paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0005265 | CASE: Using REF we get that MATH . So we get for every MATH that MATH . It follows that MATH. Moreover, we have for all MATH that MATH implying that MATH and hence MATH . CASE: This follows immediately from REF. CASE: Using the estimate MATH, this is an easy consequence of REF. |
math/0005265 | One implication is trivial. We will prove the other one. Therefore suppose that there exists a non-zero element MATH such that MATH. Let MATH denote the MATH-weak closure of MATH in MATH. Then MATH is a MATH-weakly closed left ideal in MATH so there exists a projection MATH in MATH such that MATH. Choose MATH. By the p... |
math/0005265 | Let MATH. Then MATH. Thus MATH and MATH. By REF , we get that MATH and MATH . By the left invariance of MATH, we know that MATH, thus MATH, so MATH. So we have proven that MATH is a two-sided ideal in MATH. Using the techniques of the proof of REF, we arrive at the conclusion that MATH is MATH-weakly dense in MATH. |
math/0005265 | Because MATH belongs to MATH, we have that MATH belongs to MATH and MATH . On the other hand, since MATH, the left invariance of MATH implies that the element MATH belongs to MATH and MATH . Therefore MATH belongs to MATH and MATH . Since MATH is a unitary corepresentation, we get that MATH . Combining this with REF , ... |
math/0005265 | By REF, we get for every MATH that MATH . REF tells us that MATH belongs to MATH and MATH . Combining these two facts, we see that MATH belongs to MATH and MATH . |
math/0005265 | Take an orthonormal basis MATH for MATH. Choose MATH. Then the MATH-weak lower semi-continuity of MATH implies that MATH so the previous lemma implies that MATH . By the lower semi-continuity of MATH, this implies that MATH for all MATH from which we conclude that MATH belongs to MATH. By result REF, We have moreover f... |
math/0005265 | We know that there exists a bounded net MATH in MATH such that MATH converges strongly-MATH to MATH and MATH is a bounded net that converges strongly-MATH to MATH. Then MATH is surely a bounded net that converges strongly-MATH to MATH. It follows easily from the previous lemma that MATH belongs to MATH and MATH . There... |
math/0005265 | The element MATH belongs to MATH. Hence the previous lemma implies that MATH belongs to MATH, implying that MATH belongs to MATH. |
math/0005265 | By REF we get that MATH belongs to MATH implying that the element MATH belongs to MATH. Because MATH, the result follows. |
math/0005265 | Choose MATH and MATH. Notice that MATH. Using REF , we see that MATH . Let us now get hold of some interesting elements: CASE: Using REF , we get the existence of a net MATH in MATH such that CASE: MATH for all MATH, CASE: The net MATH converges strongly to REF. CASE: REF guarantees that the MATH-algebra MATH is a non-... |
math/0005265 | Choose MATH and MATH. The previous lemma guarantees the exstence of a directed set MATH, a net MATH and a net MATH in MATH such that CASE: MATH for all MATH, CASE: The net MATH converges strongly to REF. By result REF, we have for every MATH that MATH belongs to MATH and MATH . From this it follows that MATH is a bound... |
math/0005265 | Let us first prove the first statement. Choose MATH and MATH. Then we have for all MATH that MATH which belongs to MATH. This implies that MATH belongs to MATH. Let MATH. Using the inequality MATH, the above considerations imply that MATH belongs to MATH. Now we turn to the second statement. Therefore choose MATH, MATH... |
math/0005265 | Choose MATH, MATH, MATH such that MATH, MATH and MATH. REF tells us that MATH is the closed linear span of elements of the form MATH . Take a basis MATH of MATH. There exists MATH such that MATH for all MATH. Since MATH is in standard form, we can find an element MATH such that MATH for all MATH. From the previous lemm... |
math/0005265 | Choose MATH, MATH, MATH and MATH. Then MATH . Choose an orthonormal basis MATH for MATH. Choose MATH and MATH. By REF , we know that the net MATH converges to MATH . In the next part we will show that each of the sums in this net belongs to MATH. Therefore fix MATH. Because MATH belongs to MATH and MATH belongs to MATH... |
