paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9903087 | By passing to a cover if necessary, we may assume that MATH admits a lift MATH. The formula follows directly from REF in the proof of REF , since the subspace MATH is trivial in this case. |
math/9903087 | We may assume that MATH is connected and that it admits a lift MATH. We compare the two measures at some point MATH. Again we go back to REF from the proof of REF . Let MATH and MATH. The splitting MATH is both MATH-orthogonal and MATH-orthogonal. By the moment map condition, the restriction of MATH to MATH is given by... |
math/9903087 | We may assume that MATH is the exponential of a Hamiltonian space MATH. The symplectic quotients MATH and MATH coincide. Since MATH, the NAME are related by MATH. Since MATH, we have MATH. Hence, REF follows from the well-known statement for Hamiltonian MATH-manifolds. |
math/9903087 | View MATH as a conjugacy class of MATH for the semi-direct product MATH defined by MATH. The adjoint action of MATH on MATH is given by MATH. The automorphism MATH induces an isometry from the orthogonal complement of MATH onto MATH. Therefore, MATH . |
math/9903087 | In this case, MATH is the MATH eigenspace of MATH. Therefore MATH. |
math/9903087 | By REF the NAME volume is equal to MATH, divided by the Riemannian volume of the diagonal subgroup MATH. Since the induced Riemannian metric of MATH is twice the Riemannian metric of MATH, its volume is MATH. |
math/9903087 | The identification of NAME measures and Riemannian measures follows from REF , since fusion does not change the NAME measure REF . Hence the NAME MATH is the push-forward of the Riemannian measure on MATH under the moment map MATH. Its NAME coefficients are obtained from the calculation, MATH where we have used conjuga... |
math/9903087 | This follows directly from REF , together with REF for the NAME coefficients of convolutions of invariant measures. |
math/9903087 | It suffices to show MATH generated by elements MATH and MATH. Let MATH be a maximal torus of MATH, and MATH a regular element generating MATH. Choose MATH such that MATH is not contained in a proper closed subgroup MATH containing MATH. This is possible since MATH is a regular conjugacy class. Then MATH generate MATH a... |
math/9903087 | Let MATH be the fusion product of MATH copies of MATH, and MATH the moment map. By REF from REF, all components of MATH meet the principal orbit type stratum of the corresponding component of MATH. Let MATH be a fusion product of MATH copies of MATH. The principal stabilizer is equal to MATH on MATH and to MATH on MATH... |
math/9903087 | Write MATH with MATH and MATH. Let MATH be the following product structure on MATH, MATH where MATH is the diagonal embedding. Arguing as in the proof of REF , we have MATH . Integrating over MATH, the terms involving contractions MATH make no contribution, and we obtain MATH . Thus MATH . Since left convolution by MAT... |
math/9903087 | We show that MATH is the exponential (compare REF) of a Hamiltonian MATH-manifold. Let MATH be the standard symplectic structure on MATH. The defining MATH-action on MATH is Hamiltonian, with moment map MATH . To compute MATH we use the formula MATH . Putting MATH, and using that MATH, we obtain MATH. The symplectic fo... |
math/9903087 | Note first that since the component group MATH is abelian, the commutator map takes values in MATH. Let MATH be a generator. We want to show that MATH is onto MATH on the component of MATH, for any MATH. Using MATH, we may assume MATH. If MATH, let MATH be the largest non-negative integer such that MATH. Using MATH we ... |
math/9903087 | Let MATH be a positive NAME chamber. By composing MATH with an inner automorphism, we may assume that MATH takes MATH to itself, and in fact any element of MATH has a unique representative of this form. (That is, MATH is induced from an automorphism of the NAME diagram of MATH.) We have to find a NAME group element MAT... |
math/9903087 | For MATH let MATH denote projection to the MATH-th factor. Given a component MATH it is possible to choose MATH such that the principal stabilizer for MATH equals that of MATH. (The index MATH is arbitrary if MATH; otherwise pick MATH such that MATH.) Thus MATH whenever MATH. It hence suffices to show that for any give... |
math/9903097 | If MATH is defectless, then so is MATH, by the foregoing lemma. Suppose that MATH is a finite immediate algebraic extension. Hence, MATH. Since MATH is a henselian field, there is a unique extension of MATH from MATH to MATH. Since MATH is defectless, we have that MATH, showing that MATH. As every proper immediate exte... |
