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 extremal and so MATH is in the linear span of the extremal elements MATH. |
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 MATH as a sum with MATH, extremal elements. By REF , for MATH, we have MATH . Since MATH spans MATH, we conclude MATH. Thus, MATH is a well defined linear functional. It readily follows that MATH defines a bilinear form. It is symmetric by REF . It remains to establish that MATH is associative. Take MATH. Interchanging MATH and MATH in REF , we obtain MATH . On the other hand, by NAME and NAME REF, MATH . Suppose now MATH. Comparing the coefficients of MATH in the two expressions for MATH, we find MATH. But then, by symmetry of MATH, we have MATH, proving MATH whenever MATH. Similarly, MATH, whenever MATH and MATH, whenever MATH. Suppose that MATH. We have to show MATH. If MATH and MATH, then MATH by the previous equations. Hence (up to symmetry) we are left with the case MATH, MATH. Then MATH, so we must show that MATH. By NAME, MATH. Applying MATH yields MATH as needed. We conclude that MATH for all MATH. Since MATH linearly spans MATH, this proves that MATH is associative. |
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 root in MATH unless MATH, so MATH unless MATH, in which case the result is a multiple of MATH. To see that MATH is generated by the long root elements, observe that MATH is generated by the root elements of the basis MATH and (by analysis of the NAME rank REF case) that any short root element can be written as a sum of three long root elements, as is clear from the following computations in the NAME algebras of type MATH and MATH, respectively. MATH . As for the last assertion, let MATH. Under the given characteristic restriction for MATH, the proof of REF and the paragraph following REF in the proof of REF show that there exists a NAME subalgebra of MATH with eigenspace MATH. Thus MATH is a root element. If the diagram of MATH is not simply laced, short root elements of MATH are easily seen not to be extremal. Hence MATH is a long root element. |
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 such NAME algebra, MATH. Observe that each MATH is in MATH. By REF MATH for any bracketing MATH in MATH. (We consider also MATH as a bracketing.) Since MATH, this implies that MATH for any two of these bracketings, see REF . In the language of CITE, each MATH is a crust of a thin sandwich, and by CITE the associative algebra generated by MATH is nilpotent. Hence all products of some fixed length, say MATH, of the MATH are MATH. This yields that MATH is a nilpotent NAME algebra with bracketings of length MATH being MATH. In particular, MATH is finite dimensional. To see that there is a universal such NAME algebra, let MATH be the free NAME algebra generated by MATH elements MATH. Now factor out by the ideal MATH generated by all MATH where MATH is a bracketing in the MATH. Now MATH is a universal NAME algebra spanned by MATH extremal elements with MATH identically MATH. This completes the proof of the lemma. We now show how this implies REF . Suppose that MATH is generated by MATH extremal elements where the values of MATH need not all be MATH. Then the dimension of MATH is at most MATH. In particular MATH is finite dimensional. For a nonzero bracketing in MATH, its length is understood to be the total number of the MATH which appear, counting multiplicities (for example, the length of MATH is REF). Choose a basis MATH of MATH and let MATH be the linear span of MATH. Then MATH. Consider the surjective homomorphism MATH determined by MATH. We claim that MATH linearly spans MATH, that is, MATH. Any element of MATH is a linear combination of terms MATH with MATH bracketings in the MATH. Suppose that MATH is properly contained in MATH. Then there exists a nonzero bracketing MATH of minimal length with MATH. We may write MATH as a linear combination of the MATH and terms MATH as in REF with all MATH of the same length as MATH. (Note that in the free NAME algebra MATH, the linear span of the bracketings of a fixed length MATH trivially intersects the linear span of all bracketings of lengths distinct from MATH.) Among these terms MATH there is a term MATH as in REF with MATH. Set MATH. Then MATH. Hence MATH and MATH with MATH a bracketing of shorter length than MATH, a contradiction with the choice of MATH. Combining these two lemmas proves REF . |
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 by REF monomials of length REF on the list. All monomials of length REF which involve only REF letters are reducible, since MATH are extremal. There are MATH monomials of length REF with three different letters. With antisymmetry, all MATH and all MATH may be expressed. By NAME this holds also for the remaining ones. All monomials of length REF which involve only three letters are reducible, see the identity for MATH in REF . There are MATH monomials of length REF with different factors. With NAME we have the following equations MATH . The elements in the first two columns are on the list. Hence all monomials beginning with MATH, MATH, or MATH may be expressed as a linear combination of the given REF basis elements. For the monomials beginning with MATH, we calculate: MATH . The products of the form MATH may be expressed as follows. MATH . Hence also the monomials of length MATH beginning with MATH may be expressed as a linear combination of the monomials on the list. We obtain monomials of length REF by multiplying a letter from the left to a monomial MATH of length REF. This yields four possibilities: First MATH, which obviously is reducible. Second MATH, which is of the form MATH and hence it is reducible. Third MATH. There is no obvious way to rewrite this. Fourth MATH, which yields the previous case by interchanging MATH and MATH. We are left