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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 the skew-adjoint operator MATH. Hence, its top exterior power is MATH. Together with REF this proves the Proposition. |
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 conjugation invariance and the orthogonality relations for irreducible characters. The formula for the NAME of MATH follows from REF. |
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 and therefore MATH generate MATH. |
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. An open dense subset of the symplectic quotient MATH fibers over MATH with fiber MATH. Hence the volumes are related by a factor, MATH. To compute MATH, we calculate the NAME coefficients MATH for MATH. Since MATH vanishes on MATH for MATH, we only need to consider MATH. We have MATH. The NAME coefficients MATH are given by REF as usual. We claim that MATH is given by a similar formula but with MATH in place of MATH: MATH . Indeed, MATH by the formula for MATH, since MATH restricts to an irreducible character of MATH. It follows that MATH . These NAME coefficients contribute with a factor 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 MATH is a differential operator, the description of the support and singular support of MATH follows. The interpretation of MATH at the group unit MATH follows as in the proof of REF by a reduction to the Hamiltonian case, using REF . |
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 form MATH can be written in the form MATH . The term MATH appearing in the formulas for both MATH and MATH is the pull-back of the normalized volume form MATH on MATH under the quotient map MATH. Thus MATH . We want to identify this expression with MATH. Introduce polar coordinates by the map MATH where MATH is viewed as the unit sphere in MATH. Then the form MATH is a multiple of the normalized volume form on MATH: MATH . Using again MATH, it follows that MATH. This shows that MATH is the exponential of the subset MATH of the Hamiltonian MATH-manifold MATH. Multiplying the moment map MATH by the central element MATH one obtains MATH. The volume form MATH is obtained from REF , since MATH. |
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 see that MATH is onto MATH on the component of MATH if and only it is onto MATH on the component of MATH. Let MATH. Iterating this procedure, we obtain a finite sequence MATH such that MATH is onto MATH on the component of MATH. The sequence terminates when MATH. We have thus reduced the problem to the case MATH. Since MATH is equivariant, it suffices to show that some maximal torus MATH is in the image for this component. REF below shows that one can find MATH such that MATH preserves MATH and MATH has no eigenvalue equal to MATH. Given MATH, choose MATH with MATH. Then MATH satisfies MATH . |
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 MATH such that MATH has no eigenvalue equal to REF. Then any MATH preserving MATH and acting as MATH on MATH will have the required property. If MATH, one can take MATH to be any NAME element, compare CITE. Suppose MATH. By decomposing MATH with respect to the action of MATH, we may assume that MATH contains no proper MATH-invariant ideal. We consider two cases: CASE: Assume that MATH is not simple. Let MATH be a simple ideal of MATH. Then MATH, where MATH is the order of MATH. Let MATH be a NAME element of MATH. Then MATH has the required property. CASE: We are left with the case of MATH simple. Recall that MATH for the NAME algebras MATH and MATH, MATH for MATH, and MATH in all other cases. For the NAME algebras MATH and MATH, and MATH, one can take MATH to be the longest element of the NAME group. Indeed MATH in both cases (see CITE, planche I REF , planche V REF ). For the NAME algebra MATH (MATH), consider the standard presentation of the root system as the set of lattice vectors in MATH of length MATH. In the standard basis MATH of MATH, a system of simple roots is given by MATH. The automorphism MATH exchanges MATH and MATH, hence MATH is the linear map changing the sign of the last coordinate in MATH. Take MATH to be a cyclic permutation of the coordinates. Then MATH has no fixed vector. It remains to consider the case of MATH. If MATH is induced by a diagram automorphism of order MATH, the map MATH has eigenvalues the third roots of unity. The longest NAME group element MATH acts as MATH on MATH, hence MATH has no eigenvalue equal to MATH. Finally, if MATH is induced by the automorphism exchanging MATH, we can take MATH to be cyclic permutation of the coordinates as above, the other diagram automorphisms of order MATH are obtained from this by conjugation with a third order automorphism. |