math/0005265 | Define the unitary element MATH such that MATH for all MATH. Then the following holds CASE: MATH for all MATH, CASE: MATH , CASE: MATH for all MATH. Call MATH the norm closure of MATH, then MATH is a MATH-weakly dense sub-MATH -algebra of MATH such that MATH belongs to the multiplier algebra MATH (for once, the tensor ... |
math/0005265 | Define MATH to be the projection on MATH. Also define MATH, then MATH implying that MATH is a partial isometry in MATH. Define MATH to be the final projection of MATH. Thus, since MATH, we get that MATH . Because MATH, we have that MATH, implying that MATH. Hence MATH . Arguing as in the proof of REF, we conclude from ... |
math/0005265 | By REF and the previous lemma, we know that MATH . Since MATH is an isometry, this implies that MATH and hence MATH. Using the previous lemma once more, we get that MATH so that MATH is unitary. Because MATH, the proposition follows. |
math/0005265 | We may assume that MATH and MATH. We have for MATH that MATH . Therefore the right invariance of MATH implies that MATH. Using REF, we conclude from this that MATH belongs to MATH. By right invariance of MATH we have moreover for all MATH that MATH implying that MATH. |
math/0005265 | Let MATH denote the densely defined closed linear map from within MATH into MATH such that MATH is a core for MATH and MATH for all MATH. Recall that the pair MATH,MATH is by definition the polar decomposition of MATH, that is, MATH. Choose MATH and MATH. Let MATH. Then MATH belongs to MATH and MATH. Moreover, MATH whi... |
math/0005265 | Apply the previous proposition with MATH and some NAME MATH for MATH. In this case , MATH and MATH is the modular operator for MATH in this NAME. Then we get for MATH and MATH that MATH and the corollary follows. |
math/0005265 | Call MATH the modular operator of the pair MATH, MATH. By REF we know that MATH for all MATH. Applying the previous proposition to the pair MATH, MATH (in which case MATH and MATH), we get that MATH. If we apply the previous proposition to the pair MATH, MATH in which case MATH, we get that MATH. So we get for all MATH... |
math/0005265 | By the previous proposition, we have for MATH that MATH from which we conclude that MATH belongs to MATH. By REF we also get that MATH . Therefore REF implies the existence of a n.f.s. weight MATH on MATH such that MATH for all MATH. REF tells us that MATH for all MATH. Hence, MATH. |
math/0005267 | Partially ordering the up-sets and (separately) the down-sets in MATH by set inclusion, let MATH be a minimal up-set in MATH containing the vertices of some path from MATH to MATH. (By assumption, MATH is an up-set containing such a path, so MATH exists.) Let MATH be a minimal down-set in MATH such that MATH contains t... |
math/0005267 | Let MATH be the poset on ground set MATH obtained by deleting the edge joining MATH and MATH from the NAME diagram of MATH. Since MATH is a path from MATH to MATH in MATH, by REF there is a path MATH from MATH to MATH which is an induced subposet of MATH. Then, the cycle MATH is as desired. |
math/0005267 | Let MATH and MATH be two unequal upward paths from MATH to MATH in MATH. Without loss of generality, the two paths differ in their first step, that is, MATH. Clearly MATH and MATH are incomparable. Let MATH. Then MATH is nonempty because MATH. Let MATH be a minimal (in MATH) element of MATH. Let MATH and MATH be upward... |
math/0005267 | If MATH has a diamond as an induced subposet, then by REF we can find a subdivided diamond MATH which is an induced cyclic subposet of MATH. Clearly MATH has height at least MATH. Suppose now that MATH has no diamond as an induced subposet. Let MATH be a cycle with height at least MATH such that MATH. Let MATH and let ... |
math/0005267 | CASE: If MATH is the diamond as in REF , then there are at least two unequal upward paths from MATH to MATH. By REF , MATH has a cycle (namely, a subdivided diamond). CASE: Suppose that MATH is a subdivision of the MATH-crown displayed and labeled in REF . Since by REF has height at least MATH, without loss of generali... |
