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math/9909116 | From the NAME REF, we have to estimate the norms of MATH for MATH. The crucial ingredient here is REF , which gives the following system of equations for the components of a unit length vector MATH in MATH: MATH . The solution is a special case of REF : MATH and moreover this is valid for any choice of unit MATH and MATH. These formulae easily yield the refined NAME inequalities and their equality cases. |
math/9909116 | If MATH is an eigenspinor with eigenvalue MATH, then MATH lies in the kernel of the NAME operator given by the NAME connection MATH, which is a metric connection on spinors. Hence we have the following refined NAME inequality for MATH, wherever it is nonzero: MATH . We next consider the conformal Laplacian of MATH where MATH: the conformal Laplacian is invariant on scalars of weight MATH and so this power is natural in view of the conformal weight MATH for the NAME operator. Using the NAME formula and the elementary identity MATH with MATH, we obtain the following equalities on the open set where MATH is nonzero: MATH since MATH. This is nonpositive by REF. Notice that this gives a local version of the NAME inequality, with equality iff MATH is the projection of MATH onto the kernel of NAME multiplication, for some MATH-form MATH. If the eigenvalue MATH is nonzero, then differentiating and commuting derivatives shows in fact that MATH. The case MATH is distinguished by conformal invariance and the fundamental solutions MATH give examples with MATH. In order to globalize, we consider the NAME quotient for the first eigenvalue MATH of the conformal Laplacian: MATH . We can estimate the integral in the numerator by setting MATH on the open set where MATH is nonzero and writing MATH . Taking MATH, integrating over MATH and letting MATH gives REF. The equality case is also easy to establish. |
math/9909116 | The starting point is the trivial formula MATH where the summation is over all virtual weights (it does not matter whether we include MATH as MATH). However, by the cancellation REF , almost all of the non-effective weights cancel. Examining the cases, we find that MATH . If we now apply this formula to the trivial representation, where MATH and MATH, we readily obtain the statement of the proposition. |
math/9909116 | For each MATH, MATH since the partial traces act by scalars on MATH. The relative dimensions MATH may be computed as follows. Let MATH denote the residue at MATH of the rational function within parentheses. Then: CASE: if MATH is odd, MATH CASE: if MATH is even, MATH . Proof of the lemma. NAME 's dimension formula (see for instance CITE) gives MATH where MATH is the set of positive roots of MATH, hence MATH . Unless the dominant weight MATH of MATH is equal to MATH, MATH is one of the MATH virtual weights MATH. Hence MATH so that MATH if MATH is even, and MATH if MATH is odd. Applying the cancellation REF and analyzing each case in turn completes the proof. Proof of REF (continued). It follows from the lemma that MATH by the residue theorem. It is straightforward to check that this residue is the coefficient of MATH in the desired rational expression of REF . |
math/9909116 | We compute the generating function MATH . This yields the stated formula. |
math/9909116 | We give the proof of NAME and NAME, which is by complete induction on MATH: clearly REF holds for MATH and we have an inductive formula for MATH. Introducing the temporary notation MATH we have MATH since MATH commutes with MATH. The result follows by observing that MATH by equivariance of MATH, where MATH is defined using the action of MATH on MATH. This gives, finally, MATH which completes the proof. |
math/9909116 | Note that MATH and so MATH . Now, by definition, we have MATH and so MATH . Finally, observe that MATH . Adding one half of this onto REF completes the proof. |
math/9909116 | Positivity holds because MATH, while the inequality follows from the identity MATH and the fact that MATH. |
math/9909118 | REF is well-known. For REF , notice that the relations in REF imply that MATH . The second equality above follows by using the NAME - NAME formula. The calculation of MATH is now straightforward. The other equations are proved similarly. REF follows immediately from REF . |
math/9909118 | Observe that the left - hand side of the equation is an antisymmetric polynomial in MATH and hence is divisible by the right hand side. Hence, by comparing degrees, we can write, MATH . But it is easy to see that the coefficient of MATH on the left hand side is MATH thus proving the proposition. |
math/9909118 | Since, MATH we get MATH which becomes the formula in REF on putting MATH. REF follows trivially. |
math/9909118 | For MATH, MATH, we know by REF that the matrix MATH is invertible. Let MATH denote the inverse of this matrix. For MATH, MATH, MATH, set MATH . Clearly MATH satisfies MATH . The second formula in REF is now clear from REF . To prove REF , assume that MATH is a submodule of MATH and let MATH. By REF , we can choose MATH, MATH, MATH, such that MATH where MATH is a polynomial in the elements MATH, MATH, MATH and MATH, MATH. Applying MATH to MATH repeatedly we find that MATH. Repeating the argument we find that MATH thus proving the Proposition. |
math/9909121 | We argue for the case MATH, the case of MATH being similar. Recall that in type MATH the coroot lattice is all vectors with integer components and zero sum with respect to a basis MATH, that MATH for MATH and that MATH. The elements of the coroot lattice contributing to some MATH are: MATH . One observes that the inverses of the permutations in the above card shuffling description for a given MATH contribute to MATH where MATH . The total number of such permutations for a fixed value of MATH is MATH, the number of interleavings of MATH cards with MATH cards preserving the relative orders in each pile. Since MATH, and MATH is a sum of MATH group elements, the proof is complete. |
math/9909121 | For the first assertion, from REF, the coefficient of MATH in MATH is MATH . The second assertion is similar and involves two cases. |
math/9909121 | The proof proceeds in several cases, the goal being to show that the inverse of the above processes generate MATH with the probabilities in REF . We give details for one subcase - the others being similar - namely even MATH when MATH satisfies MATH. (The other case for MATH even is MATH). The inverse of the probabilistic description in the theorem is as follows: CASE: Start with an ordered deck of MATH cards face down. Successively and independently, cards are turned face up and dealt into one of MATH uniformly chosen random piles. The even piles are then flipped over (so that the cards in these piles are face down). CASE: Collect the piles from pile REF to pile MATH, so that pile REF is on top and pile MATH is on the bottom. Consider for instance the permutation MATH given in REF-line form by MATH. Note that this satisfies MATH because the top card has a negative value (that is, is turned face up). It is necessary to count the number of ways that MATH could have arisen from the inverse description. This one does using a bar and stars argument as in CITE. Here the stars represent the MATH cards, and the bars represent the MATH breaks between the different piles. It is easy to see that each descent in MATH forces the position of two bars, except for the first descent which only forces one bar. Then the remaining MATH bars must be placed among the MATH cards as MATH consecutive pairs (since the piles alternate face-up, face-down). This can be done in MATH ways, proving the result. |
math/9909121 | Using the definition of MATH, one sees that MATH . Now let MATH. Then the expression for MATH simplifies to MATH . This proves the first assertion of the theorem. The second assertion follows from the first by viewing partitions diagramatically and taking transposes. The third assertion follows from the second and REF . The fourth assertion follows from either the first or second assertions together with the well-known fact that the generating function for partitions with at most MATH parts of size at most MATH is the q-binomial coefficient MATH. |
math/9909121 | REF calculates MATH for MATH. Thus it is only necessary to calculate the left hand side of the quantity in Conjecture REF for each conjugacy class of MATH. For the identity conjugacy class this amounts to counting polynomials of the form MATH with MATH. By the principle of inclusion and exclusion, the number of such polyomials with distinct roots is MATH where MATH is the number of cube roots of MATH in MATH. The number of solutions with exactly two of MATH equal is MATH and the number of solutions with MATH is MATH. Thus the total number of such polynomials is MATH agreeing with the coefficient of the identity in MATH. For the conjugacy class corresponding to a tranposition, one must count monic degree REF polynomials factoring into a quadratic and a linear factor, with product of the roots equal to REF. This is simply the number of degree REF irreducible polynomials, that is, MATH, since the degree REF factor is then determined. This answer agrees with the answer in the first step. Since the only class left is the class of REF and because the total number of semisimple conjugacy classes of MATH is MATH, the proof is complete. |
