paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9905001 | We proceed by induction on MATH. For MATH, choose a system of parameters MATH and a neighbourhood MATH of MATH where MATH and MATH have regular representatives (which abusing notation we call MATH also) and MATH. Then the claim is clear, because MATH, and the virtual transform of MATH at MATH is MATH itself, so we can take MATH. Suppose now MATH and apply the induction hypothesis to MATH. We obtain the existence of MATH, MATH, MATH, MATH, and the polynomials MATH as in the claim. By making a linear substitution in MATH we may assume that the last point MATH of MATH lies in the direction of MATH. MATH is open in MATH, and is the blowing-up of MATH in MATH, which can be described as the subvariety of MATH given by the equation MATH, where MATH are projective coordinates of MATH. The exceptional divisor MATH has equations MATH. We define now MATH to be the open subset determined by MATH, and the isomorphism MATH as MATH. Let MATH be the projections of MATH on its two factors. Then it is easy to see that MATH generate the ideal of MATH. Because of the induction hypothesis we know that for any MATH and MATH the virtual transform of MATH at MATH relative to MATH is MATH . This virtual transform must have multiplicity at least MATH at MATH, therefore MATH and the virtual transform transform of MATH in MATH is given locally by MATH which because of the commutativity MATH can be written in terms of the local parameters MATH: MATH . This allows us to define the polynomials MATH after which the claim is satisfied. |
math/9905001 | By induction on MATH. The case MATH is clear, because then MATH is the closed subset of MATH determined by the ideal MATH. Suppose now MATH and the claim is true for MATH. Let MATH be a cluster and MATH the open neighbourhood given by REF . It will be enough to see that MATH is closed in MATH, because MATH can be covered by a finite number of such open neighbourhoods. Define MATH . Because of the induction hypothesis and the fact that MATH is continuous, MATH is closed in MATH. Moreover MATH if and only if MATH and the virtual transform of MATH at MATH has multiplicity at least MATH, that is, by REF , MATH . These equations define a closed subset MATH, therefore MATH is closed in MATH. |
math/9905001 | Let MATH be a cluster and MATH the open neighbourhood given by REF . As in REF , it will be enough to see that MATH is closed in MATH. We also know from the proof of REF that MATH is defined by a finite number of polynomials MATH, MATH. Now for MATH, MATH for all MATH, so the polynomial MATH must be identically zero. This implies in fact that MATH . Now MATH is the affine coordinate ring of MATH, which is a MATH - vector space admiting the basis MATH. Therefore the MATH define a NAME - closed subset in MATH, and it is immediate to see that this is in fact MATH. |
math/9905001 | Choose MATH such that MATH for all MATH. Then MATH and the claim follows because the dimension of the fibers of the (non flat) family MATH is upper semicontinuous. |
math/9905001 | The morphism MATH defines a closed subset MATH, and we have to see that MATH. As MATH is closed and MATH it will be enough to see that MATH is the closure of MATH. Take MATH, with MATH. The family determined by MATH corresponds to an ideal sheaf MATH. By definition, MATH, so there is an open neigbourhood MATH of MATH in MATH and a section MATH with MATH. As MATH is open, we may assume MATH, therefore MATH. So we have a morphism MATH whose image contains MATH and has every other point in MATH, therefore MATH lies in the closure of MATH. As this holds for every MATH and MATH is closed, the claim is proved. |
math/9905001 | It is clear that MATH. On the other hand, as MATH is consistent, there are curves going sharply through MATH which miss MATH (compare CITE), so MATH. Abusing slightly notations, we will call MATH; as the length of MATH does not depend on MATH we have MATH and the schemes MATH form a flat family. |
math/9905001 | CASE: MATH is consistent in MATH, but not consistent in MATH, otherwise MATH. Therefore the proximity inequality at MATH must be satisfied for MATH but not for MATH. This means that MATH . Therefore, if MATH we must have MATH, and for any cluster MATH we have MATH. So MATH for all MATH, so MATH against the hypotheses. We conclude that MATH, and it only remains to be seen that the case MATH is not possible. But in this case MATH in MATH, so again MATH for all MATH and MATH. CASE: We are given two systems of multiplicities MATH with MATH and MATH. We have to see that for any cluster MATH there is a deformation of MATH whose general member is of the form MATH with MATH. For that, let MATH be a smooth irreducible curve MATH with MATH and consider the morphism MATH . Consider also an auxiliary system of multiplicities and the associated morphism MATH . For MATH, MATH. Therefore MATH for all MATH. Moreover, REF shows that MATH, so MATH. On the other hand, MATH does not satisfy the proximity inequality at MATH, and unloading multiplicity on this point gives MATH . Note that MATH may be non - consistent, but still MATH and MATH. All together, we have MATH and we are in the conditions of REF . Therefore there is a point MATH in the exceptional divisor MATH of blowing up MATH such that MATH, with MATH . Let MATH be the only point proximate to MATH. If MATH then unloading gives MATH so MATH is the family we are looking for. If MATH is not proximate to MATH then we only know that MATH. As MATH, though, by REF we can say that for any free point MATH, MATH. As the free points are dense in MATH, this implies MATH. Now it is enough to see that MATH, because then MATH. If MATH is consistent for MATH free, then MATH because of REF ; if it is not, then the equivalent consistent system obtained by unloading is MATH, whereas for the cluster MATH the equivalent consistent system is MATH. This implies that MATH, so again MATH. CASE: Follows from an easy unloading calculation. |
math/9905001 | We will proceed by induction on MATH. For MATH, there is nothing to prove. For MATH, we have either MATH or MATH so we can apply the induction hypothesis and MATH . On the other hand, a straightforward computation shows that MATH, therefore MATH. In these contitions, REF tells us that MATH thus finishing the proof. |
math/9905001 | By induction on MATH. The cases MATH are clear, so assume MATH and the claim to be true for MATH and MATH. This means that MATH, MATH are closed subvarieties and the morphism MATH is the restriction of MATH. So we have also a closed subvariety MATH and MATH so by the definitions, MATH is the strict transform of MATH under the blowing-up MATH (compare CITE), MATH is a subvariety of the exceptional divisor, hence of MATH, and MATH is the restriction of MATH. |
math/9905001 | Write MATH with MATH. It is enough to define MATH and MATH . |
math/9905001 | It is clear that MATH; we will prove the claim by induction on MATH. For MATH the result is obvious. For MATH we have MATH. By semicontinuity it is enough to see that there are no plane curves of degree MATH containing MATH with MATH general in MATH. Now for MATH, unloading gives MATH, and the result follows from the induction hypothesis. |
math/9905001 | By semicontinuity it is enough to see that there are no plane curves of degree MATH containing MATH with MATH general in MATH. But for MATH, unloading gives MATH, and the result follows from REF . |
math/9905001 | The proof runs by induction on MATH, the case MATH being clear from REF . Assume MATH. For MATH general in MATH, consider the line MATH passing through MATH in the direction of MATH. If MATH we can assume (MATH being general) that MATH does not lie on MATH; for MATH this is automatic. For every curve MATH therefore MATH is a component of MATH, so it is a fixed part of MATH. The residual linear system is MATH with MATH because of REF . As this is true for general MATH we have now MATH. But MATH is not consistent in MATH. The equivalent consistent system is MATH . Observe that MATH. Finally MATH because of REF in the first case and because of the induction hypothesis in the second. |
