paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0002117 | Immediate from REF since MATH and MATH. |
math/0002117 | Let MATH be the fraction field of the NAME algebra MATH and consider the MATH-representations on MATH and MATH given respectively by MATH and MATH. The NAME transform REF extends uniquely to an algebra isomorphism MATH. Moreover MATH identifies the MATH-finite parts of MATH and MATH. MATH is the MATH-finite part of MAT... |
math/0002117 | Using REF we find the commutative diagram MATH . Then commutativity of the middle square gives MATH. |
math/0002117 | To prove this is it easiest to start with the geometry of MATH and MATH. In REF, we introduced MATH and MATH expressly for the purpose of extracting a square root of the function MATH. In fact, MATH occurs naturally in the geometry of MATH. To begin with, MATH and so MATH is a smooth Lagrangian submanifold of MATH with... |
math/0002117 | The only point that is not immediate is the surjectivity of MATH and MATH in REF . But this follows since, as MATH and MATH are MATH-orbits, MATH and MATH are the MATH-finite parts of the function fields MATH and MATH. |
math/0002117 | Let MATH be a non-zero MATH-sided ideal in MATH. Then MATH is in particular MATH-stable with respect to the representation REF . Let MATH be a subspace carrying a non-zero MATH-irreducible representation. Then MATH lies in MATH or MATH. This follows since, by REF , MATH is isomorphic as a MATH-representation to a subsp... |
math/0002117 | Define MATH by MATH. Then MATH extends to an algebra anti-involution MATH where MATH and MATH. This follows from the definition of MATH in the proof of REF . Now MATH naturally extends to an anti-involution of MATH and then to a MATH-invariant anti-automorphism MATH of MATH such that MATH. Then MATH preserves the MATH-... |
math/0002117 | The equivalence REF is immediate from REF . The implication MATH follows easily from REF . We get REF as follows. Given REF , we know by REF that MATH admits a square root MATH. But the equation MATH in MATH forces MATH. So MATH. So MATH. Finally, the equivalences with REF follow from REF . |
math/0002117 | By decomposing MATH as a MATH-representation, we find that the (unique) MATH-stable direct sum complement to MATH in MATH is MATH. So in particular, MATH is the complement to MATH in MATH. But MATH and so MATH. We find that MATH generates MATH, and so MATH. |
math/0002117 | Since MATH is a lowest weight vector in MATH and MATH is MATH-semi-invariant, it follows that MATH. But MATH. |
math/0002117 | The commutator MATH is a differential operator on MATH of order at most MATH and so we can write it uniquely as the sum of a vector field MATH and a function MATH. It is convenient to compute these parts individually. Using REF we find MATH . The fourth equality follows from NAME identities. Indeed, REF the operator MA... |
math/0002117 | Starting from REF we find MATH . We will explain the second equality. We start from the fact that MATH is a primitive idempotent in the NAME algebra MATH. Indeed, MATH is a decomposition of MATH into orthogonal primitive idempotents. For any primitive idempotent MATH, then the map MATH is the orthogonal projection onto... |
math/0002117 | If MATH then REF gives MATH and using REF we find MATH. If MATH then REF gives MATH and we find MATH. Thus REF is true when MATH equals MATH or MATH. Now we can prove REF for all values of MATH without further calculation by simply examining the form of MATH. Let MATH be arbitrary. Then MATH where MATH. Since MATH we g... |
math/0002117 | The result is clear once we prove that MATH where MATH. It suffices to check this for MATH and MATH. Clearly MATH is the identity on MATH. So for MATH we find MATH. Next suppose MATH. Then MATH . The second equality follows by REF because MATH and so MATH. |
math/0002117 | MATH is simple by REF since MATH does not lie in MATH. (Note MATH by REF .) The result follows by REF . |
math/0002117 | MATH follows by REF while MATH follows by REF . MATH is maximal since MATH is simple. The infinitesimal character follows by REF ; the two weights are then NAME group conjugate. |
