paper stringlengths 9 16 | proof stringlengths 0 131k |
|---|---|
math/9903077 | We use induction on the length of MATH as a bracketing of elements from MATH (that is, obtained by forming successive brackets of elements from MATH). It readily reduces to the case where this length is REF, the case of length REF being MATH. Suppose MATH with MATH. Now let MATH. As MATH is an automorphism, MATH is ext... |
math/9903077 | Take MATH to be a basis of MATH consisting of elements from MATH. Its existence is guaranteed by REF . Notice that if MATH is extremal, then MATH and so MATH is also extremal with MATH. Now suppose MATH. Define MATH by MATH . Notice from above this is MATH. Now suppose MATH are two ways of writing an element MATH of MA... |
math/9903077 | By REF , we have MATH. Since MATH is associative by REF , we find MATH for MATH. |
math/9903077 | Clearly, MATH, MATH and MATH linearly span MATH. The rest is straightforward. |
math/9903077 | Consider a NAME basis MATH of MATH with respect to a given NAME subalgebra MATH of MATH, write MATH for the corresponding root system, and MATH for the element of MATH associated with a given root MATH. Suppose that MATH is a long root in MATH. Then, for MATH, we have MATH. Moreover, for MATH, the sum MATH is not a roo... |
math/9903077 | Suppose that MATH is generated by extremal elements MATH. Denote by MATH the associated bilinear form. We first consider the case in which MATH is identically zero. Suppose that MATH is generated by elements MATH with MATH for MATH. Then MATH is finite dimensional and is nilpotent. Furthermore, there is a universal suc... |
math/9903077 | We know that MATH is non-trivial by REF . (Otherwise MATH is identically MATH and MATH is nilpotent.) Thus, MATH is a proper ideal of MATH and so, as MATH is simple, MATH. This means that MATH is non-degenerate. Suppose MATH with MATH. Then MATH is extremal and MATH for all MATH. Hence MATH. This means MATH. |
math/9903077 | By induction on the length of a bracketing (with respect to MATH) and NAME. |
math/9903077 | Since MATH is linearly spanned by the monomials in MATH, we have to show that each monomial may be written as a linear combination of the given REF elements. All monomials of length REF are on the list. There are MATH monomials of length REF with different factors. Since MATH for all MATH, all of them may be expressed ... |
math/9903077 | Let MATH be a non-trivial quadratic highest weight module for MATH of finite dimension. By REF , there is a single MATH-orbit of extremal elements in MATH. Moreover, all these (long root) elements have the same nilpotency index on MATH (because the highest weight representation is equivalent to a composition of itself ... |
math/9903077 | Let MATH be the highest weight of MATH. Suppose MATH is a weight for MATH. Then there is a path of weights from MATH to MATH, such that, for each adjacent pair MATH from the path, the difference MATH is a positive root. Since there is only one root length, each of these differences is a root whose root element is extre... |
math/9903077 | The simple root elements with respect to MATH generate the subalgebra MATH where MATH is the set of positive roots and MATH is a root space. The lemma follows from the fact that the MATH-submodule of MATH generated by the root element corresponding to the root of lowest height, coincides with MATH. |
math/9903077 | If MATH is generated by the extremal elements MATH, then the image of MATH in MATH, the underlying MATH-dimensional vector space, is generated by MATH of dimension at most MATH. Thus, irreducibility of the representation implies MATH, whence the lemma. |
math/9903077 | This follows immediately from REF . |
math/9903077 | This is immediate from the dimensions given in REF and in REF . |
math/9903077 | For MATH, using REF , we find MATH . This implies that MATH maps MATH to MATH. |
math/9903077 | Note that MATH and (by REF ) MATH. Hence MATH is invariant under MATH. Furthermore, MATH is zero on MATH by REF . We write MATH on MATH in NAME normal form with eigenvalues MATH or MATH. Now REF is obvious. Next suppose MATH. Then MATH are linearly independent. Denote by MATH the eigenspace of MATH with eigenvalue REF.... |
math/9903077 | Let MATH. Then MATH for all MATH. Hence MATH for all MATH by REF. Therefore MATH. |
