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math-ph/9906019 | Since MATH has finite statistics, MATH and hence the centre of MATH are finite dimensional. Therefore if MATH is a covariant representation, MATH is trivial on such centre. Then, since MATH implements automorphisms of MATH, it implements an action of MATH by automorphisms of MATH, preserving any factorial component. Th... |
math-ph/9906019 | Of course we may restrict to the case MATH. Since MATH is the product of MATH with MATH, such reflections are implemented by the corresponding modular involutions MATH, MATH, and MATH is equivalent to MATH, both being conjugate endomorphisms of MATH, there exists a unitary MATH intertwining MATH and MATH. Since MATH, M... |
math-ph/9906019 | By property (MATH), MATH does not depend on MATH, as observed in REF . Concerning the relation between spin and statistics, we may define a function MATH, MATH, as in the proof of REF , which is indeed a local group representation namely, if MATH verify MATH, we have MATH. Setting MATH, we get MATH and MATH. Hence, for... |
math-ph/9906019 | Let MATH belong to MATH, MATH. By REF MATH because of the above property of MATH since the integrand is constantly equal to MATH on the set MATH. Then the localization of MATH and NAME duality imply that the range of MATH is contained in MATH. Setting MATH gives a conditional expectation of MATH onto the range of MATH ... |
math-ph/9906019 | We have MATH and MATH is locally normal because both MATH and MATH are locally normal. |
math-ph/9906019 | If MATH and MATH is localized in a double cone, the commutator function MATH vanishes on a right wedge, hence MATH. Since MATH is locally normal, MATH commutes with every MATH, thus with MATH; but MATH being irreducible, it is therefore a scalar equal to its vacuum expectation value: MATH as MATH is normal and MATH-inv... |
math-ph/9906019 | One may repeat the proof of REF for each of the one-parameter light-like unitary translation groups. |
math-ph/9906019 | MATH is a simple MATH-algebra since it is the inductive limit of type III factors (that are simple MATH-algebras). Therefore MATH is one-to-one and the statement will follow if we show that MATH if cyclic for MATH. To this end we may use a classical NAME argument. If MATH is orthogonal to MATH, and MATH, then for all M... |
math-ph/9906019 | We begin with the case where MATH is irreducible and assume for convenience that MATH. Notice then that MATH is finite-dimensional and, by covariance, globally MATH-invariant with MATH in the subgroup of boosts because these transformations preserve MATH. Therefore MATH is a finite-dimensional subspace of MATH globally... |
math/9906004 | Let MATH denote the NAME graph of MATH with respect to some finite generating set for MATH. Let MATH denote the almost invariant subset MATH of MATH and let MATH denote the almost invariant subset MATH of MATH. Recall from the start of this section, that if we identify MATH with the MATH - cochain on MATH whose support... |
math/9906004 | As MATH and MATH are translates of MATH or MATH, it suffices to prove that MATH is small if and only if it lies in a bounded neighbourhood of each of MATH, MATH, MATH, MATH. If MATH is small, it projects to a finite subset of MATH which therefore lies within a bounded neighbourhood of the image of MATH. By lifting path... |
math/9906004 | We need to show that MATH is transitive and that if MATH and MATH then MATH . Suppose first that MATH and MATH. The first inequality implies that MATH is small and the second implies that MATH is small, so that two of the four sets MATH are small. The assumption of REF implies that one of these two sets must be empty. ... |
math/9906004 | REF are obvious from the definition of MATH and the hypotheses of REF . To prove REF , we observe that if MATH and MATH, then MATH and MATH must both be small. This implies that MATH itself is small, so that MATH or MATH must be small. But this contradicts the hypothesis that MATH is a non-trivial MATH - almost invaria... |
math/9906004 | To prove the first part, we let MATH denote the set of all translates of MATH and MATH by elements of MATH, let MATH be the involution on MATH and let the relation MATH be defined on MATH by the condition that MATH if MATH is empty or the only small set of the four sets MATH. REF show that MATH is a partial order on MA... |