math/9909121 | It is enough to show that MATH . Elementary algebra shows that the terms corresponding to a given value of MATH on both sides of this equation are equal. |
math/9909121 | The semisimple conjugacy classes MATH of MATH such that MATH are simply the number of ways (disregarding order) of picking MATH elements of MATH whose product is the identity. After choosing a generator for MATH viewed as a cyclic group of order MATH, one sees that the number of such semisimple conjugacy classes is the number of ways of expressing MATH as the sum mod MATH of MATH integers from the set MATH. By REF and the fact that there are MATH semisimple conjugacy classes, the probability that MATH is MATH . By REF the coefficient of the identity in MATH is equal to MATH . The result follows by arguing as in REF . |
math/9909121 | This is an elementary NAME inversion running along the lines of a result in CITE. |
math/9909121 | The left hand side is equal to MATH where the last step uses REF on page REF. Note by REF that this expression is precisely the cycle structure generating function under the measure MATH, multiplied by MATH. To complete the proof of the theorem, it must be shown that the right hand side gives the cycle structure generating function for degree MATH polynomials over a field of MATH elements with constant term MATH. Let MATH be a fixed generator of the multiplicative group of the field MATH of MATH elements, and let MATH be a generator of the multiplicative group of the degree MATH extension of MATH, with the property that MATH. Recall NAME 's correspondence between degree MATH polynomials over MATH and size MATH aperiodic necklaces on the symbols MATH. This correspondence goes by taking any root of the polynomial, expressing it as a power of MATH and then writing this power base MATH and forming a necklace out of the coefficients of MATH. It is then easy to see that the norm of the corresponding polynomial is MATH raised to the sum of the necklace entries. The result now follows from REF . Note that there is no MATH term because the polynomial MATH can not divide a polynomial with constant term REF. |
math/9909121 | From REF , when MATH is odd the coefficient of MATH in MATH is the same as the formula for MATH in CITE (stated there for type MATH but which is the same in type MATH as it is defined only in terms of descents sets - for which types MATH are equivalent - and not in terms of cyclic descents). The paper CITE showed, using delicate combinatorial techniques of CITE, that Conjecture REF holds for type MATH. Thus it suffices to show that the mass on a class MATH of MATH obtained by picking one of the MATH semisimple conjugacy classes of MATH at random and applying MATH is equal to the mass on a class MATH of MATH obtained by picking one of the MATH semisimple adjoint orbits of MATH on its NAME algebra at random and then applying MATH. To do this, recall first that the semisimple conjugacy classes of MATH are monic degree MATH polynomials MATH with non-zero constant term invariant under the involution sending MATH to MATH. The conjugacy classes of MATH correspond to pairs of vectors MATH where MATH, MATH and MATH (respectively, MATH) is the number of positive (respectively, negative) MATH cycles of an element of MATH, viewed as a signed permutation. From REF, the map MATH can be described as follows. Factor MATH uniquely into irreducibles as MATH where the MATH are monic irreducible polynomials and MATH. Note that all terms in the second product have even degree. This is because by REF all MATH invariant under the involution MATH have even degree, except possibly MATH. However MATH must each appear with even multiplicity as factors of the characteristic polynomial of an element of MATH and hence only contribute to the first product. The class of MATH corresponding to MATH is then determined by setting MATH and MATH. Next, recall that the semisimple orbits of MATH on its NAME algebra are monic degree MATH polynomials MATH satisfying MATH. Arguing as in the previous paragraph, the description of the map MATH is similar. One factors MATH uniquely into irreducibles as MATH where the MATH are monic irreducible polynomials and MATH. Then MATH and MATH. Thus the theorem will follow if it can be shown that the number of monic, degree MATH irreducible polynomials satisfying MATH is equal to the number of monic, degree MATH irreducible polynomials satisfying MATH. From REF , the first quantity is MATH . This formula agrees with the second quantity, as computed in CITE. |
math/9909121 | Suppose that the identity of the theorem is true. Then taking coefficients of MATH on both sides would yield the equation MATH . Set MATH (which is an integer since MATH is assumed even). Taking the coefficient of MATH on the left hand side of this equation and dividing by MATH gives by REF the probability that a MATH chosen according to the MATH probability measure is in a conjugacy class with MATH positive MATH-cycles and MATH negative MATH-cycles for each MATH. By REF , doing the same to the right hand side of the equation gives the probability that when one factors a uniformly chosen random degree MATH polynomial over MATH as MATH (with MATH are monic irreducible polynomials and MATH), that one obtains MATH and MATH. The theorem now follows from the fact that in even characteristic the map MATH of Conjecture REF has the same description as in the proof of REF . |
math/9909121 | For the first equation, consider the coefficient of MATH on the left hand side. It is the probability that a uniformly chosen unimodal permutation on MATH symbols has MATH cycles of shape MATH. The coefficient on the right hand side is MATH. These are equal by REF . To deduce the second equation, observe that setting all MATH in the first equation gives that MATH . Taking reciprocals and multiplying by the first equation yields the second equation. |
math/9909121 | The result follows from the claim that if MATH has a NAME series around REF and MATH and MATH has a NAME series around REF, then the MATH limit of the coefficient of MATH in MATH is MATH. To verify the claim, write the NAME expansion MATH and observe that the coefficient of MATH in MATH. |
math/9909121 | Symmetric function notation from REF is used. Thus MATH are the power sum, complete, elementary, and NAME symmetric functions parameterized by a partition MATH. From REF, the number of MATH in MATH with a given cycle structure and descent set MATH is the inner product of a NAME character MATH and a NAME character MATH. From the proof of REF, MATH where MATH is the number of standard tableaux of shape MATH with descent composition MATH. Thus the sought number is MATH . Expanding these NAME functions using REF on page REF, using the fact that the MATH are an orthogonal basis of the ring of symmetric functions with known normalizing constants (page REF), and using the expansions of MATH and MATH in terms of the MATH's (page REF) it follows that MATH . |
math/9909121 | From the first equation in the proof of REF , it suffices to prove that MATH . For this it is enough to define a MATH to MATH map MATH from the MATH type MATH characteristic REF shuffles to unimodal elements of MATH, such that MATH preserves the number of MATH-cycles for each MATH, disregarding signs. To define MATH, recalling REF observe that the MATH shuffles are all ways of cutting a deck of size MATH, then flipping the first half, and choosing a random interleaving. For instance if one cuts a REF card deck at position REF, such an interleaving could be MATH . Observe that taking the inverse of this permutation and disregarding signs gives MATH . Next one conjugates by the involution transposing each MATH with MATH, thereby obtaining a unimodal permutation. Note that this map preserves cycle structure, and is MATH to MATH because the first symbol (in the example MATH, can always have its sign reversed yielding a possible shuffle). |
math/9909121 | From the above bijection, the MATH inverse shuffles correspond to multisets of primitive necklaces on the symbols MATH in a cycle length preserving way. Thus the sought probability is MATH times the product (over shapes MATH) of the number of ways choosing a multiset of size MATH from the primitive necklaces on MATH symbols and of shape MATH. The number of such necklaces is MATH, because from CITE, this is the probability that an inverse MATH shuffle on MATH symbols gives the cycle MATH. |
math/9909121 | Let MATH be the unique minimal length coset representative of the coset MATH. Then MATH is given explicitly by reordering the images under MATH of the elements within each MATH to be in increasing order. Consider MATH in some MATH. Suppose that MATH is the MATH smallest element in MATH. If the MATH-orbit of MATH is MATH, then the MATH-orbit of MATH consists of the MATH-th smallest elements of each of MATH, and hence has size MATH. By hypothesis, the MATH orbit of MATH also has size MATH, implying the result. |