math/9903097 | By assumption, MATH induces an embedding of MATH in MATH. Further, we know by ramification theory REF that MATH is separable-algebraically closed. Thus, MATH. Using NAME 's Lemma, one shows that the inverse of the isomorphism MATH can be extended from MATH to an embedding of MATH in MATH. Its image is separable-algebra... |
math/9903097 | If MATH is separable and MATH is equivalent to MATH, then also MATH is separable; thus, we can assume that the restriction of MATH to MATH is the identity. Take a finite purely inseparable extension MATH; we have to show that it is linearly disjoint from MATH. As MATH is the identity on MATH and MATH is purely insepara... |
math/9903097 | Choose elements MATH such that the values MATH are rationally independent over MATH and the residues MATH are algebraically independent over MATH. Then by the foregoing lemma, MATH. This proves that MATH and the rational rank of MATH are finite. Therefore, we may choose the elements MATH such that MATH and MATH to obta... |
math/9903097 | The first assertion is well-known. The second assertion follows from the first, keeping in mind that MATH. We leave the straightforward proofs to the reader. |
math/9903097 | Take finitely many polynomials MATH. From REF we know that for all MATH close enough to MATH, the values MATH of the non-zero among the polynomials MATH, MATH, are fixed. Since by REF the set MATH has no maximal element, we can then take MATH so close to MATH that for every fixed MATH, the values of all non-zero elemen... |
math/9903097 | If MATH is not torsion modulo MATH or MATH is transcendental over MATH, then we set MATH. Otherwise, we set MATH if REF of the foregoing lemma holds, and MATH if REF holds (by REF , only one of the two cases can hold at a time). If the polynomial MATH is separable over MATH, then we set MATH. Otherwise, we proceed as f... |
math/9903097 | Since MATH is separable, we can choose a separating transcendence basis MATH of MATH. We set MATH and MATH. We proceed by induction on MATH. If the extension MATH is not immediate, then we choose MATH according to the assertion of the foregoing lemma. Otherwise, we set MATH. Since every extension MATH is separable, we ... |
math/9903097 | By REF , value group and residue field of MATH are finitely generated. We choose MATH such that MATH. Since MATH is a finitely generated separable extension, it is separably generated. Therefore, we can choose MATH such that MATH is separable-algebraic (MATH). Now we can choose some MATH such that MATH. Since MATH is s... |
math/9903097 | We only prove the first assertion; the proof of the second assertion is similar. Take MATH, and choose a finite NAME extension MATH of MATH such that MATH and that MATH is uniformizable with respect to the MATH's. By our assumption on MATH, MATH is algebraic over MATH. Hence by REF of REF , there is a finite subextensi... |
math/9903097 | For the proof, we set MATH. By REF , we know that MATH . Thus, MATH is an isomorphism from MATH onto the multiplicative subgroup of MATH generated by MATH. We denote this group by MATH. Now let MATH. Take MATH to be any of these elements and write MATH with polynomials MATH. Write MATH as sums of monomials (in such a w... |
math/9903097 | Let MATH and write MATH with polynomials MATH, MATH. We apply REF to these finitely many polynomials and choose MATH according to this lemma. Then by REF , for every MATH we can find MATH such that MATH and MATH. Thus, we can write MATH where MATH are polynomials with coefficients in MATH and MATH. Note that also the f... |
math/9903097 | Take any separable-algebraic extension MATH such that MATH lies in the completion of MATH. Further, take MATH. Let MATH be any of these elements. We extend MATH to the algebraic closure of the completion. Since MATH is separable, we can write the minimal polynomial of MATH in the form MATH where MATH are automorphisms ... |
math/9903097 | It suffices to prove the assertion for every finitely generated separable extension MATH within the completion of MATH. As MATH is finitely generated and separable, we can choose a transcendence basis MATH such that MATH is separable-algebraic. By induction on the transcendence degree, using REF and transitivity REF , ... |
math/9903097 | By REF , MATH is henselian generated and there are MATH as in the assertion of that theorem such that MATH. By the foregoing corollary it follows that MATH is uniformizable. By REF , MATH is uniformizable. Now our assertion follows by transitivity. |
