paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9910102 | The spectrum consists precisely of the MATH such that MATH; this amounts to MATH. Now this holds when MATH, MATH or MATH for some MATH. These give respectively MATH, MATH and MATH, which after simplification yield the proposition. |
math/9910102 | We compute the determinant: MATH which completes the proof of the lemma, using the easily verified equations MATH valid for all scalar MATH and MATH. |
math/9910102 | Let us write temporarily MATH and MATH for MATH in MATH. Then we have MATH . Now, first MATH as claimed; then by induction, using REF , we have for MATH as claimed. |
math/9910102 | Solving MATH gives MATH. The result then follows from the factorization of MATH as a product of MATH's given by REF . |
math/9910102 | In the CASE: MATH is clearly MATH, because MATH. Now assume that MATH, and define two continuous maps MATH such that MATH. Then MATH is a free semigroup, because MATH and MATH have disjoint image-interiors. In the CASE: MATH is obviously all MATH, because MATH. Clearly MATH-orbit of MATH, is dense in MATH, so its closure MATH is MATH. In the last case: we first show that MATH. Since MATH, every point in MATH is specified by an infinite sequence of MATH's applied to some point. Since the MATH's are contracting, the choice of that point is unimportant - hence we may choose MATH, and this proves the claim. If MATH, then clearly MATH and we are done. Otherwise, set MATH, which is a countable set of isolated points. Clearly MATH and MATH are disjoint (else MATH would be in MATH). Finally MATH accumulates on the infinite orbits under MATH, hence on MATH. Let MATH. Then MATH disconnects MATH, and MATH disconnects each of the connected components of MATH, etc., so MATH is disconnected. Take any point MATH; then for all MATH we may write MATH for some MATH and some word MATH of length MATH. Taking some MATH close to MATH and letting MATH tend to infinity gives a sequence in MATH converging to MATH. Since MATH is clearly closed, it is perfect, so is a NAME set. Finally MATH, where MATH denotes the NAME measure. |
math/9910102 | We compute the determinant: MATH which completes the proof of the lemma, using the easily verified equations MATH valid for all scalar MATH. |
math/9910102 | Let us write temporarily MATH and MATH for MATH in MATH. Then we have MATH . Now, first MATH as claimed; then by induction, using REF , we have for MATH as claimed. |
math/9910102 | Solving MATH gives MATH. The result then follows from the factorization of MATH as a product of MATH's given by REF . |
math/9910102 | It suffices to note that for all MATH we have MATH; this follows by induction on MATH. |
math/9910102 | Consider the NAME graph MATH associated to the action of MATH on MATH, the MATH-th level of the tree MATH. The vertex set of MATH is MATH, and its edges are described by the action of MATH. Note first that the axiom is MATH. We construct MATH from MATH. Split MATH as MATH. By virtue of the definition of MATH given in REF , the MATH-edges within MATH are in bijection to the MATH-edges in MATH, while the MATH-edges within MATH are in bijection with the MATH-edges in MATH, and there are MATH-loops at all MATH. Moreover there are ``parallel edges" labeled MATH between MATH and MATH for all MATH. Now consider any MATH-edge in MATH, say between MATH and MATH. In MATH, it gives rise to a MATH-edge between MATH and MATH. Consider then a MATH-edge in MATH between MATH and MATH. In MATH, it gives rise to the following subgraph: a MATH-edge from MATH to MATH; two edges, labeled MATH and MATH, from MATH to MATH; a MATH-edge from MATH to MATH; and two loops, labeled MATH at MATH and MATH. This is precisely the substitutional rule for MATH, completing the proof. |
math/9910102 | We prove the claim for MATH only, as the same reasoning applies for MATH. Consider the NAME graph MATH associated to the action of MATH on MATH, the MATH-th level of the tree MATH. The vertex set of MATH is MATH, and its edges are described by the action of MATH. Note first that the axiom is MATH. We construct MATH from MATH. Split MATH as MATH. By virtue of the definition of MATH given in REF , the MATH-edges within MATH are in bijection to the MATH-edges in MATH, while the MATH-edges within MATH are in bijection with the MATH-edges in MATH, and there are MATH-loops at every MATH. Moreover there are ``parallel triangles" labeled MATH between MATH, MATH and MATH for all MATH. Now consider any MATH-edge in MATH, say between MATH and MATH. In MATH, it remains a MATH-edge, but now between MATH and MATH. Consider then a MATH-edge in MATH between MATH and MATH. In MATH, it gives rise to the following subgraph: a MATH-triangle between MATH, MATH and MATH; a MATH-edge between MATH and MATH; MATH-loops at MATH and MATH; and a MATH-triangle between MATH, MATH and MATH. Actually the MATH-edges form triangles so these subgraphs overlap at the MATH-triangles and MATH-loops. This justifies the substitutional rule for MATH-triangles, completing the proof. |
math/9910103 | To show the existence of MATH, MATH giving positivity we may modify the proof of REF so as to make some specific estimates, that is, we show that if a solution to REF exists, then the sequences MATH, MATH may be taken to grow at most linearly. Let MATH, MATH be the maximum eigenvalues of MATH, MATH. Let MATH exceed the largest absolute value of any other eigenvalue, and let MATH be the largest absolute value of the reciprocal of any eigenvalue. Consider MATH. Using the above-mentioned (see REF ) two invariant complex vector-space (column vectors) decompositions MATH we note that the contribution of the maximum eigenvector in MATH will be at least MATH for some positive MATH. When we multiply it by MATH we get MATH. The largest magnitude of any other term will be some MATH. We want the former terms to dominate the sum of all the others, say to be MATH times the largest, where MATH is the dimension of the matrices. Take logarithms, and we want MATH or rearranged equivalently as MATH . Then some arithmetic progression where the ratio of MATH to MATH exceeds MATH will give the domination. Consider denominators in the matrix entries which have as divisor