math/9906004 | The given splitting of MATH over MATH corresponds to an action of MATH on a tree MATH such that MATH-has a single edge, and some edge of MATH-has stabiliser MATH. Let MATH denote the fixed set of MATH, that is, the set of all points fixed by MATH. Then MATH is a (non-empty) subtree of MATH. As MATH normalises MATH, it ... |
math/9906004 | Our first step will be to show that MATH and MATH must be commensurable. Without loss of generality, we can suppose that MATH is small. The other small set can only be MATH, as otherwise MATH or MATH would be small which is impossible. It follows that for each edge of MATH, either it is also an edge of MATH or it has (... |
math/9906004 | The preceding lemma showed that the hypotheses imply that MATH equals MATH and also that the images MATH and MATH of MATH and MATH in MATH are almost equal or almost complementary. By replacing one of MATH or MATH by its complement if needed, we can arrange that MATH and MATH are almost equal. We will show that in most... |
math/9906004 | Recall that MATH is associated to a splitting of MATH over MATH. It follows that MATH is associated to the conjugate of this splitting by MATH. Thus MATH and MATH are associated to splittings of MATH which are each conjugate to one of the two given splittings. If MATH and MATH are each translates of MATH or MATH, the n... |
math/9906004 | Let the MATH splittings MATH of MATH be over subgroups MATH with associated MATH - almost invariant subsets MATH of MATH, and let MATH. We will start by supposing that no two of the MATH's are conjugate. We will handle the general case at the end of this proof. We will apply the second part of REF to MATH . Recall that... |
math/9906004 | The idea of the proof is much as before. We let MATH denote the almost invariant subset MATH of MATH, and let MATH denote MATH. We want to apply the first part of REF . As before, the assumption that MATH implies that MATH is almost nested. However, in order to apply REF , we also need to know that for any pair of elem... |
math/9906004 | The first inclusion is clear. The second is proved in essentially the same way as the proof of the first part of REF . Let MATH be an element of MATH, and consider the case when MATH ( the other case is similar). Recall that this means that the sets MATH and MATH are both small. Now for each edge of MATH, either it is ... |
math/9906004 | REF tells us that the given set is contained in the union of a finite number of double cosets MATH. If MATH, we claim that the double coset MATH is itself the union of only finitely many cosets MATH, which proves the required result. To prove our claim, recall that MATH is commensurable with MATH. Thus MATH can be expr... |
math/9906004 | Consider an element MATH in MATH. REF tells us that MATH and MATH are commensurable subgroups of MATH. Let MATH denote their intersection and let MATH denote the intersection of the conjugates of MATH in MATH. Thus MATH is of finite index in MATH and MATH and is normal in MATH. Now consider the quotient MATH. Let MATH ... |
math/9906004 | We will consider how MATH can intersect the normaliser of a subgroup of finite index in MATH. Let MATH denote a subgroup of MATH of finite index. We denote the image of MATH in MATH by MATH. Then MATH is an almost invariant subset of MATH. We consider the group MATH, which we will denote by MATH. Then MATH acts on the ... |
math/9906004 | The previous lemma tells us that there is a subgroup MATH of finite index in MATH such that MATH normalises MATH. Let MATH denote the almost invariant subset MATH of MATH, and let MATH denote the almost invariant subset MATH of MATH . Suppose that the index of MATH in MATH is infinite. Recall from the proof of the prec... |
math/9906004 | Suppose that there exists a subset MATH of MATH which is MATH - almost equal to MATH, such that MATH. We have MATH as MATH and MATH are MATH - almost equal. So, it is enough to show that for every MATH, either MATH or MATH is MATH - finite. Suppose that MATH. Consider MATH which is a union of a finite number of right c... |
math/9906004 | Let MATH be the NAME graph of MATH with respect to a finite system of generators and let MATH. As in the proof of REF , for a set MATH of vertices in a graph, we let MATH denote the maximal subgraph with vertex set equal to MATH. We will show that exactly one component of MATH has infinite image in MATH. Note that MATH... |