math/9903077 | One inclusion comes from REF . Let MATH. Assume that MATH. Then there exists MATH such that MATH. Hence MATH in characteristic MATH. This is a contradiction. |
math/9903077 | It is enough to show for MATH and MATH that MATH. We have MATH. Hence MATH. |
math/9903077 | Let MATH. Take MATH. If MATH, then MATH by REF . If MATH, suppose MATH. Then MATH. Hence MATH is a non-solvable NAME subalgebra of MATH which is a contradiction. So MATH. This means that MATH for all MATH and MATH. The lemma follows as MATH linearly spans MATH, compare REF . |
math/9903077 | Suppose that MATH is a solvable ideal of MATH and let MATH. For MATH, either MATH, in which case MATH by REF , or MATH and then MATH by REF . |
math/9903077 | Put MATH and MATH. Then, by REF , for MATH, MATH . For the last equality, use that MATH is associative and MATH. We obtain MATH. Also MATH, as MATH. Consequently, MATH . |
math/9903077 | Since MATH, the expression MATH is well-defined for MATH. For MATH, MATH, MATH, whence MATH. |
math/9903077 | Recall that MATH, so if MATH is contained in the center MATH of MATH, there is nothing to prove. Suppose, therefore, MATH. We show that MATH. By REF , there is a nonzero element MATH with MATH. (If MATH, any element of the intersection will do, otherwise, take MATH, and MATH such that MATH, whose existence is guarantee... |
math/9903077 | As MATH, we have MATH for MATH. Let MATH and MATH. Write MATH, MATH with MATH, MATH. We calculate MATH. Hence MATH for all MATH, and MATH. Since MATH is linearly spanned by MATH, each MATH is linearly spanned by the projections. Hence MATH is linearly spanned by extremal elements (with form MATH restricted to MATH). Fi... |
math/9903077 | Assume that MATH is a direct sum of simple ideals MATH. By REF , the decomposition is an orthogonal one with respect to MATH. Suppose MATH for some MATH. Then the form MATH restricted to MATH is trivial. But then MATH is nilpotent by REF , a contradiction. This means that the form MATH restricted to each MATH has a tri... |
math/9903077 | In REF , we have already proved that MATH. Set MATH and let MATH be as in REF . Since MATH, we have MATH if and only if MATH, so the corollary is a direct consequence of REF . |
math/9903077 | All equations can be verified in a straightforward manner, by use of the definition of MATH. In the calculations we use the NAME identity, associativity of MATH (for example, MATH, MATH when MATH, and MATH when MATH) and the rewriting rule for MATH of REF . |
math/9903077 | Note that REF ' are equivalent by REF . For MATH, MATH, we have MATH and MATH. Hence REF implies REF . If REF holds, then MATH is extremal and a short calculation shows MATH. Notice MATH here. This yields the result. |
math/9903077 | By REF we have MATH. By associativity of MATH (compare REF ), we see MATH, which proves REF ' of REF , as required. |
math/9903077 | Denote by MATH the NAME algebra generated by MATH. Let MATH with MATH for some MATH. Since MATH is invariant under MATH for MATH, it is also invariant under MATH. Hence MATH. We apply MATH to this vector and see that MATH is in MATH. Hence also the difference, which is MATH, is in MATH. Since MATH, we conclude that MAT... |
math/9903077 | Every element of MATH is a long root element by REF . Suppose that MATH is generated by the extremal elements MATH. Write MATH (as usual) and MATH. It suffices to show that MATH coincides with MATH. Put MATH. Notice that, for MATH, also MATH. Hence, by the argument of the proof of REF with MATH instead of MATH, we obta... |
math/9903080 | We may assume that the pair MATH, MATH is a direct sum of several blocks of the form MATH and MATH, and that for any MATH the family MATH. We suppose that MATH, it is easy to consider the remaining case separately. Start with supposing that there are only blocks of the form MATH, MATH. Then the only things we need to p... |
math/9903080 | In this prove we assume that MATH is a complex manifold, so that MATH is a complex vector space for any MATH. If MATH is a MATH-manifold, one should substitute MATH instead of MATH in the arguments below. In conditions of REF if MATH in a neighborhood MATH of the point MATH, then vectors MATH span a vector subspace MAT... |