math/0005267 | We prove the claim of REF by induction over the cardinality of MATH. The claim is vacuous when MATH. We now suppose that the claim is true for an acyclic poset MATH when MATH for fixed MATH, and consider an acyclic poset MATH with cardinality MATH. Let MATH be a leaf in MATH. Without loss of generality, we assume that ... |
math/0005267 | Notice that for any MATH, MATH . Since MATH, we deduce MATH. Thus we obtain MATH which completes the proof. |
math/0005267 | Since MATH, we see that the system MATH is stochastically monotone. To see that it is not realizably monotone, suppose that there exists a system MATH of MATH-valued random variables which realizes the monotonicity of MATH. By applying REF repeatedly, we (almost surely) have MATH. But then (after perhaps taking care of... |
math/0005267 | We have already seen that a poset MATH of Class B has either the diamond or a crown as an induced subposet. If MATH has an induced MATH-crown, then MATH is in Class B by definition. If MATH has an induced subposet which is either the diamond or a MATH-crown for some MATH, then, by REF , MATH is non-acyclic and therefor... |
math/0005267 | Let MATH be a non-acyclic poset and MATH be either the diamond or a MATH-crown for some MATH. We will construct a stochastically monotone system MATH of probability measures on MATH which is not realizably monotone, by dividing the construction into two cases. CASE: Suppose that MATH has a diamond MATH as an induced su... |
math/0005267 | If MATH, then MATH is both a down-set and an up-set in MATH. If MATH, then MATH and there is a unique predecessor of MATH; otherwise, the uniqueness of the path is contradicted. Let MATH be the predecessor of MATH. Clearly, MATH belongs to the cover graph of MATH. Suppose that MATH covers MATH in MATH. We claim that MA... |
math/0005267 | By REF and its trivial consequence REF , MATH clearly implies REF - REF . We proceed to the converse. Since any up-set MATH in MATH is the disjoint union of the components MATH of the subgraph of MATH induced by MATH and MATH are all up-sets in MATH, to prove MATH it suffices to show REF for every up-set MATH which ind... |
math/0005267 | Suppose that the hypotheses in REF hold. If MATH, then MATH and the inequality clearly holds. Otherwise, MATH for every MATH. Since the path MATH is upward and MATH covers MATH in MATH, by REF MATH is a down-set in MATH. Therefore we have MATH for all MATH, as desired. REF is reduced to REF by considering the dual MATH... |
math/0005267 | We first claim that for every MATH, there are incomparable elements MATH of MATH in MATH which satisfy both the hypotheses of one of REF and also MATH . We will show this by induction over the cardinality of MATH. If MATH, then MATH is a minimal element in MATH and indeed MATH. Suppose that the claim holds for any MATH... |
math/0005267 | Suppose that there exists an induced subposet MATH of MATH which is poset-isomorphic to one of the posets REF - REF . If there is an acyclic poset MATH which has MATH as an induced subposet, then MATH is also an induced subposet of MATH; by REF , this is impossible. We have thus shown that REF . To prove REF , observe ... |
math/0005267 | Let MATH be the collection of all subsets MATH of MATH such that the subposet via induced cover subgraph of MATH on the ground set MATH is a poset of the form REF for some MATH. Note that if a subposet MATH via induced cover subgraph is of the form REF then MATH is an induced subposet. By REF , MATH has an induced cycl... |
math/0005267 | Suppose that a poset MATH satisfies REF . We will make an inductive argument over the cardinality of MATH. But if MATH is acyclic, then the argument is vacuous. In particular, MATH with cardinality at most MATH is acyclic. Now let MATH be a connected non-acyclic poset with cardinality MATH. By REF , there exists a pair... |
math/0005267 | Suppose first that a non-acyclic poset MATH is not enlargeable to an acyclic poset. Then, by REF , MATH has an induced subposet MATH which is one of the posets REF - REF . If MATH is the diamond, then by REF MATH has a cycle with height at least MATH. Thus REF implies that monotonicity equivalence fails for MATH. If MA... |