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 given MATH, MATH, the fiber MATH meets each component of MATH. Interpret MATH as a fusion product of MATH copies of the double MATH, with MATH as its moment map. By REF , the restriction of MATH to any component MATH is surjective, and by another application of this Lemma each fiber of MATH meets MATH. If MATH is REF-connected, this completes the proof since each MATH contains a unique component of MATH. (Recall CITE that the fibers of the moment map for a compact, connected, group-valued Hamiltonian space are connected, provided the group is REF-connected.) If MATH is not REF-connected, we construct a finite cover MATH, with identity component MATH the universal cover of MATH: Choose MATH such that MATH preserves a maximal abelian subalgebra MATH. Any such MATH has finite order MATH, hence it defines an action of MATH on MATH by automorphisms. We define MATH, with covering map MATH where MATH is the image of MATH. Let MATH be the projection. The product of commutators, MATH is the moment map for the MATH-action. The fiber MATH is covered by the union of MATH, as MATH ranges over pre-images of MATH. Again, REF shows that the restriction of MATH to any component MATH is surjective, and MATH meets MATH. This proves the Lemma since MATH is connected. |
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 extension would contain a proper finite immediate extension, it follows that MATH does not admit any proper immediate algebraic extension. |
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-algebraically closed and contains MATH. Hence, MATH. Further, MATH contains MATH, which by REF contains MATH. As MATH is algebraic, so is MATH. Therefore, if MATH is algebraic, then MATH is algebraic over MATH and hence separable-algebraically closed. Since MATH, it follows that MATH. |
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 inseparable, MATH is also the identity on MATH. Hence, MATH. It follows that MATH where the last equality holds since MATH is separable by assumption. Hence, equality must hold everywhere, showing that MATH, that is, MATH is linearly disjoint from MATH. |
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 obtain REF . Assume that this is an equality. This means that for MATH, the extension MATH is algebraic. Since MATH is finitely generated, it follows that MATH is finite. By the fundamental inequality, this yields that MATH and MATH are finite extensions. Since already MATH and MATH are finitely generated by the foregoing theorem, it follows that also MATH and MATH are finitely generated. |
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 elements MATH, MATH, are distinct. Having picked such an element MATH, we choose an element MATH such that MATH. Then REF holds by the ultrametric triangle law. |
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 follows. Set MATH; this must be divisible by the characteristic MATH of MATH. In REF is torsion modulo MATH by assumption, and it follows that for MATH, the value MATH is still not torsion modulo MATH. As MATH is not divisible by MATH, we find that MATH is separable over MATH. In REF is torsion modulo MATH by REF , and it follows that there is some MATH such that MATH. Then for MATH we have that MATH, and that MATH is separable over MATH. |
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 obtain a separating transcendence basis MATH of MATH. We set MATH and MATH. As MATH and in view of the fact that MATH we find that for precisely MATH many values of MATH, MATH will be rationally independent modulo MATH. Collecting all of these MATH and calling them MATH we thus obtain that MATH is a maximal set of elements in MATH rationally independent modulo MATH. Similarly, we find that for precisely MATH many values of MATH, the residues MATH will be transcendental over MATH. Collecting all of these MATH and calling them MATH we thus obtain that MATH form a transcendence basis of MATH. |
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 separable-algebraic over MATH, by NAME 's Lemma there exists an element MATH in the henselization of MATH such that MATH and that the reduction of the minimal polynomial of MATH over MATH is the minimal polynomial of MATH over MATH. Then MATH lies in the absolute inertia field of MATH. Now the field MATH has the same value group and residue field as MATH, and it is contained in the henselization MATH of MATH. As henselizations are immediate extensions and the henselization MATH of MATH can be chosen inside of MATH, we obtain an immediate algebraic extension MATH. On the other hand, we observe that MATH is a defectless field since MATH is trivial on MATH. By construction, MATH is without transcendence defect, and the same is true for MATH since this property is preserved by algebraic extensions. Hence we know from REF that MATH is a defectless field. Now REF shows that the extension MATH must be trivial. Therefore, MATH is contained in MATH, which in turn is a subfield of the absolute inertia field of MATH. This shows that MATH is inertially generated. If MATH, then we do not need the elements MATH and MATH. If MATH is a rational function field, then we can choose MATH such that MATH, and we do not need MATH. In both cases, we find that MATH, which yields that MATH is henselian generated. |