some algebraic prime MATH. The prime MATH is fixed, but we will do this for all prime divisors in MATH. (For the definition of ``algebraic prime", see the end of REF.) For simplicity extend the coefficient field and assume we can diagonalize the matrices (the case of a standard NAME form can be treated similarly). The maximum denominator in MATH is MATH for some constant MATH, which for instance can be worked out from the determinant. Then consider the matrix entries in MATH. They will be sums of constants from the diagonalizing matrices times MATH powers of the eigenvalues MATH of MATH, that is, MATH. The eigenvalues, when factored, only involve nonnegative powers of MATH, since they are algebraic integers. The terms in this sum for eigenvalues not divisible by MATH must add up to be an integer at the prime MATH: otherwise, no very large powers MATH could make the total an integer. For the other terms, as soon as MATH exceeds MATH plus the degrees of constants arising from diagonalization process, we will have algebraic integers. |
math/9910103 | There is a MATH such that MATH, and let MATH be a column vector basis for the free abelian group MATH. Define a matrix MATH by MATH . The MATH matrix MATH maps MATH bijectively onto MATH. Let MATH be a MATH matrix extending the inverse of this bijective map. Then MATH is a MATH matrix. The mapping MATH on MATH has image in MATH, and hence we may define a MATH matrix map MATH by MATH . Now one immediately verifies that MATH, MATH, MATH, MATH, the equations of shift equivalence. Over an extension field, we may triangularize MATH, and find that MATH over this field becomes the sum of all nonzero generalized eigenspaces, since on them MATH is an isomorphism, but on the zero generalized eigenspace of dimension at most MATH, MATH is zero. Therefore MATH is nonsingular. |
math/9910103 | Form the matrices MATH, MATH, using REF , but now with the respective NAME inverses in place of inverses. In doing so, we can multiply by suitably chosen higher powers of MATH or MATH, and thereby force the MATH's and MATH's to have their row and column spaces contained in those of MATH, and then we must have the same inverse formulas for them as in the nonsingular case, provided that we use the NAME inverse. |
math/9910103 | Since MATH is finite, there is a MATH and a MATH such that MATH then MATH for all MATH and thus MATH for all MATH. Choose MATH such that MATH and put MATH. This gives MATH so MATH is idempotent. If MATH and MATH are idempotents, then MATH, so the idempotent is unique. If it is called MATH, then if MATH, then MATH, so MATH is a semigroup. |
math/9910103 | We first show that to any given modulus such as MATH, we can put a matrix modulo MATH into diagonal form using row and column operations (elementary matrices, having one non-zero off-main diagonal entry) modulo MATH. Each such operation lifts to a similar operation over MATH, so it preserves the given set, and moreover, these lifted operations will preserve the norm of the matrix. The reason that this works is that MATH is a principal ideal domain, even though MATH may not be. This means that every ideal is modulo MATH generated by some element - for this it suffices to factor MATH into primes, use the fact that finite extensions of the MATH-adic integers are principal ideal domains, and then use the NAME remainder theorem CITE to assemble primes. We can determine the group MATH generated by row operations modulo MATH; it is a subsemigroup of the finite semigroup MATH generated by a given finite list of generators. The criterion for being in MATH is that a matrix is in MATH where MATH . We can determine (list) the finite set of such reductions by determining a finite list of generators for the group of units of MATH, that is, units of MATH and combinations of prime factors of MATH, as well as the class group, the prime factors of MATH, and their images in the class group. CITE gives methods for finding an integral basis - we can take some rational basis, compute its discriminant, and then find integral bases locally at primes which divide the discriminant (in effect by finding extensions of the ring of MATH-adic integers). CITE further gives MATH-adic methods for determining what the prime ideals are which lie over given rational primes MATH. We can factor an element into primes, by factoring its rational norm into primes CITE, and successively attempting division by the algebraic primes over MATH. Also from CITE we have algorithms for bounding the norms of representatives of the class group, so that to compute the class group, it remains to tell when two given elements generate the same ideal class, which is discussed by CITE. These authors CITE also give an algorithm for finding generators for the group of units in algebraic number fields. Necessity of this condition follows by the first paragraph. We now show sufficiency, that is, that every such diagonal matrix modulo MATH actually arises from a matrix of MATH. To do this, we start with a diagonal matrix the product of whose entries modulo MATH is some MATH. Then we alter it by multiples of MATH so as to modify the determinant by an arbitrary multiple of MATH. To do this, make the entries MATH for each MATH equal to MATH and the entry MATH equal to any MATH. This adds the single product MATH to the determinant. |
math/9910103 | We find the eigenvalues of MATH, diagonalize MATH over MATH, factoring ideals into primes, using standard algorithms, for example, CITE. Define MATH and MATH as in the paragraph before the proposition. The effect of NAME action and the families of intersections of these spaces can be considered by taking NAME bases MATH for MATH. We have ordered the intersections MATH with bases MATH by inclusion, and have put NAME conjugates next to each other. Then the subspace generated by all bases succeeding any given basis is preserved, and we have a block-triangular structure corresponding to it, and a larger block-triangular structure, whose blocks are the sets of NAME blocks of those from the former structure. The latter will be defined over MATH as required. Since the elements of MATH strictly increase filtration, any MATH-fold product of elements of MATH is zero, where MATH is the filtration length, that is, the elements of MATH are the matrices in the algebra which are zero on the main-diagonal blocks, and so the quotient maps isomorphically into the sum of the general linear groups on MATH with basis MATH. Each MATH, by induction, and thus each MATH is spanned by the union of the MATH's contained in it. But we note that the general linear group on the span of MATH will preserve all subspaces MATH, and their images will span each of the required summands, so together they will span the sum. Finally, the larger filtration mentioned above gives the MATH'MATH . |
