH, then MATH for some MATH, vectors MATH and distinct elements MATH, MATH, MATH in MATH. Then MATH so that MATH is MATH-bounded. We also deduce that MATH is the NAME decomposition of the positive definite kernel MATH and MATH is the NAME decomposition of MATH. Since MATH we deduce that MATH and, by REF , in CITE we deduce that MATH and MATH are disjoint kernels. Since REF are obvious implications, the proof is complete. |
math/0007183 | Indeed, in this REF becomes MATH for all MATH, where MATH is the unit of the group MATH. Also, if MATH is a hermitian kernel then it is MATH-invariant if and only if MATH . Let MATH be arbitrary. Then MATH and hence MATH is MATH isometric. |
math/0007183 | CASE: Let MATH be the similarity such that MATH for MATH. We first notice that MATH is also an involutory similarity (with the terminology from CITE), that is MATH . Then, we consider on MATH the positive inner product MATH, MATH. Since MATH is boundedly invertible, there exists a selfadjoint and boundedly invertible operator MATH such that MATH, MATH. Therefore, for arbitrary MATH and MATH we have MATH . Thus, MATH and letting MATH it follows that MATH is a fundamental symmetry on the NAME\ĭn space MATH such that MATH. CASE: If MATH is a fundamental symmetry on the NAME\ĭn space MATH such that MATH, for all MATH, then MATH is a symmetric projective representation with respect to the NAME space MATH. |
math/0007183 | CASE: We use the same notation as in the proof of REF . Thus, MATH is the NAME space obtained by the completion of the quotient space MATH with respect to MATH, where MATH is the isotropic subspace of the inner product space MATH. Let MATH be the NAME operator of MATH with respect to MATH and let MATH be the projectively invariant NAME decomposition of the kernel MATH described in the proof of REF . Since MATH is MATH-bounded, it follows that each MATH extends to a bounded operator MATH on MATH. Since MATH is MATH-invariant, we deduce that MATH for all MATH, which implies that MATH . This relation and REF imply that MATH for all MATH. Let MATH be the polar decomposition of MATH and let MATH be the symmetry introduced in REF . Using REF, we deduce that MATH therefore the representation MATH is fundamentally reducible. CASE: We consider the elements involved in the proof of REF for a fundamental symmetry MATH on MATH for which MATH, MATH. Therefore MATH for all MATH, and then MATH . CASE: Just set MATH. |
math/0007183 | This follows from REF . |
math/0007183 | This follows from REF . |
math/0007183 | We can obtain a NAME decomposition of the kernel MATH by adapting the NAME space construction from the positive definite case. Let MATH be a fundamental symmetry on MATH and let MATH be the associated NAME space structure on MATH. We then consider the NAME space MATH where MATH is a one-dimensional NAME space generated by the unit vector MATH and MATH is the MATH-fold tensor product MATH. Then, the operator MATH is a symmetry of MATH. Let MATH be the orthogonal projection of MATH onto its symmetric part, where MATH and MATH is an element of the permutation group MATH on MATH symbols. We notice that MATH for all MATH, therefore MATH . It follows that the compression of MATH to MATH, the boson NAME space, gives a symmetry on MATH. So, MATH becomes a NAME\ĭn space by setting MATH, MATH and MATH is the positive definite inner product on MATH obtained from MATH. We now use the NAME decomposition that gives the NAME space (see CITE or CITE for more details). Thus, for MATH we define the map MATH where MATH and MATH. Then, we define MATH and we can easily check that MATH is a NAME decomposition of MATH. We now define MATH and each element of MATH admits a representation of the form MATH, where MATH, MATH, MATH, MATH, MATH are complex numbers and MATH, MATH, MATH are elements in MATH. For each MATH we can define a map on MATH by the formula MATH . In order to show that this map is well-defined it is enough to prove that the relation MATH implies MATH. This follows from the fact that MATH is a linearly independent set in MATH REF . Next we have MATH so that the family MATH satisfies the canonical commutation relations. We also notice that the operators MATH are isometric on the common domain MATH with respect to the indefinite inner product on MATH. |
math/0007183 | Let MATH be a GNS data of MATH. Then for arbitrary MATH, MATH defined as in REF is a linear bounded mapping from MATH into MATH such that MATH is dense in MATH. Also, MATH and hence MATH is a NAME decomposition of MATH. Let MATH be a NAME decomposition of MATH. Define MATH which is dense in MATH. Each element of MATH admits a representation of the form MATH, where MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH. We can define a map on MATH by the formula MATH . We have to prove that this map is well-defined. To this end, it is enough to show that MATH implies MATH . From MATH we deduce that for any MATH we have MATH . On the other hand, for an arbitrary MATH, we have MATH whence letting MATH, the last sum is zero. This implies that for arbitrary MATH, MATH . But MATH is dense in MATH, therefore MATH and hence MATH is well-defined on MATH. Also, MATH is a linear operator on MATH. We easily check that MATH for MATH. Also, from the relation MATH and the linearity of MATH, it follows that MATH for all MATH and MATH. Therefore, MATH is a closable representation of MATH. For MATH and MATH, we have MATH so that the domain of MATH contains MATH and REF holds. Therefore, for every MATH the operator MATH is closable and MATH is a hermitian representation of the algebra MATH on the NAME\ĭn space MATH. Also, for MATH, MATH . Clearly, MATH, therefore, MATH is a GNS data associated to MATH. It is easy now to see that the association of the GNS data MATH, as in REF, to an arbitrary NAME decomposition MATH is a two-sided inverse to the correspondence defined as in REF. |