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 subextension MATH of MATH such that MATH is uniformizable with respect to the MATH's. Take MATH to include generators of MATH over MATH and the uniformization coefficients of MATH in MATH. By hypothesis, there is a finite NAME extension MATH of MATH such that MATH and MATH is uniformizable with respect to MATH. Then MATH is a finite NAME extension of MATH. By construction, we have that MATH is a finite extension of MATH and that MATH. Hence by REF , MATH is uniformizable with respect to MATH. |
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 way that in either polynomial, two different monomials differ in at least one exponent). Then by REF , the value of MATH is equal to the least of the values of its monomials, say, to the one of the first. Similarly for MATH. So we can write MATH where the denominator has value REF and the summands appearing in it all have value MATH. Since MATH, also the numerator has value MATH, and the same must thus be true for all its summands. The only obstruction is that some of the MATH's may appear with negative exponents in some summands, and that, if MATH is non-trivial on MATH, some of the MATH or MATH may have negative value. We collect all summands of the form MATH which appear in the numerator or denominator of REF , and all products of the form MATH which appear in these summands. We do the same for all elements MATH. In this way, we obtain finitely many elements MATH, all of them having non-negative value, and corresponding elements MATH. We note that MATH, and as MATH is a convex subgroup of MATH and MATH is non-negative, it follows that MATH is non-negative in MATH, for the induced order. After adding some suitably chosen elements of the form MATH or MATH (depending on whether MATH or MATH is positive) to the MATH's if necessary, we obtain elements MATH which also generate MATH. At this point, we apply REF to the non-negative values MATH. Pulling the result back through the above isomorphism, we find generators MATH of MATH for which the values MATH in MATH are positive, and such that every of the MATH's can be written as a (unique) product of the MATH with non-negative exponents. Let MATH be the minimum of the values of the coefficients MATH appearing in the monomials MATH. We take MATH such that MATH. If MATH, which in particular is the case if MATH is trivial on MATH, then we can take MATH. Now we set MATH. It follows that MATH. Since the values MATH in MATH were positive, all values MATH are positive. This remains true if MATH is replaced by any MATH such that MATH. As we can read off from REF , every MATH can be written as MATH, where MATH lie in MATH, with MATH. We set MATH, MATH for MATH and MATH for MATH. Next, we set MATH and put MATH and MATH, and we are done. |
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 first fraction is an element of MATH since its value is equal to MATH and MATH by assumption. Now we set MATH, MATH, MATH and MATH for MATH, and we are done. |
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 in Aut-MATH and all MATH are distinct from MATH. Since MATH lies in the completion of MATH, it follows that there is some MATH such that MATH for all MATH, and that MATH. Since MATH, the latter implies that MATH. On the other hand, MATH and MATH, so we can choose MATH such that MATH. Thus, MATH and MATH. Therefore, the reduction MATH of the minimal polynomial MATH of MATH over MATH is the polynomial MATH which has MATH as a simple root. We set MATH. If MATH was MATH then we set MATH, MATH, MATH, MATH, MATH, MATH, and we are done. |
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 , we find that MATH is uniformizable. By the foregoing lemma, the same holds for MATH. Now our assertion follows by transitivity. |
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 and MATH, then we employ REF in the place of REF . This is possible since MATH implies that MATH and thus, being equal to the residue field of a separable-algebraically closed field, MATH is itself separable-algebraically closed and hence henselian under every valuation. |