math/9903080 | First of all, one can proceed as in REF up to the moment we concluded MATH. Under the conditions of the amplification we conclude that MATH, thus MATH. REF implies that MATH, thus MATH. This shows that in fact MATH. We can conclude that for MATH in an appropriate open subset MATH the pair of bilinear pairings on the ve... |
math/9903080 | For translation-invariant brackets the tensor fields MATH and MATH are constant, thus to describe the bracket we need to describe the pairing on one cotangent space. On the other hand, we know that to satisfy REF this pairing should be isomorphic to MATH, thus any pair which satisfies the lemma is isomorphic to one giv... |
math/9903080 | It is easy to write the diffeomorphism explicitly as MATH, MATH, MATH with an appropriate function MATH. |
math/9903080 | Indeed, MATH defines a structure with constant coefficients (compare with REF), thus a flat one. Any other function MATH which satisfies REF will define a non-isomorphic bihamiltonian structure, thus a non-flat one. |
math/9903080 | Actually it is possible to describe all analytic functions MATH such that MATH admits a MATH-family MATH, at least if we are allowed to restrict our attention to smaller open subsets. If MATH is a MATH-family, then MATH, MATH, gives a MATH-family. Call such families simple families. By REF a simple family may exist on ... |
math/9903080 | Let MATH. Consider the symplectic leaf of MATH passing through MATH. The codimension of this leaf is MATH, fix a transversal manifold MATH of dimension MATH, and coordinate functions MATH, MATH, on this manifold. Obviously, if MATH is close to MATH and MATH is close to MATH, then the symplectic leaf of MATH passing thr... |
math/9903080 | To simplify notations assume MATH. Let MATH. Since MATH for any function MATH, we see that MATH if MATH or MATH. Since MATH is linear in MATH, MATH for any MATH as far as MATH. On the other hand, the same identity is true for MATH by continuity in MATH. |
math/9903080 | Put MATH for MATH. This makes REF applicable for MATH too. For any MATH and MATH . Repeating this process MATH times, one gets MATH. Moreover, MATH. |
math/9903080 | For any MATH and MATH one gets MATH again. Repeating this several times, one can decrease MATH until it becomes REF or REF (depending on MATH being even or odd). If MATH is even, use MATH, if MATH is odd, use MATH. Thus MATH is always REF, moreover, MATH. |
math/9903080 | To simplify notations, suppose MATH (the case of general MATH is absolutely parallel). Then MATH is defined uniquely up to a change MATH. Additionally, given MATH, REF determines MATH up to a change MATH. Moreover, the NAME series for MATH provides one solution to the recursion relations REF. Since the change of the fo... |
math/9903080 | Since the NAME series for MATH in MATH converge, it is enough to show that the NAME coefficients for MATH depend on MATH. Fix numbers MATH, MATH. Let MATH, and MATH . Obviously, this is a MATH-Casimir family. It is easy to show that for generic values of MATH the first MATH . NAME coefficients MATH of MATH are independ... |
math/9903080 | Indeed, if a structure is NAME at MATH, then it is also NAME at MATH for MATH in an appropriate open subset of MATH. It is easy to show that by decreasing this subset MATH one may assume that at any point MATH the sizes of NAME blocks of the pair of pairings on MATH are the same. Functions MATH, MATH, MATH, given by th... |
math/9903080 | A tiny modification of the above proof together with REF imply this statement immediately. |
math/9903080 | Let MATH be the determinant of the upper-left minor of MATH of size MATH. We need to show that MATH, MATH. Let us show that MATH, MATH, MATH. Use induction in MATH. Plugging the identity MATH into MATH shows that the step of induction will work as far as MATH. On the other hand, due to obvious symmetry MATH of brackets... |
math/9903080 | It is enough to consider the case when no MATH vanishes. Indeed, if we leave all the variables MATH except MATH fixed, then MATH is quadratic in MATH without the linear term. Thus MATH implies MATH. On the other hand, if MATH, the matrix breaks into two blocks, and the derivatives with respect to other variables can be... |
math/9903080 | Since this function is invariant with respect to translation MATH, it is enough to show this for the bracket MATH. When one calculates MATH, only the factor MATH of MATH matters, and by REF MATH vanishes. Similarly, for MATH only MATH matters, and it also vanishes. |