math/0007183 | Note that REF is equivalent to MATH and then apply REF . |
math/0007183 | Let MATH, MATH, be two NAME decompositions of MATH that are unitarily equivalent, that is, there exists a unitary operator MATH such that MATH. Let MATH, MATH, be the corresponding GNS data for MATH as in REF . Then, MATH . Also, for MATH and MATH, MATH which implies that MATH. Finally, MATH therefore MATH and MATH are unitarily equivalent GNS data for MATH. Conversely, let MATH, MATH, be two unitarily equivalent GNS data for MATH and let MATH, MATH, be the NAME decompositions of MATH associated to these GNS data by REF . Therefore, there exists a unitary operator MATH such that MATH, MATH for all MATH and MATH. It follows that MATH which shows that MATH and MATH are unitarily equivalent NAME decompositions of the kernel MATH. Now, an application of REF concludes the proof. |
math/0007183 | This is a consequence of REF . |
math/0007183 | The implications MATH are direct consequences of REF . For MATH we use the proof of REF in order to deduce that there exists MATH such that MATH. Then REF shows that MATH for all MATH. Also, in this case, MATH is linear in the first variable (hence, antilinear in the second variable). If we define MATH for MATH, then MATH is a linear functional on MATH and MATH . Now all the required properties of MATH follow from the corresponding properties of MATH. |
math/0007183 | Let MATH be a minimal NAME dilation of MATH and define MATH as in REF. Then MATH . Since MATH it follows that MATH is a NAME decomposition of MATH. Let us note that, by the defintion of MATH, MATH and hence, letting MATH, it follows that the NAME decomposition MATH is invariant. Conversely, let MATH be an invariant NAME decomposition of the hermitian kernel MATH, that is, there exists a hermitian representation MATH of the multiplicative semigroup with involution MATH, such that MATH . Define MATH and MATH. Since MATH is linear it follows easily that MATH is also linear, hence a selfadjoint representation of the MATH-algebra MATH on the NAME\ĭn space MATH. Then, taking into account that MATH for all MATH, it follows MATH and since MATH we thus proved that MATH is a minimal NAME dilation of MATH. We leave to the reader to show that the mapping defined in REF is a two-sided inverse of the mapping associating to each invariant NAME decomposition MATH the minimal NAME dilation MATH as above. |
math/0007183 | In the following we let MATH be the unitary group of MATH. Then MATH has the involution MATH and acts on MATH by MATH. MATH . We use the NAME 's off-diagonal technique. Briefly, assume that MATH is completely bounded. By REF, there exist completely positive maps MATH and MATH such that the map MATH is completely positive. Define MATH, which is a completely positive map. We can check that MATH. First, let MATH, MATH. Then MATH, so that MATH. In particular, for MATH, MATH or MATH. Therefore, MATH are positive maps. The argument can be extended in a straightforward manner (using the so-called canonical shuffle as in CITE) to show that MATH are completely positive maps. MATH . Since MATH is completely positive, the kernel MATH is positive definite and satisfies MATH. Also, MATH . By REF , there exists a NAME decomposition MATH of MATH and a fundamentally reducible representation MATH of MATH on MATH such that MATH . Let MATH be a fundamental symmetry on MATH such that MATH for all MATH. Then MATH is also a representation of MATH on the NAME space MATH. Also, for MATH, MATH . Since MATH is the linear span of MATH and MATH is linear, MATH can be extended by linearity to a representation MATH of MATH on MATH commuting with MATH and such that MATH holds for all MATH. Also, MATH and using once again the fact that MATH is the linear span of MATH, we deduce that MATH is dense in MATH. CASE: Note first that REF is equivalent with the following: there exists a NAME space MATH, a MATH-representation MATH, and a bounded operator MATH such that MATH where MATH is a symmetry on MATH commuting with MATH for all MATH, and MATH is dense in MATH. Assuming this, define for MATH, MATH . Then MATH is in MATH and MATH. MATH becomes a NAME\ĭn space by setting MATH, MATH. Also, for MATH, MATH so that MATH is a NAME decomposition of MATH. Let MATH be the NAME decomposition of MATH and define, as in the proof of REF , the hermitian kernels MATH . It was checked in the proof of REF that MATH for all MATH and MATH. For MATH define MATH . Then MATH are linear maps on MATH and for MATH, MATH, we get MATH . Since MATH and MATH are antilinear in the second variable and MATH is the linear span of MATH, it follows that MATH for all MATH. This implies that MATH are disjoint completely positive maps such that MATH. The implication MATH follows from REF . |
