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math/9911082 | It is sufficient to prove REF with additional assumption that MATH and respectively, MATH. Thus let us assume the corresponding one. For brevity, we set MATH, MATH. First we shell prove that MATH is complete. Suppose MATH be any NAME 's sequence in MATH. Then for every positive number MATH and for every MATH there is a number MATH such that MATH . Hence, in particular, MATH . Thus we obtain that the sequence MATH is uniformly convergent on any compact in the upper half plain MATH. Then there is a holomorphic function MATH in MATH which is the uniform limit of MATH on every compact in MATH. Therefore, by REF it follows MATH . Hence, accordingly to the choise of MATH we obtain MATH . In particular, MATH and MATH. Thus we obtain that MATH is complete and hence it is a NAME space. Second, we shell prove that MATH is complete. Note MATH is a linear subspace of the NAME space MATH. So, to obtain the completeness of MATH it is enough to prove that MATH is a closed subspace of MATH. We claim that MATH is a closed subspace of MATH. Suppose MATH is such a sequence of elements of MATH that is convergent in the space MATH. Let MATH be such that MATH. Then for every positive number MATH there is a number MATH such that MATH . Fix a MATH, MATH. Then it follows by MATH that there is MATH, MATH, such that MATH . Hence MATH and therefore MATH. Thus the assertion which we claim above is proved. |
math/9911082 | Let MATH, MATH, be an arbitrary non zero element of MATH. Then MATH . Hence MATH and in particular the holomorphic in the upper half plane MATH function MATH is bounded in the band MATH . Then by the NAME principle we obtain that MATH is convex in MATH, MATH. Thus MATH is convex in MATH. So, the first assertion of REF is proved. In order to prove the second assertion of this Lemma we set MATH . From the first assertion of REF it follows that MATH is an open convex subset of Euclidean plane MATH. Further, by MATH it follows that there is MATH, MATH, such that MATH (in fact this inequality holds for every MATH, but we do not use it). Hence there exists a point MATH, MATH, MATH, which does not belong to MATH. Then there exists a line that distinguishes the set MATH and the point MATH. Hence, there are two numbers MATH such that MATH . So, REF is proved. |
math/9911082 | First we prove that if MATH then there are two numbers MATH such that MATH . Suppose MATH and let MATH, MATH be an arbitrary non zero function that belongs to MATH. Then by NAME REF, there exist two numbers MATH, MATH such that MATH . Further, by MATH and MATH, MATH, it follows MATH . We set MATH, MATH and obtain MATH . Thus we prove that if MATH then the condition on the function MATH which is stated in REF is fulfilled. With view to show MATH, suppose that the function MATH satisfies REF and in addition there exist two numbers MATH and MATH such that MATH . Then for such numbers MATH, MATH we set MATH, where MATH. Then a direct computation shows MATH. So, MATH. Thus the proof of REF is completed. |
math/9911082 | Suppose MATH and let MATH, MATH be an arbitrary non zero function that belongs to MATH. In particular, from the inclusion MATH it follows MATH. Therefore, by REF the first condition on the function MATH is fulfilled. Furthermore, according to REF MATH . Moreover, by REF , there exist two numbers MATH, MATH such that MATH . Hence, MATH which yields MATH. Thus, as we assert in REF , both conditions on function MATH are fulfilled. Further, suppose that the function MATH meets REF and in addition the following two conditions are fulfilled, too: CASE: there exist two numbers MATH, MATH such that MATH . CASE: MATH. We claim that MATH. NAME, the function MATH defined by MATH belongs to MATH. One can verify this by a simple direct computation and we omit the details. Thus REF is proved. |
math/9911082 | Let MATH be such that MATH . In particular, from the inclusion MATH, it follows MATH. Then by REF , there exist two numbers MATH, MATH such that MATH . Hence MATH. We set MATH . So, MATH. In particular, MATH is a well-defined holomorphic function in the upper half plane MATH. We prove that the function MATH has the following properties: CASE: MATH, MATH - which is obvious and we omit the details; CASE: MATH is convex in MATH - indeed, MATH is convex in MATH by REF and hence, according to the previous item, MATH is convex, too; CASE: MATH - this follows immediately from both REF and the first item; CASE: MATH is a decreasing function in MATH - we deduce this from both the second and the third items: for arbitrary MATH and MATH, MATH, and for MATH by the second item it follows MATH and we obtain by the third item MATH, that is, MATH decreases in MATH; CASE: MATH is bounded from above on MATH - it follows immediately from the previous item that MATH, MATH. Further from MATH it follows that MATH is finite and hence by the first item MATH is finite, too, which yields the boundedness of MATH on MATH; CASE: MATH tends uniformly to MATH as MATH, with respect to MATH, MATH, where MATH-is any. We obtain this by the corresponding property of the function MATH, namely according to REF , MATH has the same property. We set MATH. According to the fifth item MATH. Moreover, MATH because MATH is a non zero element of MATH. Note, according to the first item, the assertion of REF can be stated in terms of the function MATH as follows MATH . In order to prove REF , we have to obtain that for every positive number MATH there exists a number MATH such that MATH, MATH. Let MATH be an arbitrary number such that MATH. It follows by the sixth item of the above listed properties of the function MATH that there exists a number MATH, MATH, such that MATH . Further, we fix MATH, MATH, in such a way that MATH . We intend to show that MATH, MATH. Let MATH be such that MATH. Let MATH, MATH, be any. We set MATH . Then it follows from the fifth item that there exists MATH, MATH such that MATH . We construct the rectangle MATH and define the holomorphic function MATH in the upper half plane MATH by MATH . By the choise of MATH, MATH, MATH it is easy to verify that the modulus MATH of the function MATH is less or equal to MATH on the boundary of the rectangle MATH. Indeed: CASE: for MATH such that MATH, MATH, we obtain MATH CASE: for MATH such that MATH, MATH, we obtain MATH and further there are two cases CASE: if MATH then we have to continue with the following inequalities in which we use the notation MATH CASE: if MATH then it is obvious that MATH, that is, in both cases MATH; CASE: for MATH such that MATH, MATH we obtain MATH . Thus, the modulus MATH of the function MATH is less or equal to MATH on the boundary of the rectangle MATH. Then, by the maximum modulus principle, it follows MATH and hence, in particular, MATH. Therefore, MATH . Then MATH because of the choise of MATH. So, MATH. Thus the claim MATH, MATH, is proved. The proof of REF is completed. |
math/9911083 | Since MATH it follows that MATH lies in MATH for every MATH. Hence MATH. So we may set MATH. |
math/9911083 | Denote by MATH the images in MATH of MATH respectively. Since MATH is the identity on MATH there is a MATH-vector space isomorphism MATH induced by MATH which respects the alternate bilinear form MATH on MATH. If MATH is MATH then MATH respects the quadratic form MATH. Since NAME 's extension theorem holds for MATH (see CITE), we may extend MATH to a MATH-orthogonal transformation MATH of MATH. Using the standard generators for MATH we may lift MATH to an automorphism MATH of MATH. If MATH then MATH for some MATH. Since MATH has enough inner automorphisms, we may assume that MATH extends MATH. To be more precise: pick MATH whose images under MATH constitute a basis for MATH. Since the alternate bilinear form MATH on MATH is nondegenerate we can pick MATH such that MATH. Hence conjugation by MATH fixes MATH for MATH and sends MATH to MATH. So we can correct MATH by an inner automorphism of MATH to ensure that MATH on MATH. Now suppose that MATH is odd. As NAME 's extension theorem holds for MATH (see CITE), we may extend MATH to a transformation of MATH that respects MATH. But then MATH respects MATH too: this is trivial in the exponent MATH case, as MATH is then zero. In the exponent MATH case, we may assume by REF that MATH contain MATH and that MATH fixes MATH. So MATH respects MATH by REF below. As in the MATH case we can now lift MATH to an automorphism of MATH which extends MATH. |
math/9911083 | Each element MATH of MATH has canonical form MATH. Then MATH. |
math/9911083 | If MATH then MATH is not abelian. In MATH, all order MATH elements are central. The other three groups really are exceptions: take MATH. |
math/9911083 | Observe that MATH is a NAME MATH-subgroup of MATH, and so all NAME MATH-subgroups of MATH have NAME subgroup MATH. Now set MATH equal to MATH. By assumption the NAME subgroup of MATH is that of MATH, namely MATH. But MATH is contained in a NAME MATH-subgroup of MATH. We conclude that MATH. Therefore MATH is itself a NAME MATH-subgroup of MATH, and so MATH for some MATH. Take MATH and MATH. |
math/9911083 | Set MATH equal to MATH, which contains MATH. Since the centre of MATH is cyclic and MATH is maximal in MATH, it follows that MATH is a NAME MATH-subgroup of MATH. Now let MATH be MATH, the centralizer of MATH in MATH. Since MATH has exponent MATH, so does MATH. As MATH centralizes MATH, we deduce that MATH is a maximal subgroup of MATH. But the only exponent MATH maximal subgroup of MATH is MATH, which has centre MATH. So MATH normalizes MATH. Now set MATH. Then MATH is a NAME MATH-subgroup of MATH, and MATH is too since MATH lies in MATH. So MATH for some MATH. Take MATH and MATH. |
math/9911083 | If MATH comes from MATH, then it certainly satisfies both conditions. Conversely, observe that the conditions combined with the NAME formula mean that MATH. |