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 NAME over MATH. On the other hand, MATH is trivial on MATH. Hence, REF tells us that MATH has NAME over MATH. Now our assertion follows by transitivity. The proof for normal-uniformization is similar. |
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 contained in the henselization of MATH, where MATH. But MATH by REF , and MATH is henselian. Hence by REF , MATH is uniformizable. Since MATH is trivial on MATH and since MATH, we know that MATH is a convex subgroup of MATH, and that MATH. Consequently, our choice of the MATH's yields that MATH form a maximal set of rationally independent elements in MATH modulo MATH. Hence by REF , MATH is uniformizable. By transitivity, the same holds for MATH. This proves our assertion. |
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 has normal-uniformization over MATH. Now our assertion follows by transitivity. |
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 group and MATH is algebraic by hypothesis, REF implies that the extension MATH and hence also its subextension MATH are immediate. For the case of MATH with MATH non-trivial, it also implies that MATH and MATH. Hence, our assumption that MATH is algebraic implies that MATH. So let us now assume that MATH and that our assertion is true for transcendence degree MATH. We take a separating transcendence basis MATH of MATH. Then we pick a subset MATH such that MATH for MATH. It follows that MATH is a separable function field of transcendence degree MATH and that MATH is a separable function field of transcendence degree MATH. As MATH is a subextension of MATH, MATH is a torsion group, MATH is algebraic, and MATH will have rank at most REF if MATH has rank REF. But the fact that MATH is a torsion group implies that MATH has the same rank as MATH. This shows that MATH has rank REF if MATH has rank REF. Similarly, if MATH with MATH and MATH has rank REF, then MATH and MATH will have rank at most REF. But the fact that MATH is a torsion group also implies that MATH has the same rank as MATH. Hence in this case, MATH has rank REF if MATH has rank REF. We have shown that also MATH satisfies the assumptions of our proposition. As MATH, our induction hypothesis yields that MATH has NAME over MATH and that MATH has NAME over MATH. Hence by transitivity, MATH has NAME over MATH. The proof for normal-uniformization is similar: instead of separable-algebraic closures we use algebraic closures. |
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. Hence by REF , the extension MATH has NAME over MATH. By REF , MATH has NAME over MATH. Now our assertion follows by transitivity. |
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 over MATH. Now our assertion follows by transitivity. |
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 hand, MATH is algebraic over MATH by construction, and therefore, MATH is equal to the algebraic closure of MATH. As MATH, this shows that MATH. Since MATH is the identity on MATH and MATH are algebraically independent over MATH, it induces an isomorphism on MATH and hence also on MATH. Passing to an equivalent place if necessary, we can assume that MATH is a place of MATH. Since MATH, REF now shows that MATH has normal-uniformization over MATH, and hence also over MATH. As the rank of MATH is equal to the rank of MATH and thus smaller than the rank of MATH, our induction hypothesis (or REF , if MATH is trivial) yields that MATH has normal-uniformization over MATH. Now our assertion follows by transitivity. |
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 orders of MATH and REF respectively. Furthermore, by our choice of MATH as the normal bundle, we have MATH and MATH so that MATH is of (at most) order REF in MATH. Locally, NAME 's REF are now of the form MATH . Comparing relative orders, the first equation implies that MATH is of order MATH with respect to MATH. The second equation then implies that MATH is of order REF. We also know that, as MATH, MATH approaches MATH. Hence, in this limit, MATH approaches the fibrewise vector field MATH given by MATH . Indeed, this equation defines the limiting vector field MATH globally and we note that the convergence is MATH for any MATH. To show that the flow of MATH is fibrewise quasiperiodic, we recall that in each fibre MATH there are coordinates MATH such that MATH and MATH . So, the flow of MATH in each fibre is given by the sets of O.D.E's MATH for MATH . This flow is clearly quasiperiodic and the proof of the lemma is done. |