math/9905001 | By REF we may assume MATH. But then MATH so MATH, and REF concludes. |
math/9905001 | By REF we may assume MATH. We will distinguish two cases according to the parity of MATH. CASE: Write MATH. As MATH is consistent in MATH, and MATH, we must have MATH, so by semicontinuity it is enough to see that there are no plane curves of degree MATH containing MATH with MATH general in MATH. It is clear that MATH does not contain MATH, and MATH, therefore by REF MATH . By semicontinuity applied to the tautological flat family on MATH then, it is enough to see that there are no plane curves of degree MATH containing MATH with MATH general in MATH and MATH. But then MATH and still MATH so the result follows by REF . CASE: Write MATH. By semicontinuity it is enough to see that there are no plane curves of degree MATH containing MATH with MATH general in MATH. But for MATH, unloading gives MATH, and again MATH and MATH so the result follows by REF . |
math/9905001 | Consider MATH. Let MATH be the closed variety of MATH where MATH are proximate to MATH. It is easy to see that MATH is an irreducible variety and MATH. Then the claim may be equivalently stated as MATH where MATH. By semicontinuity it is enough to see that MATH and unloading gives that MATH is equivalent to MATH in MATH. Therefore MATH . It only remains to be seen that for a general cluster MATH, MATH has maximal rank. Because of REF we can assume that MATH has level MATH, and it is enough to see that MATH. Let MATH, MATH. The proof runs by induction on MATH. The case MATH has already been settled in REF , so suppose MATH. We will distinguish two cases according to the parity of MATH. CASE: Write MATH. As MATH is consistent in MATH and MATH we must have MATH, so by semicontinuity it is enough to see that there are no plane curves of degree MATH containing MATH with MATH general in MATH. It is clear that MATH does not contain MATH, and MATH, therefore by REF MATH . By semicontinuity applied to the tautological flat family on MATH then, it is enough to see that there are no plane curves of degree MATH containing MATH with MATH general in MATH and MATH . An easy computation shows that we are still in the numerical conditions of the claim, and MATH, so we can apply the induction hypothesis, except in the case that MATH. In this case, either we are in one of the exceptions above, or MATH, MATH, MATH and the claim follows from REF . CASE: Write MATH. By semicontinuity it is enough to see that there are no plane curves of degree MATH containing MATH with MATH general in MATH. But for MATH, unloading gives MATH, so the result follows by the induction hypothesis again. |
math/9905001 | REF and the fact that all singularities of general curves sit at the proper base points of MATH are standard. For a proof of REF, compare CITE. Compare also NAME 's remark on REF. |
math/9905001 | This is an easy application of the residual exact sequence of the NAME method. There is a unique point MATH where the length of the component of MATH supported at MATH is bigger than that of MATH. Let MATH be a general straight line through MATH, and consider the residual exact sequence MATH . We have to prove that MATH . As going through MATH imposes independent conditions to curves of degree MATH and MATH is general, MATH, therefore MATH and MATH. As by REF, it will be enough to see that MATH contains the residual scheme MATH. Let MATH be the ideals locally defining MATH, MATH and MATH, and let MATH be a local equation of MATH. MATH implies that MATH for some MATH with MATH, therefore MATH and MATH, as wanted. |
math/9905001 | Because of REF , MATH has no fixed part and MATH. By REF then, we only have to see that if MATH is composed of the curves of a pencil then MATH is a point with multiplicity MATH. Let MATH be composed of MATH curves of degree MATH in a pencil. Then MATH and MATH because MATH imposes independent conditions to curves of degree MATH. Therefore MATH and MATH. A pencil of lines has a unique base point MATH so curves in MATH are composed of MATH lines through MATH and MATH is as claimed. |
math/9905001 | Consider a scheme MATH where MATH is a tacnode scheme of order MATH supported at MATH and MATH is a cusp scheme of order MATH supported at MATH, all of them having their points in general position. We claim that the linear system MATH of curves of degree MATH containing MATH is nonempty and that a generic curve in it has a tacnode of order MATH at MATH, a cusp of order MATH at MATH and no other singularities. Specializing the cusp schemes MATH to tacnode schemes MATH of order MATH we obtain a scheme MATH in the conditions of REF , therefore of maximal rank and length MATH. So by semicontinuity MATH is also of maximal rank. The bound on MATH assures that MATH imposes independent conditions to curves of degree MATH. So by REF a general curve in MATH is irreducible and has no other singularities but the ones in MATH, MATH, which are tacnodes and cusps of the desired orders. |
math/9905001 | Analogous to the proof of REF. |
math/9905001 | The cluster of infinitely near singular points of a MATH - singularity, with MATH even, is a weighted unibranched cluster with MATH free points with multiplicities MATH, so we are in a particular case of REF . If MATH is odd, then the cluster of infinitely near singular points of MATH is a weighted unibranched cluster with MATH points, the last of which being satellite and the others free (that is, MATH), with multiplicities MATH. For simplicity, we assume MATH (or MATH), as the cases with MATH small require special care. Let MATH be the open subset where only MATH and MATH are satellites (we specialize the third point to be proximate to the first). For MATH, unloading gives MATH with MATH, so the length of MATH is the same for MATH in MATH or in MATH, therefore MATH. As MATH is dense in MATH, by semicontinuity it is enough to see that MATH has maximal rank for MATH general in MATH and the result follows from REF , because the multiplicity of the last point is now REF so we can assume MATH general in MATH. |
math/9905007 | This follows from REF and CITE. |
math/9905007 | REF is in CITE, see also CITE. MATH follows from REF , and REF is trivial, by defining MATH from MATH. |
math/9905007 | We want to prove MATH . Multiplying to the left by MATH and to the right by MATH, this amounts to showing MATH as a map from MATH into MATH. To this end, pick a MATH such that MATH where MATH and MATH here refers to MATH-class in MATH. It suffices to verify the commutation on these MATH since they generate MATH. Pick projections MATH such that MATH . We may and will assume that MATH. Note that MATH and MATH . If MATH is any real-valued additive character on MATH (no positivity assumed), there is a unique linear functional MATH on MATH corresponding to MATH, which restricts to a scalar multiple of the standard trace on each MATH, such that MATH for each projection MATH in MATH. To show our commutation, we just need to show MATH . But this follows from the computation where we use MATH to denote MATH . Here the second equality follows from the definition of MATH, and the third from MATH and the fact that MATH commutes with MATH and MATH, and MATH commutes with MATH so MATH is a projection in a tensor product complement of MATH. |
math/9905007 | If MATH, MATH, choose again projections MATH and MATH such that MATH . Here we have identified MATH with a subgroup of MATH through MATH. Thus, for any real valued additive character MATH on MATH we have again MATH establishing the lemma. |
math/9905007 | By definition of MATH we have MATH . But MATH so MATH and hence MATH . Replacing MATH by MATH in the reasoning above, we obtain the other identity. |
math/9905007 | Suppose that MATH. Then as in the proof of REF we argue that MATH is properly outer where we use the fact that MATH for MATH. Thus we conclude that MATH is properly outer on MATH. |
math/9905007 | Without the condition MATH, this follows from CITE. To obtain this condition we consider each cycle MATH separately, and thus we may suppose that MATH is one cycle, that is, MATH except MATH. In this case we find MATH projections MATH, MATH, MATH in MATH such that MATH where MATH. Then we set MATH for MATH. It follows that MATH . |