math/9910103 | If MATH can be defined over a subfield of MATH, then the NAME group of that field must fix MATH; conversely if the NAME group fixes MATH, it will also fix the complementary sum of generalized eigenspaces, hence it will fix a projection operator to the subspace whose kernel is the complementary sum of generalized eigenspaces, and from the columns of a matrix for this operator, the subspace can be defined. Given an endomorphism of MATH over MATH which arises from a mapping over MATH, the endomorphisms of all other MATH are uniquely determined as its NAME conjugates. This means we have a one-to-one linear mapping from endomorphisms of MATH over MATH fixing MATH (and these by NAME conjugacy fix every MATH), into the general linear group of MATH over MATH. In fact the image lies in the general linear group over MATH since over it, we can define a projection operator to MATH. This mapping is also an epimorphism, since, given any MATH-linear mapping MATH of MATH to itself, there are NAME conjugates defined on the other MATH (the NAME operator is unique up to the subgroup fixing MATH, which also fixes MATH). We can take the sum of MATH and its NAME conjugates on the other MATH, and the sum will be a NAME mapping of MATH, and therefore defined over MATH. |
math/9910103 | Note that for our matrix representation the norm conditions on MATH will give norm conditions on MATH, since the latter gives the main-diagonal blocks in a block-triangular representation, and the product of their determinants is the determinant in MATH. The condition that the determinant is a fixed algebraic integer MATH times products from a finite list of primes and units will translate into a finite list of similar conditions at each main diagonal block, based on the prime factorizations of MATH. Additively, write an element which is to have determinant involving certain primes, and satisfy congruences, as MATH where MATH is in the NAME radical. The congruences will say, for some MATH, a Boolean combination of congruences MATH hold. If we take all possibilities MATH for MATH, this will be a Boolean combination of congruences MATH. |
math/9910103 | We can eliminate the other prime factors of moduli and the denominators by multiplying by a power of the defining matrix MATH large enough to cancel off the denominators. That is, if we have a solution mapping MATH at a particular generalized eigenspace which satisfies congruences for all primes except those which divide the eigenvalue MATH, then MATH will produce a solution at all the other primes, which is congruent to zero modulo any set power of the primes in MATH, and therefore exists over MATH. And if any solution does exist, multiplication by a large power of MATH must produce one which is congruent to zero modulo high powers of the primes in MATH; hence it is one that can be found in this way. The last statement follows by invertibility of the matrix MATH restricted to any eigenspace, at all primes not dividing the eigenvalue, so that there will be arbitrarily large powers of MATH congruent to the identity. |
math/9910103 | This result is a consequence of the preliminary discussion and the propositions above. First reduce the problem to the case of nonsingular matrices by the method in REF. This reduces the problem to one of finding a matrix MATH which preserves certain subspaces, has certain primes in its determinants, and satisfies congruences, going from MATH to MATH. We find such a matrix MATH over the rational numbers; the proposed solution must differ from it by multiplying with a matrix MATH meeting corresponding conditions (we multiply by a constant MATH to arrange that MATH have integer entries). We find the NAME radical of MATH and the simple components of the quotient by it, and restate the congruences in terms of those simple components. They are determined in terms of certain combinations of generalized eigenspaces, as general linear groups over algebraic number fields. Again we conjugate, and obtain a new finite set of congruences on a tuple of matrices over algebraic number rings (no longer necessarily fields, because they are images of integer matrices) of the same general nature as the originals, (MATH), (MATH), (MATH), and a determinant condition (iii-MATH). We use REF to ensure that the congruences involve moduli relatively prime to the eigenvalues and can allow these eigenvalues as denominators. By REF , we can solve them. |
math/9910103 | The first assertion is by CITE, and the second is by CITE. The third assertion follows because the space of vectors in the dimension group MATH such that some fixed multiple is arbitrarily divisible by a given algebraic prime is sent to the corresponding subspace of the other dimension group, and this set is the sum of the generalized eigenspaces for all eigenvalues divisible by the prime. If these spaces are intersected over all primes dividing the NAME - NAME eigenvalue, we get, by our hypothesis, only the NAME - NAME eigenspace. The fourth assertion follows because the eigenspace of MATH will consist precisely of those vectors in the dimension group which are divisible by arbitrary powers of primes occurring only in MATH, so it must be preserved by any isomorphism of dimension groups. In addition, vectors in this MATH-dimensional space which are not divisible by primes other than those in MATH will be unique up to multiplication by units and primes dividing MATH, so they will be preserved by any isomorphism, up to such multiplication. Next note that the fourth statement and the first part of the fifth statement are equivalent whenever we have an isomorphism from the dimension group to itself induced by an integer matrix. The reason is that since row and column NAME - NAME eigenvectors are preserved, this integer matrix in a basis corresponding to generalized eigenvectors becomes block diagonal, and the block for the NAME - NAME eigenvectors must be the same element for the row eigenvectors as for the column eigenvectors, and by REF it involves only primes dividing MATH. Now consider integer matrices MATH and MATH inducing mappings each way between two different column dimension groups, with NAME - NAME eigenvectors MATH and MATH normalized over the algebraic number ring. We have MATH and MATH, where MATH and MATH are algebraic integers since MATH and MATH are integer matrices. But MATH arises from a map of the dimension group to itself, so it divides a power of MATH, hence so do MATH and MATH. |