math/0007186 | Follows from REF . |
math/0007186 | Straightforward. |
math/0007186 | REF follows from the fact that MATH is a NAME subsheaf of MATH and the definition of the NAME map. REF follows from REF are obvious. |
math/0007186 | Let MATH be functions such that MATH, MATH, form a local basis in MATH and all the MATH commute and form a local basis in MATH. Let us set MATH. This rule defines a local connection. Let us prove the invariance of MATH, that is, MATH . But for MATH, MATH, MATH this equation is equivalent to the relation MATH which holds by the NAME identity for the NAME bracket MATH. Let us proof that MATH is torsion free. Since the torsion is MATH linear, it is enough to prove that MATH for all pairs MATH of our basis. But this follows from the definition of MATH and from the fact that all MATH pairwise commute. Similarly, from the fact that all MATH pairwise commute follows that MATH is flat. That MATH preserves MATH is obvious. Since MATH is equivalent to MATH, it is obvious that MATH has the property: for any MATH such that MATH one has MATH . Now, let us prove the existence of a global connection. Let MATH is an open covering of MATH such that on each MATH there is a MATH-connection MATH. Then the differences MATH form on MATH a NAME cocycle MATH, MATH. MATH satisfy the following properties. It follows from REF that MATH . Since all MATH are torsion free, MATH are symmetric. Since all MATH preserve MATH, one has MATH for MATH. In addition, MATH considered as elements from MATH, MATH, belong to MATH where MATH consists of endomorphisms of MATH preserving MATH invariant. Since all the properties above are MATH linear, it is possible to find tensors MATH satisfying the same properties and such that MATH. Then MATH is a globally defined connection. Flatness of MATH on MATH along MATH follows from the fact that for all MATH for MATH, which follows from REF . MATH is torsion free because all MATH are symmetric. So, MATH satisfies the proposition. |
math/0007186 | Since MATH is a MATH valued REF-form, it follows from the NAME identity that it is a closed REF-form. From flatness of MATH along MATH follows that the MATH component of MATH does not contain terms of the view MATH with MATH. It follows that MATH is a polarization of MATH. It is easy to see that MATH being restricted to MATH defines a connection on MATH, MATH, and MATH where MATH is the MATH-degree MATH component of the MATH valued REF-form MATH. Now REF follows from the definition of the NAME classes. Since NAME classes are integer valued elements of MATH, that is, belong to MATH, the independence of MATH on the parameter MATH is obvious. |
math/0007186 | First of all, we apply the NAME method, REF , to find MATH satisfying REF. According to REF , MATH must obey the equation MATH . Look for MATH as the limit of the sequence, MATH, where MATH, MATH, and MATH. As in REF, using REF and the fact that MATH, such MATH can be calculated recursively: MATH . So, REF are proven. REF follows from REF . Let us prove that MATH for all MATH. That MATH follows from the fact that MATH and from REF. Suppose that MATH for MATH. Then MATH because MATH preserves MATH. On the other hand, MATH because of REF, therefore from REF follows that MATH as well. So, we have that MATH being the limit of the convergent sequence MATH satisfy REF, and REF. REF obviously follows from REF . |
math/0007186 | Again, we apply the NAME iteration procedure. According to REF , we look for MATH as a limit, MATH, there MATH can be calculated recursively: MATH . Put MATH. As in REF , one proves that such MATH and MATH satisfy REF. Now observe that MATH and if MATH then MATH. By induction we conclude that MATH for all MATH. So MATH as well, which proves REF. Let us prove REF. We have MATH. Since by REF MATH, MATH as well. It follows that MATH and MATH, because MATH is a MATH linear map and MATH. |
math/0007186 | Let MATH be a REF-form representing MATH and having MATH as its polarization (see REF ). Construct the star-product for MATH and MATH as in REF . According to REF, the NAME curvature corresponding to that star-product is equal to MATH and its NAME class is equal to MATH. REF follows from REF. |
math/0007186 | First of all, let us prove that MATH is a polarization, that is, MATH is integrable. Since MATH is involutive, the NAME complex MATH is meaningful, where MATH is a deformation of the differential MATH from the NAME complex for MATH. Let MATH be a neighborhood such that the complex MATH is exact. It is enough to prove that if MATH is a function on MATH satisfying MATH then there is an extension MATH such that MATH. Suppose that we have already found a series MATH such that MATH mod MATH. Then MATH mod MATH. Applying MATH to the both sides of this equation and dividing by MATH we obtain MATH mod MATH. Hence, MATH. So, MATH is a closed REF-form in the NAME complex corresponding to MATH, and one can find a function MATH such that MATH. It is clear that if we put MATH, we obtain MATH mod MATH. In this way we construct step-by-step the series MATH such that MATH and prove that MATH is a polarization of MATH. That MATH is strong is obvious because of the upper-semicontinuity argument. |
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