math/0002117 | Both MATH and MATH are faithful as right or left modules over MATH since the algebra MATH has no zero-divisors. Since MATH identifies with MATH, the left and right annihilators in MATH are MATH and MATH. |
math/0002117 | Let MATH be the algebra generated by MATH and MATH. Then MATH and this is the MATH-grading. Now MATH, regarded as a space of multiplication operators, lies in MATH. Consequently MATH, regarded as a space of functions, lies in MATH. The problem then is to show that MATH is MATH-stable. We know REF that MATH is generated... |
math/0002117 | By REF we know that MATH is a NAME module for MATH with lowest weight vector MATH of weight MATH. Similarly MATH is a NAME module for MATH with lowest weight vector MATH of weight MATH. This follows because for MATH we have MATH, and for MATH we have (by REF ) MATH. The weights MATH exponentiate to characters of MATH a... |
math/0002117 | of REF . MATH and MATH are faithful simple MATH-modules by REF . Since MATH and MATH carry different MATH-representations, they are the only non-trivial MATH-submodules of MATH. Neither MATH nor MATH is MATH-stable, since multiplication by MATH moves each into the other. Thus MATH is simple for MATH. Faithfulness is au... |
math/0002117 | Clearly MATH lies in MATH. The converse follows since MATH is a localization of MATH and so any differential operator on MATH extends to one on MATH. |
math/0002117 | The map is injective because any differential operator on MATH is uniquely determined by its values on MATH. This is true since any vector space basis of MATH is a set of local étale coordinates on MATH. To prove surjectivity we need to show that if MATH then MATH extends to a differential operator MATH on MATH. We may... |
math/0002117 | The first part is clear. Now MATH is a filtered MATH-invariant algebra automorphism of MATH which is trivial on MATH. By considering the induced action of MATH on MATH, we see easily that MATH lies in MATH. We have MATH and so MATH. |
math/0002117 | Let MATH. Since MATH is multiplicity-free as a MATH-representation, there is a unique MATH-stable complement, call it MATH, to MATH in MATH. Then MATH is a MATH-stable vector space grading and MATH acts on MATH by multiplication by MATH. So MATH . The pairing MATH is MATH-invariant. Suppose MATH. We know by REF that MA... |
math/0002117 | Everything is immediate except non-degeneracy. Now MATH because MATH is simple REF and MATH is a MATH-sided ideal in MATH which does not contain MATH. |
math/0002117 | We need to show that MATH is a simple bimodule over MATH; we already know that MATH is a simple ring. Suppose MATH is a MATH-bisubmodule of MATH and MATH with MATH. Then MATH. Now MATH is non-degenerate on MATH and so there exists MATH such that MATH. Then MATH since MATH is a supertrace. Thus MATH is not MATH-orthogon... |
math/0002117 | The map MATH is MATH-invariant, equivariant with respect to MATH and MATH, and also MATH intertwines MATH and MATH. So our results on MATH transfer over to MATH via MATH. This proves the first paragraph. Since MATH is a filtered superalgebra product, we have MATH. Now proving REF reduces to showing that if MATH is not ... |
math/0002117 | Let MATH. Then we have MATH by REF . Now the result is immediate because of REF . |
math/0002117 | The graded star product is defined by MATH where MATH and MATH are NAME homogeneous. This is strongly MATH-invariant by REF . |
math/0002117 | Define MATH by MATH. Then REF imply REF . We claim MATH. We may assume MATH and MATH. Then REF gives MATH since the grading of MATH is MATH-orthogonal. Now MATH since MATH is a supertrace and MATH and MATH are even. But MATH and MATH by the parity relation REF . This proves our claim. Now REF of MATH follow immediately... |
math/0002117 | We can easily compute the restriction to MATH of MATH. We find MATH where MATH, MATH and MATH was defined in REF . Now if MATH and MATH then MATH. |
math/0002117 | We have MATH and so MATH is MATH-abelian. Suppose MATH and MATH for all MATH. We can write MATH where MATH and MATH. Since MATH is graded it follows easily that MATH for all MATH. But then MATH since MATH is maximal NAME abelian. It follows by induction on MATH that MATH. |