math/9906004 | Observe that REF shows that, by changing MATH up to commensurability, and changing MATH, we may assume that the translates of MATH are almost nested with respect to MATH and nested with respect to MATH or MATH. If we do not have almost nesting for all translates of MATH, then there is MATH outside MATH such that none o... |
math/9906004 | Let MATH be a non-trivial MATH - almost invariant subset of MATH, let MATH be an element of MATH and let MATH, so that MATH-has stabiliser MATH. Let MATH denote the intersection MATH which has finite index in both MATH and in MATH because MATH lies in MATH. Thus MATH and MATH are both almost invariant subsets of MATH. ... |
math/9906004 | The proof of REF goes through because of our assumption that MATH. The mistake in CITE occurs in the proof of REF where it is implicitly assumed that if MATH crosses MATH, then it crosses MATH strongly. Since we have assumed that the two intersection numbers are equal, the argument is now valid. |
math/9906007 | If MATH is a relative equilibrium then MATH for some MATH. Since MATH is quadratic homogeneous, we have MATH for all MATH. Hence MATH. On the other hand, since MATH is definite, the ray MATH is transverse to the level set MATH, hence MATH. Contradiction. |
math/9906007 | Since the action of MATH preserves the one-form MATH, the action of MATH on MATH is Hamiltonian. The action of MATH on MATH is also Hamiltonian: MATH is a corresponding moment map. Consequently MATH is a symplectic orbifold diffeomorphic to MATH; from now on we identify MATH and MATH. The Hamiltonian action of MATH on ... |
math/9906007 | We use open sets in our definition of the category. Let MATH be a generic component of the moment map for the action of MATH on MATH so that MATH is precisely the set of critical points of MATH. The function MATH is NAME. Therefore MATH decomposes as a disjoint union of the unstable orbifolds MATH of the gradient flow ... |
math/9906007 | The proof is a standard computation that uses the local normal form of the moment map of NAME and of CITE. Similar computations are carried out in CITE and in CITE. We use the version of the normal form theorem described in CITE which we now recall without proofs: Let the symbols MATH, MATH, MATH, MATH, MATH, MATH and ... |
math/9906008 | There exists a constant MATH such that MATH for all conjugacy classes MATH in MATH and circuits MATH in MATH representing MATH. Choose MATH such that MATH. We conclude that MATH for all conjugacy classes MATH. |
math/9906008 | Since MATH is indivisible, its initial edge and its terminal edge are contained in MATH. Suppose that both endpoints of MATH are contained in MATH. By REF , one of them is contained in a contractible component of MATH. Let MATH denote this endpoint. By REF , MATH is necessarily a zero stratum, and we have MATH. This im... |
math/9906008 | Let MATH be a homotopy inverse of MATH. Since the restriction of MATH to MATH is hyperbolic, there exist numbers MATH such that MATH for all circuits MATH in MATH. There exists a path MATH such that MATH is an immersed circuit in MATH, satisfying MATH and MATH. We can find some constant MATH, depending only on MATH and... |
math/9906008 | We will show by induction that MATH . For MATH, the bounded cancellation lemma implies that MATH so the claim holds for MATH. Assume that the claim is true for some MATH. Again, the bounded cancellation lemma tells us that MATH by induction. Hence, we conclude that MATH . |
math/9906008 | Given MATH, choose an exponent MATH according to REF , for MATH. Express MATH as a concatenation of paths MATH such that MATH and MATH. There exist preimages MATH such that MATH is their concatenation. We claim that MATH for all MATH. Suppose otherwise, i. e., MATH for some MATH. Because of our choice of MATH, REF impl... |
math/9906008 | Fix some MATH. Let MATH be a nontrivial circuit in MATH. We will distinguish several cases, and in each case we will show that there exist numbers MATH, MATH and MATH such that there exists a collection MATH of subpaths of MATH with the following properties: CASE: For every integer MATH and for every MATH, we have MATH... |
math/9906008 | By CITE, there exists some number MATH such that for all MATH there exist numbers MATH and MATH such that MATH. Now MATH has the desired property. |
math/9906008 | Fix some MATH such that MATH. Although there may be infinitely many paths MATH whose endpoints are vertices and whose MATH-length is at most MATH, there is only a finite number MATH of intersections MATH of such paths with MATH. Choose MATH according to REF , with MATH and MATH as above. We will show that MATH is the d... |