math/9903080 | Associate to a point MATH of the infinite NAME lattice an infinite MATH-diagonal matrix MATH in the same way we did it in REF. Consider a matrix equation MATH, here MATH is a two-side-infinite vector. Since this equation may be written as the recursion relation MATH this matrix equation has a two-dimensional space of s... |
math/9903080 | Reduce this statement to one of Amplification REF. In our case MATH, and, by submersion condition, MATH. Thus the only thing one needs to show is that MATH. This momentarily follows from the definition of the action dimension. |
math/9903084 | MATH since MATH. If MATH is noncrossing, then the limit, as MATH, of REF is MATH. On the other hand, assume that MATH is crossing. As in CITE, the number of elements of MATH, which is a NAME number, is less than MATH and for each MATH, MATH. Thus the absolute value of REF is less than MATH. In particular, it converges ... |
math/9903084 | More generally, MATH . For MATH as in REF , MATH. Applying REF and using the estimate in the proof of that Lemma, we see that the above REF is less than MATH, where MATH. It was shown in REF that MATH (note that our use of MATH and MATH is the opposite of theirs). Therefore MATH . In particular, MATH unless MATH is non... |
math/9903084 | By REF , MATH. By the Theorem, for MATH. The second equality follows from the first one by the use of NAME inversion on the lattice of noncrossing partitions. |
math/9903084 | By REF , MATH . Here the last equality follows from applying the first equality to MATH. |
math/9903084 | If MATH contains a singleton, and the process is centered, then MATH . In fact each term is MATH, for any MATH. On the other hand, if MATH contains an inner singleton, so does any partition MATH with at most MATH crossings, for any MATH. Thus in the sum in REF , only the partitions MATH with MATH enter, and so by the a... |
math/9903084 | The only noncrossing partition with no inner singletons for which no consecutive elements lie in the same class is MATH. |
math/9903084 | For MATH the NAME function on the lattice of noncrossing partitions, MATH . |
math/9903084 | First note that since the MATH's are orthogonal and identically distributed, MATH . MATH . MATH's are centered, so MATH. Thus only those partitions MATH contribute to the sum REF for which MATH has no single-element classes. In particular MATH and so MATH (since MATH). Then the sum REF is MATH where MATH is the partiti... |
math/9903084 | First consider the process on the interval MATH. In the notation from the beginning of the section, MATH and MATH. MATH since MATH. In the limit MATH, by the Theorem the only term that survives is the one for MATH (that is, MATH), and that term is MATH. Therefore MATH . For the full NAME process, the result follows fro... |
math/9903084 | For a free NAME element MATH, MATH, so MATH. Therefore MATH and so MATH, which implies MATH. |
math/9903084 | MATH where MATH are the blocks of MATH. By REF , once again the only term that survives in the limit MATH is the one for MATH, and that term is MATH. |
math/9903084 | The proof will be by induction on the level in the above partial order on the classes of the partition MATH. For the product measure MATH one can sum over the indices corresponding to the minimal classes, which are precisely the classes that are intervals, without disturbing the rest of the partition. If any of these c... |
math/9903084 | Once again, the proof is by induction on the level of the class of MATH in the partial order. Since the diagonal measures of the free NAME process are equal to the process itself, each minimal (interval) class can be shrunk to a one-element class. Again, consider a class which used to be at a distance of MATH from the ... |
math/9903084 | Any MATH-tuple of indices determines a partition MATH by MATH. REF restricts MATH precisely to those for which MATH, and the union of all collections of indices corresponding to such MATH's is precisely the set given by REF . This decomposition of the set of indices gives the representation REF. The second equality fol... |
math/9903084 | The proof of REF shows that MATH . That proof is purely combinatorial, and works for noncommutative stochastic measures as well as for the commutative ones. Therefore the first equality follows from REF above. Moreover, the proof also works for the lattice of noncrossing partitions, provided we use the appropriate NAME... |