math/9910103 | By REF, there is a unimodular matrix MATH sending the NAME row eigenvector of MATH to the NAME row eigenvector of MATH and the NAME column eigenvector of MATH to the NAME column eigenvector of MATH (and we can choose signs for positivity). By CITE this gives a positive mapping on dimension groups. Since the row eigenvectors are perpendicular to the sum MATH of all non-Perron-Frobenius generalized eigenspaces, MATH, and also MATH as noted in REF. Write any vector MATH as a direct sum according to REF , MATH. This splitting can introduce certain fixed primes MATH in the denominator. Note that the matrix MATH is unimodular and integer restricted to the integer vectors in MATH (and similarly for the matrix MATH), because each determinant is the product of its determinant on this space and its determinant on the NAME - NAME eigenspace, and because it is an integer matrix preserving this subspace. Multiplication by MATH is multiplication by MATH on MATH (see REF ), and the same is true for MATH. For MATH to be in the dimension group means for all sufficiently large MATH, MATH has integer entries. Any prime MATH which does not divide MATH will not occur in the denominator of the expression MATH. Consider those primes MATH which divide MATH. We claim that they cannot occur in denominators of MATH. Restricted to vectors MATH, the matrix MATH is unimodular, so modulo any powers of those primes it lies in a finite group, MATH. Thus we can choose arbitrarily large MATH so that MATH is congruent to the identity. But then in MATH, the denominators in MATH have vanished, being multiplied by MATH and those in MATH remain. So MATH is not in the dimension group, a contradiction. Therefore in MATH both terms are integer for sufficiently large MATH (with a symmetrical argument the other way) which verifies the conditions in REF for isomorphism of ordered dimension groups. |
math/9910103 | Let MATH; it will be a MATH-ideal which spans MATH over MATH. Now define an integer matrix MATH which expresses the action of MATH on MATH, that is, form an additive basis MATH for MATH, let MATH, MATH. This matrix will have an eigenvalue MATH, and we claim that at the corresponding eigenspace, the image of the trace is isomorphic to MATH. This is because the action of MATH on MATH has been forced to be that of MATH on MATH, and because the trace reflects this module structure, by means of the short (nearly exact) sequence. Finally we claim that we can conjugate MATH over MATH to a matrix whose powers are eventually positive; then those powers will be nonnegative matrices whose image of trace is the same. To get eventual positivity, given that MATH is the largest eigenvalue (the largest of its NAME conjugates), it is necessary and sufficient that its row and column eigenvectors for this eigenvalue be positive, by a limit argument somewhat like that in REF . Let MATH be row and column eigenvectors at the eigenvalue MATH, with signs chosen so that their inner product is positive. Multiply each by a large integer, and then take relatively prime integers approximating its components. Such a pair of vectors can be mapped over MATH to any vectors whose entries are relatively prime integers having the same inner product, by CITE, in particular, to ones which are positive, if MATH. If MATH we use the same result and get a congruence condition, but that is compatible with positivity. |
math/9910103 | The first three statements are a reformulation of CITE, except for the relationship with the trace, which we next show. REF , and MATH imply that the image of the trace is a module over MATH and a subset of MATH. Using the standard basis for MATH, it is generated by MATH as a module over MATH since MATH, MATH generate the dimension group. The trace mapping is an epimorphism if we pass to rational coefficients (that is, tensor dimension groups with MATH), just because its image is nonzero (consider MATH as a NAME - NAME column eigenvector) and closed under field operations in MATH. Its kernel is zero since the dimension group with rational coefficients is also a MATH-dimensional vector space over MATH (for instance, by CITE). Thus the trace mapping is an isomorphism to its image as asserted in the second part of REF . REF (stated as REF) in fact allows us to replace the condition in REF above that MATH and MATH be nonnegative with the condition that instead they are integral eventual positive (IEP), that is, that they are in MATH and have respective powers with strictly positive entries. Next we show that dimension group isomorphism in our sense implies shift equivalence if the irreducible characteristic polynomials of MATH, MATH are equal. The only difference with the isomorphisms used in CITE is that there the actions of MATH, MATH are the same, that is, the matrices themselves represent the field element acting on this module. But the field element MATH, MATH represent are roots of the same irreducible characteristic polynomials, and are the unique NAME - NAME roots of these polynomials so they must be the same field element. Equivalence to REF : REF is, properly understood, a rephrasing of REF , given the isomorphism in the first paragraph of the proof. We will clarify the kind of module structure which is involved. Assuming REF , the images of the traces generate the fields MATH, but the rings MATH and MATH may be