with the monomials of the form MATH or MATH, where MATH is one of the monomials of length REF on the list. This yields the following REF monomials MATH (MATH is in row MATH and column MATH): MATH . Our intended basis vectors are MATH. We may express the other REF elements as follows: Finally, we show that all monomials of length REF are reducible. We multiply the monomials of length REF on the list with a letter. Note that all these monomials of length REF are of the form MATH. Multiplication from the left with MATH or MATH yields reducible monomials. Next, we deal with MATH. With NAME we may pass from MATH to MATH and MATH. From MATH, we pass to MATH and MATH. Now we multiply first with MATH, then with MATH from the left. This yields products of the form MATH, MATH and MATH, which are all reducible. (Look for patterns of the type MATH.) The last monomial we have to reduce is of the form MATH. We pass from MATH by NAME to MATH and MATH. The products MATH and MATH are reducible. (In the second one, the two letters MATH are at distance MATH.) As a consequence all monomials of length MATH are reducible and may be written as a linear combination of REF vectors on the list. |
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 with conjugation by an element of MATH). Consequently, each element MATH satisfies MATH. In particular, for every long root MATH, the corresponding NAME basis element MATH is nilpotent of index REF. It follows that if MATH is a weight of MATH, then MATH is not. Since the set of weights for MATH is a convex subset of the coset of the root lattice containing the highest weight, only small weights can occur. Thus, the result can be readily established by use of NAME, compare CITE, and the following argument for the case of a single root length. |
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 extremal. Now take the fundamental MATH-triplet containing this root element. NAME means that the non-trivial irreducible sub-representations in MATH of the corresponding subalgebra isomorphic to MATH all have dimension MATH. But MATH, MATH are in the same representation, so they belong to a MATH-dimensional module. But then, they are conjugate by an element of the corresponding subgroup of type MATH. This establishes that MATH has a basis of weight spaces which are all conjugate, which can be taken as a definition of minuscule. |
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. Then MATH and MATH. This proves the lemma. |
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 guaranteed by the hypotheses that MATH and MATH, and apply REF .) By REF (see also REF), if the characteristic of MATH is not REF or REF, the element MATH for any MATH, satisfies MATH. In particular, there is a nonzero element MATH with MATH. If MATH, then we are done. Otherwise, there is MATH with MATH. If MATH, then MATH, and we are done. So, assume MATH. By REF, as MATH has characteristic not MATH or MATH, we have MATH. Since MATH, we have MATH and since MATH, we have MATH, whence MATH, proving that MATH. This establishes MATH. Thus, if MATH, it contains a nonzero sandwich, and so MATH. But then, in view of REF and by induction on the dimension, MATH is solvable. |
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). Finally, for MATH, MATH, we have MATH. Since MATH is linearly spanned by extremal elements, this shows that the decomposition MATH is an orthogonal one with respect to MATH. |
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 trivial radical. Hence MATH. As for the converse, assume MATH. Suppose that MATH is a minimal nonzero ideal of MATH. By REF , MATH is not Abelian. Let MATH be the orthoplement of MATH with respect to MATH. As MATH is associative, MATH is an ideal of MATH. We claim MATH. For, MATH would imply MATH by minimality of MATH and hence MATH as MATH is not Abelian, so MATH, yielding the contradiction MATH. Hence, if MATH is a nonzero ideal of MATH, it also is an ideal of MATH, and so coincides with MATH. This means that MATH is simple. Moreover, by REF , MATH is generated by extremal elements with form MATH. Notice that MATH, so by induction on the dimension, we find that MATH is a direct sum of simple ideals. |
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 MATH. The graph MATH is connected, so we obtain that MATH, whence MATH. |
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 obtain that MATH is linearly spanned by MATH. Subgroups of the algebraic group MATH generated by connected algebraic subgroups are connected algebraic subgroups (see REF ), so MATH is a closed algebraic subgroup of the connected linear algebraic group MATH. Clearly, the derivative MATH of the embedding MATH at the identity of MATH is the embedding MATH of the NAME algebra of MATH in the NAME algebra of MATH. As MATH linearly spans MATH, we have MATH and so MATH is surjective. By REF, this implies that MATH is dominant, that is, MATH is dense in MATH. But MATH is closed as well, so MATH. This establishes that MATH is generated by at most MATH long root subgroups. The converse is handled by the previous lemma. |