math/0002118 | CASE: MATH is simple implies MATH since MATH is a MATH-sided ideal in MATH which does not contain MATH. CASE: Let MATH be a non-zero two-sided ideal in MATH. Pick MATH with MATH. Then REF implies there exists MATH such that MATH. It follows that MATH where MATH lies in MATH. So MATH contains MATH. But then MATH contain... |
math/0002118 | Suppose MATH is simple. Then MATH is non-degenerate on MATH (by REF ) and hence (since MATH is MATH-invariant) MATH is non-degenerate on MATH. Then (by REF again) MATH is simple. Conversely, assume MATH is simple. Let MATH be a non-zero MATH-sided ideal in MATH. To show MATH is simple, it suffices to show that MATH is ... |
math/0002118 | The second sentence is clear. Now REF is equivalent to MATH . We have MATH since MATH is a superalgebra. Because of REF , showing REF reduces to showing that MATH is orthogonal to MATH if MATH. So suppose MATH is not orthogonal to MATH. Then there exist MATH, MATH and MATH such that MATH. Then MATH has a component in M... |
math/0002118 | Identical to the proof of CITE. |
math/0002118 | If MATH and MATH then MATH. Then MATH because of the parity REF . |
math/0002118 | Given MATH, we define a product MATH on MATH as follows: MATH is MATH-bilinear and if MATH and MATH with MATH where MATH then MATH. This is the only possible way to extend MATH to a graded star product. The properties of MATH imply that MATH is in fact a perfectly graded, MATH-invariant star product. The converse is cl... |
math/0002118 | Same as the proof of CITE. |
math/0002119 | First suppose that MATH is reachable. Then there is a firing sequence MATH. Therefore, as above, there exist MATH such that MATH. Hence MATH. For the converse, suppose MATH. Then MATH . The proof is by induction on MATH. For the base step put MATH then MATH. The correspondence between markings and their associated poly... |
math/0002119 | Let MATH be a field. First observe that MATH. Let MATH be a NAME basis for MATH. Then MATH if and only if there exists MATH such that MATH and MATH reduce to MATH by MATH. |
math/0002124 | MATH. By choosing MATH small enough as a function of MATH, the NAME fillings contribute negligible volume so this property is retained by MATH. |
math/0002124 | According to CITE a non-oriented minimizer among all nonzero codimension one cycles always exists and is smooth provided the ambient dimension is at most MATH. Let MATH be this minimizer. For a contradiction, assume MATH. The NAME surgeries in section MATH were confined to MATH, so the surfaces MATH, MATH persist as su... |
math/0002124 | We actually show that any homotopically essential loop obeys this estimate. The long collar condition MATH REF implies that any arc in MATH with end points on MATH can be replaced with a shorter arc with the same end points lying entirely within MATH. It follows that any essential loop in MATH can be homotoped to a sho... |
math/0002126 | Consider the derivation MATH of degree MATH given by MATH. On the polilinear form of weight MATH acts by MATH. Define the homotopy MATH to be MATH on the forms of weight MATH and MATH on the forms of weight MATH. We define map of complexes MATH by MATH where MATH . This is a map of (total) complexes. Indeed, using iden... |
math/0002126 | We check all the required properties. First MATH . Then MATH . Also MATH and MATH . |
math/0002126 | Let MATH be the differential graded algebra of MATH matrices over the algebra MATH can define an action MATH of MATH on MATH by MATH, where MATH, MATH, MATH. Put now MATH . It is easy to see that MATH, MATH satisfy REF, and hence we can twist the action MATH by MATH, MATH to define a new action MATH, as in REF. Conside... |
math/0002126 | We use the notations of the proof of REF . There we established that MATH. We need only to verify that MATH is well defined on the quotient complex MATH. But since MATH, and MATH is easily seen to be MATH on MATH, the result follows. |
math/0002126 | We have: MATH . Next, if we write MATH we have: MATH . Also, if MATH is compact the algebra MATH has a unit given by the function MATH . Then MATH . Finally, we have MATH . |
math/0002126 | The first identity is clear, the second follows from the fact that MATH is a constant map, taking the value REF. |