math/9903097 | By REF , MATH is henselian generated. That is, there is some MATH such that MATH. Since MATH is separable-algebraically closed, REF shows that REF holds. Therefore, REF shows that MATH is uniformizable. By REF , the same holds for MATH. Hence by transitivity, MATH is uniformizable. If MATH such that MATH has rank REF a... |
math/9903097 | For given MATH, we take MATH to be the normal hull of MATH over MATH; then MATH is a finite NAME extension. Since MATH is assumed to be perfect and MATH to be algebraic, MATH is a separable-algebraic extension. By REF , for MATH we have that MATH. Thus, REF shows that MATH is uniformizable. This proves that MATH has NA... |
math/9903097 | For given MATH, we take MATH to be the normal hull of MATH over MATH. We set MATH. As MATH is zero-dimensional, we obtain that MATH. Hence, MATH is a rational NAME place of MATH. By REF , MATH is henselian generated: there are MATH such that MATH is a maximal set of rationally independent values in MATH, and MATH is co... |
math/9903097 | We choose MATH such that MATH is a transcendence basis of MATH. We take MATH to be the algebraic closure of MATH. We extend MATH to MATH. Then MATH induces an isomorphism on MATH. Passing to an equivalent place if necessary, we can assume that MATH. Hence by REF , MATH has normal-uniformization over MATH. By REF , MATH... |
math/9903097 | We give the proof for NAME. We proceed by induction on the transcendence degree. The case of transcendence degree REF is covered by REF : For given MATH, we take MATH to be the normal hull of MATH over MATH, which is a finite NAME extension of MATH. Then we apply REF to MATH. We observe that since MATH is a torsion gro... |
math/9903097 | By REF , we can choose a separating transcendence basis of MATH which contains elements MATH such that MATH is a maximal set of rationally independent elements in MATH. We set MATH. Then MATH is separable. Further, MATH is a torsion group by REF . By the same theorem, MATH. By assumption, MATH is algebraic over MATH. H... |
math/9903097 | We choose MATH such that MATH is a maximal set of rationally independent values in MATH. We set MATH. Then by REF , MATH and MATH is a torsion group. Hence by REF , MATH has normal-uniformization over MATH. By construction, MATH and MATH satisfy the assumptions of REF . This yields that MATH has normal-uniformization o... |
math/9903097 | Since the rank of MATH is finite (compare REF), we can proceed by induction on this rank. Assume that MATH such that MATH is a place of MATH of rank REF, with MATH possibly trivial. We take MATH such that MATH is a transcendence basis of MATH. Then we set MATH and MATH. Since MATH is algebraic, so is MATH. On the other... |
math/9903100 | We restrict ourselves to a neighborhood of MATH on which our normal bundle is trivial and there we split things into fibre (vertical) and base (horizontal) components, that is, let MATH and MATH where MATH has only horizontal terms, MATH has only fibre terms and MATH has only mixed terms. The forms MATH and MATH have o... |
math/9903100 | The proof will come in two steps, both dependent on NAME 's generalizations of NAME 's work. Fix MATH. First we will use REF 's to find a MATH-invariant vector field MATH on MATH whose zeroes correspond to closed orbits of MATH near MATH. Then we will extend the method of proof of NAME 's second result in CITE to find ... |
math/9903104 | Set MATH and compute MATH. Since MATH, where MATH is the inclusion map of MATH into MATH regarded as a MATH sector, and MATH, we have MATH, and this contains the identity only once. So MATH is an irreducible MATH sector. We can similarly prove that if MATH, then MATH. We next set MATH as a MATH sector, which is also ir... |
math/9903104 | We compute MATH, the dimension of the intertwiner space between MATH and MATH, by using CITE. This number is then equal to MATH . This gives the conclusion. |
math/9903104 | By a similar argument to the proof of the above lemma, we know that MATH is also irreducible. CITE gives MATH which gives the conclusion. |
math/9903104 | MATH and MATH map sectors localized in bounded intervals to soliton sectors localized in right unbounded and left unbounded half-lines, respectively. Hence MATH is localized in a bounded interval. By the above corollary, we may assume that MATH for some MATH, hence MATH must have trivial monodromy with MATH, that is, M... |
math/9903104 | We compute MATH . The only sector which can be contained in MATH and MATH is the identity by the above proposition. So the above number is MATH. Since the square sums of the statistical dimensions for MATH and MATH are the same, it completes the proof. |
math/9903104 | By a direct computation. |