math/9911083 | Pick any outer automorphism of order prime to MATH, and lift it to an automorphism MATH of the same order. Let MATH be the semidirect product MATH. |
math/9911083 | See CITE for a proof that MATH is a NAME group. The groups MATH and MATH are treated in REF below. So we may assume that MATH satisfies the hypotheses of REF and hence those of REF . Let MATH be a finite group with extraspecial NAME MATH-subgroup MATH, and let MATH be a MATH-invariant. We show that the conditions of REF are satisfied. Let MATH be an element of MATH. If MATH is elementary abelian but not maximal in MATH, then corestriction from MATH to MATH is zero: for corestriction from any group MATH to MATH is zero. We may therefore assume that MATH, MATH both contain MATH. Write MATH for MATH and MATH for MATH. By REF we deduce that MATH normalizes MATH. Moreover, since MATH is invariant under MATH, we may in fact assume that MATH stabilizes MATH. If MATH has odd exponent MATH we deduce further by REF either that MATH normalizes MATH and can be taken to act trivially on MATH, or that MATH. So we can now apply REF and deduce that conjugation MATH extends to an automorphism MATH of MATH. Then MATH, which means that MATH as MATH is a proper subgroup of MATH. |
math/9911083 | Let MATH be MATH. Then the upper triangular matrices with ones on the diagonal form a NAME MATH-subgroup isomorphic to MATH. We may take MATH and MATH. Pick MATH. Then MATH equals MATH, a maximal elementary abelian subgroup. Let MATH be the dual basis for MATH. Taking the first NAME class is a MATH-linear monomorphism from MATH to MATH. Let MATH be an ordinary representation of MATH with character MATH, the sum of all MATH linear characters. These linear characters restrict to MATH as scalar multiples of MATH, each scalar multiple being the image of MATH characters. So by the NAME sum formula, the total NAME class MATH restricts to MATH as follows: MATH . This equals MATH. Set MATH equal to MATH in MATH. Then MATH lies in MATH, since MATH is an invariant of MATH. But MATH is distinct from MATH, so MATH is not stable. |
math/9911083 | MATH is a self-centralising, maximal elementary abelian subgroup. By REF, the NAME summand MATH of MATH is a transfer summand of MATH. |
math/9911086 | Given the data MATH for all MATH, MATH, MATH, the NAME formula, MATH (here MATH) yields the solution of the problem MATH for each fixed MATH. In particular, MATH in REF represents MATH . By symmetry of the NAME 's function MATH with respect to MATH and MATH, MATH we also determined MATH . Moreover, using MATH with MATH (instead of MATH) in REF then determines MATH . In other words, we managed to lift the data from MATH to MATH. Next, the explicit REF for MATH yields MATH with MATH defined in REF. Hence one concludes MATH . Thus, the data MATH, MATH uniquely determine MATH by the unique continuation property CITE applied to REF. The singularity structure of REF then determines MATH and hence MATH. Similarly, taking MATH and MATH independently, determines MATH, MATH, and hence MATH. Thus, REF is proved. |
math/9911088 | A vector MATH is in the stabilizer algebra of MATH iff the vector field induced on MATH vanishes at MATH, iff the symplectic gradient of MATH vanishes at MATH, iff the differential MATH vanishes at MATH, iff MATH vanishes at MATH, which means MATH is in the perp of the image of the differential MATH at MATH. |
math/9911088 | If MATH is an extremal point, the image of the differential can't be onto - some directions from MATH lead outside the image of MATH. So the perp to the differential is positive-dimensional, and generates a positive-dimensional subgroup of MATH, the connected component of the stabilizer group. And any positive-dimensional compact group contains a circle group. |
math/9911088 | We have already explained how to identify MATH and MATH with coadjoint orbits of MATH. Therefore by REF of moment maps listed above, their product has a Hamiltonian action of MATH. (In fact it is a coadjoint orbit for this big group.) Consider the action of the diagonal MATH, that is, conjugating both Hermitian matrices by the same unitary matrix. Then by REF this action is Hamiltonian, and we can compute its moment map as the transpose of the inclusion MATH, composed with the MATH moment map, which was just inclusion of the coadjoint orbit. The transpose of diagonal inclusion MATH is summation MATH; so the moment map for MATH's action on MATH just takes a pair of Hermitian matrices to their sum. But now we have convexity, as our map MATH is just the map used in the alternate description of NAME 's convexity theorem. If MATH is a strictly decreasing list of real numbers, that means it's in the interior of the positive NAME chamber, therefore in the image of the symplectic slice, which for MATH is the set of pairs MATH whose sum is diagonal with decreasing entries (already a familiar set to people studying NAME 's problem). Then we can use some element of MATH to conjugate MATH into the symplectic slice, on which MATH is the moment map for the action of MATH. Since MATH is on the boundary of the image of MATH, we can apply the corollary from subsection REF, and determine that MATH is invariant under some circle in MATH (and not just the scalar matrices, which fix all pairs MATH). Being invariant under conjugation by a nonscalar diagonal matrix forces each of MATH and MATH to be block diagonal. |
math/9911088 | To see what the problem is, let's trace through the definition of MATH. Since MATH and MATH are graded, that is, have MATH-actions, MATH and MATH have MATH actions. The map between the rings being grading-preserving is equivalent to it being MATH-equivariant. Consequently, the map backwards on MATH is MATH-equivariant. The problem comes when we try to rip out MATH and MATH. Of course there is no problem when in restricting the map to MATH. But the image of this may not land inside MATH; there may be points of MATH that hit MATH. Once we rip them out, then there's no problem. |
math/9911088 | NAME already seen that the first space is the symplectic quotient of MATH by the diagonal action of MATH, at the level MATH. By the symplectic vs. GIT equivalence, this is the geometric invariant theory quotient of a certain product of flag manifolds by MATH. The NAME theorem, or rather the small part of it presented here, explains how to embed MATH into projective space in order to be able to apply the symplectic vs. GIT theorem. The individual coordinate rings are MATH (likewise MATH); the coordinate ring of the product is, by NAME embedding, MATH . Then to take the GIT quotient we take MATH of the invariant subring. |
math/9911088 | If the tensor product has an invariant vector, then the coordinate ring of the GIT quotient is nontrivial in its degree REF piece. Since this ring is a subring of a ring with no zero divisors, it itself has no zero divisors, and therefore is nontrivial in all degrees. This is enough to conclude that MATH of it, the GIT quotient, is nonempty. By the identification in the theorem, the symplectic quotient is therefore nonempty. So there are three Hermitian matrices with the desired spectra adding to zero. |
math/9911088 | The condition says the symplectic quotient is nonempty, so running the theorem in reverse, the GIT quotient is nonempty. Consequently the ring of invariants is nontrivial. Therefore in some graded piece MATH it is nontrivial, giving the desired result. |
math/9911094 | Proof. - The right inequality follows directly from the definition of MATH and the fact that we can bound by MATH the number of monomials of MATH. Thus we only consider the left inequality. Let MATH denote the coefficient of MATH with respect to the monomial MATH. Applying REF we obtain for all MATH: MATH . We have MATH. We integrate both sides of the last inequality on MATH and we deduce MATH . The statement follows then by induction and the fact that MATH. |
math/9911094 | REF are immediate from the definition of MATH. REF : Let us consider the case MATH. Set MATH. First we compute MATH for the exponent MATH of a monomial of MATH. Applying REF we obtain MATH . The polynomial MATH has at most MATH monomials and so MATH . The case MATH follows in a similar way. REF : In case MATH, we apply directly REF : MATH . In case MATH, NAME Lemma implies that MATH. |
math/9911094 | Proof. - Let MATH be a non-archimedean absolute value, and set MATH for the corresponding prime ideal of MATH. Then MATH for some MATH. We set MATH . Clearly MATH, and MATH for every MATH, that is MATH. We have MATH for MATH, and also MATH for all MATH. Thus MATH . On the other hand MATH . |
math/9911094 | Proof. - Set MATH. We have MATH and so MATH by REF . On the other hand we have MATH for MATH and thus MATH by application of REF . |
math/9911094 | Let MATH be the NAME matrix associated to the generic polynomial system MATH. This is a non-singular square matrix of order MATH, where MATH denotes the cardinality of the set MATH . Here MATH, and MATH is a vector such that each point in MATH is contained in the interior of a cell in a given triangulation of the polytope MATH. Every non-zero entry of MATH is a variable MATH. In fact, each row has exactly MATH non-zero entries, which consist of the variables in some group MATH. We refer to CITE for the precise construction. Thus MATH is a multihomogeneous polynomial of total degree MATH and height bounded by MATH. This polynomial is a non-zero multiple of the sparse resultant MATH CITE. The assumption that MATH is primitive implies that MATH lies in MATH, and so MATH. Let MATH be a unimodular triangulation of MATH, so that MATH is a triangulation of MATH. For every MATH, the set of integer points in MATH is in correspondence with a subset of those of MATH. Moreover, for a generic choice of MATH we loose - at least - the set of integer points in a facet of codimension REF. Thus MATH and so MATH. Applying REF we obtain MATH . We conclude MATH as MATH. |