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 prove is that MATH, and MATH. The first statement is obvious. The following lemma follows immediately from the explicit description of the pair MATH: For the pair MATH of skewsymmetric pairings there exists a family of vectors MATH polynomially depending on MATH such that MATH for any MATH and MATH, and the degree of MATH in MATH is MATH. This family is defined uniquely up to multiplication by a constant, and it spans a MATH-dimensional vector subspace. Any other polynomial family MATH such that MATH for any MATH and MATH may be written as MATH for an appropriate scalar polynomial MATH. Denote the family MATH for the NAME block MATH by MATH. Due to this lemma one can write MATH, thus MATH. Since MATH is even, this shows that MATH, thus finishes proof of the proposition in the case when there are no NAME blocks. Consider now the general case. First of all, MATH for a generic MATH, thus MATH (for a generic MATH) is in the null-space of the linear combination MATH. Since for a block of the form MATH and generic MATH this combination has no null-space, it is obvious that MATH is in the sum of components of the form MATH. Since removing a component of the form MATH decreases MATH by MATH, does not change MATH, and may only decrease MATH, one can see that conditions of the proposition are applicable to the sum of components of the form MATH, but the equality on MATH is sharpened by at least MATH. However, we have seen that it is not possible to sharpen this inequality more than by MATH, which proves that MATH contains no NAME components. |
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 MATH satisfying MATH. There is an open subset MATH where MATH has a constant corank MATH. Obviously, there is MATH such that the point MATH is in the closure of MATH. Restrict our attention to this value of MATH. Let MATH be in MATH, and MATH, MATH, and MATH. Then the span MATH of vectors MATH considered for all possible MATH satisfies MATH, thus MATH. The vector space MATH is equipped with two skewsymmetric bilinear pairings MATH, MATH given by values of MATH, MATH (see REF ) at MATH. Obviously, MATH is in the kernel of MATH. By the conditions of REF , there is at most one independent NAME function near MATH, thus MATH. Obviously, this is the same MATH as in REF , thus MATH implies MATH. Hence the pair MATH, MATH is isomorphic to MATH for an appropriate MATH, thus MATH is odd. |
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 vector space MATH is isomorphic to a direct sum of MATH . NAME blocks. What remains to prove is that the dimensions of these blocks do not depend on MATH in an appropriate open subset of MATH. Fix a vector space MATH. For a sequence MATH denote by MATH the set of pairs of skewsymmetric bilinear pairings which are isomorphic to MATH. In particular, MATH is not empty iff all MATH are odd and MATH. Moreover, MATH is a MATH-orbit. It follows that if MATH intersects the closure of MATH, then MATH is contained in this closure. Fix a neighborhood MATH of MATH, let MATH be such sequences that there are points MATH in MATH where the pair of pairings is in each of MATH, MATH. Suppose that MATH are of maximal possible dimension among MATH, then the points MATH in MATH where the pair of pairings is in any one of MATH, MATH, form an open subset. Obviously, at least one of these subsets has MATH in its closure. |
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 given by the brackets REF. Obviously, MATH satisfies the conditions of REF , and the bracket MATH has exactly one independent NAME function MATH. |
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 a flat bihamiltonian structure only. Given an open subset MATH with two analytic functions MATH, and an open subset MATH with an analytic function MATH, let MATH . Consider an analytic bihamiltonian structure MATH with a MATH-family MATH which is not simple. Then there exists an open subset MATH with two analytic functions MATH, and an open subset MATH with an analytic function MATH such that MATH and MATH . In particular, MATH is locally isomorphic to an open subset of MATH. Write MATH. Being a NAME function for MATH, MATH depends on MATH only, similarly MATH depends on MATH only. If MATH does not depend on MATH, then MATH, thus MATH is simple. Similarly one can exclude the case when MATH does not depend on MATH. Restricting to a smaller open subset, one can assume that MATH, MATH, thus one can consider MATH and MATH instead of coordinates MATH and MATH on MATH. In particular, we may assume that MATH, MATH. By REF , given a point MATH, MATH may be written as MATH, here MATH is a scalar polynomial in MATH, and MATH is a vector-valued polynomial of degree REF in MATH. Thus MATH, denote the zero of MATH by MATH. We conclude that for each MATH there is MATH such that MATH if MATH. Restricting to an appropriate open subset of MATH, we may assume that MATH depends analytically on MATH and MATH. If MATH is constant, then MATH implies that MATH is linear, thus MATH is simple. Thus, decreasing MATH again, we may assume that either MATH or MATH give a local coordinate system on MATH. Assume that MATH is a local coordinate system. The condition MATH implies MATH . Thus MATH-form MATH vanishes, in other words, MATH. One can conclude that in the coordinate system MATH one has MATH, or MATH with an unknown function MATH. REF leads to MATH . This leads to a formula for MATH in coordinates MATH and MATH: MATH . By REF, MATH for an appropriate function MATH. If MATH, then MATH is linear, thus MATH is simple. Hence decreasing MATH we may assume that one can write MATH for an appropriate function MATH. Thus MATH locally isomorphic to MATH. Moreover, REF implies that MATH is also a coordinate system on an open subset of MATH. Exchanging MATH and MATH, we see that our assumption that MATH is a local coordinate system is always satisfied. There is a way to associate to an open subset MATH and a function MATH a homogeneous bihamiltonian structure MATH of type REF with a family of function MATH which is quadratic in MATH. In conditions of REF there is a diffeomorphism of an open subset of MATH with an open subset of MATH and MATH such that the diffeomorphism sends one bihamiltonian structure to another, and the family MATH to the family MATH. A change of MATH of the form MATH leads to an isomorphic bihamiltonian structure with the isomorphism sending MATH to MATH, MATH. Indeed, given the functions MATH and MATH, let MATH. Let MATH be the subset of MATH where MATH and MATH provide