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 a MATH-invariant function MATH on MATH whose critical points correspond directly to zeroes of MATH. Combining these results we get the following as desired: MATH CASE: In CITE, the following local decomposition is proved. Let MATH be a MATH vector field on a manifold MATH that generates a flow MATH with a compact nondegenerate periodic submanifold MATH of period REF. Suppose that MATH is a MATH vector field MATH-close to MATH in some neighborhood of MATH. Then there exists a MATH-invariant function MATH (close to REF), and unique MATH-small REF and V, of MATH and MATH respectively, such that for MATH and MATH defined by MATH we have the following decomposition: MATH . The following conditions also hold: CASE: MATH. CASE: MATH. CASE: MATH with respect to a MATH-invariant metric on MATH, MATH. We now outline a simple proof of this proposition for the case MATH. Consider the following subsets of MATH, the NAME space of MATH vector fields on MATH. Let MATH and MATH . Clearly MATH is a closed subspace of MATH, and MATH is also a NAME space given the norm MATH . Let the map MATH be defined by MATH where MATH and P REF are as above. Note that MATH. One can show quite easily that MATH is an isomorphism. So, by the inverse function theorem for NAME spaces, for all MATH sufficienly MATH-close to MATH we get unique MATH and MATH such that MATH . Given this decomposition, the existence of the MATH-invariant function MATH, that ensures MATH, can be established using the implicit function theorem as it is in CITE. With a little extra effort this proof yields a MATH version of a result of NAME that establishes the existence of a closed orbit for vector fields MATH-close to the NAME field on MATH, CITE. It is a simplification of the proof by NAME, CITE, which is possible only because of its stronger closeness hypothesis, (see CITE). REF yields a decomposition of MATH near MATH, that is, there is a unique embedding MATH of MATH, and a MATH-invariant vector field MATH on MATH such that on MATH we have MATH . This is a useful splitting because if MATH then MATH for all MATH, since MATH. Thus, we get MATH which means that MATH is a closed orbit of MATH on MATH . Since MATH is close to REF, MATH is also a closed orbit of MATH and we get a direct correspondence between zeroes of MATH and periodic orbits of MATH on MATH. In looking to apply this decomposition to our situation we note, for small MATH, that MATH is MATH-close to MATH for any MATH. Consequently there exists a diffeomorphism MATH such that MATH is close to the identity. Letting MATH, MATH, MATH, and MATH, REF gives unique MATH and MATH such that MATH . Now if MATH for MATH we get MATH . This implies that MATH is a closed orbit of MATH on MATH and we get the first desired correspondence between zeroes of a MATH-invariant vector field MATH on MATH and closed orbits of MATH near MATH. CASE: Expanding on the work of NAME we constuct a MATH-invariant function MATH whose critical points correspond to zeroes of MATH. Define the following REF-form on MATH where MATH. As suggested by the notation, MATH is exact. From the definition of MATH we have MATH . The map MATH is homotopic to the identity map on MATH. By the homotopy invariance of the NAME cohomology we get MATH and MATH . The second term of REF is exact since MATH . Here the term involving MATH vanishes after integrating over the closed curve. We still must show that the first term in REF is exact. Overall, we have the following situation MATH and MATH where MATH, MATH and MATH are the obvious projections and MATH is inclusion. Again by homotopy invariance, we have MATH for some REF-form MATH on MATH. We also know that MATH . Hence MATH and we see that MATH for some REF-form MATH on MATH. This means we can write MATH where MATH satisfies MATH. Consequently MATH . This completes the proof of the claim. MATH is a MATH-invariant form. MATH . Using the definition of MATH we have MATH and MATH . Hence MATH . The previous two claims establish the existence of a MATH-invariant function MATH whose differential is given by REF. It is worthwhile to note here the following two facts. First, our construction of MATH does not depend on the relation of the embeddings MATH and MATH to the vector fields MATH and MATH. In fact, any MATH and MATH will yield a MATH-invariant function MATH in the same manner. This freedom of choice will be used in what follows. Also, just as we used the exponential map to obtain the diffeomorphism MATH from the section MATH, we can consider MATH as being defined by a small smooth section MATH of MATH, that is, MATH . We still need to show that the critical points of MATH correspond to zeroes of MATH. Using our expression for MATH and the decomposition REF we get MATH . Substituting this into the definition of MATH yields MATH . The term involving MATH vanishes since MATH on MATH and we are left with