math/9903084 | The first part is obtained by taking expectations of both sides in REF and applying the multiplicativity property of REF . The second part follows from the first part. |
math/9903084 | The boundary conditions follow directly from the definitions. MATH where MATH. By an argument similar to the one in the proof of REF , each inner singleton contributes a factor of MATH, and so MATH. |
math/9903084 | By a repeated use of the Lemma, MATH . For MATH, using the boundary conditions, the last term in REF is MATH . Thus, continuing REF , MATH . The result follows from the definitions of MATH and MATH. |
math/9903084 | MATH . Therefore MATH and so MATH . |
math/9903084 | Since the diagonal measures for the free compensated NAME process are MATH, MATH, by REF MATH . Therefore MATH . |
math/9903084 | By REF , MATH . |
math/9903084 | By REF , the diagonal measures of MATH are MATH. Therefore by the Theorem MATH . |
math/9903084 | First, let MATH be a free NAME process, MATH. Then MATH. The joint distribution of MATH and MATH is determined by the condition that MATH is freely independent from MATH. Thus we may assume that in fact MATH is freely independent from MATH. In that case MATH is freely independent from MATH, and MATH. Then the result fo... |
math/9903084 | As in REF , MATH . Here in MATH, MATH acts on the subset MATH while MATH acts on its complement. The rest of the proof proceeds as in REF . |
math/9903085 | REF : obvious. CASE: Let MATH be a finite cover of MATH. For every finite MATH denote by MATH the (non-empty, finite) collection of all MATH-essential elements of MATH. Clearly, whenever MATH, one must have MATH. The compactness (or rather finiteness) considerations lead one to conclude that MATH thus finishing the pro... |
math/9903085 | Since the unit spheres MATH, MATH, are contained in MATH in a canonical way both as uniform subspaces and MATH-subspaces, it follows that the compactifications MATH, MATH, are compact MATH-subflows of MATH. This establishes REF . Now assume that both systems MATH, MATH do not have the concentration property. Then there... |
math/9903085 | Let MATH denote a MATH-invariant mean on the unit sphere MATH, that is, a positive functional of norm MATH, taking the function MATH to MATH and such that MATH for all MATH and all MATH, where MATH. For every bounded linear operator MATH on MATH define a function MATH by MATH . Then MATH is bounded (by MATH) and NAME w... |
math/9903085 | Let MATH be a MATH-fixed point in the NAME compactification MATH. For every MATH set MATH, where MATH is the unique continuous extention of MATH to MATH. Clearly, MATH is a MATH-invariant mean on MATH. |
math/9903085 | It is enough to apply the following result of NAME REF : the left regular representation MATH of a LC group MATH is amenable if and only if MATH is amenable. It is instructive to look at the direct proof as well. Let MATH be a MATH-fixed point in MATH. For every NAME set MATH and each MATH set MATH, where MATH denote, ... |
math/9903085 | It is enough to make an obvious remark: a MATH-invariant mean on MATH is invariant with respect to the action of every group MATH represented in MATH by unitary operators. Since MATH is infinite-dimensional, one can find a non-amenable discrete group MATH of the same cardinality as is the density character of MATH, and... |
math/9903085 | If MATH is infinite-dimensional, the statement follows from Corol. REF or REF. If MATH, the unitary group MATH possesses no non-zero invariant vectors, and there is no concentration property in a trivial way. |
math/9903085 | Since the left regular representation of MATH in MATH (defined by MATH) is well known (and easily checked) to be strongly continuous CITE, so is the subrepresentation of MATH in MATH. Now it follows from a result of Teleman CITE that the action of MATH on the NAME spectrum of MATH is continuous, that is, MATH is a topo... |
math/9903085 | By REF , the action of MATH upon MATH is continuous for every NAME space MATH, that is, MATH forms a compact MATH-space. According to a result by NAME and NAME that we cited in the Introduction REF , if a compact group MATH acts by isometries on the unit sphere MATH of MATH, then the pair MATH has the concentration pro... |