a local coordinate system. Plugging into REF, one obtains functions MATH, MATH, MATH, MATH defined on MATH. Functions MATH and MATH provide a local coordinate system near any point of MATH. On an open subset MATH one has MATH, MATH, thus putting MATH into REF instead of MATH defines a homogeneous bihamiltonian structure on MATH. As we have seen in the proof of REF , an open piece of MATH is isomorphic to an open piece of MATH, moreover, the families MATH are preserved by this isomorphism. By REF, a change of the form MATH together with the change MATH-MATH would lead to a parallel translation of the surface MATH, and to the required change of functions MATH. In particular, a change in MATH only will not change MATH, and will change the family MATH by an additive constant only. In the conditions of REF the family MATH on MATH is a MATH-family. By construction of bihamiltonian structure on MATH, the functions MATH, MATH, and the leading coefficient MATH of the quadratic family MATH are NAME functions for MATH, MATH, and MATH correspondingly. Fix MATH. Then MATH is a vector-function which is quadratic in MATH. Moreover, at MATH this vector-function vanishes, thus MATH, here MATH is a vector-function of degree REF in MATH. Extend MATH to become a homogeneous function MATH of homogeneity degree REF, here MATH. This function MATH is in the null-space of MATH for three values REF and MATH of MATH. However, the null-space for a pair of pairing which is isomorphic to MATH depends linearly on MATH (compare with REF). We conclude that MATH is in the null-space for MATH for any MATH, thus MATH is a MATH-family indeed. The bihamiltonian structure MATH of REF is flat on any open subset iff MATH. By arguments of REF, MATH is flat on an open subset iff there is a local dependence between MATH, MATH and MATH of the form MATH. If MATH, then MATH, thus MATH, thus MATH is flat. By REF this proves the ``if" part. If a dependence MATH exists, then MATH, or MATH here MATH, MATH . Taking derivative MATH of REF, and dividing by the cube of REF, one obtains MATH. Since MATH and MATH are independent, and MATH, MATH, we conclude that MATH . Thus MATH, MATH, MATH, MATH. Hence MATH by REF, and MATH for appropriate MATH. Together with MATH this shows that MATH. This finishes the proof of REF . |
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 through MATH intersects MATH in exactly one point, and this point depends smoothly on MATH and MATH. Thus there is exactly one NAME function MATH for MATH which coincides with MATH when restricted to MATH. Obviously, it satisfies the conditions of the proposition. |
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 form MATH corresponds to a change of the form MATH in the proof of REF , we conclude that there is a locally defined solution to the recursion relations REF for any initial data MATH which is a NAME function for MATH. Next, proceed by induction in MATH. To do the step of induction, it is enough to prove the following statement: given a MATH-Casimir family MATH near MATH such that MATH, and given any function MATH of one variable defined in a neighborhood of MATH one can find another MATH-Casimir family MATH such that MATH. Putting MATH finishes the proof in the case MATH. |
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 independent, which finishes the proof. |
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 the anchored NAME scheme provide a mapping MATH, MATH. Decreasing MATH yet more, we may assume that the differential of this mapping is of constant rank MATH (recall that components of MATH are independent). Fix a point MATH and MATH, MATH. Let MATH. Let MATH, MATH. By REF , MATH is in the null space of pairing MATH on MATH, and MATH for any MATH. An immediate check shows that if MATH, MATH, then MATH, MATH . Similarly, if MATH, then all vectors MATH are in the subspace MATH of MATH. The dimension of this subspace is MATH (it is the same subspace which appears in a similar context in REF ). In general case, taking a decomposition of MATH into a sum of indecomposable components, one can see that all vectors MATH are in the sum of NAME blocks of MATH, moreover, they are in a direct sum of subspaces MATH for these blocks. This shows that MATH, here MATH, MATH, are NAME blocks of MATH. The restriction on the action dimension shows that MATH, thus MATH has no NAME blocks, which finishes the proof of the theorem. |
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 REF and the determinant MATH, it is enough to check MATH for MATH. Moreover, if we know MATH for MATH, then we know it for MATH, thus MATH. Since all these expressions are polynomials in MATH, and MATH, one would be able to conclude that MATH. Thus the only relations to check are MATH for MATH such that MATH. This leaves only MATH and MATH, which are easy to check (using one step of induction for the latter one). |
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 calculated when we consider two blocks separately. Now the case when some MATH vanish can be proved by induction using the following obvious The multiplication mapping MATH is a submersion at MATH iff MATH and MATH are mutually prime. In the case when all MATH the matrix MATH is similar to a MATH-diagonal matrix with diagonal entries MATH, above-diagonal entries REF, and below-diagonal entries MATH. Denote by MATH the set of MATH-diagonal MATH matrices with the above-diagonal entries being REF. Denote by MATH the mapping MATH of taking the characteristic polynomial. Denote the diagonal entries of MATH by MATH, MATH, the below-diagonal entries by MATH, MATH. Now the proposition is an immediate corollary of the following The mapping MATH restricted on the subset MATH, MATH, is a submersion. To prove this lemma, denote the characteristic polynomial of the upper-left principal