MATH . Since MATH we know that MATH is a REF-form on MATH where MATH is taken with respect to a fixed MATH-invariant metric MATH. Fixing a MATH we can consider REF as defining a family of linear maps MATH parameterized by the embeddings MATH and MATH and hence the sections MATH and MATH, (see REF ). Here MATH is the dual space to MATH at MATH and MATH, as defined in REF. We know that MATH and MATH are small in their MATH-norms. If we can show that MATH is nondegenerate then for our small sections, MATH and MATH, MATH will also be nondegenerate. This would imply that MATH and we would have the desired correspondence between the zeroes of MATH and the critical points of the MATH-invariant function MATH on MATH. MATH is nondegenerate. With MATH and MATH we have MATH and MATH is given by MATH . REF-form MATH is almost conserved by MATH since MATH is close to the flow of MATH. So we have MATH and MATH . All that is required now is to show that MATH is nondegenerate. The kernel of MATH restricted to MATH is MATH the projection of MATH onto MATH. But MATH is MATH-close to MATH and hence to MATH. The nondegeneracy of MATH on MATH implies that it is also nondegenerate on MATH and both the claim and REF follow. For each MATH we have constructed a MATH-invariant function MATH whose critical submanifolds correspond to distinct periodic trajectories of MATH near MATH. Now, MATH generates a circle action on each MATH without fixed points. This yields MATH as an immediate lower bound for the number of closed orbits on low energy levels. If all the MATH are equal then the circle action generated by MATH on MATH is free. In particular, MATH is a MATH bundle over MATH obtained from a vector bundle. For such bundles we have that MATH as modules over MATH, CITE. Using NAME theory we then get MATH . In fact, even if the MATH aren't equal, and MATH is a twisted projective space bundle, it was shown by NAME in CITE that MATH . Relation REF still holds in this case and so we always have MATH . Summing over the MATH and using the fact that MATH and MATH we get at least MATH distinct closed orbits of MATH and hence MATH. The proof of REF is now complete. |
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 irreducible. We compute MATH which contains the identity only once. So we have MATH. The rest is now easy. |
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, MATH, which in turn gives MATH for all MATH. The non-degeneracy assumption gives MATH as desired. |
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. Hence MATH is a globally defined harmonic function on the compact manifold and hence a constant MATH. This shows the claim. |
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 set MATH we have MATH with MATH a holomorphic function on the intersection. For the pull-backs we obtain MATH . This implies (using REF ) MATH . But note that MATH and MATH. The claim follows from REF . |
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 and MATH to MATH (MATH and MATH may be the same). Then the NAME MATH and MATH are MATH and MATH for some column vectors MATH and a matrix MATH, where MATH is a zero vector of suitable size. We choose MATH to calculate MATH. Putting MATH and MATH, we have MATH for some half integers MATH, some column vectors MATH, and some matrix MATH. (Note that MATH.) Then the NAME potential function of the labelled singular link MATH presented by MATH is given by MATH . Similarly we see that MATH . Here MATH is a singular link diagram with MATH double points presenting MATH and MATH is given as follows. We arrange MATH so that four arcs adjacent to each double points are different after inserting kinks if necessary. Then MATH is of the form MATH where MATH is a MATH matrix of the form MATH . (The author does not know whether the matrix above can be derived from NAME 's free differential calculus applied to MATH.) It is not hard to see that the total degree of MATH is at least MATH putting MATH. Therefore the total degree of MATH is at least MATH. This shows that MATH vanishes if MATH and so MATH is a NAME invariant of type MATH, completing the proof. |