math/0002126 | Let us chose another trivialization of the bundle MATH, and let MATH be a transition matrix between the two bases of the fiber MATH. Then we have a new map MATH, related to MATH by MATH . Let MATH denote the action corresponding to the map MATH. Consider now the pull-back MATH as a map MATH, where we consider forms on ... |
math/0002126 | Since MATH is horizontal, it can be written as a sum of the expressions of the form MATH, where MATH is a function, MATH - projection, and MATH is a form on MATH of the type MATH. Then MATH is also of the type MATH. |
math/0002126 | The curvature of the connection MATH is a horizontal form MATH on MATH, given by MATH . Hence MATH has only components of the type MATH and MATH. The statement of the lemma will then follow from the fact that MATH also has only components of the type MATH and MATH. But this follows from REF . |
math/0002127 | By definition, MATH if and only if MATH; this fact combined with the commutation relation MATH establishes the equivalence of MATH and MATH. Furthermore, these two facts show that if MATH is invariant under the action of MATH, then so is MATH, whence MATH is as well. Finally, suppose that MATH is invariant under MATH. ... |
math/0002127 | We prove by contradiction. Let MATH. Hence, MATH for some MATH. Let MATH, for MATH, MATH. Suppose that MATH is not divisible by MATH, and MATH is a set of non-zero measure such that both MATH and MATH are subsets of MATH. Let MATH; we have, MATH a contradiction. If. It suffices to show that MATH for some MATH. Again, l... |
math/0002127 | Now we shall construct a second wavelet set to form an interpolation pair. Take the same multiplicity function, MATH and construct a MATH similar to MATH, except we will take the interval MATH and shift it right MATH. The result is MATH . This new MATH also satisfies REF 's theorem, so there is a wavelet set MATH: MATH... |
math/0002127 | Fix MATH and MATH. We shall first give a brief outline to show that MATH satisfies the consistency REF . For a complete proof of this fact see CITE. Let MATH, then MATH where MATH and MATH, and MATH are in either MATH or MATH. Consistency equation: MATH. A similar argument is used for MATH to prove that MATH satisfies ... |
math/0002127 | The above is one way of constructing MATH and MATH. Now, we will construct a second generalized scaling set MATH and a second wavelet set MATH that have the same multiplicity function by moving the interval MATH to the right MATH. MATH . Consider the wavelet sets MATH and MATH, then we have the following: MATH . This g... |
math/0002127 | We shall prove by contrapositive. Suppose that MATH, then there exists a MATH that is not divisible by MATH. We shall show that there exists a positive integer MATH such that for MATH, MATH, by showing that MATH. Note that MATH converges to MATH in measure. Let MATH be such that MATH. Choose MATH, a set of non-zero mea... |
math/0002129 | Let MATH be a compact NAME. To show that MATH is hereditarily indecomposable we take disjoint closed sets MATH and MATH and open sets MATH and MATH such that MATH and MATH. We must exhibit three closed sets MATH, MATH and MATH such that MATH, MATH, MATH, MATH, MATH and MATH. Choose a continuous function MATH such that ... |
math/0002129 | For each MATH define MATH by MATH . Observe that MATH and that, for each individual MATH, the sets of values MATH and MATH are equal. It follows from this that the coefficients of MATH, MATH, , MATH in MATH and MATH are the same and hence that the polynomials are the same. |
math/0002129 | Let MATH be such a polynomial and assume that there are MATH and MATH such that MATH. Completing the square gives us MATH. Now because MATH we know that MATH so that we can write MATH for some nonnegative MATH. We find that MATH; write MATH and MATH. Observe that for each MATH either MATH or MATH and that MATH. We cove... |
math/0002132 | Let MATH be a lifting of MATH. Ad-MATH is an automorphism of MATH preserving the invariant scalar product and sending MATH to MATH for all MATH. Thus, Ad-MATH for suitable numbers MATH and MATH. |
math/0002132 | It is known that MATH are distinct positive roots and MATH. Hence, the collection MATH does not depend on the reduced presentation. The vector MATH is a singular vector in MATH. If MATH is another reduced presentation, then the vectors MATH and MATH are proportional. Since MATH is a free MATH-module, we have MATH in MA... |