math/9911094 | For MATH we denote by MATH the set of MATH-roots of REF in MATH. Let MATH such that MATH. Then MATH . Set MATH. Let MATH such that MATH, that is MATH. Then for any MATH we have MATH . Let MATH such that MATH. From the previous expression we derive that for each MATH there exists MATH such that MATH. The set MATH is NAME dense in MATH, and so MATH is also dense. Thus we can take MATH such that MATH, and therefore MATH . The other inequality is straightforward. |
math/9911094 | First we consider the case when MATH is a REF - dimensional variety. We may assume without loss of generality that MATH is irreducible, that is MATH for some MATH. Set MATH. Then MATH where MATH and MATH denote generic polynomials in MATH variables of degree MATH and MATH respectively. Then MATH and so MATH. Analogously, MATH and so MATH. Now we consider the general case. Set MATH, MATH and MATH . Then MATH is a REF-dimensional variety of degree MATH for MATH in a NAME open set MATH of MATH. Let MATH. By CITE there exist MATH such that MATH where MATH stand for the specialization of MATH into MATH. Applying the morphism MATH linking the MATH-Chow form with the usual one we obtain MATH and so MATH. We consider the case MATH. Any NAME closed set of MATH intersects MATH in a set of MATH-measure REF, and so the previous relation holds for almost every MATH. Therefore MATH . The case MATH follows analogously from REF-dimensional case and the previous lemma. |
math/9911094 | We assume without loss of generality that MATH is irreducible. Set MATH and MATH. The case MATH follows directly from CITE: in this case, there exists a partial monomial order MATH such that MATH where MATH denotes the initial polynomial of MATH with respect to MATH. In particular MATH is the sum of some of the terms in the monomial expansion of MATH. The general case MATH reduces to the previous one: we choose standard coordinates MATH of MATH such that the projection MATH verifies MATH for MATH. Let MATH denote the canonical projection. Then MATH, MATH and MATH. We have that MATH for MATH by the theorem of dimension of fibers. Thus MATH, and in particular MATH where MATH denotes the canonical inclusion MATH. We have MATH and so MATH is a NAME form of MATH CITE. Now we estimate the height of MATH. Let MATH be a number field of definition of MATH, and set MATH for some polynomial MATH. From the proof of CITE, there is a non-zero coefficient MATH of MATH such that MATH for all MATH. Clearly MATH also holds for all MATH. Thus MATH for MATH, while MATH for MATH. Let MATH. From CITE we obtain MATH. Hence MATH by application of REF . In case MATH we have analogously MATH, and so MATH . |
math/9911094 | We assume again without loss of generality that MATH is irreducible. Let MATH be a number field of definition of both MATH and MATH, and set MATH for some MATH-matrix MATH of maximal rank and MATH. Then let MATH be the linear map MATH defined by the transpose of the matrix associated to MATH. Set MATH, and let MATH, MATH denote the projective closures of MATH and MATH respectively. For MATH we let MATH, and we set MATH for the homogenization of the associated linear form. Then MATH if and only if there exists MATH such that MATH lies in the linear space determined by MATH. Equivalently MATH lies in the linear space determined by MATH. We conclude that MATH . Let MATH. Then MATH . Here we have applied REF and the proof of REF , using the fact that the number of monomials of MATH is bounded by MATH. In case MATH we obtain analogously MATH, and hence MATH . |
math/9911094 | Proof of REF . - Let MATH be the injective map MATH. Then MATH decomposes as MATH where MATH denotes the canonical projection. Thus MATH . |
math/9911094 | Proof. - Set MATH and MATH. By CITE there exists MATH such that MATH. Then - as in the proof of REF - there exists a non-zero coefficient MATH of MATH such that MATH for all MATH and MATH for all MATH. Now let MATH. From REF we obtain MATH since MATH has degree MATH in each group of variables. Then MATH by straightforward application of REF . The case MATH follows in an analogous way. |
math/9911094 | We just consider the case when MATH is archimedean, as the other one follows similarly. From the preceding result we obtain MATH for some MATH. For the final estimate we apply iteratively this inequality and we set MATH. |
math/9911094 | Proof. - We apply the previous result to MATH, using the fact that MATH . |
math/9911094 | Proof. - We proceed by induction on MATH with respect to the product order of MATH. The cases MATH or MATH are both trivial. Thus we assume MATH. Let MATH be the decomposition of MATH into irreducible components. In case MATH we have that MATH and so MATH with MATH. In case MATH we have either MATH or MATH. The first case is trivial. In the second case we have MATH . To obtain this, we proceed as in the proof of REF , applying REF instead of REF . Then we apply the inductive hypothesis and we obtain MATH . |
math/9911094 | Set MATH. The case MATH is trivial, and so we assume MATH. We also assume that MATH are the vectors of the canonical basis of MATH. The map MATH induces an isomorphism between MATH and the affine toric variety MATH. The projection map MATH defined by MATH restricted to MATH is the inverse map of MATH. For MATH we set MATH and we let MATH be the associated linear form. Set MATH and MATH. We have MATH and so MATH. Then MATH and MATH by successive application of REF . Finally MATH and so MATH. |
math/9911094 | Proof. - It is enough to consider the case when MATH is irreducible. Let MATH be a MATH-characteristic polynomial of MATH. For MATH we set MATH so that MATH. We observe that MATH . In particular, this variety is non-empty, and so we infer that MATH. This implies that MATH as MATH is an irreducible polynomial. On the other hand MATH is also irreducible, as it is multihomogeneous and MATH. We conclude that MATH and MATH coincide up to a factor in MATH. |
math/9911094 | Proof. - We have that MATH in MATH and so MATH is a monic equation for MATH in MATH, for MATH. Thus the projection MATH is finite. For the second assertion, set MATH . This is a polynomial of degree MATH. It is monic with respect to MATH, as MATH and MATH. We have MATH in MATH. Now let MATH be the monic minimal polynomial of MATH. Let MATH be a group of MATH variables and set MATH for the generic polynomial of degree MATH in the variables MATH. Then MATH is an equation for MATH over MATH. Since MATH is a REF-dimensional variety of degree MATH and MATH separates its points, we infer that MATH, and so MATH. Finally we obtain MATH . |
math/9911094 | We keep notations as in Subsection REF. Set MATH and MATH. We have then MATH by REF . Then MATH . From the previous expression we also obtain that the coefficients of MATH are some of the coefficients of MATH, and so MATH for every absolute value MATH of MATH. Let MATH. Then MATH by REF , and REF . In a similar way we obtain MATH for MATH. |
math/9911094 | Let MATH, and let MATH be a new variable. Then MATH is again an integral inclusion. Set MATH. We have then MATH and so MATH . Set MATH with MATH. The last identity implies then MATH. Set MATH, and let MATH denote the group of MATH-roots of REF. Then MATH for MATH, and so MATH . From REF we get MATH. For MATH, we then obtain MATH . Analogously, for MATH we take MATH such that MATH, and we obtain MATH. |
math/9911094 | Set MATH for the quotient field of MATH and MATH. Then MATH is a finite MATH-algebra of dimension MATH and MATH can be uniquely extended to a MATH-linear map MATH. The fact that MATH is a torsion-free MATH-algebra implies that the canonical map MATH is an inclusion. We will only consider the case MATH. For the case MATH we refer to REF . Whenever it is clear from the context, we will avoid explicit reference to the ring in which we are considering a given element of MATH. Let MATH be any polynomial such that MATH in MATH. We have that MATH is a non-zero divisor in MATH, and so it is invertible in MATH. Then MATH in MATH and therefore MATH for all MATH. Then we set MATH . NAME trace formula implies that MATH, and so MATH in MATH. Let MATH denote the Jacobian determinant of the complete intersection MATH with respect to the group of variables MATH. This is a non-zero divisor because of the Jacobian criterion, and so it is also invertible in MATH. Let MATH be the adjoint polynomial of MATH and set MATH . We have MATH, and so MATH . In particular MATH in MATH, and we have the expression MATH . Clearly MATH . Next we analyze the total degree of MATH. Let MATH be the monomial expansion of MATH with respect to MATH. Then MATH as MATH is a MATH-linear map. We have the estimates MATH and MATH, from where we get MATH by REF . Thus MATH . For the rest of the proof, we will use several times the following basic estimates: MATH . Finally we estimate the local height of MATH. Let MATH. We have MATH and so MATH . Therefore MATH by REF . We recall that MATH by REF and so MATH . Then MATH by REF . Hence MATH by application of Identity REF . We have MATH as each MATH is a different monomial in MATH. Thus it only remains to estimate the local height of each MATH. Let MATH be any non-zero coefficient of MATH. Then MATH by REF and the fact that MATH. From REF we obtain MATH . This implies that the right hand side of REF is non-negative. So the inequality also holds for MATH, and thus for MATH. The case MATH is treated analogously. We remark that the election of MATH is independent of MATH, and so it can be done uniformly. |