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 recall that we are assuming that there are MATH double points outside the region appearing in the equality above. Taking the total degree MATH part, we have MATH since MATH vanishes if MATH from REF . By the way, we have from the definition MATH and MATH . Therefore the required formula follows from REF . From the fourth relation of CITE we have MATH . Taking the total degree MATH part of both hand sides, we have MATH . So we have MATH . Here we use REF and the fact that MATH vanishes for a split link in the fourth equality. (Proof of REF ) Since MATH (which is the fifth relation of CITE), MATH and we have REF . Here MATH is the trivial knot with label MATH. The relation REF follows from the fact that MATH vanishes for split links (the sixth relation of CITE). We use NAME 's third relation CITE: MATH . We take the total degree MATH part. (Recall that we assume that there are MATH double points outside.) Since MATH for MATH and any singular link MATH with MATH double points, we have MATH . Now since MATH is a type MATH invariant, all its values in the equality above are the same and equal to MATH. So we have MATH . Now we have from the definition of MATH which vanishes from REF and the proof is complete. |
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 look at MATH. If MATH contains no end point of a chord, then MATH from REF . If MATH contains one end point, then from REF we can reduce the number of circles. If MATH contains more than one end point, we use REF . We assume that MATH is labelled MATH in REF . The second term there contains two end points and the others contain one or less. So we can reduce the number of end points. Repeating this process we have chord diagrams with one or no end point, which can be calculated as described above. So the proof is complete. |
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 will satisfy the NAME equation if MATH for all MATH, MATH and MATH. Now MATH . Similarly, MATH . Comparing coefficients then yields the theorem. |
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 then they must be scalar. Embed the rational function field MATH into the NAME power series ring MATH in the usual way. Considering the cases MATH and MATH, MATH yields that MATH and MATH have eventually constant coefficients of opposite sign. Hence, MATH where MATH and MATH. But then the functions MATH are also all polynomial. Hence MATH and MATH must be scalars, as required. |
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 suppose that MATH and let MATH. Then MATH . Using this identity, the second equation follows easily from the first equation. This yields all solutions of the first form. A similar analysis of the case MATH yields all solutions of the second form. The restriction on MATH implies that these are the only possibilities. |
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 REF , for each MATH, MATH is the completion of MATH with the topology on MATH defined by using MATH as basic open neighborhoods of MATH. Now we show that this topology on MATH is the same as the one defined by using MATH for MATH as basic neighborhoods of MATH. If MATH, it is clear. Now we assume MATH with MATH. By REF we have MATH . On the other hand, MATH because MATH for MATH. Thus MATH . Therefore, MATH is the completion of MATH with the given topology. Furthermore, an element MATH of MATH with MATH corresponds to the NAME sequence MATH with MATH. From this, the MATH-graded associative algebra MATH is the same as the MATH-graded associative algebra MATH. |
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 that MATH and MATH are equivalent MATH-modules. |
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 acts as a scalar MATH on MATH. From MATH, we get MATH for MATH. Then MATH (We call MATH the lowest weight of MATH.) Consequently, any submodule of MATH is graded. Let MATH be any nonzero submodule of MATH. We shall prove that MATH, which implies that MATH because MATH. Let MATH be a nonnegative integer such that MATH. Since MATH, by REF MATH, hence MATH . Since MATH is a completely reducible MATH-module, from REF there is an irreducible MATH-submodule MATH of MATH such that MATH. Then MATH because MATH is a nonzero MATH-submodule of MATH and MATH is irreducible. Consequently, MATH. On the other hand, MATH. Thus MATH is an irreducible MATH-module, hence MATH. If MATH, we have MATH. If MATH, we have MATH, so that MATH . Then MATH. Thus MATH. Therefore MATH is irreducible. REF : any MATH-graded continuous MATH-module MATH is a direct sum of irreducible graded MATH-modules. Let MATH be the sum of all irreducible graded submodules of MATH. We must prove that MATH. Since MATH is semisimple, MATH as a MATH-module is completely reducible. By REF the MATH-submodule generated by MATH is completely reducible, so that MATH. Suppose that for some nonnegative integer MATH, MATH for MATH. Write MATH, where MATH is a MATH-submodule of MATH. Then MATH hence MATH . On the other hand, MATH. Thus MATH for MATH, hence MATH. Since MATH is continuous, MATH, so that MATH is a MATH-module. Now we claim that MATH. If MATH, then let MATH be an irreducible (MATH-) MATH-submodule of MATH. Then MATH on MATH for some MATH. If MATH for all MATH, by REF , MATH generates an irreducible graded MATH-submodule of MATH. By the definition of MATH, MATH. It is a contradiction. If MATH for some MATH, then MATH, where MATH is the lowest weight of some irreducible MATH-graded continuous MATH-module because MATH . This contradicts REF . Thus MATH, hence MATH. Therefore MATH by induction. |