MATH minor by MATH. The lemma is an immediate corollary of The mapping MATH restricted on the subset MATH, MATH, is a bijection onto the subset of mutually prime polynomials MATH. This lemma is a direct discrete analogue of the inverse problem for NAME - NAME equation by the spectrum with fixed ends and normalizing numbers (compare CITE). In fact zeros of MATH determine the spectrum, and values of MATH at these points determine the normalizing numbers. Indeed, extending the sequence MATH by MATH, MATH, one can see that this sequence is uniquely determined by the recurrence relation MATH . From this relation one can immediately see that if MATH, MATH, do not vanish, then MATH and MATH are mutually prime. On the other hand, given mutually prime MATH and MATH, one can uniquely determine MATH and two numbers MATH and MATH from the above relation, and MATH. This finishes the proof of the proposition. |
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 solutions if MATH for any MATH. If MATH is in the periodic NAME lattice, then the equation MATH is invariant with respect to the shift MATH of coordinates of MATH. This shift induces a linear transformation MATH of monodromy in the MATH-dimensional vector space of solutions. As in REF, denote by MATH an element of MATH with REF on even positions, REF on odd positions. If MATH for any MATH, then MATH, and MATH is a polynomial of degree MATH in MATH with the leading coefficient MATH. Indeed, the recursion REF induces a linear transformation MATH, MATH. In an appropriate basis MATH can be written as MATH, and each matrix MATH has determinant MATH. Moreover, MATH is of degree REF in MATH with the leading term MATH. Thus MATH is a polynomial in MATH of degree MATH with the leading term being MATH, which finishes the proof. The function MATH defined on the open subset MATH, MATH, of MATH is a NAME function for the NAME bracket MATH. We do not prove this standard statement about the periodic NAME lattice. As in the case of REF , the proof is reduced to a check of a finite number of identities. The following lemma is obvious: On the open subset MATH, MATH, of MATH the NAME bracket REF has symplectic leaves of codimension REF given by the equations MATH, MATH. This shows that MATH in REF . To demonstrate REF the only thing which remains to be proved is that at a generic point MATH the differentials MATH for different MATH and the differential of MATH span a MATH-dimensional vector subspace of MATH. It is enough to show that for a generic MATH the differentials of MATH for different MATH span a MATH-dimensional vector subspace of the hyperplane MATH in MATH. The leading coefficient in MATH of MATH is REF, thus the function MATH defines a mapping MATH. Again, it is enough to show that the restriction of this polynomial mapping to MATH is a submersion for a generic MATH and MATH. On the other hand, multiplication of MATH by the same non-zero constant does not change MATH, thus if we prove this statement for one MATH, is it true for any MATH. Thus it is enough to demonstrate this statement for MATH, MATH. Again, it is enough to show that the restriction of MATH to an open subset of MATH is a submersion. However, if MATH, then MATH thus the restriction of MATH to MATH is a surjection, thus is a submersion in a generic point. This shows that REF is applicable, thus the bihamiltonian structure is indeed flat indecomposable at a generic point. |
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 to MATH as MATH. |
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 noncrossing. |
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 argument following that equation we see that the limit defining MATH is MATH. |
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 partition obtained by expanding the MATH-th point of the set on which MATH operates into MATH points. That is, MATH . (Note that MATH.) Therefore the sum REF is less in absolute value than MATH . We can always choose MATH's to be self-adjoint, while the MATH's are so by assumption. Hence MATH and so converges to MATH as MATH. |
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 from the free independence of the increments corresponding to disjoint intervals. |
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 classes contains at least MATH elements, the result is MATH, while each two-element class gives MATH, which is a scalar and can be factored out. In the partition resulting from factoring out the minimal two-element classes, all the classes which were at the distance of MATH from the minimum now become minimal. |
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 minimum, and after the above shrinking, covers only a number of singletons: MATH . NAME though the free NAME process is not centered, it is easy to see that each inner singleton contributes a factor of MATH. Thus all the singletons covered by the MATH class contribute MATH. Note that MATH is precisely the number of classes covered by the MATH class, that is, the number of edges emanating down from it in the partial order incidence graph. The result easily follows by induction. |
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 follows from REF . |
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 function. Therefore the second equality holds as well. |
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 follows from REF . In general, MATH . Here in the above REF , for partitions MATH, in the partition MATH we let MATH act on MATH while MATH acts on MATH. This sum can probably be estimated directly using REF numbers, but instead we'll use the method of REF: the result follows from the fact that the cumulants grow no faster than an exponential, and the fact that the limit exists for the free NAME process. |