math/9911094 | Set MATH for MATH. Also set MATH and MATH. The fact that MATH satisfies REF implies that the inclusion MATH is integral. We note that the sets of free and dependent variables of MATH have cardinality MATH and MATH respectively. Also the set of dependent variables of MATH is contained in that of MATH for MATH. For MATH we denote by MATH the degree of MATH in the dependent variables MATH of MATH with respect to the integral inclusion MATH. For MATH, the previous observation implies that MATH. Applying the Division REF , we will construct inductively polynomials MATH: first we take MATH such that MATH . For MATH we assume that MATH are already constructed and we set MATH . Then MATH is a non-zero divisor and MATH in MATH. We apply again Division Lemma to obtain MATH such that MATH . Continuing this procedure until MATH, we get MATH in MATH. Let us analyze degrees. First we consider the case MATH. Again we proceed by induction. First we have MATH and MATH . Now let MATH. Then MATH and MATH where MATH . Hence MATH . For MATH we have MATH and therefore MATH. Then for all MATH: MATH . Next we consider the case MATH. In this case MATH is a REF-dimensional variety and so MATH . Let MATH. Then MATH and MATH . We have also MATH. We conclude for all MATH: MATH . Finally we estimate the local height of these polynomials. In the rest of the proof we will make repeated use of the following degree bounds: MATH . As usual, we consider only the case MATH, the case MATH can be treated analogously. From Division Lemma we obtain MATH for some MATH. Let MATH and set MATH. Then there exists MATH such that MATH . Applying the inductive hypothesis we obtain MATH . For MATH we apply REF : there exists MATH such that MATH . We set MATH. Then MATH . This last inequality follows from the facts that MATH for MATH and MATH as MATH. To conclude the proof, observe that for MATH, REF guarantees that the obtained estimate for MATH differs from the one for MATH by a positive term. Thus, the same estimate holds for MATH, MATH. |
math/9911094 | Let us first consider degrees. We assume without loss of generality MATH. From the preceding result we obtain MATH . Next we consider the local height estimates. Let MATH. We have MATH for some MATH. Applying REF , MATH for some MATH. Therefore MATH where MATH is defined as MATH. The case MATH follows anagolously. |
math/9911094 | A direct computation shows that for MATH . |
math/9911094 | Equation MATH is equivalent to a MATH-linear system of MATH equations in MATH unknowns, which can be solved applying NAME rule. The integer MATH is the determinant of a non-singular submatrix of the system. |
math/9911094 | Proof. - Let MATH be generic linear combinations of MATH. Then MATH and MATH are coprime polynomials, and so there exist MATH with MATH and MATH such that MATH . Expanding this identity there exists MATH with MATH such that MATH . Thus the above NAME identity translates to a consistent system of MATH-linear equations. The number of equations and variables equal MATH and MATH respectively. This system can be solved by NAME rule. The integer MATH is the determinant of a non-singular MATH-submatrix of the matrix of the linear system. |
math/9911094 | Proof of REF . - We assume MATH and MATH. Let MATH denote the group of MATH-roots of REF, for a prime MATH. For MATH and MATH we set MATH . Also, for MATH and MATH we set MATH . We will assume that for a specific choice of MATH, MATH and MATH there exists MATH such that MATH is a radical ideal of dimension MATH for MATH and MATH. We also assume that MATH is a linear change of variables, and that MATH satisfies REF for MATH with respect to MATH. This is guaranteed by the fact that these conditions are generically satisfied: there exists a hypersurface MATH of the coefficient space such that MATH implies that MATH satisfy the stated conditions with respect to the variables MATH CITE, CITE. As MATH is NAME dense in MATH, it follows that these coefficients can be chosen to lie in MATH for some MATH. Moreover, MATH can be chosen such that for MATH a primitive MATH-root of REF and MATH, MATH and MATH are linearly independent and MATH does not divide the discriminant of MATH. We refer the reader to REF, where we give a self-contained treatment of this topic. Set MATH and MATH. For MATH set MATH . Then MATH satisfy the hypothesis of REF . Let MATH and MATH be the non-zero element and the polynomials satisfying NAME identity we obtain there. Now, for MATH, set MATH so that MATH holds. Finally set MATH. By REF there exists MATH such that MATH. We define MATH for MATH. Then MATH as MATH and MATH is a MATH-linear map. Aside from the degree and height bounds, we will show that since MATH, MATH and MATH. Let us first analyze degrees and local heights. As MATH, MATH. Now let MATH and let MATH such that MATH. We have MATH and so MATH by REF . From REF MATH . Therefore MATH again by REF and the fact MATH. We have MATH and so MATH . We have MATH and so the previous estimate also holds for MATH. Now let MATH and MATH. Analogously we have MATH as MATH. Then MATH and MATH, which in term implies that MATH and MATH as desired. The global height estimate follows then from the expression MATH . |
math/9911094 | We note first that MATH is a radical ideal, and so MATH is also radical. We readily obtain from the definition of MATH that MATH, and so MATH. We can write MATH with MATH. Therefore MATH also lies in MATH for all MATH. A direct computation shows that for MATH for some MATH with MATH. Set MATH . The previous argument shows the inclusion MATH. On the other hand, MATH for some MATH. Set MATH. Then MATH and so for every MATH we have that MATH modulo MATH, and hence modulo MATH. For MATH, MATH which implies MATH as desired. |
math/9911094 | Set MATH and MATH. First we observe that MATH is a linear bundle over MATH: the projection MATH is surjective, and the fibers are affine spaces of dimension MATH. This follows from the assumption that the MATH have no common zeros. This implies that MATH because of the theorem of dimension of fibers. Namely MATH is a complete intersection, and in particular the ideal MATH is unmixed. Set MATH for the primary decomposition of this ideal. We will show that MATH is prime for all MATH, and then that MATH. First we have that MATH where MATH does not depend on MATH. Therefore MATH is a domain, that is MATH is prime. We have MATH, and so there exists MATH such that MATH . In particular MATH is prime. The fact that MATH ensures that MATH runs over all MATH, and so MATH is radical. The expression MATH implies that MATH contains the dense open set MATH. In particular MATH is not contained in any of the hypersurfaces MATH and so MATH for all MATH. This implies that MATH, and so MATH is prime. |
math/9911094 | Set MATH. We have MATH and MATH. First we consider the case MATH. This occurs, for instance, when MATH, since then MATH. Let MATH be the canonical projection. Then MATH is a proper subvariety of MATH, and thus it is contained in a hypersurface of degree bounded by MATH. This can be seen by taking a generic projection of this variety into an affine space of dimension MATH CITE. Let MATH be a defining equation of this hypersurface. Then MATH as MATH is prime, and we have MATH. Thus MATH and therefore MATH for MATH such that MATH. Next we consider the case MATH. We adopt the following convention: for an ideal MATH and for MATH any new group of variables, we denote by MATH and MATH the extension of MATH to the polynomial rings MATH and MATH respectively. We assume for the moment MATH. Then MATH and so the extended ideal MATH is a REF-dimensional prime ideal. We have then that MATH is a radical ideal, as MATH CITE. Our approach to this case is based on Shape REF . We will determine a polynomial MATH such that MATH implies that the shape lemma representation of MATH can be transferred to a shape lemma representation of MATH. Let MATH be a group of MATH variables and set MATH for the associated generic linear form. Consider the morphism MATH and let MATH be the variety defined by MATH in MATH, that is MATH. The NAME closure MATH is then an irreducible hypersurface. We set MATH for one of its defining equations. If MATH is the extension of MATH to MATH, the polynomial MATH can be equivalently defined through the condition that MATH is a generator of the principal ideal MATH. Namely, MATH is a characteristic polynomial of the MATH-dimensional variety MATH defined by MATH in MATH. Let MATH denote the polynomials arising in Shape Lemma applied to MATH. From the proof of this lemma we have that MATH and so MATH. Set MATH and MATH. Then MATH . We have that both MATH and MATH are prime ideals of MATH with trivial intersection with the ring MATH. Thus they coincide with the contraction of MATH and MATH to MATH respectively, and so MATH . Define MATH as any of the non-zero coefficients of the monomial expansion of MATH with respect to MATH. Let MATH such that MATH. Then MATH and so MATH is squarefree. Then MATH is radical, which implies in turn that MATH is a radical ideal of MATH as desired. It remains to estimate the degree of MATH. To this end, it suffices to bound the degree of MATH with respect to the group of variables MATH. We recall that MATH was defined as a defining equation of the hypersurface MATH. The map MATH is linear in the variables MATH and MATH, and so MATH . This implies that MATH. Finally we consider the case MATH for MATH. Let MATH be groups of MATH variables each, and set MATH for MATH. Set MATH, MATH and MATH. The extended ideal MATH verifies MATH and thus falls into the previously considered case. Thus there exists MATH with MATH such that MATH for MATH implies that MATH is a radical ideal of MATH. This implies in turn that MATH is a radical ideal of MATH, as MATH . We can assume without loss of generality that MATH lies in MATH. We conclude by taking MATH as any non-zero coefficient of the monomial expansion of MATH with respect to the variables MATH and MATH. |
