paper stringlengths 9 16 | proof stringlengths 0 131k |
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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 tha... |
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 distributi... |
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 MAT... |
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 ... |
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/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 re... |
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... |
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 t... |
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 quant... |
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 ... |
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 ... |
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.... |
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 b... |
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 fa... |
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 ... |
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... |
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 MAT... |
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 co... |
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, M... |
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 M... |
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 s... |
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... |
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/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 MAT... |
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 exis... |
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... |
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 obt... |
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 ... |
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 identi... |
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 s... |
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 R... |
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 equa... |
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.... |
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... |
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/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 t... |
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 h... |
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 case... |
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 isomorphi... |
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/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 fo... |
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 exi... |
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 ... |
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 ... |
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 d... |
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 wit... |
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 previou... |
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 o... |
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 projective... |
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... |
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 a... |
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... |
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 M... |
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 NAM... |
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 positi... |
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 hold... |
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 t... |
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. ... |
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... |
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 t... |
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