math/0005267 | Suppose first that MATH. Let MATH and MATH denote the distribution functions of MATH and MATH, respectively. Let MATH be fixed, MATH, and MATH. If MATH, then REF obviously holds. If MATH, then we have MATH for all MATH such that MATH. By REF , the section MATH is a down-set for every MATH, which implies, by REF , that ... |
math/0005267 | Let MATH be a stochastically monotone system of probability measures on MATH. Let MATH be a random variable uniformly distributed on MATH. Then we can construct a system MATH of MATH-valued random variables satisfying REF via MATH . By REF , the system MATH satisfies REF ; thus, MATH is realizably monotone. |
math/0005268 | The dimension statement is straightforward. To prove the relation, we pull MATH apart along MATH, and study the corresponding moduli spaces (see REF for more discussion on such matters). Let MATH denote the complement MATH, given a cylindrical-end metric modeled on the product metric MATH, where MATH is given its stand... |
math/0005268 | This is a combination of REF when MATH, and REF in the remaining case. |
math/0005268 | This follows from REF (see also the proof of REF in the perturbed case). |
math/0005268 | According to REF , the space MATH is identified with the space of degree MATH line bundles over MATH with non-trivial MATH. The forgetful map MATH which takes a degree MATH divisor, thought of as a complex line bundle with section, to the underlying complex line bundle gives the surjection to this locus. |
math/0005268 | When MATH, this is proved in REF, where it appears as REF . The remaining case is covered by REF . |
math/0005268 | This is a dimension-counting argument. Suppose we have a sequence MATH, for some increasing, unbounded sequence MATH of real numbers. By local compactness, there is a subsequence which converges in MATH to a pair of configurations MATH and MATH over MATH and MATH respectively. By the usual compactness arguments (see CI... |
math/0005268 | The compactness of MATH and MATH follows from the usual compactness arguments, together with the facts that the NAME functional is real-valued, MATH, and there are no other critical manifolds. Compactness of MATH follows from this, together with a straightforward dimension count (see the discussion above in the proof o... |
math/0005268 | Recall that MATH consists entirely of reducibles all of which are smooth, according to REF ; thus, gluing theory identifies the moduli spaces MATH for large MATH with MATH. (See also CITE, where this result appears as REF .) |
math/0005268 | The moduli space MATH comes equipped with an obstruction bundle MATH, defined by MATH. (whose MATH-theory class canonically extends over all of MATH). The dimension count in REF guarantees that each solution in MATH extends uniquely to a smooth reducible over MATH. Thus, gluing theory gives that MATH where MATH denotes... |
math/0005268 | Gluing shows that MATH . According to REF , MATH either has degree MATH, or MATH is empty. The latter case would force MATH (for the given genus and self-intersection number). To rule out this latter case, we need only look at an example where the irreducible term is non-zero. Let MATH be a ruled surface MATH over MATH... |
math/0005268 | This a standard argument from NAME theory. A general discussion of compactness results for the anti-self-duality equation can be found in CITE (see especially REF); so we sketch the argument here only briefly. A sequence MATH converges in MATH, after passing to a subsequence, to some solution MATH to the NAME equations... |
math/0005268 | If a sequence MATH converges to an ideal point MATH then there is a divergent sequence MATH of real numbers so that MATH where MATH is the map induced by translation by MATH on the first coordinate. Since each path has finite energy, continuity of the restriction maps (see CITE) guarantees that MATH . |
math/0005268 | Gluing describes the end of MATH as a fibered product MATH where the superscript denotes based versions of the moduli spaces. This gives the space of ideal solutions a disk-bundle neighborhood in MATH. |