math/9911094 | Set MATH for the minimal MATH such that MATH. Then MATH, and by the previous result there exists MATH with MATH such that MATH implies that MATH. On the other hand, for MATH we take a polynomial MATH of degree bounded by MATH such that MATH implies that MATH is a radical ideal of dimension MATH. Then we take MATH and so MATH . Finally, MATH implies there exist MATH such that MATH and MATH. |
math/9911094 | Proof. - Let MATH be a NAME form of MATH and MATH be the characteristic polynomial of MATH associated to MATH given by REF . Set MATH and let MATH be its expansion with respect to MATH. Also set MATH for the discriminant of MATH with respect to MATH. Observe that as MATH is multihomogeneous of degree MATH in each group of variables MATH, the degree of MATH in each of these group of variables is bounded by MATH. Now let MATH such that MATH is a MATH-dimensional variety of cardinality MATH, and MATH be a NAME forms of MATH. Set MATH. Then applying CITE, there exists MATH such that: MATH where MATH is a characteristic polynomial of MATH. This implies MATH is a squarefree polynomial and so MATH. We take MATH as any non-zero coefficient of the expansion of MATH with respect to MATH. Therefore MATH . The condition MATH implies that MATH, and so MATH. |
math/9911094 | This follows readily from the previous result. We take MATH as the polynomial corresponding to the variety MATH and we set MATH. We have MATH and so MATH . We conclude by taking MATH such that MATH and MATH. |
math/9911094 | Let MATH be the coefficient matrix which realizes the degree and height of MATH and set MATH . Let MATH be minimum such that MATH. Let MATH be the matrix formed by the first MATH rows of MATH and let MATH be a staircase matrix equivalent to MATH. The polynomial system MATH is then equivalent to MATH, that is MATH for MATH. Also we have MATH, and so MATH . We have also that each coefficient of MATH is a subdeterminant of MATH. Thus MATH and so, applying REF , MATH . |
math/9911094 | Proof. - Let MATH and set MATH for MATH. Then MATH and so MATH. Applying REF we obtain MATH and MATH . |
math/9911094 | Let MATH be a coefficient matrix which realizes the degree and height of MATH. We have then MATH, MATH and MATH. We set MATH for MATH. Then MATH is a radical ideal of dimension MATH for MATH and MATH for some MATH. Also let MATH be integers with MATH such that MATH satisfies REF with respect to the variables MATH for MATH. Set MATH and MATH, and set MATH for the affine map MATH. For MATH we then set MATH . Thus MATH are in the hypothesis of REF with respect to MATH and we let MATH be the polynomials satisfying NAME identity we obtain there. Finally, for MATH, we set MATH . We have MATH. Now we analyze the degree and the height of these polynomials. We will assume MATH as the remaining cases have already been considered in REF . Set MATH for MATH. We have MATH and so MATH. We have also MATH and so MATH as MATH for MATH. Now let MATH and set MATH. We have MATH . Then MATH by REF and the facts that MATH and MATH for MATH. Next, applying REF , we obtain MATH as MATH and MATH for MATH. By REF there exists MATH such that MATH with MATH. Then MATH . Analogously MATH for MATH. Hence MATH . Finally we apply REF to obtain MATH such that MATH. Thus MATH and the corresponding height estimates are multiplied by MATH. |
math/9911095 | REF follows from the definition and REF , and REF is a consequence of the projection formula for MATH-modules. |
math/9911095 | The canonical morphism MATH induces a monomorphism MATH of MATH-modules. Applying MATH we obtain a sequence of morphisms MATH of MATH-modules. The morphism MATH is nontrivial by the definition, and the morphism MATH is a monomorphism by the left exactness of MATH. Thus the composition MATH is nontrivial. Hence it induces a canonical nontrivial morphism MATH of MATH-modules. |
math/9911095 | Since MATH is stable under the action of MATH, we have MATH for any MATH. In particular, MATH for any MATH, and hence MATH for any MATH. By the definition we have MATH and MATH for any MATH. Hence the assertion follows from REF . |
math/9911095 | We have MATH and MATH by REF. |
math/9911095 | For MATH define an object MATH of MATH by MATH where MATH has degree MATH (see REF for the notation). For MATH we set MATH. By MATH we have MATH. Set MATH. Then REF are obvious. Let us show REF . Applying MATH to the distinguished triangle MATH we obtain a distinguished triangle MATH . By REF we have MATH . REF is proved. |
math/9911095 | REF are obvious from REF . REF follows from REF . Assume that MATH satisfies the assumption in REF . Then we have MATH and MATH. Hence REF follows from REF . |
math/9911095 | REF is a consequence of the definition of MATH, and the first statement in REF follows from REF and the definition of MATH. By MATH we have only to show MATH for MATH. We have MATH by MATH and MATH. Since MATH, we obtain MATH. Hence MATH . |
math/9911095 | The first statement is obvious by REF. Since MATH and MATH are stable under the action of MATH and MATH respectively, we have MATH for any MATH. In particular, we have MATH for any MATH. Hence we obtain MATH . Therefore, the relation MATH is equivalent to the system REF. |
math/9911095 | Since MATH and MATH are fixed by the action of MATH and MATH respectively, We have MATH . By MATH we obtain MATH. Hence by MATH and MATH, we have MATH for any MATH. Moreover, we have MATH for any MATH by REF. |
math/9911095 | For MATH we set MATH . Take MATH such that MATH for any MATH, and let MATH such that MATH. We can assume that MATH for any MATH satisfying MATH. Then MATH is the (unique) element of MATH with minimal length. Let us first show: MATH . It is sufficient to show MATH. Since MATH is the element of MATH with minimal length, we have MATH, and hence MATH. By REF we have MATH. Hence by MATH we have MATH. Thus MATH. By MATH we have MATH, and hence MATH. REF is proved. We next show MATH . For any MATH we have MATH and hence MATH by the choice of MATH. Thus we have MATH . Hence MATH. Here, we have used the well-known fact that for MATH we have MATH if and only if MATH. Similarly, we have MATH for any MATH by the definition of MATH and hence MATH. Thus we have MATH . Hence MATH. REF is proved. Note that MATH and that MATH is a quotient of the ordinary NAME module MATH. Hence by our assumption and by REF we obtain MATH with respect to the standard partial order on MATH by a result of CITE concerning the composition factors of NAME modules. In particular, we have MATH. Hence we obtain the desired result by REF. |
math/9911103 | Choose a bounded, almost everywhere smooth NAME cross-section MATH, which can then be used to define the NAME cocycle MATH corresponding to MATH, and such that MATH. As in REF, for MATH, there is an index idempotent, MATH where for MATH, MATH . Then as in REF, one sees that MATH, MATH are smoothing operators and MATH is a parametrix for MATH for all MATH. The MATH-index map is then MATH where MATH is the idempotent MATH . Let MATH. We adapt the strategy and proof in CITE to our situation. MATH where MATH denotes a bounded measurable section. Notice that the phase term MATH appearing in the expression for the cocycle MATH compare REF, is exactly cancelled by the twisted product of the integral kernels MATH, compare REF. Notice also that the right hand side of REF is independent of the choice of the section MATH, since upon changing MATH to MATH, we obtain MATH where MATH which is exactly as in the case when the multiplier is trivial. Observe that if MATH is a smooth local section, then there is a unique element MATH such that MATH. Moreover, we have the equality MATH (and MATH otherwise), where MATH denotes the MATH-equivariant REF MATH-cocycle on MATH representing the pullback MATH of the group REF-cocycle via the classifying map MATH. Since MATH is mainly supported near the diagonal as MATH, and using the equivariance of MATH, one sees that MATH where we have identified MATH with a fundamental domain for the MATH action on MATH. The proof is completed by applying the local higher index REF, to obtain the desired cohomological REF for MATH. |
math/9911103 | By REF , one has one has MATH where MATH is smooth and MATH is a finite orbifold cover. Here MATH is the lift of the map MATH (since MATH in this case) which is the classifying map of the orbifold universal cover (and which in this case is the identity map) and MATH degree REF cohomology class on MATH that is the lift of MATH to MATH. We next simplify the right hand side of REF using the fact that MATH and that MATH . We obtain MATH . When MATH, is the area REF-cocycle, then MATH is merely the restriction of the area cocycle on MATH to the subgroup MATH. Then one has MATH . The corollary now follows from REF above together with the fact that MATH, and MATH. |
math/9911103 | We need to consider the expression MATH . Let MATH denote the symplectic form on MATH given by: MATH . The so-called `symplectic area' of a triangle with vertices MATH may be seen to be MATH. To appreciate this, however, we need to use an argument from CITE, pages REF. In terms of the standard basis of MATH (given in this case by vertices in the integer period lattice arising from our choice of basis of harmonic one forms) and corresponding coordinates MATH the form MATH is the two form on MATH given by MATH . Now the `symplectic area' of a triangle in MATH with vertices MATH is given by integrating MATH over the triangle and a brief calculation reveals that this yields MATH, proving the lemma. |
math/9911103 | Since MATH, then MATH for some smooth, compactly supported function MATH. Now by definition, MATH, and since MATH is closed under the smooth functional calculus by the result of CITE, it follows that MATH. |