math/9905007 | For each MATH in REF we find a projection MATH such that MATH. However this only shows that MATH, and MATH will never be faithful. To overcome this difficulty we replace MATH by a smaller projection and we set MATH as in CITE. Note that we have not assumed that MATH. |
math/9905007 | First we take a sequence MATH of projections in MATH which represents MATH in REF and satisfies that MATH and for MATH we have that MATH . For any MATH with MATH it follows from the density of MATH in MATH that there exists a function MATH such that MATH . If MATH then MATH since MATH. Thus for a sufficiently large MATH, it follows that MATH in MATH. Then we find a subprojection MATH of MATH in MATH such that MATH . For an increasing MATH and a decreasing MATH we use the above argument to replace MATH by a smaller projection satisfying the required properties. |
math/9905007 | By using the projections MATH in REF we obtain that MATH are almost mutually orthogonal when MATH and we must show that they are equivalent as projections in MATH. To show that they are equivalent, we regard MATH as homomorphisms of MATH into MATH and compute by REF - REF . REF MATH . With a similar computation for MATH we obtain that MATH that is, MATH in MATH. We can repeat this process. Thus we find a C*-subalgebra MATH of MATH such that MATH is a factor of type MATH which almost contains the MATH projections MATH, the identity of MATH is close to MATH when the distance is measured with the traces MATH, and the large portion of MATH is MATH-invariant. From this one can deduce the NAME property for MATH as in CITE. By reinterpreting MATH in terms of MATH and MATH, one obtains the NAME property of MATH. |
math/9905007 | By REF it suffices to show that there is a MATH such that MATH and MATH for MATH. (The existence of MATH is clearly necessary for the density. For the sufficiency, let us extract the apposite argument from CITE: Let MATH and put MATH. Then MATH. By NAME theorem, there exists a real polynomial MATH approximating MATH, and we may assume that each MATH has the form MATH. If MATH and MATH we may approximate each MATH arbitrarily well on MATH by MATH by choosing MATH large and replacing MATH by a finite number, and thus we may approximate the first order polynomials MATH arbitrarily well by polynomials MATH for MATH. Thus MATH is uniformly small for MATH, and since MATH and MATH, the sufficiency follows.) We may use MATH to establish REF and MATH to establish REF in the Proposition. In REF , let MATH and MATH, where MATH are integers, and let MATH for a sufficiently large MATH. |
math/9905007 | Note that MATH follows automatically. If MATH for an automorphism MATH of MATH, then there is an isomorphism MATH of MATH onto MATH such that MATH . By setting MATH and MATH, all the properties follow easily. |
math/9905007 | In view of REF it suffices to show this when MATH is a cocycle perturbation of MATH, that is, MATH with MATH a one-cocycle for MATH. We define an isomorphism MATH of MATH onto MATH by MATH . Since MATH, MATH, MATH naturally extends to an isomorphism MATH of MATH onto MATH. By setting MATH and MATH, all the properties follow easily, perhaps except for MATH. For this we shall show that MATH. Supposing that MATH is represented on a NAME space MATH, we represent MATH on MATH by MATH for MATH, MATH, and MATH. Then MATH is generated by MATH and MATH, MATH and MATH is generated by MATH, MATH, which naturally identifies with MATH. Then MATH is given by MATH and MATH is implemented by the unitary MATH defined by MATH which is a multiplier of MATH. Thus we obtain the commutative diagram: MATH . Thus the assertion follows. |
math/9905007 | We have shown REF Suppose REF . Note that MATH and MATH are stable AF algebras and MATH are all isomorphisms such that MATH is commutative. Then by a standard intertwining argument we obtain isomorphisms MATH such that MATH is commutative and MATH, MATH, and MATH. Since MATH is purely infinite and MATH has only one trace MATH up to constant multiple, MATH must scale MATH, that is, MATH with MATH. The same is true for MATH. Hence by REF, MATH, MATH, MATH, MATH have the NAME property and hence MATH and MATH (respectively, MATH and MATH) are outer conjugate, that is, MATH and MATH for unitaries MATH in MATH and automorphisms MATH of MATH, respectively. Moreover we can assume that MATH and MATH. Since MATH we obtain MATH which implies, with MATH and MATH, MATH . Let MATH (respectively, MATH) be the spectral projections for MATH, MATH in MATH (respectively, MATH), that is, MATH . Since MATH and MATH, we obtain that MATH . Since MATH forms an approximate identity for MATH, there is a unitary MATH in MATH such that MATH, MATH. Thus replacing MATH by MATH we obtain an isomorphism MATH of MATH onto MATH such that MATH where MATH is a unitary in MATH. Since MATH commutes with MATH. Let MATH . Then MATH commutes with MATH, and satisfies that MATH . We define a unitary MATH by MATH . Then it follows that MATH . Hence we obtain, replacing MATH by MATH and MATH . Then the extension of MATH to the multiplier algebras maps MATH onto MATH and MATH . (compare CITEEF). Thus MATH is conjugate to MATH. |
math/9905007 | We have shown REF Assume that REF holds. As in the proof of REF we obtain an isomorphism MATH of MATH onto MATH such that MATH for some unitary MATH. Since MATH, MATH is isomorphic to MATH, MATH, which is again isomorphic to MATH, we obtain an isomorphism MATH of MATH onto MATH such that MATH, extending MATH. The image of MATH in MATH is MATH; and MATH preserves this class, that is, MATH where MATH is a minimal projection in MATH. Thus we find a unitary multiplier MATH of MATH such that MATH is the identity on MATH. Then there is an automorphism MATH of MATH such that MATH, from which it follows that MATH . Using MATH, which is a one-cocycle for MATH, we obtain that MATH . This shows that MATH is cocycle conjugate to MATH. |
math/9905007 | By general theory we obtain that REF . From a part of the proof of REF we obtain REF . Assume that REF holds. Denoting by MATH the compact operators on MATH, we consider the systems MATH and MATH. Since MATH is a hereditary C*-subalgebra of MATH and MATH is a stable simple AF algebra, we have that MATH. In the spectral subspace MATH we find a unitary MATH CITE. Note that MATH is generated by MATH and MATH. Define an automorphism MATH of MATH by MATH. Then, as MATH is isomorphic to MATH, which implies that MATH. In the same way we obtain a unitary MATH in MATH and that MATH . Hence, by using the NAME property for MATH and MATH, we obtain an isomorphism MATH of MATH onto MATH such that MATH for some unitary MATH. Then MATH extends to an isomorphism MATH of MATH onto MATH such that MATH for MATH. It follows that MATH. |
math/9905010 | Immediate from the definition. |
math/9905010 | Notice all simple objects in one orbit give the same link invariant, so we could as well compute MATH by taking the sum only over a representative MATH of each orbit class, and replacing the factor MATH with MATH where MATH is the stabilizer of MATH and thus MATH is the number of elements in the orbit of MATH . Because the functor is a ribbon functor we can replace the invariant associated with MATH with the invariant associated with MATH if we replace the label MATH on MATH with its image under the functor. This image is the direct sum of MATH many simple objects, each with quantum dimension MATH . Decomposing the invariant into a sum of invariants, each with the link labeled by a single one of these simple objects, we obtain MATH . |
math/9905010 | See the end of REF. |