math/0002132 | MATH is a singular vector in MATH. It has to have weight components of the form MATH for suitable MATH and MATH. Since MATH is generic, we have MATH and MATH is of the required form for a suitable MATH. |
math/0002132 | Since MATH, we have that Ad-MATH and Ad-MATH for any MATH orthogonal to MATH. Hence MATH and Ad-MATH. |
math/0002132 | See REF . |
math/0002132 | The first equation holds since MATH commutes with the comultiplication. Now MATH . |
math/0002132 | It is sufficient to check the equation for the residues of both sides at MATH, and for the limit of both sides as MATH. The residue equation MATH is true since the NAME operator commutes with the comultiplication. The limit equation MATH is a corollary of REF . |
math/0002132 | The Corollary follows from the following simple Lemma. For MATH, the operator MATH is the product in a suitable order of all operators MATH with MATH and MATH. |
math/0002132 | If MATH is a reduced presentation, then MATH and MATH . |
math/0002132 | For MATH we have MATH, and REF is equivalent to REF . |
math/0002132 | We have MATH. Hence MATH . The last equality follows from REF . |
math/0002132 | Let MATH be a minuscle dual fundamental weight. We have MATH and MATH according to REF-cocycle property. Now MATH . |
math/0002136 | Due to the symmetric and dual nature of the NAME diagrams, both arguments will be similar, and I only need discuss one. So without loss of generality, let MATH denote the MATH . NAME diagram, consisting of REF row with MATH boxes. We can use the NAME diagram (see REF ) to count the dimension of each of the representati... |
math/0002136 | Both expressions are simplifications of MATH. |
math/0002136 | Both expressions can simplify to MATH. |
math/0002136 | From REF , it is clear that MATH implies MATH. The cited proposition, applied to the present situation, states that if MATH is a minimal idempotent, then MATH is a minimal idempotent of MATH. For the present purposes, it is not necessary to deal with minimal idempotents, but notice that since MATH is REF-dimensional, t... |
math/0002136 | CASE: MATH B. MATH C. MATH D. If MATH, MATH . If MATH, MATH . |
math/0002138 | This proof is adapted from CITE. Let MATH be a continuously differentiable function with compact support such that MATH . Set MATH where MATH where MATH is a infinitely differentiable function with MATH=REF when MATH and MATH when MATH and MATH is a positive real number. The properties of MATH and MATH are the same as ... |
math/0002139 | For notational simplicity, we will do the case of pair correlation (MATH). Our test function MATH can then be written as MATH for some MATH, say MATH supported inside MATH. Let MATH be smooth, compactly supported and such that MATH is constant on MATH, where it equals MATH. Set MATH. For further notational simplicity a... |
math/0002139 | Write MATH for MATH, MATH, MATH. Then MATH . Since MATH, we may count only the contribution of those MATH (MATH distinct) for which all the components MATH are divisible by MATH. There are MATH such MATH-tuples. If MATH then since MATH we have MATH and so we find MATH . Since MATH and MATH with MATH, this gives MATH wi... |
math/0002139 | Without loss of generality we will assume that MATH. For each pair of coprime integers MATH with MATH, denote by MATH the interval MATH . Then MATH is MATH-approximable if and only if it lies in infinitely many of the intervals MATH with MATH. That is for all MATH, MATH lies in MATH . Thus we need to compute the measur... |
math/0002147 | Let MATH, MATH. It is clear that MATH and MATH are disjoint, and MATH has two connected components, one of them contains zero and the other one is not bounded. Thanks to NAME 's theorem, for any MATH there exists a holomorphic function, MATH, (with pole in zero), such that: CASE: MATH on MATH, CASE: MATH on MATH, where... |
math/0002147 | To prove REF notice that REF implies: MATH . Taking into account MATH and MATH, MATH, we have MATH . From MATH and MATH, MATH, we obtain MATH . Using the above three inequalities and REF in page REF we conclude the proof of the first assertion in this proposition. To obtain REF , consider MATH. From REF , there is MATH... |