math/9911103 | Let MATH denote the smooth NAME kernels of MATH respectively, and MATH denote the smooth NAME kernels of MATH respectively. Then one has MATH since each term in the summand vanishes by symmetry, and we have used the fact that the fundamental domain MATH is compact in order to interchange the order of the summation and integral. |
math/9911103 | We need to show that MATH is a MATH-algebra which is weakly closed. We will establish that it is has a semifinite trace a bit later on. Let MATH. Since MATH, it follows that MATH. Since MATH it follows MATH. Clearly the identity operator is in MATH. Finally, if MATH and MATH converges weakly to MATH, it follows that for all MATH, MATH converges weakly to MATH and also to MATH. By uniqueness of weak limits, we deduce that MATH. |
math/9911103 | The proof follows by straightforward calculations as done above. The reader is warned that the righthand side of the inequality in REF is not necessarily finite. |
math/9911103 | Let MATH be a sequence of finite subsets of MATH which is an exhaustion of MATH, that is, MATH. For all MATH, define MATH by MATH . Then in fact MATH by definition, and using the previous lemma, we have MATH . By hypothesis, MATH, therefore MATH as MATH, since MATH is an increasing exhaustion of MATH. This proves that MATH. The second part is clear from the definition, once we identify MATH with the algebra of sequences MATH . |
math/9911103 | By the Lemma above, it follows that MATH are bounded operators commuting with the given twisted action, that is, MATH. Since the NAME kernels of MATH are smooth MATH by the Lemma above, it follows that MATH. Let MATH denote the word metric with respect to a given finite set of generators, and MATH the Riemannian metric on MATH. Then it is well known that MATH for some positive constant MATH. By REF above, one has, MATH for some positive constants MATH, and a similar estimate holds for MATH. Setting MATH observe that one has the estimate MATH for some positive constants MATH, since the volume growth rate of MATH is at most exponential. Therefore one has MATH for all positive integers MATH. By the Lemma above, it follows that MATH. |
math/9911104 | If the reduced rooted splice diagram is not minimal, the root vertex has just one edge emanating and that edge has far weight MATH. In terms of the NAME polygon this means that for some MATH there is a segment of slope MATH touching the MATH-axis (MATH is the near weight at the vertex adjacent to the root). Then MATH has the form MATH . Composing MATH with any of the automorphisms MATH will reduce the degree of MATH. |
math/9911104 | The specific form of MATH described in the theorem is easily seen if MATH is affine, so assume it is not. Since the automorphism MATH does not change the splice diagram, it does not change the degree of MATH (see REF). By REF , a normal form representation for the automorphism MATH involves exactly one factor from MATH. We can thus write it as MATH, with MATH and MATH, but with no restriction on non-triviality of MATH. We can modify MATH and MATH to make MATH triangular, say MATH with MATH. Denote MATH. By REF , MATH all have the same degree MATH say. Note that MATH has the same number of points at infinity as MATH. If MATH had a point at infinity other than MATH then it would have a monomial MATH with MATH. Taking the largest such MATH, it would follow that MATH has a monomial MATH and thus has degree larger than MATH. Thus MATH has just one point at infinity at MATH, so MATH fixes this point and is therefore in MATH. The same argument applied to MATH with MATH shows MATH so MATH. Thus MATH, as claimed. The statement about the degree of MATH follows by noting that, if the slope of the top segment in the NAME polygon is MATH and if MATH then a monomial MATH corresponding to the lowest point on this top segment of the NAME polygon for MATH leads to a monomial MATH for MATH. Since MATH, this completes the proof. |
math/9911112 | We write here the minimal choice of MATH, which we use in REF: MATH . For NAME algebras of classical series, the statement of the lemma with the above MATH follows from the explicit formulas for the entries MATH of the matrix MATH given in REF. For exceptional types, the lemma follows from a case by case inspection of the matrix MATH. |
math/9911112 | Denote by MATH the subalgebra of MATH generated by MATH. Let MATH be the inverse matrix to MATH from REF. The following formula for the universal MATH - matrix has been proved in CITE: MATH where MATH (here we use the notation REF), MATH, and MATH acts as follows: if MATH satisfy MATH, then MATH . By definition, MATH is obtained by taking the trace of MATH over MATH and then projecting it on MATH using the projection operator MATH. This projection eliminates the factor MATH, and then taking the trace eliminates MATH (recall that MATH acts nilpotently on MATH). Hence we obtain: MATH . The trace can be written as the sum of terms MATH corresponding to the (generalized) eigenvalues of MATH on the vectors MATH of the NAME basis MATH of MATH for the operators MATH (and hence for MATH). The eigenvalues of MATH on each vector MATH are given by REF . Suppose that MATH and MATH are given by REF . Then the eigenvalue of MATH on MATH equals MATH . Substituting into REF and recalling REF of MATH we obtain that the corresponding term MATH in MATH is the monomial REF. |
math/9911112 | Recall REF for MATH. We have: MATH; MATH is a generalized eigenvector of MATH; and MATH is a linear combination of tensor products MATH, where MATH has a lower weight than MATH. Therefore the diagonal matrix element of MATH on MATH equals the generalized eigenvalue of MATH on MATH. On the other hand, as explained in the proof of REF, the monomial MATH is equal to the diagonal matrix element of MATH corresponding to MATH. Therefore the diagonal matrix element of MATH corresponding to MATH equals the eigenvalue of MATH (considered as an element of MATH) on MATH. In particular, if MATH is the highest weight vector, then the corresponding monomial MATH is the highest weight monomial MATH. Therefore we find that the diagonal matrix element of the non-normalized MATH - matrix corresponding to MATH equals the eigenvalue of MATH on MATH. By REF the diagonal matrix element of the normalized MATH - matrix equals MATH. Therefore the eigenvalue of MATH on MATH equals the scalar function MATH. Therefore we obtain that the diagonal matrix element of the normalized MATH-matrix MATH corresponding to the vector MATH is equal to the eigenvalue of MATH on MATH. According to REF , MATH. Therefore, if MATH is given by REF , we obtain from REF that this matrix element is given by REF . |
math/9911112 | These formulas follow from the relations given in REF and the formula MATH. |
math/9911112 | The statement of the lemma follows from the fact that the matrix MATH is non-degenerate. |
math/9911112 | The monomials in MATH encode the common eigenvalues of MATH on MATH. It follows from REF that the monomials in MATH encode the common eigenvalues of MATH, and MATH, on MATH. Therefore we obtain that the restriction of MATH to MATH has a filtration with the associated graded factors MATH, where MATH is a MATH - module with MATH, and MATH is a one - dimensional MATH - module, which corresponds to MATH. By our assumption, the modules MATH over MATH are pairwise distinct. Because MATH commutes with MATH, there are no extensions between MATH and MATH for MATH, as MATH - modules. Hence the restriction of MATH to MATH is isomorphic to MATH. |
math/9911112 | This follows from the formula MATH. |
math/9911112 | The fact that multiplication by MATH makes the diagram commutative follows from REF . The uniqueness follows from the fact that MATH and the multiplication by MATH are injective maps. |
math/9911112 | The proof follows from a combination of REF . First, we observe that it suffices to prove the statement of REF for fundamental representations MATH. Indeed, then REF will be true for any tensor product of the fundamental representations. By REF , any irreducible representation MATH can be represented as a quotient of a submodule of a tensor product MATH of fundamental representations, which is generated by the highest weight vector. Therefore each monomial in a MATH - character of MATH is also a monomial in the MATH - character of MATH. In addition, the highest weight monomials of the MATH - characters of MATH and MATH coincide. This implies that REF holds for MATH. Second, REF is true for MATH. Indeed, by the argument above, it suffices to check the statement for the fundamental representation MATH. But its MATH - character is known explicitly (see REF ): MATH and it satisfies the required property. For general quantum affine algebra MATH, we will prove REF (for the case of the fundamental representations) by contradiction. Suppose that the theorem fails for some fundamental representation MATH and denote by MATH its MATH - character MATH. Denote by MATH the highest weight monomial MATH of MATH. Recall from REF that we have a partial order on the weight lattice. It induces a partial order on the monomials occurring in MATH. Let MATH be the highest weight monomial in MATH, such that MATH can not be written as a product of MATH with a monomial in MATH, MATH, MATH. This means that MATH . In REF we will establish certain properties of MATH and in REF we will prove that these properties can not be satisfied simultaneously. Recall that a monomial in MATH is called dominant if does not contain factors MATH (that is, if it is a product of MATH's in positive powers only). The monomial MATH is dominant. Suppose MATH is not dominant. Then it contains a factor of the form MATH, for some MATH. Consider MATH. By REF , we