math/9906008 | Given MATH, choose an exponent MATH according to REF , with MATH. Express MATH as a concatenation of paths MATH such that MATH and MATH. There exist preimages of MATH such that MATH is their concatenation. We claim that MATH for all MATH. Suppose otherwise, i. e., MATH for some MATH. Because of our choice of MATH, REF ... |
math/9906008 | We will proceed by induction up through the filtration of MATH (as in the previous sections, we equip MATH with the metric constructed in REF). The restriction of MATH to MATH is a homotopy equivalence, and we claim that MATH is of exponential growth. If it were a zero stratum, this would imply that MATH, but MATH. If ... |
math/9906015 | To abreviate, we denote MATH. Then, MATH where MATH and MATH denote the partial derivatives with respect to MATH and MATH respectively. But, according to the above computation, MATH and MATH. Therefore, MATH where the last determinant has to be considered with respect to any oriented orthonormal frame of MATH. |
math/9906015 | Let MATH be an arbitrary extension of MATH. Then the map MATH given by MATH is a diffeomorphism. In particular, it is transverse to the submanifold MATH. By the Transversality Theorem, it follows that for almost any MATH, the map MATH given by MATH is also transverse to MATH. Since MATH has codimension REF, this implie... |
math/9906015 | Let MATH be an extension of MATH such that MATH for any MATH, being MATH. Then we can extend MATH to MATH by putting MATH . As in REF , it follows that MATH defines a smooth MATH-form on MATH. Moreover, since MATH is closed on MATH, MATH and by NAME Theorem, MATH where MATH denotes a small ball centered at MATH in the ... |
math/9906015 | Let MATH. For any MATH, we have that MATH, where MATH. Thus, we can define MATH and MATH, so that MATH is a right-handed orthonormal frame of MATH. Now, we can consider REF-forms on MATH defined by MATH, for any MATH. Since MATH, by taking differentials we see that MATH. Moreover, we have that MATH . But from the fact ... |
math/9906015 | Suppose that this is not true. Then, for each MATH, there are MATH and pairs MATH such that MATH. By taking subsequences if necessary, we can suppose that MATH and MATH. If MATH, we arrive to MATH, in contradiction with REF . Otherwise, let MATH. If we denote by MATH a frame for MATH, we have for any MATH, MATH . Since... |
math/9906015 | Let MATH be an orthonormal oriented frame of MATH and consider the map MATH given by MATH . By the Transversality Theorem, we have that for a residual subset of curves MATH and vector bundles MATH with the corresponding MATH . NAME topologies, the curve MATH meets the hypersurface MATH transversely at a finite number o... |
math/9906018 | CASE: Let MATH, MATH both be as in REF. So for each MATH, MATH for some MATH, MATH. Hence, if MATH, MATH and each is in MATH as defined by MATH (as MATH). As this holds for every MATH, all generators of MATH as defined by MATH are in MATH as defined by MATH. As the situation is symmetric we finish. CASE: Similar proof.... |
math/9906018 | Remember that (by CITE) there is MATH, a generating sequence for MATH. Let for MATH, MATH exemplify MATH, MATH; by CITE, without loss of generality CASE: MATH. Without loss of generality for MATH if MATH, MATH if MATH. We define MATH by: MATH and for MATH: MATH . Clearly CASE: MATH. MATH . Let MATH, so MATH and hence f... |
math/9906018 | CASE: Trivially the second and third terms are equal (see REF ). Let MATH be defined as in the second term, so MATH. So by REF without loss of generality MATH, so MATH. Using REF's notation, MATH exemplify MATH. CASE: By REF . CASE: This follows by the proof of REF, but as I was asked, we repeat the proof of REF with t... |
math/9906021 | Consider the spectral representation associated with a vector MATH and perform a straightforward computation: MATH . |
math/9906021 | The proof is a direct computation. For details, we refer to REF . |
math/9906021 | The proof is standard and we provide it for the sake of completeness. See, for example, CITE for detailed exposition of similar results and further references. Throughout the proof, we assume that the function MATH is sufficiently smooth to justify integration by parts (local MATH is sufficient). Clearly this is the ca... |
math/9906021 | We have MATH since MATH and MATH satisfy the boundary conditions. Next note that MATH . We used the NAME inequality in the last step. |