math/9905010 | It is well known that MATH is isomorphic to the group MATH by the map sending MATH to the homomorphism MATH on MATH such that MATH acts on the classical representation MATH by MATH (this is clearly a homomorphism descending to MATH and is injective by the faithfulness of the left regular representation. That the domain and range have the same dimension is shown in CITE). Thus by REF we can construct a bijection MATH from MATH to the set of fundamental weights MATH meeting the two characterizations of the proposition. This gives the first two sentences of the proposition. For the rest, we argue as follows. We first show that the map MATH defined in the proposition satisfies the conditions of the proposition. We then use this to give the characterization of the range of MATH in the last sentence of the proposition. Finally this will allow us to show that MATH is a homomorphism. The map MATH is certainly an isometry of the weight space taking the weight MATH (that is, the object MATH) to MATH . The image of the elements of the NAME alcove are those weights MATH such that MATH for MATH and MATH . Since MATH for MATH and MATH such MATH are exactly the elements of the NAME alcove. Thus MATH is an isometry taking the NAME alcove (and the simplex) to itself. To see that MATH note that for any element MATH of the quantum NAME group MATH where MATH is another element of the quantum NAME group with the same sign, because this is true for generating reflections. Thus in REF MATH . This confirms MATH . Now MATH certainly takes a neighborhood of the weight MATH to a neighborhood of the weight MATH and since it is an isometry of the simplex it also connects these neighborhoods intersected with the simplex. Conversely, if MATH is an extreme point of the simplex and MATH is an isometry of a neighborhood of the weight MATH intersected with the simplex to a neighborhood of MATH intersected with the simplex, then MATH extends to an isometry from an entire neighborhood of the weight MATH to a neighborhood of the weight MATH which takes the hyperplanes MATH to the hyperplanes MATH for all MATH and for some MATH together with the hyperplane MATH . Therefore MATH is an isometry taking MATH to MATH and the hyperplanes MATH to MATH . Such an isometry is necessarily a composition MATH with MATH an element of the classical NAME group and MATH a linear isometry permuting the simple roots MATH (such isometries of the NAME chamber correspond to automorphisms of the NAME graph). The map MATH corresponds to the base MATH and therefore by REF , the associated MATH is in the range of MATH . Since the composition of isometries MATH of the alcove takes MATH to MATH we must have that MATH and we see that every corner of the simplex which is locally isometric to the corner MATH is in the image of MATH . To see that MATH is a homomorphism, let us first consider the case MATH . Then if MATH we have MATH which is again an extreme point locally isometric to MATH and thus is MATH for some MATH . Of course in general each MATH such that MATH differs from MATH by an element of the root lattice, and thus MATH . In particular MATH and MATH . Thus MATH is an endomorphism. Finally notice that the map multiplication by MATH where it is defined forms a commuting square with the maps MATH acting respectively on the NAME alcove at level MATH and the NAME alcove at level MATH and thus the map MATH takes on the one hand MATH to MATH and on the other MATH to MATH and thus MATH is an endomorphism. |
math/9905010 | Suppose MATH is long, and MATH . Since MATH is the coefficient of MATH in the expansion of MATH in terms of simple roots, we have MATH for some positive integers MATH. To conclude the existence of a unique MATH as in the statement of the lemma, it suffices to show that MATH is a base Since the coefficient of MATH in the expansion of MATH is MATH every positive root can be written in terms of the original base with the coefficient of MATH being either MATH or MATH . In the former case the root is already a nonnegative combination of MATH in the second MATH writes the positive root as a sum of two terms, each with all nonpositive coefficients in the new base. That the second condition implies the first is a straightforward inverting of the above argument. For the second assertion, notice that the homomorphism is well-defined on MATH because MATH so MATH is an integer for all roots MATH . A count of such MATH from REFCITE (For MATH all fundamental weights, for MATH for MATH for MATH and MATH for MATH and for MATH) shows that we need only check that no MATH gets sent to the trivial homomorphism. This is confirmed by computing inner products MATH using REFCITE. |
math/9905010 | By REF , the weights occurring in the decomposition of the truncated tensor product of two weights lie in the product of their cosets in MATH and thus are annihilated by any homomorphism which annihilates the factors (since the reflections that generate the quantum NAME group preserve these cosets). Thus MATH is closed under the truncated tensor product. Of course it is closed under the duality relation, since duality corresponds to inverse in MATH . In MATH the truncated tensor product and dual on simple objects corresponds to product and inverse in MATH so closure is immediate. |
math/9905010 | See end of REF. |
math/9905010 | See end of REF. |
math/9905010 | Of course since MATH on MATH can be identified with group multiplication on MATH the only closed subsets of MATH will be subgroups and hence of the form MATH . So assume MATH and that MATH is degenerate. We will show that MATH is in the range of MATH which suffices for the theorem. In that case if MATH then MATH so MATH for any MATH with MATH and likewise for MATH . If MATH is not a corner of the NAME alcove then MATH by REF so MATH so MATH is not degenerate. Likewise in the nonsimply-laced case if MATH for some short simple root MATH then by REF MATH so MATH since MATH so MATH is not degenerate. Now if MATH is in the NAME alcove for some long root MATH then by REF MATH so MATH . Also MATH for all MATH so MATH can only be a multiple of MATH if CASE: MATH or CASE: MATH . If MATH is a corner and is orthogonal to all short simple roots, then MATH is MATH for MATH long, where MATH . If MATH then MATH is in the NAME alcove, MATH and MATH . For MATH to be degenerate requires this quantity to be equal to MATH (since it is negative, it is not MATH), so that MATH and we conclude MATH is in the range of MATH by REF . Thus the only possible degenerate objects which are not in the range of MATH are weights MATH dual to long roots for MATH . We argue first that for each such MATH there is a long positive root MATH with MATH such that either MATH is in the NAME alcove with MATH or MATH is a long root and MATH is in the NAME alcove, and second that this contradicts degeneracy. To see the existence of such a MATH observe from the NAME diagrams CITE[pg. REF] that for every fundamental weight MATH dual to a long root, either MATH or for one of the subdiagrams into which the removal of MATH divides the diagram, the weight MATH adjacent to MATH is dual to a long root and satisfies MATH for MATH the highest root associated to this subdiagram. In the first case MATH is in the image of MATH in the second MATH will do, and in the third we choose MATH . Of course in the third case MATH for MATH and MATH because the decomposition of MATH into simple roots contains exactly one simple root adjacent to MATH . So MATH is in the NAME chamber. If MATH then MATH is in fact in the NAME alcove. If not then MATH . Except for the case MATH where all corners are in the range of MATH and there is nothing to prove, MATH is of the form MATH for some MATH so MATH indicates that the subdiagram contained that MATH . Further inspection of the NAME diagrams indicates that the MATH for which the only subdiagram meeting the desired conditions contains this MATH are MATH of MATH and MATH of MATH . In the first case MATH and thus MATH is in the NAME alcove. In the second use MATH is a positive root, MATH is a root, and MATH is in the NAME alcove, so MATH meets the desired condition. Thus we need only check that the existence of such MATH contradicts degeneracy. In the first case MATH because of course MATH . In the second MATH . Thus the only degenerate objects for MATH are those in MATH . Now if MATH it is not true that MATH . However, the only elements of MATH and hence of MATH are elements of MATH and weights MATH with MATH short. Thus the argument involving MATH above suffices to show that only those in MATH can be degenerate. |