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 proof: every element MATH of the above intersection is MATH-essential. REF : if MATH is a finite cover of MATH, then at least one of the sets MATH, MATH, is MATH-essential in MATH, and it follows that MATH is MATH-essential in MATH. REF : emulates a proof of REF. There exists a point MATH whose every neighbourhood is essential: assuming the contrary, one can cover the compact space MATH with open MATH-inessential sets and select a finite subcover containing no MATH-essential sets, a contradiction. We claim that MATH is a common fixed point for MATH. Assume it is not so. Then for some MATH one has MATH. Choose an entourage, MATH, of the unique compatible uniform structure on MATH with the property MATH. Since MATH is uniformly continuous, there is a MATH with MATH for all MATH. Since MATH, we conclude that MATH is MATH-inessential (with MATH), a contradiction. CASE: Let MATH be a MATH-equivariant compactification of MATH. There exists a unique continuous MATH with MATH, and it follows easily that MATH is a MATH-equivariant mapping, that is, MATH for every MATH. The image of a MATH-fixed point MATH under MATH is a MATH-fixed point in MATH. REF : trivial. REF : fixed points in the NAME space MATH of the commutative MATH-algebra MATH correspond to MATH-invariant multiplicative means (states) on MATH. CASE: If MATH is a finite cover of MATH, then there is a MATH containing a MATH-fixed point, and such a MATH is clearly MATH-essential. REF : this is the only implication where we assume MATH to be a group. Without loss in generality and replacing MATH with its separated reflection if necessary, one can assume that MATH is a separated uniform space (that is, MATH): indeed, the NAME compactifications of a uniform space MATH and of its separated reflection are canonically homeomorphic. Thus, we can identify MATH (as a topological, not uniform space!) with an everywhere dense subspace of MATH. If now MATH is a finite cover of MATH, then the closures of all MATH taken in MATH cover the latter space, and so there is a MATH with MATH containing a MATH-fixed point MATH. We claim that MATH is MATH-essential in MATH. To prove this, we need a simple fact of general topology. Let MATH be subsets of a uniform space MATH satisfying the condition MATH for some entourage MATH. Then the closures of MATH, MATH in the NAME compactification MATH have no point in common: MATH. [Let MATH be a uniformly continuous bounded pseudometric on MATH subordinated to the entourage MATH in the sense that MATH whenever MATH and MATH. For each MATH and MATH set MATH. The real-valued functions MATH are uniformly continuous (indeed MATH-Lipschitz-REF) and bounded on MATH, and therefore extend to (unique) continuous functions MATH on MATH. If there existed a common point, MATH, for the closures of all MATH in MATH, then all MATH would vanish at MATH and consequently for any given MATH there would exist a MATH with MATH, and in particular MATH, for all MATH. However, for every MATH, there is a MATH with MATH.] Now assume that MATH for some MATH, where MATH is a finite subfamily of MATH. Every MATH has a uniformly continuous inverse MATH, and there is an entourage MATH with MATH for all MATH and MATH. Let MATH, that is, MATH for some MATH. Then MATH, that is, MATH, and consequently MATH. We conclude: MATH for each MATH, and therefore MATH as well. The above observation from uniform topology implies that MATH. Since extensions of MATH to MATH are homeomorphisms, MATH, and consequently MATH, a contradiction because the intersection contains MATH. |
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 exist finite covers MATH of MATH, MATH, a finite collection MATH of elements of MATH, and a MATH having the property that for every MATH, MATH: MATH . (Here and below it is convenient to assume that the MATH-neighbourhood of MATH in the sphere, MATH, is formed with respect to the geodesic distance.) For every MATH and each MATH set MATH . The collection MATH covers MATH, and it is easy to see that for each MATH and each MATH one has MATH whenever MATH is sufficiently small, for example MATH. |
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 with constant MATH: MATH . Therefore, MATH. Set MATH . The following properties of the mapping MATH are obvious. CASE: MATH is linear. [If MATH and MATH, then for every MATH one has MATH, and consequently MATH.] CASE: MATH is positive. [If MATH, then MATH for all MATH, therefore MATH.] CASE: MATH. [MATH.] CASE: MATH is MATH-invariant. [Let MATH and MATH. Then MATH . Therefore, MATH . REF imply that MATH is also bounded of norm MATH (compare REF ), and by REF MATH is a MATH-invariant mean on MATH, as required. |
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, as usual, the characteristic function of MATH, and the dot stands for the muptiplication of (equivalence classes of) functions. Since the mapping MATH is MATH-Lipschitz on MATH, so is the function MATH. Being also bounded, MATH. Denote by MATH the unique continuous extension of MATH to the NAME compactification of the sphere, and set MATH. Then MATH is a finitely-additive left-invariant normalized measure on NAME subsets of MATH, vanishing on locally null sets, and consequently MATH is amenable. |
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 to realize MATH as MATH. (For example, take as MATH the free group of rank equal to the density character of MATH.) Now one can apply REF . |
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 topological MATH-space. The uniformly continuous mapping of compactification MATH has everywhere dense image and is MATH-equivariant. It only remains to prove the maximality of MATH as a MATH-equivariant compactification of MATH. Let MATH be a compact MATH-space, and let MATH be a continuous MATH-equivariant mapping. It determines a morphism of MATH-algebras, MATH, from MATH to MATH via MATH . The dual continuous mapping MATH between the NAME spaces of the corresponding MATH-algebras is MATH-equivariant and its restriction to MATH is easily seen to coincide with MATH. The proof is thus finished. |