math/9906021 | An interplay of the scales in space and in the spectral parameter plays an important role in the analysis. Let us assume that MATH . Take sufficiently large MATH such that MATH . By NAME 's formula we have MATH . In the above computation we used the definition of MATH and the fact that the function MATH satisfies MATH ... |
math/9906021 | Recall that for every self-adjoint operator there is an associated spectral measure of maximal type, MATH such that for every MATH and any measurable set MATH implies MATH . A vector MATH is of the maximal type if for any measurable set MATH given that MATH . We will show that for any MATH of positive MATH-dimensional ... |
math/9906021 | Suppose that MATH has positive MATH measure. By the remark after the proof of REF implies that MATH for every MATH and finitely supported MATH such that MATH . Proceeding as in the proof of REF , we can find a vector MATH such that MATH for any MATH in some set of positive MATH measure. This is not possible by NAME (se... |
math/9906021 | The argument repeats the proof of REF , except that we do not need REF . The analog of REF is proven directly by the observation that MATH and MATH . |
math/9906021 | Without loss of generality, assume MATH . We first establish the existence of a NAME set MATH, for which the following three properties are true: CASE: MATH. CASE: The restriction of the spectral measure MATH to MATH is MATH-H. CASE: There exists a constant MATH, such that for each MATH and MATH, the corresponding gene... |
math/9906021 | By the results of the NAME theory, the spectral measure MATH is supported on the set of the energies MATH for which the decaying solution REF satisfies the boundary condition (namely, MATH coincides with the NAME boundary condition). Moreover, these decaying solutions, which we will denote by MATH, are exactly the gene... |
math/9906024 | Straightforward. |
math/9906024 | CASE: Immediate (by the MATH - completeness of MATH). CASE: Clause REF should be clear (remember REFMATH). For clause REF note that MATH, so in the game MATH the condition MATH is chosen by the anti-generic player. So if the conclusion fails, then for some MATH - name MATH for an ordinal we have MATH. Thus the anti-gen... |
math/9906024 | Let MATH be a strictly increasing continuous sequence cofinal in MATH. By induction on MATH choose MATH such that CASE: MATH is above (in MATH) of all MATH for MATH, CASE: MATH, MATH, and MATH, CASE: if MATH then MATH and MATH. (The choice is clearly possible as MATH is MATH - generic.) Let MATH be an upper bound of MA... |
math/9906024 | REF Straightforward (use the MATH - completeness of MATH). CASE: If MATH is MATH - generic, then our assertion is clear (remember clause REF ). So suppose that we are in the second case (so MATH). Let MATH be a strictly increasing continuous sequence cofinal in MATH. For MATH and MATH let MATH, MATH (clearly MATH). Sin... |
math/9906024 | By REF, the forcing notion MATH is MATH - complete, so we have to concentrate on showing clause REF for it. So suppose that MATH is large enough, MATH and MATH, MATH, MATH and MATH, and MATH satisfies MATH. Furthermore, suppose that MATH is a MATH - semi diamond and MATH is a MATH - candidate. We may assume that for ea... |
math/9906024 | Assume MATH is as in REF, and MATH. Put MATH and MATH. CASE: If MATH, MATH, then MATH and MATH. CASE: If MATH, MATH, and MATH, then MATH is stronger than both MATH and MATH. CASE: If MATH is MATH - increasing and MATH, then MATH has a least upper bound MATH, and MATH is a least upper bound of MATH. CASE: By the definit... |
math/9906024 | CASE: Follows from REF . CASE: Assume MATH. Let MATH, MATH and MATH be as in REF , MATH. Let MATH (for MATH) and put MATH. Clearly MATH and MATH for each MATH (remember MATH). By the proof of REF, the condition MATH is MATH - generic for MATH over MATH . To show that MATH for MATH, it is enough to do this for MATH. So,... |
math/9906024 | Assume MATH is as in REF, MATH and MATH. We are going to show that the condition MATH is MATH - generic for semi-diamonds. So suppose that MATH and MATH are as in REF. For MATH, let MATH be such that MATH, MATH, MATH, MATH. Note that MATH. Let us describe the winning strategy of the generic player in the game MATH. For... |
math/9906024 | First let us argue that MATH is MATH - lub - complete. So suppose that MATH are such that MATH, MATH. We claim that MATH is a condition in MATH. Clearly MATH is a tree, and MATH. By clause REF (for MATH's) we see that MATH, and in a similar way we justify that MATH satisfies other demands as well. Now suppose that MATH... |