different, in which case we work with the full-rank subring MATH. REF is immediate, and REF follows since algebraic primes dividing MATH are those primes which can divide elements of the dimension groups to arbitrary powers. Conversely, suppose we are given REF . The equality of fields asserted in CITE is taken in the sense of ``equality of MATH and MATH as subfields of the real numbers", which gives embeddings of MATH and MATH into the real numbers. Thus it also embeds the modules which can be considered as subsets of MATH. The isomorphism of modules as additive groups acted on multiplicatively (that is, the action MATH) by subrings of MATH having full rank (in this case MATH) means there is some element of the quotient field mapping one to the other: if the isomorphism of REF maps some element MATH to MATH (considered as images in the real numbers) then the ratio MATH is independent of the choice of MATH by definition of the isomorphism in REF , and we multiply by this ratio to get the isomorphism in REF . |
math/9910103 | Consider an isomorphism of dimension groups. The eigenvectors generate the MATH-dimensional spaces of vectors such that some multiples of those vectors are in the dimension group and are divisible by arbitrary powers of the respective eigenvalues. Hence any dimension group isomorphism must preserve those subspaces. Moreover, we claim that a dimension group isomorphism must send normalized eigenvectors to normalized eigenvectors of the image. The rational multiples of a normalized eigenvector MATH with rational eigenvalue MATH which lie in the dimension group are the elements of MATH: a vector MATH lies in the dimension group MATH, since MATH is a multiple of an integer vector by negative powers of MATH. And if MATH then there is a MATH such that MATH, so that MATH in lowest terms, MATH. It follows that dimension group isomorphism implies the existence of an isomorphism of MATH which sends each eigenvector to a multiple of the other eigenvector by a number which divides a power of MATH. Such a mapping must preserve the action of multiplication by MATH, given that the characteristic polynomials are equal, because this multiplies each eigenvector by its eigenvalue, and the eigenvalues are the same. So the mapping will be a shift equivalence. Let MATH be the diagonal matrix whose main diagonal entries are the multiples just mentioned. Then the isomorphism MATH of dimension groups will be, specifically, MATH if it exists. For if we multiply MATH and its inverse on the left by a large enough power of MATH or MATH, respectively, as in REF , we see that the resulting matrix products must be integer matrices. Moreover, these multiples are the matrices stated in the theorem. |
math/9910103 | This follows directly from REF . |
math/9910103 | Assuming, as we will show, that the image consists of torsion elements relatively prime to MATH, the inclusion gives a natural mapping. If we multiply any element in MATH by a power of power of MATH, we can get an element of MATH, so this mapping is an epimorphism. We also claim that if we multiply any element of MATH, say MATH, MATH by a power of MATH, we will get an element of MATH. This is because MATH using the left two factors. Note that since MATH and MATH are both torsion-free and MATH-divisible, their quotient has no MATH-torsion. Hence every element annihilated by a power of MATH lies in the kernel. (We will show that the image consists of torsion elements relatively prime to MATH.) Let MATH be in the kernel of this mapping. Then MATH, so that for some MATH, MATH and is zero in the original group. This identifies the quotient. The left hand group, the quotient of a free abelian group by a full-rank subgroup, is finite, so some fixed MATH works for the whole kernel. |
math/9910103 | MATH written as MATH will have as denominators only primes dividing MATH. If MATH includes MATH then we have a torsion group whose torsion involves only primes in MATH. We first argue that locally at each prime MATH in it, MATH consists of those vectors dual to the eventual MATH-adic row space MATH of MATH (see REF - REF ). That is, REF holds. The dimension group is the group of vectors MATH such that for some MATH, MATH. This is the group of vectors such that MATH such that for MATH, we have MATH. This is the group of rational vectors whose products with the row space of MATH is integer. This construction also goes through if we localize at any prime. To say that a vector has MATH-integer product with the row space of MATH for some MATH then implies that it has MATH-integer product with the idempotent MATH-adic limit MATH of powers of MATH, defined in REF and mentioned in REF. Conversely suppose it has MATH-integer product with the idempotent MATH-adic limit, then by MATH-adic continuity, it must have MATH-integer product with some finite power. This gives the claim. Now to show that the quotient group at the prime MATH is MATH-divisible, take a MATH-adic dual basis to MATH, which, like any MATH-adic torsion-free module, must be a free module (the MATH-adic integers are a principal ideal domain, and argue as with the ordinary integers). Approximate these vectors MATH-adically by rational vectors MATH using a MATH-adic approximation theorem such as CITE, choosing these rational vectors so that they give a MATH-adic basis. Take the free abelian group MATH generated by MATH. As soon as we have a lattice MATH including MATH and MATH, the MATH-adic dimension group consists of a sum of copies of the MATH-adic integers corresponding to MATH and a sum of copies of the MATH-adic field corresponding to the remaining vectors (in the null space of MATH - we can take additional basis vectors for it). When we divide by MATH, we are dividing out by all the MATH part MATH-adically, and by something isomorphic inside a MATH-adic field in the rest, and the result will be MATH-divisible. In fact, for any lattice MATH such that MATH is the dimension group, the quotient will be isomorphic to this, since multiplication by MATH gives an isomorphism of pairs MATH, and eventually this lattice must be large enough. Now consider the MATH-adic rank, in relation to the characteristic polynomial. By NAME 's method CITE, if the characteristic polynomial has the given form, we can factor it over the MATH-adics as a product of two polynomials, one of which is MATH modulo MATH, and the other of which has invertible constant term over the MATH-adics. We can put the matrix into corresponding block form. The former part will be MATH-adically nilpotent, and the null space will be its row space. |