math/0005268 | We must show that MATH is homologous to MATH. It suffices to verify this over the subset MATH, since the complement has codimension two, and the classes in question are one-dimensional. Over the subset, now, the claim is easy to verify. On MATH, MATH is represented by MATH, the holonomy around a representative of MATH ... |
math/0005268 | Clearly, the difference MATH is the first NAME class of the circle bundle MATH. Here, MATH denotes the moduli space based at MATH; see REF. To prove the proposition, we must verify that this bundle admits a section MATH in the complement of MATH (that is, over MATH) and that, with respect to a trivialization of the cir... |
math/0005268 | We can reduce to a corresponding statement for configurations over MATH, as follows. Let MATH be a line bundle over MATH, so that MATH. Then, pull-back induces a map MATH to the configurations where the fiber-wise holonomy of the connection is constant, and the section is covariantly constant around each fiber. The ide... |
math/0005268 | The vector space MATH is generated by homogeneous elements of the form MATH where MATH is a subset of MATH, and MATH, MATH, MATH, MATH are integers with MATH. Clearly, it suffices to prove the proposition for homogeneous generators of degree MATH. Modulo MATH, such an element is equivalent to the element MATH . Indeed,... |
math/0005268 | By REF , MATH lies in the ideal generated by MATH. Now the proposition follows from REF . |
math/0005268 | This follows immediately from the fact that MATH. |
math/0005268 | We prove that both moduli spaces MATH and MATH are empty. Suppose there were some finite energy solution to the NAME equations in a MATH structure with MATH. We know that the spinor lies entirely in one of the two summands in the splitting of the spinor bundle MATH (that is, it is a MATH- or a MATH-spinor, in the notat... |
math/0005268 | This follows exactly as in REF (for irreducible boundary values) and REF (for reducible boundary values) of CITE. The key observation at this point is to note that MATH is an isomorphism, which allows one to ``unroll" parts of the NAME deformation complex to identify it with the deformation theory of divisors in MATH. ... |
math/0005268 | The deformation theory around a solution MATH is identified the deformation theory around a corresponding divisor in the line bundle MATH with MATH; that is, with divisors in a line bundle which (topologically) pulls back from MATH. According to REF , all such divisors actually pull back from the base MATH; and indeed,... |
math/0005268 | By a standard excision argument, we have MATH which calculates MATH, given REF . Similarly, we have MATH which gives us MATH. Smoothness of MATH and MATH follows from adapting methods of CITE. |
math/0005268 | Most of this is a straightforward adaptation of CITE. We begin with the identification of the moduli spaces over MATH. As in CITE, the equations over MATH reduce to vortex equations over MATH. More specifically, the components of the moduli spaces MATH correspond to line bundles MATH over MATH with the property that MA... |
math/0005268 | We begin by proving MATH is empty. Note that MATH consists entirely of MATH-solutions, hence so must any section in MATH. Thus, a solution MATH induces a non-zero element in MATH with MATH . But MATH, according to REF . The same argument, now appealing to REF , shows that MATH is empty, and that MATH is smooth. |
math/0005268 | The proofs of REF apply directly in this perturbed context. |
math/0005268 | We begin by proving the second claim. Note that MATH comes with a tautological connection along the MATH factor, with the property that for any path MATH and connection MATH, MATH . MATH . Now, fix a path in MATH, MATH . We need to show that for all paths in the configuration space MATH we have that MATH . This follows... |
math/0005268 | Fix a compactly-supported cut-off function MATH in MATH, and consider the section MATH of MATH, thought of as a line bundle over MATH, giving MATH the MATH topology. This transversality follows from the fact that, for any MATH, as we vary MATH, the integral MATH can take on any complex value. This, in turn, follows fro... |