math/9905010 | Since we have exempted MATH is cyclic. |
math/9905010 | We note first that if MATH is in the NAME alcove, and MATH is an element of the quantum NAME group taking a weight MATH not in the NAME alcove into the NAME alcove, then the distance between MATH and MATH is strictly less than the distance between MATH and MATH . To see this, note that if MATH is in the NAME alcove, than MATH cannot lie on one of the `walls of the NAME alcove,' that is, the hyperplanes reflection about which generates the quantum NAME group (if it did, it and all its conjugates would have nontrivial stabilizers, which is not true of any point in the NAME alcove). In that case, one of the walls of the alcove lies between MATH and MATH and thus reflection about this wall brings MATH strictly closer to MATH repeating this procedure brings a sequence of weights conjugate to MATH getting strictly closer to MATH . Since there are only finitely many weights in the weight lattice a given distance from MATH this process must end after finite time. It can only end by reaching a point which is conjugate to MATH and which lies in the alcove or on the walls. Since the quantum NAME group acts transitively on the NAME alcove CITE, this gives the claim. Now let MATH and MATH be as in the statement of the lemma, so any MATH for which MATH is nonzero must be a distance at most MATH from MATH so if MATH is not in the NAME alcove but is conjugate to MATH which is, then MATH is a distance less than MATH from MATH and hence is not MATH . Thus the only contribution to MATH in REF comes from MATH and would be MATH . |
math/9905010 | We will make the argument for MATH noting parenthetically how it differs for MATH when not simply-laced. We will actually prove that MATH is nonzero, which is equivalent. Recall that since MATH is the weight MATH (respectively MATH the number of short simple roots). Thus in the sum REF , there is a contribution of MATH REF from the identity element of the quantum NAME group and a contribution for each MATH such that MATH . By the first paragraph in the proof of REF above, we saw that MATH can be written as a product of reflections each taking MATH strictly farther away from itself. Since MATH is in the root lattice, we conclude that after one such reflection its length is at least that of a short root, after two its length is at least that of a long root, and after three it must be longer than a long root. Thus if MATH is such a product of three or more reflections, MATH . A product of two reflections only increases MATH so it suffices to consider the effect of a single reflection. If MATH is reflection about one of the walls of the NAME alcove and MATH REF then MATH is either MATH or MATH depending on which wall, and MATH (MATH if the root is short). Thus MATH is at least MATH minus the number of walls of the NAME alcove to which MATH is adjacent (MATH minus the number of walls dual to short roots to which MATH is adjacent). In the simply-laced case, only corners are adjacent to MATH walls. In the nonsimply-laced case, only weights for which MATH for all short simple roots MATH are adjacent to MATH walls dual to a short root. |
math/9905010 | First, we note that MATH because they have the same inner product with the simple roots, and that MATH for all MATH . It suffices to check the second claim on simple roots. For MATH we have that MATH is another simple root of the same length, different from MATH and thus both sides of the equation are zero (the object MATH). For MATH we have MATH so MATH . By REF MATH where we have used the fact that MATH . Using the fact that MATH is an isometry and the identities in the previous paragraph gives MATH . |
math/9905010 | If MATH is invertible, NAME shows it is degenerate for MATH if and only if MATH for every MATH which by REF is true if and only if MATH annihilates all MATH . To determine whether MATH is odd or even, we need to check whether MATH which is to say whether MATH is an even or odd multiple of MATH . Notice MATH (Using the identities in the proof of the previous lemma). So we are asking whether MATH is an even or odd integer. |
math/9905010 | Since MATH the image of MATH under MATH annihilates MATH and hence MATH . Since MATH so the image of MATH under MATH annihilates MATH or equivalently the image of MATH annihilate MATH . Thus the image of MATH annihilates MATH . By REF , this means MATH (which of course is contained in MATH) will consist of degenerate units for MATH . If MATH consists entirely of even degenerate objects, then REF is met. REF are clear, and REF follows from REF . In the case where MATH notice MATH contains MATH . Thus by REF MATH consists of a union of cosets of MATH which is to say that it is of the form MATH for some MATH . Of course MATH but anything in MATH annihilates MATH and MATH so it is contained in MATH and in MATH so MATH and MATH . By REF , this is one of the NAME theories. To see these are the only cases of products of ribbon categories, suppose MATH is a closed subset which can be written as a product of two closed subsets MATH and MATH . If both were of the form MATH and not of the form MATH then both would contain MATH . Since this is not invertible except in the case MATH at level MATH when one would have to be MATH . REF prevents both from being only of the form MATH . They cannot both be of the form MATH because then the product would be of the form MATH with a MATH which was a product of groups, which only happens for MATH . Thus one must be of the form MATH and one of the form MATH for some MATH . But by REF , MATH must annihilate MATH under MATH and thus MATH . |
math/9905010 | A little thought will convince the reader that this result is almost immediate assuming that the intersection (a group of degenerate invertibles) acts freely on MATH . This is in fact the case, though one would like a more direct argument than the one below. Let MATH be the group MATH . Recall that, if MATH is the vector space of formal linear combinations of elements of MATH then one can extend the link invariant to an invariant of links labeled by MATH in such a way that it is linear in each component. Furthermore, the functor to MATH gives a linear map from MATH into the corresponding MATH which is consistent with the link invariant and a vector is in the kernel of this map if and only if it is in the kernel of the invariant for every link. In particular the NAME link gives a nondegenerate pairing on the image of MATH in MATH . NAME proves that if the NAME link is labeled respectively by MATH and MATH then the invariant is MATH unless MATH is degenerate, in which case it is MATH . The same is necessarily true for MATH and MATH and likewise for MATH in MATH . Thus the value of a NAME link labeled by MATH and a typical element MATH in MATH is zero unless MATH and MATH are both degenerate. In particular, since of course it is nonzero if MATH is degenerate, every degenerate object in MATH can be written as a product of degenerate elements in MATH and MATH so in fact MATH gives zero on the NAME link exactly when it is paired with a nondegenerate simple object. Thus a multiple of MATH gives the same functional on MATH via the NAME link as MATH . Since the NAME link labeled by MATH and MATH gives MATH and that labeled by MATH and MATH gives MATH we conclude that MATH and thus that MATH acts freely on MATH . With this in hand, if MATH then the stabilizer of MATH is the product of the stabilizers of MATH and MATH . The image of MATH and MATH in MATH is a sum of as many simple objects as there are elements in the stabilizer, and since each of these is a direct summand in the image of MATH they must each be simple. Every simple object in MATH arises this way, so every simple object in MATH is the product of an object in the image of MATH and MATH . Since MATH contains no degenerate objects, it is necessarily a product of these two subcategories. |