math/9909121 | Denoting the coordinates of MATH as MATH, recall that MATH, MATH, and that MATH. Semisimple conjugacy classes MATH of MATH are monic polynomials with constant term REF. Since the roots of an irreducible degree MATH polynomial MATH are a NAME orbit of some element of a degree MATH extension of MATH lying in no smaller extension, these roots correspond to some subset of values of MATH (there may be repetition among the MATH's) which are of the form MATH with MATH integers. For clarity of exposition, we point out that the roots of MATH are simply MATH for MATH is a generator of the multiplicative group of the degree MATH extension of MATH and was determined by the bijection in REF. The MATH-action performs a cyclic permutation of these MATH-values. Letting MATH be such that the class MATH corresponds to a polynomial whose factorization into irreducibles has MATH factors of degree MATH, it follows that there is a permutation MATH, with MATH-cycles, which satisfies the equation MATH for some MATH and such that MATH is the conjugacy class of MATH. Now suppose that some other MATH satisfies the equation MATH for some MATH. Then MATH, which implies that MATH where equality means equality of coordinates mod REF. Let MATH be the partition of MATH induced by the equivalence relation that MATH when MATH mod REF. Then MATH is contained in the parabolic subgroup MATH. Thus the set of all permutations MATH satisfying the equation MATH for some MATH in MATH is simply the coset MATH. REF implies that the unique minimal length coset representative is conjugate to MATH. |
math/9909121 | First it will be shown that the probability of the identity element agrees for both measures. Fixed points under the action of MATH (which is multiplication by MATH) correspond to solutions to the equations MATH, MATH, MATH. The number of such solutions is MATH so they can be counted by the methods of REF . Doing this, and comparing with the formula for MATH in REF proves the result for the identity element. Next we argue that the measures agree for the permutation whose REF-line form is MATH (that is, the longest element). From REF , the mass which MATH puts on this element is seen to be MATH . Now consider a MATH-stable element of MATH which is stabilized by MATH but by no other elements. Next we argue that necessary and sufficient conditions for the coordinates MATH of such a point are the inequalities MATH and MATH with MATH integers satisfying MATH. Here MATH, so it is enough to argue for the form of MATH and MATH. Observe that MATH and MATH, since the origin, though stabilized by MATH, is also stabilized by the identity, which is shorter. The form of MATH follows since MATH stabilizes MATH, but MATH does not (note that MATH since MATH and MATH since MATH does not stabilize MATH). The inequalities MATH and MATH imply without much difficulty that MATH. Rewriting MATH and MATH gives that MATH . Viewing the sums in these inequalities as base MATH expansions and using the facts that MATH and MATH, one concludes that our system of inequalities is equivalent to the system of REF inequalities MATH . To count the number of integer solutions, we first count solutions ignoring the third constraint, then subtract off solutions satisfying MATH. We only work out the case MATH mod REF, the other cases being similar. The number of solutions ignoring the third inequality is MATH . The number of solutions obtained by replacing the third inequality with MATH is readily computed to be MATH, implying that the number of solutions to the original inequalities is MATH, as desired. From REF , the element MATH assigns equal mass to the remaining two tranpositions MATH and MATH in MATH. Noting that MATH is MATH-stable if and only if MATH is MATH-stable, it follows that the construction of this section also assigns equal mass to MATH and MATH. Thus it remains to show that the construction of this section and MATH assign equal mass to the conjugacy class of all transpositions. From REF , the construction of this section assigns equal mass to the conjugacy class of transpositions as does the map MATH. The result now follows by REF . To argue for the class of REF, from REF the element MATH assigns equal mass to the permutations MATH and MATH. Noting that MATH is MATH-stable if and only if MATH is MATH-stable, it follows that the construction of this section also assigns equal mass to MATH and MATH. Now argue as in the preceeding paragraph. |
math/9909121 | The proposition follows from REF . |
math/9909122 | Given MATH, choose a compact oriented NAME surface MATH and a map MATH such that MATH. Note that MATH is a principal MATH-bundle over MATH and denote by MATH the pullback bundle of MATH. Thus MATH . There are two equivariant maps MATH and MATH, given by, MATH . By definition, the map MATH descends to MATH. Hence, by REF , MATH. This proves REF . To prove REF , choose two equivariant maps MATH and MATH. Define MATH as the maps induced by MATH and MATH. Then, by REF , MATH . Consider the induced maps MATH and MATH . They can be expressed in the form MATH and MATH where MATH denotes the obvious projection. Hence MATH and MATH are homologous, that is, MATH. Since MATH is connected, MATH is simply connected. Hence two maps MATH are homologous if and only they are homotopic. (To see this note that every map from MATH to a simply connected space factors, up to homotopy, through a map of degree MATH from MATH to MATH.) This shows that our maps MATH and MATH are homotopic. Hence MATH and MATH are isomorphic. This proves the proposition. |
math/9909122 | The form MATH is obviously horizontal. We prove that it is equivariant. Denote by MATH the structure constants of MATH. This means that MATH . Then the equivariance of MATH can be expressed in the form MATH where MATH. Hence, with MATH it follows that MATH . Moreover, with MATH, the NAME identity MATH takes the form MATH . For MATH denote by MATH the vector field MATH. Then MATH. Hence, by REF , MATH and hence, by REF , MATH . Here we have used the identity MATH . It follows that MATH and hence MATH is equivariant. This proves REF . The proof of REF relies on the following identity, for MATH, MATH . For MATH the proof of REF is a computation using MATH and MATH . For general MATH, REF follows easily by induction. With this understood, we obtain, MATH . Here the second identity follows from REF , the third identity from REF , and the last identity from MATH, that is, MATH . This proves REF . We prove REF . Let MATH be a smooth family of connections and MATH be a smooth family of equivariant functions. Think of the path MATH as a connection MATH on the bundle MATH over MATH and of the path MATH as a function MATH. Given a MATH-closed MATH-form MATH write MATH where MATH and MATH. By REF , MATH is horizontal, equivariant, and closed. Hence MATH and MATH are equivariant and horizontal, MATH is closed, and MATH for every MATH. Hence MATH . Since MATH descends to MATH for every MATH, this proves the proposition. |
math/9909122 | Choose a holomorphic coordinate chart MATH, where MATH is an open set, and let MATH be a lift of MATH, that is MATH. Then the function MATH and the connection MATH are in local coordinates given by MATH and MATH where MATH. The pullback volume form on MATH is MATH for some function MATH and the metric is MATH. Hence MATH . Here MATH, MATH, and MATH are evaluated at MATH. In the following we shall drop the superscript MATH. Then REF have the form MATH . The pullback of the energy integrand under MATH is given by MATH . This proves REF . The identity MATH can in local coordinates be expressed in the form MATH . This follows directly from the definitions and the fact that MATH. This proves the proposition. |
math/9909122 | Let MATH be any smooth path in MATH and denote its derivative by MATH . Then, by NAME 's formula, MATH . Moreover, for every MATH, MATH . Hence MATH . The last equality uses the formula MATH . This proves the proposition. |
math/9909122 | We identify the NAME sphere MATH with the unit sphere MATH via stereographic projection. Explicitly, this diffeomorphism is given by MATH . Under this correspondence the quotient MATH can be identified with the open unit ball MATH via the map MATH that assigns to MATH the diffeomorphism MATH given by MATH . If MATH converges to MATH then MATH converges to MATH, uniformly in compact subsets of MATH. Hence MATH for every MATH and every MATH, where MATH denotes the obvious inclusion and MATH denotes the standard inner product on MATH. Hence MATH for every MATH, and the convergence is uniform in MATH. It follows from a standard argument in degree theory CITE that there exists a MATH such that MATH . This identity is equivalent to REF . Now suppose that MATH is a volume form and MATH such that MATH . Then the same argument as in REF shows that MATH. Namely, there exists a MATH and a matrix MATH such that MATH for every MATH. Hence, for every MATH, MATH . Denote MATH choose MATH such that MATH, and consider the flow MATH for MATH. It satisfies MATH, MATH, and, since MATH, MATH . Hence MATH . The last expression is nonegative. Integrating from MATH to MATH, we obtain from REF that it is equal to zero for all MATH. It follows that MATH, hence MATH, and hence MATH as claimed. This proves the proposition. |