have MATH where MATH's are representation of MATH and MATH's are monomials in MATH. We have already shown that REF holds for MATH, so MATH where each MATH is a product of MATH's (in positive powers only), and each MATH is a product of several factors MATH note that MATH. Since MATH contains MATH by our assumption, the monomial MATH is not among the monomials MATH. Hence MATH for some MATH and MATH. There exists a monomial MATH in MATH, such that MATH. Therefore using REF we obtain that MATH where MATH is obtained from MATH by replacing all MATH by MATH. In particular, MATH and by our REF it can be written as MATH, where MATH is a product of MATH. But then MATH, and so MATH can be written as a product of MATH and a product of factors MATH. This is a contradiction. Therefore MATH has to be dominant. The monomial MATH can be written in the form MATH where MATH is a product of factors MATH, MATH, MATH. In other words, if MATH contains factors MATH, then all such MATH have the same index MATH. Suppose that MATH, where MATH contains a factor MATH. Let MATH be the generalized eigenspace of the operators MATH, corresponding to the monomial MATH. We claim that for all MATH we have: MATH . Indeed, let MATH (recall that MATH is obtained from MATH by erasing all MATH with MATH and MATH is a monomial in MATH, MATH, MATH). By REF, MATH belongs to the direct sum of the generalized eigenspaces MATH, corresponding to the monomials MATH in MATH such that MATH (with the same MATH as in MATH). By REF , MATH . In particular, MATH contains a factor MATH, and therefore all monomials MATH with the above property must contain a factor MATH. By our REF , the weight of each MATH can not be higher then the weight of MATH. But the weight of MATH should be greater than the weight of MATH. Therefore we obtain REF . Now, if MATH contained factors MATH and MATH with MATH, then any non-zero eigenvector (not generalized) in the generalized eigenspace MATH corresponding to MATH would be a highest weight vector (see REF ). Such vectors do not exist in MATH, because MATH is irreducible. The statement of the lemma now follows. Let MATH be any monomial in the MATH - character of a fundamental representation that can be written in the form REF. Then MATH is not dominant. We say a monomial MATH see REF has lattice support with base MATH if MATH. Any monomial MATH can be uniquely written as a product MATH, where each monomial MATH has lattice support with a base MATH, and MATH for MATH. Note that a non-constant monomial in MATH, can not be equal to a monomial in MATH if MATH. Therefore if MATH can be written in the form REF, then each MATH can be written in the form REF, where MATH if MATH, and MATH if MATH (note that the product over MATH in REF may be empty for some MATH). We will prove that none of MATH's is dominant unless MATH or MATH. Consider first the case of MATH, which has lattice support with base MATH. Then MATH . Define NAME polynomials MATH, MATH by MATH . If MATH can be written in the form REF, then MATH where MATH's are some polynomials with integral coefficients. All of these coefficients are non-negative if MATH. Now suppose that MATH is a dominant monomial. Then each MATH is a polynomial with non-negative coefficients. We claim that this is possible only if all MATH. Indeed, according to REF , the coefficients of the inverse matrix to MATH, MATH, can be written in the form REF, where MATH, MATH are polynomials with non-negative coefficients. Multiplying REF by MATH, we obtain MATH . Given a NAME polynomial MATH we will say that the length of MATH equals MATH. Clearly, the length of the sum and of the product of two polynomials with non-negative coefficients is greater than or equal to the length of each of them. Therefore if MATH, and if MATH, then the length of the LHS is greater than or equal to the length of MATH, which is greater than the length of MATH by REF . This implies that MATH for MATH. Hence MATH can be written in the form MATH . But such a monomial can not be dominant because its weight is MATH, where MATH, and such a weight is not dominant. This proves the required statement for the factor MATH of MATH (which has lattice support with base MATH). Now consider a factor MATH with lattice support with base MATH, such that MATH. In this case we obtain the following equation: the LHS of REF MATH. The previous discussion immediately implies that there are no solutions of this equation with non-zero polynomials MATH satisfying the above conditions. This completes the proof of the lemma. REF now follows from REF . |
math/9911112 | This follows from the proof of REF. |
math/9911112 | A simple computation shows that MATH. Let us show that MATH. For MATH, denote by MATH the subring MATH of MATH. We have: MATH . MATH . Let MATH, and suppose it contains MATH for some MATH and MATH. Then we can write MATH as the sum MATH, where MATH's are distinct monomials, which are products of the factors MATH (in particular, one of the MATH's could be equal to MATH), and MATH's are polynomials which do not contain MATH. Then MATH . By definition of MATH, MATH belongs to MATH, while MATH belongs to the direct sum of MATH, where MATH. Therefore if MATH, then MATH. Since MATH's are distinct, we obtain that MATH. But then MATH. Therefore MATH can be written as MATH, where each MATH is a linear combination of the MATH's, such that MATH. This proves that MATH where MATH. By repeating this procedure we obtain the lemma (because each polynomial contains a finite number of variables MATH, we need to apply this procedure finitely many times). According to REF, it suffices to show that MATH, where MATH . Denote MATH by MATH, MATH by MATH, and MATH by MATH. Note that MATH does not contain factors MATH. Let MATH be the shift operator on MATH sending MATH to MATH for all MATH. It follows from the definition of MATH that MATH if and only if MATH. Therefore (applying MATH with large enough MATH to MATH) we can assume without loss of generality that MATH. We find from the definition of MATH: MATH where MATH. Therefore each MATH can be written as a sum MATH, where each MATH and MATH . It suffices to consider the case MATH. Thus, we show that if MATH then MATH where MATH . Consider a homomorphism MATH sending MATH to MATH, MATH to MATH, and MATH to MATH. This homomorphism is surjective, and its kernel is generated by the elements MATH . Therefore we identify MATH with the quotient of MATH by the ideal generated by elements of the form REF. Consider the set of monomials MATH where all MATH, and also MATH for all MATH and MATH. We call these monomials reduced. It is easy to see that the set of reduced monomials is a basis of MATH. Now let MATH be an element of the kernel of MATH on MATH. Let us write it as a linear combination of the reduced monomials. We represent MATH as MATH. Here MATH is the largest integer, such that MATH is present in at least one of the basis monomials appearing in its decomposition; MATH is the largest power of MATH in MATH; MATH does not contain MATH, and MATH is not divisible by MATH. Recall that here both MATH and MATH are linear combinations of reduced monomials. Recall that MATH, MATH, and MATH is given by REF . Suppose that MATH. According to REF , MATH where the dots represent the sum of terms that are not divisible by MATH. Note that the first term in REF is non-zero because the ring MATH has no divisors of zero. The monomials appearing in REF are not necessarily reduced. However, by construction, MATH does not contain MATH, for otherwise MATH would not be a linear combination of reduced monomials. Therefore when we rewrite REF as a linear combination of reduced monomials, each reduced monomial occurring in this linear combination is still divisible by MATH. On the other hand, no reduced monomials occurring in the other terms of MATH (represented by dots) are divisible by MATH. Hence for MATH to be in the kernel, the first term of REF has to vanish, which is impossible. Therefore MATH does not contain MATH's with MATH. But then MATH, where MATH, and MATH. Such MATH is in the kernel of MATH if and only if all MATH and so MATH. This completes the proof of REF. |
math/9911112 | Since MATH is a ring homomorphism and both MATH, MATH are derivations, it suffices to check commutativity on the generators. Let us choose a representative MATH in each MATH - coset of MATH. Then we can write: MATH . By definition, MATH . Recall from REF that MATH equals MATH times a monomial in MATH, and from REF that MATH. Using these formulas we obtain: MATH . On the other hand, when MATH, MATH is a monomial in MATH, according to REF . Therefore MATH . The proves the lemma. |
math/9911112 | Let MATH be a finite-dimensional representation of MATH. We need to show that MATH. By REF, we can write MATH as the sum MATH, where each MATH is in the image of the homomorphism MATH, and MATH is a monomial in MATH. The image of MATH lies in the kernel of the operator MATH (in fact, they are equal, but we will not use this now). This immediately follows from the fact that MATH and MATH, which is obtained by a straightforward calculation. We also have: MATH. Therefore MATH. By REF, MATH. Since MATH is injective by REF, we obtain: MATH. |
math/9911112 | First we prove that the highest monomials are dominant. By REF, MATH . The statement of the lemma will follow if we show that a highest weight monomial contained in any element of MATH does not contain factors MATH. Indeed, the weight of MATH is MATH, and the weight of MATH is MATH. Denote MATH. Given a polynomial MATH, let MATH be its monomials (in MATH) of highest degree. Clearly, the monomials of highest weight in MATH (considered as a polynomial in MATH) are MATH, in which we substitute each MATH by MATH. These monomials do not contain factors MATH. The statement about the lowest weight monomials is proved similarly, once we observe that MATH . |