math/9905010 | Of course it suffices to see that, with MATH as defined in REF, MATH is MATH . Let MATH be a generator of MATH then MATH is even degenerate and no smaller power of MATH is degenerate. To say that MATH for some MATH is degenerate for MATH is to say that MATH since it suffices to check the degeneracy condition against a generator. Now MATH and MATH for some MATH so by REF , this is to say that MATH is an integer, which by the fact that the map MATH is a homomorphism is equivalent to saying MATH is an integer. Thus for any MATH the denominator of MATH represents the order of MATH in the quotient group MATH modulo even degenerates, and thus the statement that MATH is a generator is equivalent to MATH having maximal denominator as in the statement of the proposition. Let MATH . Notice MATH and MATH is the least natural number for which this is true. The fact that MATH is even degenerate means that MATH . Applying REF recursively shows MATH while of course MATH so MATH . In order for this to be MATH we conclude that MATH is odd and MATH is a primitive MATH-th root of unity or MATH is even and MATH is a primitive MATH-th root of unity. We claim that the link invariant of a link with linking matrix MATH and with the MATH components labeled by the vector of labels MATH or MATH depending on the parity of MATH where we mean labeling the MATH-th component by MATH is MATH . To see this, notice that because all labels are units MATH which means that switching an undercrossing to an overcrossing in a link projection with these labels multiplies the link invariant by MATH which is exactly the effect this move has on MATH . Since such moves will untie any link to a collection of unlinked framed unknots, it suffices to check that the formula is correct on these. This follows from the fact that both assign MATH to the MATH-framed unknot with label MATH . Now that we know the link invariant, MATH is either MATH or MATH depending on the parity of MATH . In the second case the MATH symmetry of the labels because of the even degenerate unit means that this sum is MATH and as noted in REF an overall factor in MATH depending on the number of components is canceled out in the three-manifold invariant. |
math/9905010 | Of course if MATH is modular then MATH and it is easy to see that the category is a product of the two corresponding subcategories. The central observation is that if MATH is any abstract graph and MATH is a choice of an object MATH for each edge MATH and MATH for each vertex MATH where MATH is constructed out of MATH as in the definition of the graph invariant in REF, then there exists a set of pairs of labels MATH and MATH for some index MATH with all the data in MATH and MATH coming from the categories associated to MATH and MATH respectively, such that for any embedding of MATH into MATH the invariant of MATH labeled by MATH is the sum over all MATH of the product of the invariants of the embedding of MATH labeled by MATH and MATH respectively. To see this, notice there are unique MATH and MATH such that MATH . If we write MATH with MATH or MATH and define MATH and MATH then there is a canonical isomorphism formed from a product of MATH-morphisms MATH because of the relation MATH . In particular there is an isomorphism MATH so we can write MATH . Now for a choice MATH of a MATH for each vertex MATH we can define MATH and MATH by MATH and MATH respectively. Consider an embedding of MATH and consider two parallel copies of that embedding of MATH one shifted from the other by the framing, and one labeled by MATH the other by MATH . On the one hand the fact that MATH means we can pass these two copies of MATH through each other to obtain two entirely disjoint copies of MATH with the same invariant, and thus the invariant of the doubled graph is the product of the invariant of the two labeled graphs separately. On the other hand, since the transformation of REF does not change the invariant, it equals the invariant of one copy of the embedded MATH with edges labeled by MATH and vertices labeled by MATH . Summing over all MATH we get the invariant of MATH labeled by MATH . Now MATH is the space spanned by all nonzero labelings of a fixed graph embedded in a handlebody MATH with boundary MATH whose image is a retract of the handlebody. The map MATH defined above then gives a linear map MATH which is easily seen to be an isomorphism (the map MATH above provides the inverse). If MATH is a MATH-framed manifold presented by an embedding of MATH into MATH and a surgery link on the complement, then the value of the functional on a vector given by labelings of graphs in MATH is (up to some normalizations) the invariant applied to the embeddings of these graphs together with the surgery link, with each component labeled by MATH . Now notice that since MATH we have MATH . Thus the image under MATH of the tensor product of vectors coming from labelings which compute MATH and MATH is the vector coming from the labeling which computes MATH so MATH after the appropriate identification of the domain spaces. |
math/9905016 | Both sets represent MATH. |
math/9905016 | Because MATH there is a MATH-cofinal family MATH in MATH on which our map constant, say with value MATH. Because the map is monotone this MATH is the ordinal that we are looking for. |
math/9905016 | Let MATH be a coherent family of functions, with values in MATH. For MATH and MATH let MATH and MATH. The sets MATH generate a MATH-ideal on the countable set MATH and one readily checks that MATH whenever MATH and MATH. Now apply REF to find MATH such that MATH for all MATH and MATH. This defines MATH by MATH iff MATH. If MATH were not almost equal to MATH then we'd find infinitely many MATH with a MATH in MATH. But then MATH, where MATH follows MATH. |
math/9905016 | Take, for each MATH, the monotone enumeration MATH of MATH and apply the equalities to find MATH such that MATH. Use these MATH to define open sets MATH; observe that MATH. It follows that MATH is infinite. We let MATH be the closed set MATH; its image MATH meets every MATH in an infinite set. For every MATH let MATH be the first index with MATH and consider the open set MATH and the infinite set MATH. Observe the following CASE: MATH, because MATH; CASE: MATH, by our choice of the MATH; and CASE: MATH, by the choice of the MATH. It follows that MATH is as required. |
math/9905016 | We fix MATH and show how to find MATH and MATH for each MATH. We transfer the almost disjoint family MATH to MATH by setting MATH and MATH. It is fairly straightforward to show that MATH is neat; one stretches the bijection MATH to find an injection MATH from MATH to MATH that maps every MATH onto a branch of MATH and different MATH to different branches. This means that we can apply REF to the embedding MATH of MATH into MATH, defined by MATH. This gives us a MATH and for every MATH a subset MATH of MATH and a function MATH such that for every subset MATH of MATH the set MATH is a representative of MATH. We can define MATH by MATH for MATH and by setting MATH for the other elements of MATH. |
math/9905016 | Let MATH and consider the lifting MATH. We define a lifting MATH in a fairly obvious way: first fix MATH such that MATH for MATH and, if need be, redefine, for the duration of this proof, the values MATH for MATH and MATH so as to get MATH whenever this is needed. Then, given MATH put MATH and put MATH . We define, for MATH, MATH . Note that we implicitly defined MATH whenever MATH and MATH when MATH. It follows that MATH and hence that MATH . We already took care of the disjointness requirement so this MATH witnesses that MATH (once we use MATH to define MATH on the rest of MATH). |
math/9905016 | Consider a potential sequence MATH of points in MATH where MATH and MATH disagree. By the disjointness condition and because the symmetric difference of MATH and MATH is always finite we can assume that MATH does not meet MATH when MATH. Let MATH and MATH. Observe that MATH and MATH determine the same open subset of MATH, so that MATH. It should be clear however that by the choice of the points MATH we have MATH, which is a contradiction. |
math/9905016 | If MATH then MATH and if MATH then MATH; now let MATH code witnesses: if MATH then MATH and if MATH then MATH. |
math/9905016 | Take MATH and fix MATH and MATH such that MATH for all MATH. Now simply let MATH and MATH; clearly MATH for all MATH. We see that MATH. |