math/9903105 | Represent the metric MATH and MATH with respect to a local frame of the (same) bundle MATH as local functions MATH and MATH. By the quantum REF we obtain for the NAME form MATH . Hence MATH or equivalently MATH is a locally defined harmonic function. But the quotient of the two metrics is a globally defined function. H... |
math/9903105 | That the map is well-defined we showed above. That it is an embedding follows from the observation that REF is up to complex conjugation nothing else as the NAME embedding with respect to the very ample line bundle MATH. |
math/9903105 | Let MATH, respectively, MATH, MATH. From REF it follows MATH and MATH . Hence, MATH is a zero of the section MATH if and only if MATH. This shows the claim. |
math/9903105 | We start from REF and divide the vectors MATH and MATH on the left hand side by their first components. This can be done because we are on MATH. We obtain MATH . Here MATH is the normalized representative which has first component REF. Using MATH and REF we obtain MATH . |
math/9903105 | First consider the section MATH. In this case MATH. Note that in view of REF it is enough to show that MATH. But by REF, hence MATH . Now take a general MATH. Recall that the complement of a zero-set of a REF is always a dense open subset. Hence the same is true for finite intersections of such sets. On the dense open ... |
math/9903105 | REF is immediate from REF follows from REF using REF . REF follows from REF , respectively, REF by taking the squared modulus. Note that MATH independently on the section MATH chosen. |
math/9903105 | Using REF we see that from REF the first equality follows. Now using REF we obtain the last equality. Note that as MATH any homogeneous representative can be chosen (but then it has be kept fixed). The ambiguity will cancel in this combination. The invariance under cyclic permutations is clear. |
math/9903108 | Let MATH and MATH be the link diagram as shown below where they are the same outside this figure. (This figure has already appeared in NAME 's paper CITE.) We also let MATH be the singular link diagram which is the same as MATH and MATH outside the figure. MATH . We assume that MATH sends MATH and MATH to MATH and MATH... |
math/9903108 | We assume that the chord diagrams appearing in the lemma have MATH double points outside the regions described in the pictures. From the well-known relation for the potential function (NAME 's first identity, which is the first relation of CITE) MATH we have MATH . Note that this also holds for singular links and recal... |
math/9903108 | We put MATH in the relation REF and connect these two arcs as follows. MATH . Then applying the relation REF , we have MATH . Now using the relations REF , we have MATH . So the required formula follows. Similarly the following connection shows the second formula. MATH . |
math/9903108 | We proceed by induction on the number of circles in the support of a chord diagram. If there is only one circle, then we use REF to change the chord diagram into a diagram without chords. Then we apply REF or REF to evaluate the diagram. Suppose that we are given a chord diagram MATH with support MATH (MATH). We first ... |
math/9903113 | MATH . |
math/9903113 | If MATH then MATH must be a scalar and the YBE trivially holds. So assume that MATH. By applying both sides of the equation MATH to MATH and comparing coefficients one sees that a homogeneous operator MATH will satisfy the NAME equation if and only if MATH for all MATH, MATH, MATH, MATH and MATH. Equivalently, MATH wil... |
math/9903113 | First note that because MATH, MATH and MATH have generating functions of the desired form, so does any linear combination. Hence it remains to show that these are the only possibilities. Let MATH and suppose that MATH is a polynomial of degree less than or equal to MATH for all MATH. It is easily checked that if MATH t... |
math/9903113 | We need to determine which pairs of functions of the form MATH satisfy the equations CASE: MATH CASE: MATH . The first equation is satisfied if MATH . On the other hand MATH cannot have any non-zero roots. For if MATH, then MATH which is impossible if MATH. So the above are indeed the only possibilities for MATH. Now s... |
math/9903113 | Clearly, MATH and MATH . Now MATH commutes with MATH because MATH is a comodule morphism and MATH commutes with MATH by the hypothesis. Using these facts we can see that MATH and MATH . Hence MATH . Since MATH is a map from MATH to MATH, it therefore satisfies the usual NAME equation. |
math/9903113 | We prove that MATH. In matrix form this is equivalent to MATH . From REF we have that MATH . Note also that (abreviating MATH by MATH) MATH . Hence MATH as required. |
math/9903113 | The result follows from REF . |