math/9906030 | Associated to the polynomial MATH over MATH is its NAME polygon, the lower boundary of the convex hull of the points MATH for MATH. The integers MATH such that MATH are vertices of the NAME polygon are called the breakpoints, and the ratios MATH, where MATH are adjacent breakpoints, are called the slopes of the NAME po... |
math/9906030 | Let MATH be elements which reduce to a basis over MATH of the MATH-vector space solutions of MATH modulo MATH. For each MATH, we wish to exhibit MATH which reduce to a basis over MATH of the set of solutions of MATH in MATH. We construct these by an inductive argument: given the MATH, note that MATH . Since MATH has al... |
math/9906030 | By the previous lemma, the sequences MATH and MATH satisfying the given relations have the form MATH where the MATH and MATH depend only on the MATH and MATH. Clearly MATH and MATH are also of this form, with the set of MATH being the union of the sets of MATH and MATH in the former case, and the set of products MATH i... |
math/9906030 | The key observation here is that a sequence MATH of elements of MATH is twist-recurrent if and only if the sequence MATH of NAME lifts of MATH to MATH is twist-recurrent. Of course one implication is obvious. For the other, note that in MATH, MATH for any MATH and any MATH such that MATH. Thus if we write MATH, we may ... |
math/9906030 | Let MATH be the distinct slopes of the NAME polygon and MATH their multiplicities. Then we can factor MATH as MATH, where MATH is the monic polynomial of degree MATH whose roots are the roots of MATH which have valuation MATH (counting multiplicities). Similarly we can factor MATH as MATH. Now observe that MATH and so ... |
math/9906030 | We first establish that MATH is algebraically closed; that is, for any MATH, the polynomial MATH has a root in MATH. Since this assertion only depends on the values of MATH, we may rechoose the MATH to ensure that MATH where MATH denotes reduction modulo MATH. Then the polynomials MATH over MATH and MATH over MATH clea... |
math/9906030 | It suffices to note that the latter set is a ring (by REF ), and has residue ring MATH (by CITE). |
math/9906030 | The most difficult part of the proof is showing that MATH is actually a ring! If MATH then MATH for certain universal polynomials MATH. Suppose that MATH and MATH are twist-recurrent. To check MATH, we must verify the condition given above for each MATH. However, for given MATH, we may replace the infinite sum over MAT... |
math/9906037 | REF holds because the extension group MATH is a finite set. REF imply that the map MATH has finite fibers; REF follows from this fact. REF is trivial. REF is REF. REF are consequences of REF (see also CITE, p. REF). To prove REF, it suffices to follow the proof of REF. Finally, to prove REF, one can adapt the proof of ... |
math/9906037 | REF enumi CASE: It is clear that MATH is abelian and satisfies REF , III. REF, and III. REF , and REF imply that MATH satisfies REF - REF . It follows from the definitions that MATH and MATH are closed under extensions in MATH, so that MATH and MATH also satisfy REF - REF . Since subobjects and quotients of torsion she... |
math/9906037 | REF enumi CASE: It is well known (for details, see REF ) that there are short exact sequences of the form MATH whenever MATH. Setting MATH, we get MATH when MATH, and an easy induction shows that MATH for all integers MATH. A similar argument shows the validity of REF for MATH. Now take a closed point MATH and an integ... |
math/9906037 | We first prove that REF , and REF are necessary. Suppose that the sequence is exact and non-split. If one of the homomorphism MATH or MATH were the zero arrow, then the other one would be an isomorphism since the cokernel of MATH is indecomposable, and the sequence would split. Similarly, neither MATH nor MATH can vani... |
math/9906037 | REF is a consequence of the structure theorem for finitely generated torsion modules over a principal ideal domain. This assertion implies that the class of sheaves with support in some fixed finite set MATH of closed points of MATH is closed under the operations of taking subobjects, quotients, and extensions. REF fol... |