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 property. It means that there exists a MATH-fixed point MATH. Denote by MATH the closure of the MATH-orbit of MATH in MATH. It is a compact MATH-subspace of MATH. Since MATH is the centre of MATH, every point of MATH is MATH-fixed. (In particular, it follows that MATH is a proper subspace of MATH.) It is a well-known and easy consequence of NAME 's lemma that every compact MATH-flow contains a minimal subflow (that is, a non-empty compact MATH-subspace such that the orbit of each point is everywhere dense in it, see for example, CITE). Denote by MATH any minimal subflow of MATH. Since MATH has no fixed points in MATH (Corol. REF), it follows that every minimal MATH-subflow of MATH is nontrivial, that is, contains more than one point. In particular, this applies to MATH. By force of the extreme amenability of the group MATH (combine the results of CITE and CITE), there is a MATH-fixed point, MATH, in MATH. It follows from the continuity of the action that MATH is also a fixed point for MATH, that is, the stabilizer MATH contains MATH. Since every point of MATH is MATH-fixed, it follows that MATH is fixed under the action of the group MATH. The stabilizers of elements of the orbit of MATH under the action of MATH are conjugate to MATH. Since MATH is normal in MATH, every such stabilizer contains MATH. Because of minimality of MATH, the MATH-orbit of MATH is everywhere dense in MATH, and we conclude: all points of MATH are fixed under the action of MATH. It implies that the action of MATH on MATH factors through an action of the projective NAME group MATH, and the latter action is continuous. Moreover, it is also minimal. Denote by MATH the set of all MATH leaving each element of MATH fixed. This is a closed normal subgroup of MATH, containing MATH, and since it is proper (in view of minimality and nontriviality of MATH), it must be contained in MATH and consequently coincide with it. It means that the action of the projective NAME group is on the compact space MATH is minimal and effective, and the statement is proved. |
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 for example, CITE, p. REF) implies that, for MATH sufficiently large, there is an orthonormal basis in MATH formed by elements MATH each at a distance MATH from MATH. Indeed, fix a MATH and denote by MATH the NAME probability measure on MATH. Then MATH therefore for any finite subset MATH . If the size MATH of MATH grows slower than MATH, for example if MATH is polynomial in MATH, then for MATH satisfying the condition MATH we can find a rotation MATH taking MATH outside of the MATH-neighbourhood of MATH in MATH. Applying this observation to an arbitrary orthonormal basis of MATH as MATH, we obtain an orthonormal system MATH with the desired property. Now extend the collection of MATH, MATH to an orthonormal basis MATH of MATH. One has MATH . If now MATH and MATH, then MATH and the latter expression in REF is MATH. The above argument establishes the following: for each pair of positive constants MATH, if MATH is so large that MATH and MATH are two rank MATH projections in MATH satisfying MATH, then the measure of the set of points MATH that are at a distance MATH from MATH is greater than MATH. Since MATH and therefore MATH one has in particular MATH whenever MATH is so large that MATH . Now observe that the set MATH contains the open MATH-neighbourhood of the set MATH . According to REF , one has MATH whenever MATH satisfies REF . Replacing MATH in both formulae with MATH yields the following: MATH whenever MATH is so large that MATH and MATH are two orthogonal projections of rank MATH satisfying MATH. And this is the desired result in slight disguise. |
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 eigenvalue MATH, MATH being as above; if MATH, then we only consider MATH and MATH. Since MATH, MATH span the same subspace of dimension MATH (respectively MATH where MATH) as the vectors MATH and MATH do (respectively, the vector MATH does), it follows that MATH whenever MATH. Now let MATH be the projection onto the one-dimensional subspace spanned by MATH. |
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 MATH, then MATH . The orthogonal sum of linear operators MATH is an isometry between MATH and MATH. REF implies that for each MATH, MATH, one has MATH and the proof is finished by applying REF . |
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 sphere MATH. Clearly, for at least one MATH the set of natural numbers MATH is infinite. We claim that the set MATH is then MATH-essential. Proceeding to a subsequence of the selected sequence of projections if necessary, we can assume without loss in generality that MATH for all MATH. For every MATH the measure of the set MATH in the latter sphere is the same as the measure of MATH, and therefore MATH for all MATH. Let a MATH be arbitrary. According to NAME 's concentration of measure property of spheres, the measure of every set of the form MATH in MATH, where MATH, is MATH. An application of REF to MATH and MATH yields that for every MATH as above the measure of the set MATH in MATH is MATH. Indeed, there is an isometry MATH with the property that for all points MATH of MATH apart from a set of measure MATH one has MATH. Consequently, the set MATH contains the set MATH. The measure of the latter set in MATH is MATH, because so is the measure of MATH in MATH and the isometry MATH between the spheres is automatically a measure-preservig map. We conclude that if MATH is sufficiently large, then the sets MATH, MATH, have a non-empty common part, and indeed the measure of its intersection with the sphere MATH is MATH. This finishes the proof. |