math/9909122 | Write MATH, denote by MATH the real inner product on MATH, by MATH the group of unitary automorphisms of MATH, and by MATH its NAME algebra. By assumption, the action of MATH on MATH is given by a homomorphism MATH and we shall denote by MATH the corresponding NAME algebra homomorphism. We prove that there exists a central element MATH such that MATH for MATH. To see this, suppose without loss of generality that MATH with its standard Hermitian structure, and consider the inner product MATH on the NAME algebra MATH of skew-symmetric matrices. The moment map is given by MATH for MATH, where MATH is a central element and MATH denotes the adjoint of the NAME algebra homomorphism MATH. Hence MATH . This proves REF . Now fix a NAME surface MATH and suppose that MATH is a solution of REF . Consider REF in local holomorphic coordinates, where the metric has the form MATH. In our situation MATH and we abbreviate MATH and MATH . Since MATH and, by REF , MATH we obtain MATH . Hence, with MATH, MATH . Now let MATH be a point at which the function MATH attains its maximum. Since MATH is compact, such a point exists in some coordinate chart, and we have MATH at MATH. Hence MATH . Since MATH is proper, there exists a constant MATH such that MATH . Hence MATH and it follows that MATH for every solution MATH of REF . To prove the last assertion just note that the same estimate holds for solutions of the perturbed REF whenever the support of the perturbation is contained in the ball MATH. This proves the proposition. |
math/9909122 | For every MATH we have MATH . The last inclusion follows from the fact that MATH is a Lagrangian subspace of MATH and the kernel of MATH is the symplectic complement of the image of MATH. The inclusion MATH shows that MATH is constant on MATH. The inclusion MATH shows that MATH for MATH and MATH. Hence MATH is in the center of MATH for every MATH. |
math/9909127 | By CITE, the contact moment map MATH is defined by MATH for any MATH and MATH the corresponding field on MATH. We know that the reduced space is a contact manifold, CITE Hence we only need to check that REF the Riemannian metric is projected on MATH and REF the field MATH projects to a unitary Killing field on MATH such that the curvature tensor of the projected metric satisfies REF . To this end, we first describe the metric geometry of the Riemannian submanifold MATH. Let MATH be a basis of MATH and let MATH be the corresponding vector fields on MATH. Since MATH is a regular value of MATH, MATH is a linearly independent system in each MATH. From the very definition of the moment map we have MATH hence MATH. As MATH acts by contact isometries, we have MATH . Note that these also imply MATH. Observe that MATH is an isometrically immersed submanifold of MATH (we denote the induced metric also with MATH) whose tangent space in each point is described by: MATH if and only if MATH. Hence, by the definition of the moment map, the vector fields MATH and MATH are tangent to MATH. Moreover, for any MATH tangent to MATH, one has MATH, hence the vector fields MATH produce a local basis (not necessarily orthogonal) of the normal bundle of MATH. The shape operators MATH of this submanifold in MATH are computed as follows (we let MATH, MATH be the NAME covariant derivatives of MATH, respectively, MATH): MATH . In particular, for the corresponding quadratic second funadamental forms we get: MATH . Consequently, one easily obtains: the restriction of the vector field MATH is Killing on MATH too. Using the NAME equation of a submanifold MATH and REF we now compute the needed part of the curvature tensor of MATH at a fixed point MATH. We take MATH orthogonal to MATH and obtain: MATH (Note that MATH are chosen to be orthonormal in MATH; this is always possible pointwise by appropriate choice of the initial MATH). Let now MATH and endow MATH with the projection MATH of the metric MATH such that MATH becomes a Riemannian submersion. This is possible because MATH acts by isometries. In this setting, the vector fields MATH span the vertical distribution of the submersion, whilst MATH is horizontal and projectable (because MATH). Denote with MATH its projection on MATH. MATH is obviously unitary. To prove that MATH is Killing on MATH, we just observe that MATH, where MATH denotes the horizontal lift of of MATH. Finally, to compute the values MATH of the curvature tensor of MATH, we use NAME formula (compare REF ) MATH where MATH are unitary, normal to MATH and the NAME MATH tensor MATH is defined as: MATH. Using NAME formula and REF, we obtain MATH hence MATH has no vertical part and MATH. Thus MATH because of REF and the fact that MATH are normal to all MATH. Hence MATH which proves that MATH is a NAME manifold. |
math/9909128 | The quickest way to see this is to use the non-degenerate pairing MATH obtained by gluing together two solid tori to make MATH (whose skein space is canonically MATH). Pairing MATH with MATH results in MATH, so that the change of basis matrix expressing the MATH, viewed as linear functionals, in terms of the dual functionals MATH is a NAME matrix MATH times a diagonal matrix whose diagonal entries are the (non-zero) MATH. Now MATH, and one can easily check that these are all distinct when MATH is prime, hence the first matrix is invertible. The second is obviously invertible, and so the MATH indeed form a basis. |
math/9909128 | Consider MATH as a thickened MATH-holed disc. The reduced skein space is spanned by a finite number of elements of MATH, which may be represented by links lying in the holed disc (consider the usual planar projection onto this disc). Given such a planar link MATH, isotop it to be near to the boundary MATH, and rewrite each of its components (thought of as an element of MATH) as a linear combination of the basis elements MATH. Since each MATH is really MATH, for MATH a single curve parallelled MATH times, this proves the assertion about spanning. The final part follows immediately from the description of the action of a NAME twist on MATH. |
math/9909128 | The standard basis vectors are certainly simultaneous eigenvectors for the twists generating MATH, as each twist just multiplies the vector by a twist coefficient MATH. But their collections of eigenvalues are distinct since the MATH are (when MATH is prime) and hence they span individual one-dimensional eigenspaces. |
math/9909129 | Over a point of MATH representing the strict lift of a MATH-pointed immersion MATH, the substack of MATH representing the same immersion together with an additional MATH markings, is dense in the fiber of MATH. |
math/9909129 | Since MATH is smooth and birationally equivalent to MATH, the general point of the divisor MATH has a unique preimage. To see the nature of the stable lift, we follow the semistable reduction recipe of CITE. We may assume that the source curve of MATH has just two twigs, that on each twig the map is an immersion, and that the maps are transverse at the attachment point. We may create a family in which the source curves are the plane curves MATH and in which the central member is MATH. Away from the central member, the strict lifts piece together to give a map to MATH, and we now proceed to compute the ``limiting member" over MATH. (Since the moduli space is separated and proper, there is a unique such limit.) In fact the map to MATH extends to every point of MATH except the point of attachment, and we can remove the indeterminacy by one blowup. The central member will now have three twigs, and a calculation shows that on the central twig the map to MATH is an isomorphism onto a fiber over a point of MATH, while on each peripheral twig it is the strict lift of the original map. The calculation also shows that the map to MATH is indeterminate at the two attachment points. Again the indeterminacy is removed by blowing up the two points, and now the central member has five twigs arranged in a chain. The map to MATH takes the newly introduced twigs onto fibers over points over MATH. On the central twig the map is the lift of the map at the previous stage; thus it is a map into the divisor at infinity. On each peripheral twig the map is the second-order strict lift of the original map. The central member is not reduced, however: its central twig has multiplicity REF and the adjacent twigs have multiplicity REF. Thus, following the recipe, we introduce the base change MATH and then normalize the resulting surface. After these operations we find that the central member is still a chain of five twigs, that the map on the central member is a degree REF cover, that the maps on the adjacent twigs are degree REF covers, and that all ramification is concentrated at the points of attachment. |