math/9906037 | The polynomial MATH represents a morphism MATH. By unique factorization, and up to a non-zero scalar, we can write MATH, for some positive integers MATH and some distinct irreducible homogeneous polynomials MATH. For each MATH, let MATH be the closed point of MATH corresponding to MATH. The homogeneous polynomial MATH ... |
math/9906037 | We can write MATH as the direct sum of its torsion subsheaf MATH and a locally free subsheaf MATH of rank MATH. Write the maps MATH and MATH as MATH and MATH in the decomposition MATH. The morphism MATH cannot be injective, so MATH cannot be zero, so MATH is injective REF , and it follows that MATH is injective. Thus M... |
math/9906037 | Let MATH be the set of pairs MATH consisting of non-zero homogeneous polynomials of degree MATH and MATH respectively. The cardinality of MATH is MATH. One can also count the number of elements in MATH by factoring out a g.c.d. MATH of MATH and MATH. For a fixed degree MATH, there are MATH possibilities for MATH up to ... |
math/9906037 | REF enumi CASE: Since the extension group MATH vanishes by REF , any short exact sequence of the form MATH necessarily splits, and the product MATH in MATH is a scalar multiple MATH. It remains to compute the corresponding NAME number. Since MATH by REF, this NAME number is equal to the number of vector subspaces of di... |
math/9906037 | The isomorphism between the category MATH and the category of MATH-modules of finite length gives rise to an isomorphism between their NAME algebras. Thus MATH is isomorphic to the NAME algebra studied in REF and III of CITE. REF therefore follows from Paragraphs III REF , III. REF, III. REF , and III REF of CITE. It i... |
math/9906037 | Following Paragraphs I REF and I REF, we define generating series in MATH by MATH . By Formulae I REF and III REF in CITE, we have in MATH . Taking the inverse images by MATH, one obtains the relations in REF . As for REF , it follows from the equality MATH (use REF in CITE and REF ). |
math/9906037 | REF follows from REF. Let MATH be the subalgebra of MATH generated by the subalgebras MATH with MATH. Then MATH is the algebra of polynomials in the indeterminates MATH with coefficients in MATH. It is easy to see that MATH belongs to MATH, which proves the algebraic independence of the elements MATH. The algebraic ind... |
math/9906037 | REF can be proved by a tedious case by case examination, using Relations REF - REF . To prove REF , one first compute MATH using REF, and then MATH using REF. |
math/9906037 | REF comes from the fact that MATH is a subcategory of MATH closed under extensions REF and from the definition of the product MATH in MATH. REF follows from REF. Finally, REF imply REF . |
math/9906037 | We will use the generating series MATH and MATH. We set MATH and compute, using Relations REF : MATH . In view of REF, we therefore have MATH after expansion of the rational functions MATH in powers of MATH. Now in the formal power series ring MATH, one has MATH where the product in the left hand side runs over all clo... |
math/9906037 | REF implies that the multiplication MATH induces an isomorphism of MATH-modules from MATH to MATH. REF implies that MATH is a subalgebra of the NAME algebra MATH, obviously equal to MATH. Since the MATH-modules MATH and MATH are free (see REF), the multiplication MATH in the NAME algebra MATH induces an isomorphism of ... |
math/9906037 | It is asserted in REF that the map MATH induced by the multiplication of MATH is surjective. The defining relations of MATH imply easily that the map MATH induced by the multiplication of MATH is also surjective. The algebra MATH is generated by the pairwise commuting elements MATH for MATH, which shows that the monomi... |
math/9906037 | By definition, MATH is the subalgebra of MATH generated by the elements MATH and MATH, for MATH and MATH. REF follows therefore from the definition of the elements MATH REF and from the fact that the scalars MATH do not vanish in the field MATH for any MATH. REF states that for all integers MATH and MATH, one has MATH ... |
math/9906041 | It suffices to examine the effect of the formula on monomials MATH and for MATH . The first two terms produce MATH if MATH else MATH . In the sum, the (typical) term for MATH is MATH . A simple calculation shows this is the image under MATH of the corresponding term in REF . |
math/9906041 | REF follow immediately from REF . It was shown in REF that MATH where MATH is the conjugate under MATH of multiplication by MATH acting on MATH. Together with REF this proves REF . |
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