math/0005271 | The set of MATH-fixed points in MATH constitutes a circle, denoted by MATH, containing MATH and MATH. Then MATH divides the sphere MATH into two hemispheres, say MATH and MATH. Let MATH be a complex MATH-vector bundle over MATH. Note that the restriction of MATH to the subspace MATH decomposes into the NAME sum of MATH... |
math/0005271 | It suffices to show that every complex MATH-line bundle MATH over MATH is trivial by REF . Choose two MATH-invariant hemispheres, denoted by MATH and MATH of MATH containing MATH and MATH, respectively, such that MATH and MATH is a MATH-invariant circle. Since MATH (respectively, MATH) is equivariantly contractible to ... |
math/0005271 | Given a MATH-vector bundle isomorphism MATH, the map MATH sending MATH to MATH gives a MATH-vector bundle isomorphism. |
math/0005271 | The construction of induced bundles is the same as that of induced representations. |
math/0005271 | For MATH and MATH, the map MATH sending MATH to MATH gives a (non-equivariant) vector bundle map. Given MATH we have MATH for some representative MATH and MATH. Then the equalities MATH imply that MATH is MATH-equivariant. On the other hand, the inverse bundle map is given by the map sending MATH to MATH for MATH and M... |
math/0005271 | The necessity is obvious since a MATH-vector bundle isomorphism MATH restricts to a MATH-vector bundle isomorphism MATH, and the sufficiency follows from the fact that MATH is functorial. |
math/0005271 | It is easy to check that the map MATH sending MATH to MATH gives the inverse of the map in the lemma. |
math/0005271 | We now consider the map MATH sending each generator MATH of MATH to MATH. Since both MATH and MATH are free abelian groups, MATH is a well-defined group homomorphism. Moreover it is injective, since the set of the elements MATH for all MATH such that MATH can be extended to an additive basis of MATH. For each MATH, eit... |
math/0005271 | By assumption there exists an element MATH such that MATH for all MATH. It follows that MATH for any character MATH of MATH, which completes the proof. |
math/0005273 | REF are clear. We only check REF . For REF , let MATH . By REF above, MATH, so we check MATH. Let MATH. If MATH, then let MATH, otherwise we must have MATH, so we can write MATH as a disjoint union MATH with MATH, MATH. In either case we have MATH, MATH, MATH. So there is a function MATH with MATH, so MATH while MATH. ... |
math/0005273 | The upper bound follows from MATH. For the lower bound: it is known that there are MATH many maximal ideals, and we have just shown that the function MATH maps them injectively to precomplete clones. |
math/0005273 | Let MATH, MATH: MATH be two REF functions such that the ranges of MATH, MATH, MATH are disjoint. Since MATH contains a function which is not almost unary, there is some MATH with MATH for all MATH in the range of MATH. Then the function MATH is a pairing function. |
math/0005273 | Choose MATH and MATH in MATH with MATH. Using unary functions, we will construct a binary function in MATH which is not in MATH. Let MATH, so MATH. So there is some MATH and a sequence MATH of elements of MATH such that all values MATH are different. Similarly, there is some MATH and a sequence MATH of elements of MATH... |
math/0005273 | Since each MATH, we can find a decomposition MATH, MATH, and a function MATH mapping each MATH to a small subset MATH such that: CASE: For all MATH, all MATH: MATH. CASE: For all MATH, all MATH: MATH. By the previous lemma, we cannot have both MATH and MATH in MATH, so without loss of generality assume MATH. Now fix an... |
math/0005273 | We will show how to construct a pairing function from MATH and MATH. Define MATH . We claim that for all distinct MATH: MATH. Indeed, if MATH, then MATH if MATH, then MATH and if MATH, then MATH . Hence MATH is a pairing function. |
math/0005273 | See CITE or CITE. |
math/0005273 | Let MATH be an independent family of subsets of MATH. Write MATH for MATH. For each MATH we let MATH the clone generated by MATH. We will now show that CASE: MATH, for all MATH CASE: Whenever MATH, then MATH already generates MATH. This will conclude the proof, because REF together with REF implies that each MATH can b... |