math/9909129 | Again we know that the general point of the divisor MATH has a unique preimage, and again we see the nature of the stable lift by following the recipe of NAME and NAME. This time we may begin with a family whose source is the product of MATH and a curve, with the map to MATH being an immersion except on the special member. The maps to MATH and MATH obtained by piecing together the strict lifts are defined everywhere but at the ramification point of the special member. At this point and at MATH we may choose local coordinates so that the special member is MATH, then extend to the family of maps MATH . The indeterminacy in the map to MATH is resolved by one blowup. On one of the two twigs of the new special member we have the strict lift; on the other we have an isomorphism onto the fiber of MATH over MATH; the attachment point maps to MATH; and the image curves are tangent there. To resolve the indeterminacy in the map to MATH, we first blow up the new attachment point. Then the new special member has a chain of three twigs, with the map on the central one being constant. There is still, however, indeterminacy at a point on this twig. One more blowup creates a fourth twig, mapping onto the fiber of MATH over MATH. Both this twig and the central one occur in the special member with multiplicity MATH. Thus when we make the base change MATH and then normalize the resulting surface we obtain two inverse images of the fourth twig, and on the central twig the normalization map is a double cover ramified at the points of attachment to the other two twigs. |
math/9909129 | Again we know that the general point of the divisor MATH has a unique preimage, and again we see the nature of the stable lift by following the recipe. We may choose local coordinates so that the family of maps to MATH is MATH . Over the origin, the indeterminacy of the map to MATH is resolved by one blowup. The map to MATH is still indeterminate at the new attachment point, and two further blowups are required. (The first blowup creates a twig on which the map is constant, and the second one creates a twig mapping to a fiber of MATH.) A similar sequence of blowups over the other branch point creates a chain or seven twigs. Two of these twigs (those on which the map is constant) occur in the special member with multiplicity REF, and two others (those created by the last blowups) occur with multiplicity REF. Thus we make the base change MATH and then normalize, creating double and triple covers with ramification concentrated at the points of attachment. |
math/9909129 | There is a birational morphism from MATH to MATH which associates to the stable lift of an immersion its image curve in MATH together with the images of the MATH markings. Denote the graph of this morphism by MATH. Then two applications of the projection formula show that the NAME invariant and the characteristic number both equal MATH . |
math/9909129 | Without loss of generality, we may assume that MATH is irreducible. By linearity we may assume that the MATH in REF come from the following basis for MATH: MATH . Note first that if any MATH, say MATH, is the identity element then, by the projection formula applied to the forgetful morphism MATH, we have MATH since the evaluation maps MATH factor through MATH. But MATH is zero unless MATH. By the irreducibility of the moduli stack, we therefore have MATH and hence that MATH. Given that MATH is at least MATH, we see that the proposition holds in this instance. Next, consider, for MATH, the class MATH where none of the MATH's is the identity. We claim that if MATH is nonzero, then it can be represented by a cycle MATH for which MATH . To establish the claim, we use induction on MATH, the case MATH being trivial. In view of the remarks above, we need only consider the cases where MATH. For MATH, or MATH, the argument is given in the proof of REF. Thus we prove the result when MATH, MATH, MATH, MATH, or MATH. CASE: Suppose MATH. If MATH is a general line, then MATH may be represented by a cycle MATH for which MATH is contained in the intersection of MATH and the set of curves with a singularity on MATH. Hence MATH and REF holds by induction. CASE: Suppose MATH. If MATH is a general point, then MATH may be represented by a cycle MATH for which MATH is contained in the intersection of MATH and the set of curves with a singularity at MATH. Hence MATH and REF holds by induction. CASE: Suppose MATH. Let MATH be a general point. Since MATH, we have that MATH is represented by a cycle MATH for which MATH is in the intersection of MATH and the union of the set of curves having a flex tangent that passes through MATH and the set of curves whose tangent cone at a singular point passes through MATH. Thus MATH. CASE: Suppose MATH. Let MATH be a general line. Using MATH in place of MATH as in REF , we have that MATH is represented by a cycle MATH for which MATH is in the intersection of MATH and the union of the set of curves having MATH as flex tangent and the set of curves whose tangent cone at a singular point contains MATH. Thus MATH. CASE: Suppose MATH. Let MATH be a general flag. Using MATH in place of MATH, we have that MATH is represented by a cycle MATH for which MATH is in the intersection of MATH and the union of the set of curves with a flex at MATH and flex tangent equal to MATH and the set of curves singular at MATH whose tangent cone contains MATH. Thus MATH. To finish the proof, note that MATH must be zero unless MATH, that is, unless MATH. But in this REF implies that MATH . Hence MATH must be zero, as desired. |
math/9909129 | An immersion MATH has a unique (unmarked) second-order stable lift, whose source is again MATH. Thus the general MATH-marked stable lift of an immersion likewise has MATH as source. Thus the image in MATH of a numerically relevant divisor is either the entire stack or a divisor representing non-immersive maps. Every non-immersion either has a reducible source or is singular at one or more points; thus it is represented by a point of one of the listed divisors. |
math/9909129 | Consider a component MATH of type REF in which the the first marking belongs to MATH. Then the evaluation map MATH factors through projection of the fiber product MATH onto its second factor, the stack MATH. Since this stack represents maps to fibers of MATH over MATH, there is a morphism which picks out the image of the fiber and thus a composite morphism MATH. If MATH is an element of MATH, the classes MATH and MATH agree. Let MATH denote MATH, and let MATH be the analogous morphism. Then MATH, where MATH is the forgetful morphism. If MATH are elements of MATH, then by the projection formula MATH . Since the relative dimension of MATH over MATH is REF, this class vanishes. Hence MATH is irrelevant with respect to the base. A similar argument applies if the second, third, or fourth marking belongs to MATH. The argument also works if any of these markings belong to MATH or MATH, since again the corresponding evaluation map factors through projection onto a factor of MATH representing maps to fibers of MATH over MATH, or representing maps to fibers of MATH over MATH. For a component of type REF or REF we may also employ the same sort of argument, since for any such component at least one (in fact two) of the four special markings must lie on a twig which either maps to a fiber of MATH over MATH, or maps to a fiber of MATH over MATH, or is contracted to a point. |
math/9909129 | First use the alternative basis to the MATH-basis where MATH replaces MATH and MATH replaces MATH. Denote the coefficients of MATH with respect to this alternative basis by MATH's and MATH's. Each basis element may be written in the form MATH, where MATH, MATH, or MATH, and MATH does not involve MATH. The NAME invariant MATH is zero unless MATH, and otherwise equals MATH where MATH denotes the fundamental class of the lift of a fiber of MATH over MATH. This in turn must be zero if MATH appears in any MATH (since the divisor at infinity MATH is disjoint from the dual divisor MATH). It also vanishes unless all MATH's equal MATH, in which case its value is MATH. Thus MATH where the coefficient MATH is MATH if the first entry of MATH is MATH, the last entry of this sum is MATH, and each second entry is MATH; otherwise the coefficient is zero. Thus MATH is the sum of terms in MATH that are quadratic in MATH and for which the sum of subscripts is MATH for some MATH. To obtain the lemma, note that MATH for all subscripts MATH appearing in the formula (which do not include the subscripts MATH or MATH). |
math/9909129 | The NAME invariant MATH is zero unless MATH, MATH (respectively), MATH, and MATH, in which case it has value MATH. Thus MATH where the coefficient MATH is MATH if the first entry of MATH is MATH, the second entry is MATH (respectively), and the last entry of each MATH is MATH; it is zero otherwise. Hence the lemma follows. |
math/9909129 | Set MATH. Apply REF and sum over all MATH. Then apply REF . |