math/9903117 | If either MATH or MATH, it is clear. Suppose MATH and MATH. Let MATH be a basis of MATH and MATH be a basis of MATH. For any MATH, let MATH be the linear homomorphism defined by MATH for MATH. Let MATH be such that MATH and MATH for MATH and let MATH be such that MATH. Then MATH. From this REF follows immediately. |
math/9903117 | We first review a well known fact. Let MATH be a MATH-graded vector space. Endow MATH with a topology by using MATH for MATH as a base of open sets in MATH. MATH is a NAME topological vector space because MATH. It is well known that MATH is the completion of the topological vector space MATH defined above. In view of R... |
math/9903117 | With the element MATH of MATH, it follows that any MATH-submodule of MATH is automatically graded. Since for any MATH, MATH is an irreducible MATH-module and MATH it follows that MATH is an irreducible MATH-graded MATH-module. |
math/9903117 | We only need to prove that MATH if MATH as a MATH-module. By the universal property of induced modules, MATH and MATH are natural quotient MATH-modules of MATH and MATH by MATH and MATH, respectively. Since MATH and MATH are equivalent MATH-modules, MATH and MATH are equivalent MATH-graded MATH-modules. Then it follows... |
math/9903117 | Let us assume REF . CASE: Let MATH be any MATH-graded continuous MATH-module such that MATH is an irreducible MATH-module and MATH. Then MATH is an irreducible MATH-module. Since MATH is finite-dimensional and MATH is an irreducible MATH-module, MATH is finite-dimensional. Then being a central element of MATH, MATH act... |
math/9903117 | First, we have MATH . By REF , for MATH, MATH . Let MATH be the projection of MATH onto MATH. (This is an algebra homomorphism.) Now we calculate the kernel of MATH. By definition, MATH if and only MATH, which is equivalent to that MATH. Then it follows from REF that MATH. Clearly, MATH. Thus, MATH. Therefore, MATH giv... |
math/9903117 | Set MATH . Then MATH is a MATH-submodule of MATH. Since MATH, MATH. By the irreducibility of MATH, we have MATH, hence MATH for MATH. By the irreducibility of MATH again, we have MATH for any MATH. Thus MATH for any MATH if MATH. Consequently, MATH is an irreducible MATH-module and MATH for any MATH if MATH. If MATH fo... |
math/9903117 | REF directly follows from REF . Since MATH is an ideal of MATH, it follows from REF that MATH is an ideal of MATH. Since MATH for MATH, MATH is an ideal of MATH. If MATH, there is nothing to prove. Suppose MATH. Since MATH, MATH, hence MATH because MATH is simple. |
math/9903117 | Let MATH. As usual, we shall use MATH for MATH. Recall that MATH. Let MATH be the projection of MATH in the subspace MATH for MATH. Now we are going to find a sequence of elements MATH of MATH such that MATH for MATH. Then we will have MATH and MATH . Fact: for any MATH, there exists MATH such that MATH. Since MATH (by... |
math/9903117 | If MATH acts irreducibly on MATH, by REF MATH. Now it suffices to prove that MATH is an irreducible MATH-module. Since MATH is a (left) ideal, for every MATH, MATH is a MATH-submodule of MATH, hence MATH if MATH. Now it suffices to prove that MATH for every nonzero MATH. Since MATH is a (right) ideal, MATH is a MATH-su... |
math/9903117 | CASE: For any fixed MATH, let MATH be any two linearly independent homogeneous elements of degrees MATH and MATH, respectively. We decompose MATH as MATH where MATH is a MATH-submodule of MATH. Define MATH by MATH for MATH. It follows directly that MATH for all MATH. From REF we have MATH if MATH. Thus MATH, hence MATH... |
math/9903117 | Clearly, MATH . Conversely, let MATH. Then MATH for MATH, that is, MATH . For each fixed MATH, let MATH be a basis for MATH and let MATH be any nonzero vector in MATH. Define linear endomorphisms MATH of MATH by MATH for MATH. Since MATH, each MATH is a MATH-endomorphism. By NAME 's Lemma we have MATH for MATH. Thus MA... |
math/9903117 | View MATH as a graded subspace of MATH. Then from the discussion right before this lemma MATH is a topological subspace of MATH. Since MATH and MATH are the completions of MATH and MATH, respectively, MATH is a topological subspace of MATH. |
math/9903117 | In view of REF , we need to prove that MATH is a topological subspace of MATH. Since MATH is the completion of MATH and MATH is a topological subspace of MATH (by REF ), it suffices to prove that MATH is a topological subspace of MATH. This is true because for MATH, MATH with respect to this very grading is a graded su... |
math/9903117 | Since MATH is an irreducible MATH-module, MATH is an irreducible MATH-module. By REF , MATH. Since MATH is a MATH-module, MATH. Thus MATH . This completes the proof. |