math/9905016 | The first equality is clear: by construction MATH and MATH, so that MATH. To prove the second equality let MATH be an infinite subset of MATH such that MATH for all MATH; because MATH and MATH are finite-to-one we can assume that MATH. But then we would have a contradiction because on the one hand MATH while on the other hand MATH. |
math/9905016 | Let MATH be the set of MATH in MATH for which MATH is infinite. Thin out MATH to get MATH infinite for all MATH in MATH. Choose MATH as per REF for MATH and MATH, so that MATH and MATH. Because the sets MATH are pairwise disjoint we can a one-to-one choice function MATH for the family MATH such that MATH for all MATH. We choose a function MATH that captures these intersections: MATH for all MATH. It follows that MATH meets MATH in an infinite set. However, by the choice of MATH and the properties of MATH the set MATH is almost equal to MATH which is disjoint from MATH. |
math/9905016 | Fix MATH such that for all MATH and MATH in MATH: if MATH and MATH then MATH. Now note that MATH, so that MATH is finite. |
math/9905016 | Assume we have MATH with MATH infinite for all MATH and MATH whenever MATH. Choose MATH as per REF for the sets MATH and MATH and choose MATH such that MATH for all MATH. We obtain a contradiction as before: MATH and MATH are disjoint, hence MATH and MATH are almost disjoint. On the other hand MATH and MATH; the intersection of the smaller sets is infinite. |
math/9905018 | CASE: Every circle centre defined by three different sites from MATH is a NAME vertex in some MATH-th and MATH-th order NAME diagram. As there are MATH different circles, the first claim follows. CASE: Consider the arrangement of bisectors MATH. Fix one bisector MATH. Without loss of generality we can assume that bisector MATH is divided into MATH line segments by the NAME circle centres MATH where MATH. Every line segment is an edge in some MATH-th order NAME diagram. As there are MATH different bisectors, REF follows. CASE: The NAME formula holds for every order. MATH . Thus MATH . This completes the theorem. |
math/9905018 | We show that MATH is a rank function for MATH. A rank function maps an element MATH from a poset to a unique level in such a way that the level corresponds with the length of any maximal chain from MATH to MATH. Let MATH, with MATH. Then every point MATH has the MATH points from MATH as its MATH nearest neighbors. Now order those points with respect to their distance to MATH. As we assumed general position it is always possible to change the choice of MATH in such a way that this order is strict. By removing at each step the furthest point still available, we get a chain of length MATH that descends to MATH. |
math/9905018 | We apply REF to MATH and MATH. MATH . We join the two sums by applying the Symmetry REF .and evaluate the result by using REF . MATH . The lemma follows from combining the two equations. |
math/9905018 | Using REF we can write MATH in terms of numbers of faces. MATH . |
math/9905018 | We prove the theorem by induction. We use inversion. Inversion changes the point-inside-circle relation in REF-dimensional space in a point-below-plane relation in REF-dimensional space. See CITE, page REF for more details and further references. The inversion map MATH is defined by MATH . It lifts points in the plane to the unit paraboloid in three-space. As every circle defined by MATH in the plane contains only three points from MATH, every hyperplane defined by MATH contains only three points from MATH as well. MATH, the number of empty circles of MATH in REFD equals the number of faces of the lower hull of MATH in MATH and MATH, the number of circles that contain all other points of MATH equals the number of faces of the upper hull of MATH. Note that all images of points in MATH under MATH are part of the convex hull of MATH. As the convex hull of a point set consisting of MATH points consists of MATH faces, if every face is a triangle, see REF , the claim follows. [induction step] MATH . |
math/9905018 | This follows directly from REF . |
math/9905018 | We prove the property by induction. MATH is zero by definition. The number of vertices MATH in the first order NAME diagram equals the number of circles of order zero, MATH. Thus, MATH [induction step] Assume we have proved that MATH . We can rewrite this, by using induction again, as MATH . Now we evaluate MATH. MATH . We fill in this expression for MATH and apply REF . MATH . This proves the claim. |
math/9905018 | Applying REF we get MATH . The claim now follows from evaluating REF using this expression. |
math/9905018 | Write MATH. Then we get from the definition that MATH where MATH, as MATH counts the empty set. MATH is the number of zero dimensional faces plus the number of MATH dimensional faces, so MATH. Applying REF gives MATH . Straightforward calculations show that MATH . So it follows that MATH . |
math/9905032 | Another identity due to NAME (see REF, or REF ) reads MATH . Substituting MATH we get REF. Substituting MATH yields MATH . Let MATH the difference of the left-hand side and the right-hand side in REF. Using REF and the recurrence relation MATH we find that MATH. Hence for any MATH it is a periodic function of MATH and it suffices to show that MATH. Clearly, the left-hand side in REF goes to REF as MATH. From the defining series for MATH it is clear that MATH which implies that the right-hand side of REF also goes to MATH as MATH. This concludes the proof. |
math/9905032 | It is convenient to set MATH. Since the operator MATH is invertible we have to check that MATH . This is clearly true for MATH; therefore, it suffices to check that MATH where MATH and MATH. Using the formulas MATH one computes MATH where MATH. Similarly, MATH . Now the verification of REF becomes a straightforward application of REF , except for the occurrence of the singularity MATH in those formulas. This singularity is resolved using REF. This concludes the proof of REF . |
math/9905032 | Using the relation MATH and the definition of MATH one computes MATH . Clearly, the relation REF remains valid for MATH. It remains to consider the case MATH. In this case we have to show that MATH . Rewrite it as MATH . By REF this is equivalent to MATH . Examine the right - hand side. The terms with MATH vanish because then MATH. The term with MATH is equal to REF, which corresponds to REF in the left - hand side. Next, the terms with MATH vanish because for these values of MATH, the expression MATH vanishes. Finally, for MATH, set MATH. Then the MATH-th term in the second sum is equal to minus the MATH-th term in the first sum. Indeed, this follows from the trivial relation MATH . This concludes the proof. |
math/9905032 | Straightforward computation using a formula due to NAME, see REF or REF . |
math/9905032 | Follows from a computation done in the proof of REF MATH and the following corollary of REF MATH . |
math/9905032 | Our argument is similar to an argument due to NAME and NAME, see the proof of REF. The recurrence relation REF implies that MATH . Consequently, the difference between the left-hand side and the right-hand side of REF is a function which depends only on MATH. Let MATH and MATH go to infinity in such a way that MATH remains fixed. Because of the asymptotics REF both sides in REF tend to zero and, hence, the difference actually is REF. |
math/9905032 | By REF , the restriction of MATH on MATH is the square of the kernel MATH. Since the latter kernel is real and symmetric, the kernel MATH is nonnegative. Hence, it remains to prove that its trace is finite. Again, by REF , this trace is equal to MATH . This sum is clearly finite by REF. |
math/9905032 | Since MATH is the square of the operator with the kernel MATH, it suffices to check that the latter operator commutes with MATH, with the above choice of MATH and MATH. But this is readily checked using REF. |
math/9905032 | By replacing MATH by MATH, we may assume that MATH. By NAME and NAME formulas, we have MATH . Choose some large MATH and split the circle MATH into REF parts as follows: MATH . REF and the equality MATH imply that the integral MATH decays exponentially provided MATH is large enough. On MATH, REF applies for sufficiently large MATH and gives MATH . Therefore, the the integral MATH is MATH of the following integral MATH . Hence, MATH and the lemma follows. |