math/9903085 | Let MATH be arbitrary and fixed. Let for every natural number MATH, MATH and MATH be two rank MATH projections in MATH. Assume for a while that, as MATH, MATH that is, for some positive constants MATH and all MATH one has MATH where MATH denotes the unit sphere in the space MATH, MATH. Then a standard argument (compare... |
math/9903085 | Let MATH be an eigenvector of MATH corresponding to an eigenvalue MATH. Then MATH, where MATH is the angle between one-dimensional subspaces spanned by MATH and MATH. The space MATH has an orthogonal basis formed by eigenvectors of MATH which can be written in the form MATH, MATH, where MATH corresponds to the eigenval... |
math/9903085 | Let the projections MATH, MATH, MATH be as in REF . For every MATH denote by MATH the (unique) isometric isomorphism from the one-dimensional range of MATH to that of MATH. (That is, the reflection across the linear span of MATH, using the notation from the proof of REF .) Since MATH, one concludes that if MATH and MAT... |
math/9903085 | Choose a sequence of orthogonal projections MATH, MATH, having the properties: CASE: MATH, and CASE: MATH for all MATH as MATH. Denote by MATH the unit sphere in the space MATH, and for each MATH denote by MATH the unit sphere in the MATH-dimensional space MATH. Now let MATH be an arbitrary finite cover of the unit sph... |
math/9903085 | If MATH admits finite-dimensional subrepresentations of arbitrarily high dimension, then the desired projections can be constructed in an obvious way. Otherwise, using the assumption of the lemma, one can assume without loss in generality till the end of the proof that MATH contains no nontrivial finite-dimensional sub... |
math/9903085 | Combine REF . |
math/9903085 | CASE: If MATH, then under our assumption MATH clearly has a non-zero invariant vector. Otherwise, suppose there exists a non-amenable subrepresentation MATH of MATH having finite codimension. If the finite-dimensional subrepresentation MATH has no non-zero invariant vectors, then it does not have the concentration prop... |
math/9903085 | CASE: every strongly continuous unitary representation of an amenable locally compact group is amenable CITE, and therefore so are all its subrepresentations, and REF applies. REF: if MATH is finite, there is nothing to prove, otherwise MATH is infinite-dimensional and REF applies together with our assumption. |
math/9903085 | CASE: If MATH has no finite-dimensional subrepresentations, then REF applies. CASE: Combine REF with the following: a unitary representation is amenable if it contains a finite-dimensional subrepresentation REF . |
math/9903085 | CASE: REF . CASE: If MATH contains a finite-dimensional subrepresentation MATH, then the desired MATH-invariant mean is obtained by integrating the (restriction of) a MATH over the unit sphere of MATH. If MATH contains no finite-dimensional subrepresentations, then, according to REF , there even exists a multiplicative... |
math/9903087 | Write MATH, where MATH and MATH. The product MATH is a lift of MATH, with symbol MATH. As a special case of CITE, the symbol of any lift MATH of MATH is given by a formula MATH where the sign depends on the choice of lift. Replacing the variables MATH with MATH, the first part of the Lemma follows. REF is a special cas... |
math/9903087 | Given MATH let MATH. From REF we obtain the following splitting of the tangent space MATH . Here MATH is the orthogonal complement of the stabilizer algebra MATH, embedded by the generating vector fields MATH, and MATH is the pre-image under MATH of the tangent space of the stabilizer group MATH. The moment map conditi... |
math/9903087 | Since MATH is connected, the lift MATH is unique. Moreover, since MATH, the form MATH, hence also MATH, vanishes in odd degrees. The formula for MATH is obtained by using REF for MATH in the proof of REF , and keeping track of the signs. |
math/9903087 | We will need the following two identities from CITE, both of which are verified by straightforward calculation: MATH . Here MATH are the contraction operators for the conjugation action, and MATH the contraction operators for the NAME algebra. From REF and the moment map condition we obtain MATH . Since MATH, REF gives... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.