math/9909129 | Apply MATH to REF . From the first term on the right we obtain terms of the form MATH and from the second term we obtain terms of the form MATH where MATH. Since MATH any nonzero terms must have MATH for MATH. Thus MATH and MATH. Similarly, MATH, so MATH and MATH, so MATH. To summarize, MATH is determined by MATH for MATH and MATH. The theorem follows by induction. |
math/9909129 | Recall that MATH where MATH (See the beginning of REF for notation.) Thus MATH where MATH. Hence if MATH we obtain the NAME invariant MATH by computing MATH times the coefficient of MATH in MATH. |
math/9909129 | It was shown in CITE and CITE that MATH has three orbits under the action of MATH: a dense orbit MATH, represented by the germ at the origin of a nonsingular curve without a flex; a three-dimensional orbit MATH, represented by the germ at the origin of a line; and another three-dimensional orbit MATH, represented by the germ at the origin of a curve with an ordinary cusp. Thus MATH has MATH orbits of the form MATH where each MATH can be one of the symbols MATH. We will show that for every nondense orbit MATH on MATH . Hence, since MATH are assumed to be in general position with respect to the action of MATH, the transversality theory of CITE shows that MATH . Thus MATH and MATH intersect transversely and all intersections must occur in the dense orbit MATH. The product of lifts MATH is the closure of the graph of function defined on a dense subset of MATH; the join MATH is likewise the closure of the graph of a function defined on a dense subset of MATH, the fiber product over MATH of MATH copies of MATH. Thus, by general position, all intersections between MATH and MATH are intersections between the graphs. Therefore every intersection point is a MATH-tuple MATH in which MATH lies over a nonsingular point of MATH and some member MATH of MATH, MATH lies over a nonsingular point of MATH and MATH, etc. Since each MATH is a point of MATH, we see that the second-order data of MATH and MATH at MATH must be identical for MATH. Thus each intersection point is a simultaneous triple contact. Establishing REF is quite easy. First, if MATH, then MATH where MATH is the number of MATH's which are not the symbol MATH. Next, since none of the curves MATH contains a line, we have MATH where MATH is the number of MATH's which are MATH. Finally, since the general member of MATH is reduced and contains no line, MATH from which REF follows. |
math/9909130 | Use the construction of REF to define inverse isomorphisms MATH . |
math/9909130 | This is a formality. CASE: Define the MATH-transverse topological bordism groups MATH by analogy with MATH, for which there is an exact sequence MATH . The forgetful maps MATH are isomorphisms, by topological transversality. Applying REF-lemma to the map of exact sequences MATH we have that the morphisms MATH are isomorphisms for MATH, exactly as in the case MATH. |
math/9909130 | This result follows from the following two lemmas, whose proofs we defer. Given a MATH-dimensional topological manifold MATH for MATH, such that MATH is the union of two manifolds along a boundary component, MATH. We may perform the construction of REF so that the result is a homology manifold MATH so that MATH and MATH for MATH all satisfy the conclusion of REF. Given a MATH-dimensional manifold MATH for MATH and a MATH-bundle MATH over MATH, whose total space is MATH we may construct a homology manifold MATH with a map MATH which satisfies the properties of REF. In particular MATH is a homology manifold such that MATH is the total space of the bundle MATH and MATH also satisfies the properties of REF. Denote MATH by MATH and MATH by MATH, so that MATH is a codimension MATH bundle subspace of MATH with MATH . By applying REF to MATH, we get MATH satisfying the above conditions. Applying REF to MATH results in MATH . By REF we may apply the construction to MATH and MATH simultaneously so that the resulting homology manifolds MATH and MATH and maps agree on their boundaries. |
math/9909130 | CASE: The functions MATH are bijections by REF and its MATH-transverse variation REF. The forgetful function MATH is a bijection by topological transversality, so that MATH . CASE: Unfortunately REF does not apply to an arbitrary map. We may get around this by factoring any map MATH through a homotopy equivalence MATH. Any map MATH is homotopic to MATH such that MATH is a NAME fibration and MATH is a homotopy equivalence. Now by REF , MATH is normally cobordant to a transverse map, and hence MATH is bordant to a transverse map. |
math/9909130 | Define MATH to be the group of obstructions in the exact sequence MATH . Consider the homology manifold normal invariant given by MATH, by REF MATH is normally bordant, say via MATH to a map MATH such that MATH is transverse to MATH. In particular MATH is a homology manifold with a normal neighborhood MATH. Thus, this transverse normal invariant defines a splitting obstruction which lives in MATH. Since we do not a priori have an understanding of MATH we must study it by comparing to the obstruction groups MATH and the surgery exact sequences for MATH and MATH. There is clearly a commutative diagram with vertical maps given by restriction as follows: MATH . The splitting obstruction MATH can be understood as a two stage obstruction as follows. First the normal invariant MATH defines an obstruction MATH. If this obstruction vanishes, then by the surgery exact sequence of CITE MATH is normally cobordant to a simple homotopy equivalence. Let MATH denote this normal cobordism, and MATH denote the corresponding pullback of MATH. We can define a homology manifold normal invariant of MATH by gluing MATH to MATH. There is now defined an obstruction MATH to this normal invariant being equivalent to a simple homotopy equivalence of MATH, that is, an s-cobordism from MATH to some map MATH. If MATH vanishes, then both MATH and MATH are defined and vanish, so that MATH is MATH-cobordant to a split map. Conversely if MATH is MATH-cobordant to a split map, then both of MATH and MATH are defined and vanish. |
math/9909130 | For MATH of codimension MATH this was proved in CITE. We prove the theorem here for codimension MATH. First we may assume that the map MATH is a homotopy equivalence by factoring the original map through a NAME fibration. This results in MATH homotopic to the original MATH so that MATH is a homotopy equivalence, MATH is a NAME fibration, and MATH has a codimension MATH subset MATH so that the normal bundle of MATH in MATH is the pullback of the normal bundle of MATH in MATH. To achieve MATH-transversality for MATH clearly it would suffice to achieve MATH transversality for MATH. Thus we may assume that MATH is a homotopy equivalence. By REF we have that MATH is bordant to a MATH-transverse map. From here the proof proceeds in two steps, given by the following lemma. If MATH is a homotopy equivalence as above and MATH is a homology manifold bordism from MATH to MATH, then REF The map MATH is homotopy equivalent to a map MATH, which factors through a homotopy equivalence to a NAME space MATH, MATH with MATH such that MATH, and MATH is NAME transverse to MATH, that is, the inverse image of MATH is a NAME space with normal bundle the pullback of the normal bundle of MATH. CASE: The map MATH is MATH-cobordant to a MATH-transverse map. |
math/9909130 | Exactly as above. |
math/9909130 | Define the generalized homology spectrum MATH as in REF , with a MATH-dimensional MATH-coefficient cycle (CITE) MATH essentially the same as a dual transverse map MATH from a closed MATH-dimensional homology manifold, with inverse images the MATH-dimensional homology manifolds with MATH . The homotopy group MATH is the cobordism group of such cycles, and is the bordism group of dual transverse maps MATH from MATH-dimensional homology manifolds. |
math/9909130 | Given a bordism MATH . Denote the connected components by MATH . We have MATH as usual. Furthermore MATH follows from the fact that MATH for any connected homology manifold with non-empty boundary (CITE). Thus we have MATH . CASE: The augmentation map MATH sends MATH to MATH, for any map MATH. If MATH is connected then MATH and MATH so that MATH implies MATH. If MATH is dual transverse MATH . Each MATH is a REF-dimensional homology manifold, which is a disjoint union of points, with resolution obstruction MATH, so that MATH. Thus if MATH is connected and MATH is dual transverse then MATH, and hence MATH. Apply this to each component of MATH. CASE: Combine REF . |
math/9909130 | See CITE (compare REF ). CASE: By REF MATH so that MATH. CASE: By REF we have that the maps of NAME MATH-sets MATH are homotopy equivalences for MATH. It follows that for MATH and hence that MATH . |
math/9909130 | This is just the exact sequence MATH induced by the cofibration sequence of spectra MATH using REF to identify MATH . |