math/9910103 | The given structure (that is, MATH up to its equivalence with any MATH) is an isomorphism invariant because the MATH-adic row spaces MATH defined in REF - REF are invariants. REF shows that a rational matrix over MATH giving an isomorphism on dimension groups must give an isomorphism on the MATH-adic row spaces, hence the dual MATH-adic null spaces of MATH. In fact REF gives, as necessary and sufficient conditions for unordered dimension group isomorphism, in effect, the existence of MATH and MATH: the MATH-adic symmetries just mean we are considering the row spaces up to isomorphism, and the MATH symmetry means that we have a rational map which is an isomorphism at all primes other than the ones considered here. The extension class of any extension of MATH by a group MATH may be computed by extending the map MATH to a mapping MATH, and letting this give a map in MATH. This is REF - CITE. To identify this class it suffices to look at the MATH-torsion subgroup of MATH for each prime MATH since the group is the direct sum of its MATH-torsion subgroups. To identify this class, take the tensor product of MATH with the MATH-adic integers, getting a localized extension of MATH by the MATH-torsion subgroup of MATH, which is MATH. But if we write all MATH-adic vectors as the direct sum MATH of the MATH-adic null space of MATH and a complementary space MATH, by REF , MATH . Thus the extension class is represented taking MATH and collapsing by MATH to give the inclusion MATH . This map is induced by the map MATH which can be taken to send the MATH-th unit vector on the left to the MATH-th vector in a basis for MATH on the right. This means taking basis vectors for the null space of MATH as forming the columns of the matrix giving the extension. |
math/9910103 | The standard unit vectors do this for any vector in the subsemigroup of strictly positive integer vectors. We claim transforms of these by integer row and column operations, permutations, and reversals of sign, take any vector to the interior of this subsemigroup - then just reverse those operations on the standard basis vectors. In fact, we get all coordinates nonzero by certain linear combinations, then reverse their signs. |
math/9910103 | Let the dimension of MATH be MATH and the rank of all sufficiently large powers MATH be MATH. By REF , we find a set of MATH vectors MATH in the eventual row space MATH, that is, the row space of MATH, or some specific higher power, a rank-MATH subspace of MATH such that the cone over MATH generated by this set includes a neighborhood of the maximum eigenvector MATH within MATH. This is sufficient to establish that all sufficiently large powers of MATH have their rows expressed as (unique) nonnegative linear combinations of MATH, since all rows of MATH divided by their lengths converge to fixed multiples of MATH and hence are eventually in the convex cone; but to be in the convex cone means that we have these convex combinations. However, we also need that it can be chosen that these convex combinations are eventually integer. For that, it suffices that the determinant of the MATH expressed as combinations of a basis for the integral vectors in the eventual row space, that is, MATH, a rank-MATH free abelian group, is MATH or MATH. This follows from the lemma and the extra assumption. Now let MATH be the matrix of MATH expressed as acting on the vectors MATH, which will be nonnegative, and positive. Then MATH is shift equivalent to MATH over the integers (maybe with negative entries), just by the inclusion mapping given by the vectors MATH. By REF and CITE (reproved in our REF paper CITE), any shift equivalence over MATH of primitive matrices can be realized by a shift equivalence over MATH. This shift equivalence will induce an isomorphism of ordered dimension groups. |
math/9910103 | The proof of REF , or alternatively, the proof of REF , shows that having a rational mapping which induces isomorphism of MATH-adic dimension groups is equivalent to having a rational mapping which induces an isomorphism of MATH-adic eventual row spaces. To make such a mapping integer, multiply by all denominators relatively prime to MATH. At MATH we must have integrality on the eventual row space MATH which is a summand of the space of all MATH-adic integer vectors. The use of a projection MATH to this subspace enables us to get a map equal to the given integer mapping defined over some algebraic number field, MATH-adic integer, and equal to the rational mapping on the eventual row space. The irrational part of this map will also consist of algebraic integers, since MATH will be an additive summand of the algebraic number ring, and can be discarded, since it must be zero on the summand. This gives a matrix of integers inducing the MATH-adic isomorphism, and thus the local isomorphism is strong. A rational mapping inducing an isomorphism of MATH-adic row spaces is equivalent to having a matrix MATH (giving the rational mapping) and a matrix MATH such that MATH (where MATH expresses the image of the row basis vectors of MATH as linear combinations of a basis for MATH); it must be MATH-adic integer and invertible since we can also map backwards by the isomorphism of row spaces. Next we show this criterion is decidable. The field generated by all eigenvalues of MATH or MATH will be sufficient to realize the MATH-adic eventual row spaces: we take this field and some prime MATH giving an embedding in an extension of MATH. Then the MATH-adic eventual row space is spanned by the generalized eigenspaces for eigenvalues which are relatively prime to MATH, since multiplication by powers of MATH will not send them to zero, but will annihilate all other generalized eigenspaces, modulo any power of MATH. Some linear combinations of basis vectors for these generalized eigenspaces must give a MATH-adic