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 gives rise to an algebra isomorphism for REF . |
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 for some MATH, then MATH . Therefore, MATH and MATH are inequivalent MATH-modules. |
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 REF ), there are elements MATH such that MATH . There exist MATH such that MATH because MATH (by REF ). Now set MATH . Then MATH. If MATH, by taking MATH, then setting MATH, we have MATH . If MATH, we have MATH, hence MATH. In this case, we simply take MATH. Suppose that we have already found MATH such that REF hold for MATH. Since MATH by the previous fact, there are MATH such that MATH . Set MATH . Then MATH . That is, MATH . Since MATH, MATH (by REF ). Thus MATH . In this way we obtain a sequence of elements MATH with the desired properties. This completes the proof. |
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-submodule of MATH, hence MATH or MATH. We must have MATH because MATH is faithful and MATH. Thus MATH if MATH. |
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. Clearly, MATH. Therefore MATH is an irreducible graded MATH-module. CASE: For MATH, let MATH where MATH is a MATH-submodule of MATH. Define MATH by MATH . Then MATH and MATH. Since MATH and MATH, MATH and MATH are inequivalent MATH-modules. |
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 MATH. Therefore MATH for all MATH. It follows that MATH. This proves REF . Since MATH are irreducible graded MATH-modules (by REF ), using REF we have MATH . Then it follows from REF that MATH. |
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 subspace of MATH . Then the proof is complete. |
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 resolvent or by using the formula MATH where MATH . The point is that MATH vanishes at MATH if MATH is even. It follows from the iterative parametrix construction in CITE that the meromorphically continued resolvent kernel of MATH has no pole at MATH, since its residue would correspond to a solution of the eigenvalue equation with MATH (hence square-integrable) behavior at infinity. It further follows that the resolvent kernel belongs to MATH away from the diagonal, so that the NAME kernel of the scattering operator at MATH, obtained as a boundary value of the resolvent rescaled by MATH, is zero, and hence the scattering operator is zero. |
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 MATH with the MATH-module MATH. As M is finite dimensional, M has a finite composition series as a MATH-module with composition factors isomorphic to MATH, for some finite number of distinct maximal ideals MATH of MATH. As a MATH-module, we have that MATH where MATH for some MATH is a MATH-submodule of MATH. As for each MATH, MATH, we conclude that the maximal ideals MATH are all in a single orbit, and that this orbit is finite. If MATH, then MATH is a MATH-submodule of MATH. Hence we have MATH, and the result follows. |
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 image of the NAME module MATH, CITE. As MATH is a simple module we have MATH. If MATH, then as MATH is simple, MATH is torsion free and we can identify MATH with MATH. The result follows now by REF . |
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 REF . |
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-module and MATH is NAME, so is MATH and by REF it follows that MATH for some MATH. Assume that MATH for all MATH. It is enough to show that MATH contains a finite dimensional submodule MATH which is a highest weight module. For then we can set MATH and use the same argument to construct MATH provided MATH. Pick MATH such that MATH. Then MATH. Since MATH we have MATH. By REF , we have MATH . By REF , MATH so MATH contains a nonzero polynomial in MATH. Therefore MATH has a nonzero eigenvector in MATH and the result follows. Assume that every finitely generated MATH-torsion and MATH-torsion MATH-module has finite dimension. Fix MATH and set MATH, MATH and MATH. Since MATH and MATH are NAME sets, MATH is MATH-torsion and MATH-torsion. Thus MATH has finite dimension. Since MATH embeds in MATH, MATH has finite codimension in MATH. Using the graded ring structure of MATH, REF, it is easily seen that MATH is the ideal of MATH generated by MATH and MATH. Hence MATH and so 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 obtain MATH and the result follows. |
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