math/9909130 | This follows from the sequence of REF MATH whose composite is REF, and which is exact for MATH. By a simple calculation MATH . Since any zero-dimensional homology manifold is a disjoint union of points MATH . By REF MATH so that MATH . Also MATH and by topological transversality MATH so that MATH . The components of the assembly map MATH are given by the inclusion MATH and the isomorphism MATH . The cokernel of MATH is thus given by the cokernel of MATH as MATH. In particular, if MATH and MATH is one of the MATH-dimensional homology spheres with MATH constructed in CITE there exists a homotopy equivalence MATH with MATH so that MATH is not bordant to a dual transverse map. |
math/9909130 | If MATH is resolvable the mapping cylinder of a resolution MATH is a MATH-cobordism MATH with a bordism MATH to a homotopy equivalence MATH which is topologically dual transverse. Conversely, suppose that MATH is a dual transverse homotopy equivalence. Without loss of generality, it may be assumed that MATH is connected, so that MATH . The composite MATH sends MATH to MATH . Since MATH is dual transverse, MATH by REF . |
math/9909130 | See REF for the canonical MATH-orientation of a MATH-dimensional topological manifold. Let MATH be the normal map from a topological manifold determined (up to normal bordism) by the canonical MATH reduction MATH of MATH, with surgery obstruction MATH the assembly of MATH . The canonical MATH-theory orientation of MATH is given by MATH . |
math/9909130 | CASE: Immediate from MATH (stably). CASE: The canonical MATH-orientation of a topological manifold MATH is given rationally by the ordinary MATH-genus MATH with MATH, MATH. If MATH is the canonical normal map MATH . CASE: Immediate from REF , the identity MATH of REF , and the fact that the simply-connected assembly map MATH sends MATH to REF, with MATH the unique map. |
math/9909130 | CASE: For any normal map MATH of MATH-dimensional homology manifolds MATH . For MATH and MATH. CASE: Immediate from REF. |
math/9909130 | This is just the NAME theorem on the level of symmetric NAME cycles. CASE: By construction, MATH and MATH are classified by MATH . By the surgery product formula (on the level of quadratic NAME cycles) the product normal map MATH is classified by MATH . By the surgery composition formula (on the level of quadratic NAME cycles) MATH so that MATH is in the canonical normal bordism class. CASE: Immediate from REF , noting that MATH with MATH the MATH-dimensional homology kernel of MATH. |
math/9909130 | CASE: This is just the rationalization of the product formula of REF . CASE: This is just REF-dimensional component of the identity of REF . |
math/9909130 | We know from REF that the morphism of abelian groups MATH is an isomorphism for MATH. It follows that the morphism of abelian groups MATH is also an isomorphism for MATH. It remains to show that MATH preserves the multiplicative structures. By REF MATH for some element MATH. We need to show that MATH . From the above we see that MATH and MATH correspond to the MATH-numbers MATH . We compare these MATH-numbers with those of MATH, which are given by MATH . We observe that by REF MATH and that the MATH-numbers satisfy the following product formula MATH with MATH, and MATH, MATH such that MATH . Conversely MATH . |
math/9909132 | By definition, the MATH's are orthogonal and are subspaces of MATH. Since the translations of the MATH's spans MATH, it suffices to show that they are contained in this direct sum. But note that MATH is in MATH, and then MATH can be written as MATH, where MATH since MATH is obtained by a projection. By the recursive definition of the MATH's, we get that MATH is in the direct sum, and REF is established. REF follows from the general fact that the muliplicity function for a representation is the sum of the multiplicity functions for orthogonal subrepresentations. |
math/9909132 | We have: MATH . |
math/9909132 | First notice that since MATH, MATH. Furthermore, MATH, indeed, MATH is the function such that MATH. Hence, MATH. Hence, MATH . |
math/9909132 | It is shown in CITE that the dimension function is integer valued, and in fact is the dimension of a subspace of MATH. Let MATH be a sequence on MATH given by MATH. It is also shown that MATH is the dimension of the subspace of MATH spanned by MATH. We have that both the dimension function and the multiplicity function describe the dimension of some subspace of MATH. The spanning vectors are different; however, the subspaces are the same. Indeed, MATH where as MATH. Hence, MATH, and so the spaces spanned by them are the same. |
math/9909132 | We need to show that MATH for all MATH. We have: MATH . |
math/9909132 | Each wavelet can be associated to a GMRA; let MATH denote the wavelet MATH, and let MATH denote the MATH core space for the wavelet MATH. We need to construct an intertwining operator between MATH and MATH. Let MATH denote the closed subspace spanned by MATH. If MATH, then MATH. Proof of Lemma. It suffices to establish the lemma for a generating vector of MATH. Let MATH, then MATH for some MATH. Let MATH denote the wavelet MATH. We have: MATH . By REF , MATH. Hence: MATH as required. Since MATH for all MATH, we have MATH. Write MATH and MATH. Define MATH by MATH. A routine computation shows that MATH. Since MATH for all MATH the subspace MATH is invariant under MATH. Indeed, MATH. As such, MATH is also invariant under MATH. Likewise, both MATH and MATH are invariant under MATH. Note that MATH maps MATH unitarily onto MATH, and leaves MATH fixed, hence MATH maps MATH unitarily onto MATH. This proves that MATH is a unitary operator. Furthermore, as noted above, all of these subspaces in question are invariant under translation, hence by REF shows that MATH commutes with translations. Therefore, MATH is the required intertwining operator. |
math/9909133 | By definition, MATH if and only if MATH. Suppose that MATH and define MATH such that MATH. Consider the following commutation relation: MATH . This calculation establishes the statement. |
math/9909133 | Suppose that MATH is invariant under MATH. Then, by REF , for MATH, MATH is invariant under MATH, whence MATH is invariant under MATH. If follows that MATH is invariant under MATH. Only If. Suppose MATH is invariant under MATH. Then, again by REF , MATH is invariant under MATH, and hence MATH. Since MATH, it follows that MATH is also invariant under MATH. |
math/9909133 | CASE: If MATH is a MSF wavelet with wavelet set MATH, then by REF , MATH. Clearly, MATH, MATH is invariant under MATH since MATH is invariant under multiplication by MATH. It follows by REF that MATH is invariant under all translations. CASE: Since MATH is invariant under MATH for all MATH, by REF , MATH is invariant under MATH. CASE: By definition, MATH. If both MATH and MATH are invariant under integral translations, it follows immediately that MATH is also invariant under integral translations. CASE: Let MATH be the collection of all operators for which MATH is invariant. An easy calculation shows that MATH is WOT (weak operator topology) closed. If MATH is invariant under integral translations, then again by REF , MATH is invariant under MATH for all MATH. Since MATH is dense in MATH, in the strong operator topology, it follows that MATH is invariant under MATH. If we take the NAME transform, then we get that MATH is invariant under multiplication by MATH. The linear span of these operators are dense in the collection MATH with respect to the WOT. It follows that MATH is invariant under multiplication by any MATH function. Next, we wish to show that MATH, where MATH. First note that since MATH forms an orthonormal basis for MATH, MATH has maximal support in the sense that if MATH, then the support of MATH is contained in the support of MATH. This immediately implies that MATH. Let MATH be a compactly supported simple function, whose support MATH is contained in MATH. Define MATH, and define MATH. Since MATH is a simple function, it is uniformly bounded by some constant MATH. Let MATH be given. Choose a MATH such that MATH, and define MATH to be MATH. Then, MATH on MATH, so that MATH. Since MATH is closed, MATH; furthermore all such MATH's are dense in MATH, whence MATH. Since MATH, MATH is a set of measure zero, and since MATH is dense in MATH, it follows that MATH. Furthermore, by REF , MATH is MATH translation congruent to MATH, hence, by REF , MATH is a wavelet set, and MATH is a MSF wavelet. |
math/9909133 | By REF , MATH is invariant under MATH for all MATH if and only if MATH is invariant under translations by all real numbers. This is equivalent to MATH being a MSF wavelet. |
math/9909133 | Let MATH. Hence, MATH for some MATH. Let MATH, and let MATH be an odd integer. Then, by a similar computation, MATH as above. |
math/9909133 | It suffices to show that MATH for some MATH. Let MATH be such that MATH is a bijection. (It can be easily shown that MATH is a surjection.) The injective property of MATH can be assured in the following manner: for each MATH, define the set MATH, then for MATH choose MATH to be REF if MATH, if not, choose MATH, else choose MATH. Let MATH. Note that by construction, MATH is MATH translation congruent to MATH. Hence, MATH where MATH and is MATH periodic. Thus, for MATH, MATH . For almost any MATH, there exists a MATH and an integer MATH such that MATH. Moreover, by hypothesis, MATH is an even multiple of MATH, since MATH is not partially self similar with respect to any odd multiple of MATH. Since MATH is MATH periodic, we have that for MATH, MATH . This completes the proof. |