math/9903118 | This is a consequence of the continuity of the map MATH. |
math/9903118 | This proof is inspired by the proof of REF , and generalizes its argument. Let MATH be the integral kernel of MATH with respect to Riemannian measure on MATH, where MATH, MATH for MATH and MATH (upper half-space model). Then MATH where MATH denotes the diagonal. This may be checked by explicit computation of the resolv... |
math/9903120 | Clearly MATH is a unit in MATH. Let MATH so that MATH. Also any element of MATH has the form MATH, where MATH and MATH. Thus MATH is (left and right) NAME in MATH and MATH. By CITE there is an antiautomorphism of MATH interchanging MATH and MATH and it follows that MATH is NAME in MATH. |
math/9903120 | It is well known that any unit in MATH has the form MATH with MATH and MATH, CITE. The result follows since MATH is not a unit in MATH unless MATH. |
math/9903120 | Since MATH, MATH is a torsion free left MATH-module. It is easy to show that MATH is simple. Conversely, assume that MATH is a finite dimensional MATH-torsion free simple left MATH-module. From the Euclidean algorithm and the fact that MATH is MATH-torsion free it is easy to conclude that MATH. Hence we can identify MA... |
math/9903120 | Let MATH be any finite dimensional simple MATH-module. The MATH-module MATH is either simple or zero, by CITE. If MATH, MATH is MATH-torsion. Let MATH. Then MATH and as MATH, for all MATH, we have MATH. Hence there is MATH such that MATH for MATH. Now MATH is a highest weight module of weight MATH, hence a homomorphic ... |
math/9903120 | Necessary and sufficient conditions for the weight spaces of MATH to be one dimensional are given in CITE, and in particular these conditions hold in REF when MATH. The statement about NAME modules now follows from CITE. By REF any finite dimensional MATH-module is MATH-torsion, and so has the form MATH by the proof of... |
math/9903120 | By CITE, MATH where MATH, MATH, MATH, MATH and MATH. It follows that MATH if and only if MATH . If MATH, then since MATH, it follows that MATH or equivalently MATH so MATH, a contradiction. This proves MATH and MATH follows by multiplying by MATH. |
math/9903120 | Let MATH be a down-up algebra as in the statement of the proposition. By REF we have that the only finite dimensional simple MATH-modules are d-torsion and of the form MATH. By construction, the dimension of MATH is MATH if and only if MATH is minimal with MATH, and the result follows. |
math/9903120 | Immediate from the Proposition and the Lemma. |
math/9903120 | Let MATH. By MATH we have MATH and so MATH. Suppose MATH for all MATH. Then MATH for all MATH. If MATH, then since MATH, MATH and MATH we have MATH . It follows that MATH. The other inclusion is proved in a similar way. |
math/9903120 | Follows easily from REF . |
math/9903120 | Consider the chain of MATH-submodule of MATH, MATH. Choose MATH such that MATH for all MATH. If MATH then there are MATH such that MATH and this implies that MATH. |
math/9903120 | For each MATH write MATH and MATH. The result follows since MATH, MATH are prime ideals of MATH. |
math/9903120 | Obviously if MATH has a finite filtration whose modules are finite dimensional highest weight modules then MATH is MATH-torsion and MATH-torsion. Let MATH be a finitely generated left MATH-module such that MATH. Then MATH is MATH-torsion and MATH-torsion or equivalently, MATH. As MATH is a finitely generated left MATH-... |
math/9903120 | Straightforward . |
math/9903120 | If MATH and MATH have type REF then using the decomposition in REF, we see that MATH generates the ideal MATH. Hence REF follows from the remarks before REF . The proofs of the remaining statements are similar. |
math/9903120 | Since MATH and MATH, relations MATH imply that MATH. On the other hand if MATH then MATH in the commutative ring MATH, so it follows that MATH. |
math/9903120 | This is proved by computation using the decomposition in REF. |
math/9903120 | In type REF this follows directly from REF . Suppose that MATH have type REF and that MATH. By REF and the fact that MATH has only trivial units we have MATH for some MATH, where MATH is defined in a similar manner to MATH. Applying MATH to the equation MATH, we get by REF , MATH. Similarly MATH. |
math/9903120 | We prove only REF . The proof of REF is similar. The fact that MATH is normal in MATH and MATH normal MATH follows from REF. A short computation shows that MATH. Using the decomposition of MATH as a MATH-graded ring in REF we see that MATH is free as a left and right MATH-module with basis MATH, MATH. From REF, we obta... |
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