basis. The required MATH-adic matrix MATH then must lie in this field MATH, since MATH, MATH, MATH do. Being MATH-adic integer means that its denominators are relatively prime to MATH. It can then be expanded as a larger matrix over MATH, using a basis for the MATH-adic integers of MATH over MATH. Given the matrix MATH, the condition that a corresponding MATH exists over MATH can be stated by linear equations in MATH. Existence of rational MATH means that all columns of MATH are rational combinations of the columns of MATH. For the expanded matrices over MATH, that means that we have any linear column combinations of the columns of MATH yielding the desired columns of MATH. In turn that means that the columns of MATH have zero inner product with all vectors which have zero inner product with the columns of MATH, which is a linear condition. We can now determine a MATH-adic basis for this linear space of matrices MATH, write the determinant of MATH as a polynomial in the coefficients of a general linear combination of basis elements, and determine whether or not it is possible for the determinant to be nonzero modulo MATH by testing each congruence class of entries modulo MATH. |
math/9910103 | Since MATH is unimodular, we have MATH and the observation follows from REF - REF . (REF may be replaced by the strictly stronger requirement that MATH and MATH have only nonnegative matrix entries.) |
math/9910104 | From REF for MATH, it is clear that MATH is a differential operator from MATH. Since at most MATH edges of MATH can terminate at the vertex MATH (corresponding to MATH), the order of MATH is at most MATH . Also, the degree of the coefficients of this differential operator is MATH which is at most MATH. Now let MATH. Then the statement follows from the discussion above. |
math/9910104 | To describe any graph from MATH, it suffices to provide the following data for each vertex MATH: CASE: the source of an incoming edge, if any (MATH choices, counting the case when there is no incoming edge); CASE: whether the vertex is connected to the vertex MATH, vertex MATH, both or neither (REF choices). Hence we have at most MATH choices for each of MATH vertices of the first type. Taking into account the possible labelings of the outgoing edges at each vertex, we get MATH . Now the statement follows from the inequality MATH . |
math/9910104 | Recall that the weight MATH associated to the graph MATH is given by the formula MATH where MATH is the pullback of the volume form from the MATH-dimensional torus MATH under MATH. We consider the preimage of a generic point MATH under MATH. Using the action of MATH, we can fix MATH and MATH, and thus identify MATH with MATH. Rewriting REF as MATH we see that each edge gives rise to a pair of quadratic equations involving the coordinates of its endpoints. Thus we obtain a system of MATH quadratic equations in MATH real variables (the real and imaginary parts of MATH). Generically, this system has at most MATH (complex) solutions. Therefore, we conclude that MATH where MATH is the volume of MATH. Obviously, MATH and the statement follows. |
math/9910104 | Without loss of generality, we may take MATH . According to REF , MATH . The expression above can be rewritten as MATH . Set MATH. It follows from the discussion after REF , that only the graphs MATH can contribute to MATH. Observe that all partial derivatives of MATH are at most REF in absolute value (this follows form REF ), and that any derivative of MATH is a monomial with the coefficient not exceeding MATH. Therefore the absolute value of the coefficient MATH in REF does not exceed MATH. Using REF , we obtain MATH . Now by REF , MATH, hence MATH where MATH. Therefore, the series MATH converges absolutely in the polydisk of radius MATH. Hence in this neighborhood the formal series REF defines a differential operator of the form MATH, where MATH is the operator of multiplication by some analytic function. The statement follows. |
math/9910104 | It suffices to show that the constants MATH from CITE satisfy estimates of the form MATH for some constant MATH. According to CITE, we have MATH where ``the wheel" MATH is the graph from MATH shown in REF . Arguing as in the proof of REF , we conclude that MATH, hence MATH. |
math/9910104 | By CITE, for MATH we have MATH . Hence for MATH as in the statement of REF , we get MATH . We now discuss the structure of MATH. The relative positions of the vertices MATH and MATH are determined by the point MATH. The integral MATH is the sum of integrals over the top (that is, MATH-dimensional) components of MATH, which correspond to the following (degenerate) configurations: CASE: Two or more points cluster at MATH. CASE: Two or more points cluster at MATH. CASE: Two or more points cluster somewhere else in MATH. CASE: One or more points cluster on MATH. For components of type REF MATH (since the angle for any edge originating from MATH is identically REF). For components of types REF which involve clusters of three or more points, MATH by CITE. Now let MATH be a component corresponding to a two-point cluster of type REF. The corresponding boundary component of MATH is MATH and we have MATH, where MATH is the preimage of the path MATH in MATH under the forgetting map. As in CITE, MATH decomposes as a product of integrals over MATH and MATH. The first integral vanishes unless two points in the cluster are connected by an edge. MATH . Relabeling the vertices and edges as above, we see that the operator MATH is a sum of terms with factors of the form MATH . The set of graphs with the above subgraph can be grouped into sets of three corresponding to a cyclic permutation of the three outgoing edges from the cluster. Each of these graphs has the same weight MATH. By the NAME identity, the sum of the three corresponding operators MATH vanishes. Hence MATH. Now we consider two-point clusters of type REF. MATH . Relabeling the edges and vertices as above, we see that the corresponding operator is a sum of terms with factors of the form MATH where MATH acts as a differential operator. But this expression is equal to MATH (where MATH is the adjoint vector field corresponding to MATH), which vanishes since MATH is invariant (MATH). This leaves only two-point clusters of type REF. MATH . Arguing as above, we conclude that MATH is a sum of terms containing factors of the type MATH. Hence MATH-is an infinite sum of operators from MATH. By REF MATH belongs to MATH. Therefore, MATH, as was to be shown. |