math/9909141 | Here we only show that MATH exists. Let MATH be the open parallelotope MATH . Then MATH is a MATH-dimensional centrally symmetric convex body with volume MATH. By NAME 's theorem (compare CITE), MATH contains a nonzero point. This is the desired point MATH. |
math/9909141 | Let MATH. We have an identity for rational functions of MATH valid for any MATH satisfying MATH for MATH. To see this, consider the MATH matrix MATH . This matrix is singular, since the first row is a linear combination of the others. Expanding by minors along the top row, and dividing by MATH yields REF. To pass from REF to REF, we need to incorporate the exponential character and sum over MATH using MATH. There is no obstruction to doing this, although we must omit terms where any linear form vanishes. We obtain the expression MATH where the inner sum is taken over all MATH with MATH for MATH. In REF, the MATH-th inner sum corresponds with the NAME sum MATH except for the terms with MATH and MATH for MATH. In other words, to make the MATH-th sum into a NAME sum, we must add MATH . Simultaneously adding and subtracting REF to REF yields REF. |
math/9909141 | The second statement follows easily from the first, so we focus on the first. By definition, we have MATH . Letting MATH and MATH, the right of REF becomes MATH . Inserting the character relations MATH we obtain MATH where MATH. The proposition follows now from the MATH-limit REF . |
math/9909141 | Without loss of generality, assume that MATH. We can represent MATH as MATH where there are MATH columns of the form MATH. The stars in the last MATH rows represent numbers that are irrelevant, since they don't affect the value of the sum. Let MATH be a matrix that carries MATH onto MATH. For MATH let MATH, where MATH is any lift of MATH. Then MATH has the form MATH . Here the row blocks have sizes MATH, MATH, and MATH, and MATH. Now let MATH be the determinant of any MATH minor from the top MATH rows of REF. This determinant will be the same up to sign as the determinant of a MATH minor containing MATH from the top MATH rows of REF. Hence MATH, and the proof is complete. |
math/9909141 | Write MATH as in REF. Permuting columns if necessary, we may assume that MATH are linearly independent in MATH. We will show that we can write an expression like REF so that the rank MATH sums on the right have type MATH for some MATH. Iterating this construction proves that we can have the rank MATH sums on the right of REF diagonal. We proceed as follows. Since MATH are linearly independent, we can find unique rational numbers MATH such that MATH . Now we want to apply the relation in REF . For each MATH, we choose MATH rational lifts MATH,,MATH, not necessarily equal to the columns of MATH, and use them to form a matrix MATH. Clearly MATH. Now define MATH . Clearly MATH. Write MATH for the matrix made from MATH by replacing the column MATH with MATH. Using the columns of MATH and MATH in REF, we find MATH where MATH is determined by the reciprocity law, and the sums are over pairs MATH satisfying MATH and MATH. We claim that the sums in REF actually have only MATH terms. This follows since the points MATH are dependent. Hence any sum such that MATH contains these points vanishes. Moreover, the sum MATH is zero, since by construction a column of MATH induces a linear form vanishing on MATH. Hence REF becomes MATH . Now consider the types of the rank MATH . NAME sums on the right of REF. If the type of MATH was MATH then the type of MATH is MATH . Hence by induction we can write MATH as a finite MATH-linear combination of diagonal rank MATH . NAME sums plus sums of lower rank, which proves REF. To complete the proof, we must show that the indices of the NAME sums on the right of REF are no larger than MATH. Indeed, REF implies that the indices on the right of REF are no larger than MATH, and so the claim follows. |
math/9909141 | By REF , we may take MATH to be diagonal. Suppose that MATH . Then by REF , there exists MATH such that MATH satisfies MATH. As in the proof of REF , write MATH, and set MATH. Now apply REF , using MATH and the columns of MATH, to write MATH as a finite MATH-linear combination of new NAME sums. These sums won't be diagonal, but we can apply the proof of REF with MATH playing the role of MATH. The resulting sums will include lifts of MATH in their columns and will be diagonal. Thus the resulting rank MATH sums will have index MATH, and the sums of lower rank will satisfy the conditions in the statement of REF . By induction on the index, this completes the proof. |
math/9909141 | First normalize and embed properly. By REF , any rank REF NAME sum is automatically unimodular and diagonal. The result follows by applying REF and descending induction on the rank. |
math/9909141 | Write MATH, where MATH. By the proof of REF , we know how to pass from a sum of type MATH to a linear combination of sums of types MATH . By iterating this, we pass from MATH to a linear combination of sums with types MATH . We will bound the number of rank MATH (respectively rank MATH) sums produced in passing from REF to REF by a constant MATH (respectively, MATH). We can then take MATH and similarly for MATH. To describe what happens in going from REF to REF, we use a geometric construction. Let MATH be the set MATH . The points in MATH correspond to types of intermediate sums in the passage from REF to REF. In particular, passing from REF to REF can be encoded by moving from MATH to MATH in MATH. Moreover, the sums of the form REF correspond to the subset of points MATH with exactly one coordinate MATH. (See REF .) Now the constant MATH will be given by MATH, as MATH ranges over all possibilities for fixed MATH, MATH, and MATH. If we allow the MATH to become continuous parameters, then a simple computation shows that the maximum occurs when MATH. With these conditions we have MATH . The constant MATH can be computed similarly. There are MATH sums of rank MATH produced for each point in MATH . One finds again that the maximum occurs when MATH, and is MATH . |
math/9909141 | This follows easily from the proof of REF . Forming REF is purely combinatorial, and makes no reference to MATH. In particular, the number of steps needed can be bounded for fixed MATH and MATH. |
math/9909141 | The proof is very similar to the that of REF , with the following twist: we begin with a diagonal sum, increase the number of distinct linear forms by one, and then make the sums diagonal again. We can keep track of the number of sums produced using the set MATH as in the preceding proof, although we must replace MATH with MATH to accommodate the extra initial step. We leave the details to the reader. |
math/9909141 | In REF, we have MATH . Thus after MATH iterations we'll have MATH . Since the index of a NAME sum is always an integer, the condition for termination is that for some MATH, we have MATH . On the other hand, by REF , we know that MATH iterations will produce no more than MATH sums of rank MATH. So fix MATH, set MATH, and let MATH . Then MATH . Now if we define MATH, we obtain REF. |
math/9909141 | Fix MATH. We proceed by induction on MATH. First, if MATH, then by REF the sum MATH is already diagonal and unimodular. Hence we may take MATH. Next, assume that the statement is true for sums of rank MATH, and let MATH be the corresponding polynomial. First we claim that without loss of generality, we need only consider the case that MATH is diagonal. Indeed, apply REF and write MATH where the rank MATH . NAME sums are diagonal, and all the NAME sums have index MATH. By REF , the sets MATH and MATH have a bounded number of elements independent of MATH. Hence we may bound the output for diagonal MATH, and then multiply this by a constant to obtain our final answer. Now we apply REF , and we must count the number of rank NAME sums produced. By REF , we know that the total number of rank MATH sums will be bounded by MATH . Furthermore, each sum of lower rank produced in the proof of REF can be written as a sum of MATH . NAME sums by induction. To find the total output, we must count these lower rank sums. We can do this as follows. Let MATH, where MATH is the constant in REF . Represent the process of reducing the diagonal sum MATH to unimodularity by the following diagram: MATH . The top row represents the bound on the number of rank MATH sums at each step of the algorithm, and the bottom row is the number of rank MATH sums. According to REF, we have produced MATH sums of ranks MATH. Adding this estimate to REF, we find that the total number of NAME sums produced will be a polynomial of degree MATH . To complete the proof, we compute the degree of MATH by induction. Indeed, using the estimate MATH and that MATH, an easy computation shows that MATH as required. |
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