math/9905032 | We shall use Debye's asymptotic formulas for NAME functions of complex order and large complex argument, see, for example, REF. Introduce the following function MATH . REF can be rewritten as follows MATH . The asymptotic formulas for NAME functions imply that MATH where MATH provided that MATH in such a way that MATH stays in some neighborhood of MATH; the precise form of this neighborhood can be seen in REF. Because we assume that MATH and because MATH, the ratios MATH, MATH stay close to MATH. For future reference, we also point out that the constant in MATH in REF is uniform in MATH provided MATH is bounded away from the endpoints MATH. First we estimate MATH. The function MATH clearly takes real values on the real line. From the obvious estimate MATH and the boundedness of MATH, MATH, and MATH we obtain an estimate of the form MATH . If MATH then because of the denominator in REF the estimate REF implies that MATH . Since MATH, it follows that in this case the lemma is established. Assume, therefore, that MATH is finite. Observe that for any bounded increment MATH we have MATH and, in particular, the last term is MATH. Using the trigonometric identity MATH and observing that MATH we compute MATH . Since, by hypothesis, MATH and MATH, the lemma follows. |
math/9905032 | First, we check REF . In the case MATH this was done in the previous lemma. Suppose, therefore, that MATH is a regular sequence in MATH and consider the asymptotics of MATH. Because the function MATH is an entire function of MATH and MATH we have MATH where MATH is arbitrary; we shall take MATH to be some small but fixed number. From the previous lemma we know that MATH . From the above remark it follows that this estimate is uniform in MATH. This implies REF for MATH. To prove the estimate REF we use NAME 's integral representation (see REF) MATH which is valid for MATH and even for MATH provided MATH or MATH. If MATH then the second summand in REF vanishes and and the first is MATH uniformly in MATH. This implies the estimate REF provided MATH. It remains, therefore, to check REF for MATH where MATH is a regular sequence. Again, we use REF. Observe, that since MATH the second summand in REF is uniformly small provided MATH is bounded from above and MATH is bounded from below. Therefore, REF produces REF estimate for MATH. For MATH we use the relation REF and the reccurrence REF to obtain the estimate. |
math/9905032 | Let MATH be a regular sequence and let the numbers MATH and MATH be defined by REF. We shall assume that MATH for all MATH. The validity of the theorem in the case when MATH for some MATH will be obvious form the results of the next section. We have MATH where the first line is the definition of MATH and the second is REF . From REF it is obvious that MATH is entire. Therefore, we can apply REF to it. It is clear that REF , together with REF , implies REF . The factorization REF follows from the vanishing MATH. |
math/9905032 | Consider the left-hand side of REF and choose for each MATH a pair of functions MATH such that MATH and MATH as MATH. Then, on the one hand, the probability in left-hand side of REF lies between the corresponding probabilities for MATH and MATH. On the other hand, the probabilities for MATH and MATH can be expressed in the form REF for the kernel MATH and by REF and continuity of the Airy kernel they converge to one and same limit given by the Airy kernel as MATH. |
math/9905032 | Assume first that MATH. We denote MATH . It will be convenient to use the following notation MATH . REF reads MATH where MATH. We have the following estimates as MATH . Substituting this into REF, we obtain REF for MATH. Assume now that MATH. Denote MATH . Introduce the notation MATH . REF reads MATH where MATH. Again we have the estimates as MATH . These estimates after substituting into REF produce REF for MATH. |
math/9905032 | First suppose that MATH. Set MATH. We shall use REF with MATH. Provided MATH is chosen small enough and MATH is sufficiently large, MATH will be close to MATH and we will be able to use NAME expansions. For MATH we have MATH and, similarly, MATH . Since the function MATH is concave, we have MATH . The constant here is strictly positive. Since MATH (see, for example, REF) we obtain MATH where we used that MATH. Finally, we note that MATH and this completes the proof of REF. The estimate REF follows directly from REF. |
math/9905032 | From REF we have MATH . Let us split the sum in REF into two parts MATH that is, one sum for MATH and the other for MATH, and apply REF to these two sums. Note that MATH here corresponds to MATH in REF ; this produces factors of MATH and does not affect the estimate. Let the MATH's stand for some positive constants not depending on MATH. From REF we obtain the following estimate for the first sum MATH where MATH . Therefore, MATH . We can choose MATH so that MATH. For the second sum we use REF and obtain MATH . Clearly, this is less than MATH for MATH. |
math/9905032 | As shown in CITE, the Airy kernel has the following integral representation MATH . REF implies that for any integer MATH . Let us fix MATH and pick MATH according to REF . Since, by assumption, MATH and MATH lie in compact set of MATH, we can fix MATH such that MATH. Set MATH . Then the inequalities MATH are satisfied for all MATH in our compact set and REF applies to the sum in REF. We obtain MATH because the number of summands is MATH and MATH is bounded on subsets of MATH which are bounded from below. Note that MATH is a NAME integral sum for the integral MATH and it converges to this integral as MATH. Since the absolute value of the second term in the right-hand side of REF does not exceed MATH by the choice of MATH, we get MATH as MATH, and this estimate is uniform on compact sets. Now let MATH and MATH. By REF the integral REF converges uniformly in MATH and MATH on compact sets and we obtain the claim of the proposition. |
math/9905032 | It is clear that REF implies the convergence of MATH to MATH in the weak operator topology. Therefore, by REF , it remains to prove that MATH as MATH. We have MATH where the MATH correction comes from the fact that MATH may not be a number of the form MATH, MATH. By REF we have MATH . Similarly, MATH . Since we already established the uniform convergence of kernels on compact sets, it is enough to show that the both REF go to zero as MATH and MATH. For the Airy kernel this is clear from REF. For the kernel MATH it is equivalent to the following statement: for any MATH there exists MATH such that for all MATH and large enough MATH we have MATH . We shall employ REF for MATH. Again, we split the sum in REF in two parts MATH . For the first sum REF gives MATH where MATH and the MATH's are some positive constants that do not depend on MATH. Since MATH we obtain MATH . This can be made arbitrarily small by taking MATH sufficiently large. For the other part of the sum we have the estimate MATH which, evidently, goes to zero as MATH. |
math/9905032 | For MATH, set MATH . As is well known (for example, CITE), MATH is a continuous map from MATH to MATH. Next, MATH evidently is a continuous function on MATH. Consequently, REF is well defined and is a continuous function on MATH. When MATH, REF agrees with the conventional definition, because then MATH . This concludes the proof. |
math/9905032 | Indeed, MATH approximates MATH in the topology of MATH. |
math/9905032 | Indeed, this is true for finite - dimensional MATH, and then we use the continuity argument. |
math/9905032 | Given a finite subset MATH, we assign to it, in the natural way, a projection operator MATH. Then, by elementary linear algebra, MATH . Assume MATH becomes larger and larger, so that in the limit it covers the whole MATH. Then the left-hand side tends to the left-hand side of REF. On the other hand the right-hand side tends to MATH by REF . |
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