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 subrepresentations. According to the NAME property of amenable representations as established by NAME REF , for every finite MATH and each MATH there is an orthogonal projection MATH in MATH of finite rank such that MATH for all MATH. Suppose it is not in general possible to choose such a MATH of an arbitrarily high finite rank. In such a case, there are a finite MATH and a MATH such that for each finite MATH and each MATH an arbitrary projection MATH satisfying REF has rank MATH, where MATH is a fixed natural number. Moreover, one can assume without loss in generality that the equality is always achieved for a suitable MATH. Notice that on the collection of all projections of fixed finite rank MATH the trace class metric and the operator metric are both NAME equivalent to the NAME distance between unit spheres in the range spaces of the projections. (The equivalence of the trace class and operator metrics follows from the obvious inequality MATH for every two projections MATH of rank MATH. For the equivalence of the operator and NAME metrics, see CITE and REF .) The space of all projections of rank MATH is thus uniformly isomorphic to the NAME MATH and forms a complete uniform space. (Compare REF .) For every MATH and MATH as above denote by MATH the non-empty set of all projections of rank exactly MATH satisfying REF . Now equip the set MATH of all pairs MATH as above with the product partial order making it into a directed set. The diameters of MATH cannot converge to zero over MATH. Otherwise the sets MATH would form a NAME prefilter having a limit point MATH, which is again a projection of rank MATH satisfying REF for every finite MATH and each MATH. In other words, MATH commutes with every MATH, that is, MATH is the space of a nontrivial finite-dimensional subrepresentation of MATH, leading to a contradiction. We conclude that for some MATH, the set of pairs MATH satisfying MATH is cofinal in MATH. It remains to notice that, if an arbitrary finite set MATH is fixed, then having at one's disposal, for every MATH, a pair of projections MATH and MATH of the same finite rank MATH satisfying MATH for all MATH and MATH, and also the condition MATH for a fixed MATH enables one to produce a new projection MATH of rank MATH satisfying MATH where MATH as MATH, thus obtaining a contradiction with the presumed maximality of MATH. Indeed, choose two sequences of projections MATH and MATH of rank MATH having the properties MATH and MATH as MATH for all MATH and MATH. Using REF , choose for every MATH one-dimensional projections MATH, MATH, MATH, such that MATH and MATH . Without loss of generality and proceeding to a subsequence if necessary, one can assume that MATH for each MATH as MATH, where MATH. Let MATH be the smallest integer MATH with the property MATH; clearly, such a MATH exists. Let MATH . The rank of the projection MATH is MATH. The space MATH is the direct sum of subspaces MATH, MATH, where MATH for MATH and MATH for MATH. For every MATH and every MATH, the NAME distance between the unit spheres MATH and MATH approaches zero as MATH, because REF the geodesic distances between MATH and MATH are bounded from below by some positive constant and at the same time do not exceed MATH, and REF the distances between MATH and MATH, MATH, MATH converge to zero as MATH. For MATH the same is true in a trivial sort of way. Since the spheres MATH, MATH are paiwise orthogonal and the same is true of the spheres MATH, one concludes that the NAME distance between the unit sphere MATH in the space MATH and the unit sphere MATH approaches zero as MATH. Since the rank of MATH is bounded above by MATH, this means MATH, as required. |
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 property and the same is true of MATH according to REF . CASE: Immediate from REF . |
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-invariant mean on MATH. |
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 case since MATH on the set where MATH; the sign is verified by evaluating at MATH. |
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 condition shows that this splitting is MATH-orthogonal and identifies MATH. In particular, REF-forms MATH and MATH are symplectic. We will show that in terms of the splitting MATH, the value of MATH at MATH is given by the formula MATH . In particular, REF shows that MATH is a volume form. Consider the splitting MATH and the corresponding decomposition MATH with MATH and MATH. (We are dropping the base point MATH from the notation.) The form MATH is determined by the moment map condition, and is given by MATH . Indeed, since MATH, it is easily verified that MATH for all MATH, as required. Using REF and the following equality of operators on MATH, MATH we obtain MATH . Thus MATH . This expression combines with MATH to a factor, MATH where we have used REF with MATH, and MATH. This proves REF. For REF, assume that MATH is non-degenerate, that is MATH. We have MATH . Comparing with REF, this shows MATH which yields REF. |
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 MATH where MATH is horizontal projection for the exterior algebra MATH. Let MATH be the contraction operators for the exterior algebra. The symbol map takes the NAME product on MATH to the product on on MATH given by application of the operator MATH on MATH (where the superscripts refer to the MATH-factor), followed by wedge product (see CITE, or CITE). Because of REF we may replace MATH with MATH everywhere, which then commutes with MATH. REF becomes MATH . Since MATH is a ring homomorphism, this is equal to MATH, proving the Theorem. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.