math/9911112 | For MATH, let MATH be a decomposition of the set of monomials in MATH with multiplicities into a disjoint union of subsets such that each subset forms the MATH - character of an irreducible MATH module. We refer to this decompostion MATH as the MATH-th decomposition of MATH. Denote MATH the collection of MATH, MATH. Consider the following colored oriented graph MATH. The vertices are monomials in MATH with multiplicities. We draw an arrow of color MATH from a monomial MATH to a monomial MATH if and only if MATH and MATH are in the same subset of the MATH-th decomposition and MATH for some MATH. We call an oriented graph a tree (with one root) if there exists a vertex MATH (called root), such that there is an oriented path from MATH to any other vertex. The graph MATH, where MATH is an irreducible MATH - module is always a tree and its root corresponds to the highest weight monomial. Consider the full subgraph of MATH whose vertices correspond to monomials from a given subset of the MATH-th decomposition of MATH. All arrows of this subgraph are of color MATH. By REF , this subgraph is a tree isomorphic to the graph of the corresponding irreducible MATH - module. Moreover, its root corresponds to a MATH - dominant monomial. Therefore if a vertex of MATH has no incoming arrows of color MATH, then it corresponds to a MATH - dominant monomial. In particular, if MATH has no incoming arrows in MATH, then MATH is dominant. Since by our assumption MATH does not contain any dominant monomials except for MATH, the graph MATH is a tree with root MATH. Choose a sequence of weights MATH as above. We prove by induction on MATH the following statement MATH: The algorithm does not fail during the first MATH steps. Let MATH be the resulting polynomial after these steps. Then the coefficient of each monomial MATH in MATH is not greater than that in MATH and the coefficients of monomials of weights MATH in MATH and MATH are equal. Furthermore, there exists a decomposition MATH of MATH, such that monomials in MATH can be identified with vertices in MATH in such a way that all outgoing arrows from vertices with MATH - weights MATH go to vertices of MATH. Finally, the MATH-th coloring of a monomial MATH in MATH is just the number of vertices of type MATH in MATH which have incoming arrows of color MATH in MATH. The statement MATH is obviously true. Assume that the statement MATH is true for some MATH. Recall that at the MATH-st step we expand all monomials of MATH of weight MATH. Let MATH be a monomial of weight MATH in MATH, which enters with coefficient MATH and coloring MATH. Then the monomial MATH enters MATH with coefficient MATH as well. Indeed, MATH is a tree, so all vertices MATH have incoming arrows from vertices of larger weight. By the statement MATH this arrows go to vertices corresponding to monomials in MATH. Suppose that MATH for some MATH. Then MATH is MATH - dominant. Indeed, otherwise each vertex of type MATH in MATH has an incoming arrow of color MATH coming from a vertex of higher weight. Then by the last part of the statement MATH, MATH. Therefore the monomial MATH is admissible, and the algorithm does not fail at MATH. Consider the expansion MATH. Let MATH be as in REF. In the MATH-th decomposition of MATH, MATH corresponds to a root of a tree whose vertices can be identified with monomials in MATH. We fix such an identification. Then monomials in MATH get identified with vertices in MATH. Let MATH be the vertex in MATH, corresponding to a monomial MATH in MATH. Denote the coefficient of MATH in MATH by MATH and the coloring by MATH. We have two cases: CASE: MATH. Then the last part of the statement MATH implies that the vertex MATH does not belong to MATH. We add the monomial MATH to MATH and increase MATH by one (we have already identified it with MATH). CASE: MATH. Then by MATH there exists a vertex MATH in MATH of type MATH with no incoming arrows of color MATH. We change the decomposition MATH by switching the vertices MATH and MATH and identify MATH with the new MATH. We also increase MATH by one. (Thus, in this case we do not add MATH to MATH.) In both cases, the statement MATH follows. Since the set of weights of monomials occuring in MATH is contained in a finite set MATH, the statement MATH proves the first part of the theorem. REF then implies the second part of the theorem. |
math/9911112 | For MATH, the statement follows from the explicit REF for MATH. The MATH - character of any irreducible representation MATH of MATH is a subsum of a product of the MATH - characters of MATH's. Moreover, this subsum includes the highest monomial. Hence if the highest weight monomial of MATH has positive lattice support with base MATH, then so do all monomials in MATH. Now consider the case of general MATH. Suppose there exists a monomial in MATH, which does not have positive lattice support with base MATH. Let MATH be a highest among such monomials (with respect to the partial ordering by weights). By REF , the monomial MATH is not dominant. In other words, if we rewrite MATH as a product of MATH, we will have at least one generator in negative power, say MATH. Write MATH in the form REF. The monomial MATH can not be among the monomials MATH, since MATH contains MATH. Therefore MATH for some MATH, which is a product of factors MATH. Let MATH be a monomial in MATH, such that MATH. Then by REF , MATH, where MATH is obtained from MATH by replacing all MATH with MATH. By construction, the weight of MATH is higher than the weight of MATH, so by our assumption, MATH has positive lattice support with base MATH. But then MATH also has positive lattice support with base MATH. Therefore all monomials in MATH have positive lattice support with base MATH. This implies that MATH, and hence MATH, has positive lattice support with base MATH. This is a contradiction, so the lemma is proved. |
math/9911112 | Let MATH contain MATH for some MATH, MATH. We write the restricted MATH - character MATH in the form REF, where each MATH is a MATH - character of an irreducible MATH module. The monomial MATH contains MATH and therefore can not be among the monomials MATH. But the graphs of irreducible MATH - modules are connected. So we obtain that MATH for some monomial MATH in MATH, and some MATH. By REF , we have MATH which is a contradiction. |
math/9911112 | Let us show first that from the highest weight monomial MATH there is only one outgoing arrow to the monomial MATH. Indeed, the weight of a monomial that is connected to MATH by an arrow has to be equal to MATH for some MATH. The restriction of MATH to MATH is isomorphic to the direct some of its MATH-th fundamental representation MATH and possibly some other irreducible representations with dominant weights less than MATH. However, the weight MATH is not dominant for any MATH and MATH. Therefore this weight has to belong to the set of weights of MATH, and the multiplicity of this weight in MATH has to be the same as that in MATH. It is clear that the only weight of the form MATH that occurs in MATH is MATH, and it has multiplicity one. By REF, this monomial must have the form MATH. Now, the graph MATH is connected. Therefore each monomial MATH in MATH is a product of MATH and factors MATH. Note that MATH is right negative and all MATH are right negative (this follows from the explicit REF ). The product of two right negative monomials is right negative. This implies the lemma. |
math/9911112 | The ``if" part of the Theorem is obvious. Let us explain the case when MATH. Let MATH be the transposition. By definition of MATH, the linear map MATH is a homomorphism of MATH - modules MATH. Therefore if MATH has a pole at MATH, then MATH is reducible. It is easy to generalize this argument to general MATH. Now we prove the ``only if" part. If the product MATH is reducible, then the product of the MATH - characters MATH contains a dominant monomial MATH that is different from the product of the highest weight monomials. Therefore MATH is not right negative and MATH is a product of some monomials MATH from MATH. Hence at least one of the factors MATH must be the highest weight monomial and it has to cancel with the rightmost MATH appearing in, say, MATH. According to REF, MATH where MATH is a product of MATH. By our assumption, the maximal MATH occurring among MATH is such that MATH. Using REF we obtain that one of the diagonal entries of MATH has a factor MATH, which can not be cancelled. Therefore MATH has a pole at MATH. This proves the ``only if" part. Moreover, we see that the pole necessarily occurs in a diagonal entry. |
math/9911112 | By REF , MATH must be antidominant. Thus, by REF , MATH for some MATH. Recall the automorphism MATH defined in REF. The module MATH is obtained from MATH by pull-back with respect to MATH. From the interpretation of the MATH - character in terms of the eigenvalues of MATH, it is clear that the MATH - character of MATH is obtained from the MATH - character of MATH by replacing each MATH by MATH. Therefore we obtain: MATH. Consider the dual module MATH. By REF, its highest weight equals MATH. Hence MATH is isomorphic to MATH for some MATH. Since MATH is a NAME algebra, the module MATH contains a one - dimensional trivial submodule. Therefore the product of the corresponding MATH - characters contains the monomial MATH. According to REF , it can be obtained only as a product of the highest weight monomial in one MATH - character and the lowest monomial in another. Therefore, MATH. In the same way we obtain that MATH is isomorphic to MATH. From REF for the square of the antipode, we obtain that the double dual, MATH, is isomorphic to MATH. Since MATH, we obtain that MATH. |
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