math/0002152 | REF are immediate. Let us see REF. For MATH, we have MATH for MATH. Thus MATH for MATH. Therefore, MATH . Let us check the fourth property. For MATH, we have MATH . |
math/0002152 | REF are easy to check. The third property is also easy to prove with the aid of REF below. The fourth property follows from the third. |
math/0002152 | Let MATH be some fixed constants. Obviously, MATH. So we only have to show MATH. It is enough to prove that for a fixed collection of numbers MATH satisfying MATH and MATH we have MATH . For any MATH, let MATH be a cube centered at MATH with side length MATH. Then MATH and so MATH. By NAME 's covering theorem, there ex... |
math/0002152 | Let MATH. To see that MATH we set MATH for all cubes MATH. Then REF holds with MATH. Let us check that the second REF is also satisfied. We have to prove that for any two cubes MATH, MATH . Notice that if MATH, then MATH because MATH are doubling. So REF follows if MATH. However, in general, MATH does not imply MATH, a... |
math/0002152 | Assume MATH for simplicity. First we show REF. If MATH, then REF holds with MATH. Moreover, for any cube MATH we have MATH . Therefore, MATH . The second term on the right hand side is estimated as REF in the preceeding lemma. For the first and third terms on the right, we apply REF . So we get, MATH . Thus MATH satisf... |
math/0002152 | First we will show that if MATH for some MATH, then MATH . We will use the characterization of MATH given by REF . REF follows by standard methods. The same proof that shows that MATH when MATH is a doubling measure works here. We omit the details. Let us see how REF follows. For simplicity, we assume MATH. We have to ... |
math/0002152 | Let MATH be the smallest cube concentric with MATH containing MATH and MATH. Since MATH, we have MATH and MATH. Then, MATH . |
math/0002152 | For any function MATH, we set MATH, with MATH and MATH. By REF , MATH. Since MATH and MATH, we have MATH. Thus MATH. |
math/0002152 | We will prove REF for MATH. The proof for other values of MATH is similar. Recall that if REF are satisfied, then REF is also satisfied subsituting MATH by MATH, for any MATH (see REF ). Let MATH. Assume first that MATH is bounded. Let MATH be some fixed cube in MATH. We write MATH. Let MATH be some positive constant w... |
math/0002152 | REF coincide. So we only have to compare REF . For any MATH, the inequality MATH follows from NAME 's inequality. To obtain the converse inequality we will apply NAME. If MATH, then MATH and so MATH. |
math/0002152 | The proofs of REF are similar to the usual proofs for MATH. Let us sketch the proof of the third property, we can follow an argument similar to the one of REF . Given MATH, it is obvious that MATH and MATH. For the converse inequality, given an atomic block MATH with MATH, it is not difficult to see that each function ... |
math/0002152 | By standard arguments, it is enough to show that MATH for any atomic block MATH with MATH, MATH, where the MATH's are functions satisfying REF of atomic block. We write MATH . To estimate the first integral on the right hand side, we take into account that MATH, and by usual arguments we get MATH . On the other hand, f... |
math/0002152 | Following some standard arguments (see CITE, for example), we only need to show that if MATH is an atomic block and MATH, then MATH . Suppose MATH, MATH, where the MATH's are functions satisfying REF of atomic blocks. Then, using MATH, MATH . Since MATH, we have MATH . From REF we get MATH . |
math/0002152 | We have to prove that MATH . We will show that there exists some function MATH such that MATH . Let MATH be some small constant which will be fixed later. There are two possibilities: CASE: There exists some doubling cube MATH such that MATH . CASE: For any doubling cube MATH, REF does not hold. If REF holds and MATH i... |
math/0002152 | This lemma is very similar to REF . We only need to show that if MATH is a MATH-atomic block and MATH, then MATH . Suppose MATH, MATH, where the MATH's are functions satisfying REF of MATH-atomic block. Since MATH, MATH where MATH. As MATH, we have MATH . From REF we get MATH . |
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