math/0005273 | Proceed by induction on the complexity of the terms. We have to thin out the set MATH finitely many times in order to make finitely many functions REF or constant. |
math/0005273 | Let MATH be the set of subterms of MATH (including MATH itself). Collect all REF functions appearing in MATH for MATH, that is : MATH . The set MATH is finite, the identity function is in MATH, and all functions in MATH are REF. We may thin out the set MATH so that the family MATH is pairwise disjoint. So since MATH wi... |
math/0005273 | REF are easy. For REF, assume that MATH, with MATH, MATH. We have to distinguish several cases: CASE: MATH, MATH. Since MATH, and MATH is canonical, we have MATH for all MATH, so MATH is symmetrical. CASE: MATH. So MATH. Pick any MATH with MATH. Then MATH, so MATH implies MATH, this means MATH. Similarly we find MATH. ... |
math/0005273 | By REF, it is enough to find a function MATH which is REF on MATH. If MATH is symmetrical and REF on MATH (and also REF on MATH, of course), then we may assume (replacing MATH by MATH for some appropriate MATH, if necessary), that MATH for all MATH. We claim that the function MATH is REF on MATH. Indeed, if MATH, then ... |
math/0005274 | We let MATH and put MATH so that we have a filtration of subspaces MATH with MATH, for all MATH by REF . Let MATH and let MATH be the minimal integer such that MATH. Since MATH implies that MATH, this setting puts us in the situation of REF, from which we conclude that MATH is a finite-dimensional vector space over MAT... |
math/0005274 | We will continue to use the notation defined earlier. First we show that MATH. Suppose that MATH. It is easy to see that MATH is invariant under MATH. Now there exits a basis MATH of MATH together with non-zero complex number MATH such that MATH, where MATH is the element of REF . Since MATH is a finite-dimensional vec... |
math/0005274 | By REF every finite irreducible MATH-module is a homomorphic image of MATH. Now the usual argument for highest weight representations implies that given a finite-dimensional irreducible MATH-module MATH the MATH-module MATH contains a unique maximal submodule, from which the bijection then follows. |
math/0005274 | Note that MATH is singular if and only if MATH. We compute MATH . But then MATH, since otherwise REF would imply that MATH. However, MATH together with REF implies that MATH . Now REF gives MATH . Now if MATH, then REF gives MATH and MATH. But then MATH by REF . Hence MATH so that by REF we have MATH. Now if MATH, we h... |
math/0005274 | The lemma follows immediately from the following two equations: MATH . |
math/0005274 | If MATH and MATH, then by REF MATH contains no proper singular vector and hence is irreducible. Suppose that MATH and MATH. In this case consider the submodule of MATH generated by the singular vector MATH. This module is precisely MATH above and hence MATH is freely generated over MATH by MATH and MATH. We claim that ... |
math/0005274 | By REF MATH is a singular vector in MATH of MATH-weight MATH. Consider MATH, the MATH-submodule generated by MATH. Then we have MATH, where MATH is the irreducible MATH-submodule generated by MATH. Note that the map MATH extends uniquely to an epimorphism of MATH-modules from MATH to MATH. In particular it is a MATH-mo... |
math/0005274 | By REF MATH is a singular vector of MATH of MATH-weight MATH. Let MATH denote the MATH-submodule generated by MATH. Consider first the case MATH. As in the proof of REF MATH is a free MATH-module generated over MATH by MATH. We compute MATH . This implies that the set MATH generates MATH over MATH and so MATH is a MATH... |
math/0005274 | By REF MATH and MATH are singular vectors in MATH. Consider MATH and MATH, the MATH-submodules generated by MATH and MATH, respectively, and let MATH. Then we have MATH and MATH, where MATH and MATH are the irreducible MATH-submodules generated by MATH and MATH, respectively. Let's first compute MATH. Since the MATH-we... |
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