math/9910104 | In view of REF , it suffices to establish the following identity (in MATH) MATH . Since both sides are germs of analytic differential operators, it is enough to verify that MATH for all MATH. Then MATH . |
math/9910104 | A bi-invariant differential operator on MATH is precisely an element of the center MATH of MATH, hence of the form MATH for some MATH. Let MATH and MATH be the delta-distributions supported at MATH and MATH respectively. Then if MATH, by CITE (see also CITE), there exists a germ MATH, such that MATH . Hence by REF , we get MATH and we see that MATH is a local fundamental solution for MATH. |
math/9910104 | If a bi-invariant differential operator on a real NAME supergroup MATH has a nonzero body, it can be represented as MATH, where MATH and MATH has a nonzero body. Then we can find a germ MATH, such that MATH . Local solvability of MATH now follows as in the proof of REF . |
math/9910106 | Recall every framed link admits a projection (every link admits a projection according to CITE, and by adding a suitable number of full twists to the projection one can make the projection have any given framing), so choose one such. Notice the moves in REF do not change the framed link, but change the winding number of the indicated component by MATH A sequence of such moves gives a projection in which every component has winding number one or zero, and the observation preceding the proposition indicates these components must have respectively even or odd framings. For the second claim, Trace CITE has shown that two link projections are regular isotopic (that is, connected by a sequence of regular isotopy moves and planar isotopy), if and only if their links are isotopic and each component has the same writhe and winding number. The writhe of a component of a projection is the signed sum of the crossings of the component with itself, which is of course the framing if it is a projection of a framed link. That every framed link admits a height function is clear, by choosing any real function with no critical points and perturbing slightly as necessary. That two such are connected by a sequence of the height function moves appears in CITE. |
math/9910106 | CASE: The first sentence is clear, as is the fact the the spin NAME moves do not change the framed three-manifolds. So suppose MATH and MATH are two spin MATH-handlebodies with the same spin boundary and signature presented by links MATH and MATH . We will show that MATH and MATH can be connected by spin NAME moves. Since they each have the same signature, gluing MATH to MATH along MATH we get a closed spin manifold with signature zero. Hence by CITE there is a spin five-manifold MATH which this closed manifold bounds. Choose a NAME function MATH such that MATH . Each MATH for MATH not a critical point is a spin four-manifold. If MATH is an index MATH critical point then MATH is MATH connect summed with MATH with the same spin structure outside the region of the connect sum. In between these two manifolds we can change MATH by replacing the MATH-handle with a MATH-handle attached along a contractible loop in the boundary. The spin structure on the boundary MATH induces a spin structure on the boundary MATH of the MATH-handle MATH which extends to the interior. In this fashion MATH can be replaced by a new spin five-manifold in which there are no MATH-handles but the NAME manifolds (that is, MATH of noncritical values) are the same. Similarly we can replace all the MATH-handles with MATH-handles. Arrange MATH so that all MATH-handles have NAME values less than those of all MATH-handles. Because the NAME manifolds are simply-connected, each MATH-handle as it is attached connect sums MATH to the NAME manifold, which corresponds to spin NAME move II applied to MATH . Likewise by flipping MATH we see each MATH-handle corresponds to move II applied to MATH . Thus a sequence of Moves II applied to MATH and MATH gives links representing the same spin four-manifold with boundary MATH . NAME 's argument that these can be connected by a sequence of moves I goes through unchanged. CASE: If MATH and MATH have the same spin three-manifold boundary, then their signature differs by a multiple of MATH . Thus the union of one of their links with sufficiently many copies of the NAME surface link results in two links which present the same framed manifold, and REF applies. |
math/9910106 | Of course it suffices to take an arbitrary instance of spin NAME move I and decompose it as a sequence of moves MATH and II. Let MATH be an even link with a component MATH and let MATH be another component to be band connect summed with MATH as illustrated in NAME move I in REF . Choose a presentation of MATH with the winding number of MATH being one as in REF , and such that the band between MATH and MATH along which the connect sum is to be applied does not overlap any component of the link. Choose MATH crossings of MATH with itself such that flipping the parities of these MATH crossings (that is, over to under or vice versa) makes MATH a MATH-framed unknot (this step relies on the winding number condition). Apply move II MATH times to create a NAME link for each of these crossings, then apply move MATH twice at each crossing as in REF . The effect of these moves is to make MATH a MATH-framed unknot (of course the disk it bounds intersects MATH in many places), and in this new link, the band connected sum of MATH with MATH along the same band is an instance of move MATH. After applying this band connected sum, the vicinity of each of the MATH crossings looks like the left-hand side of REF , and undoing the two instances of move MATH and the instance of move II at each crossing results in the right side of REF , which is a projection of the original band connect sum of MATH with MATH . |
math/9910106 | Since MATH is simple, MATH must be simple. Of course MATH is different from MATH so MATH and MATH are distinct. For any link MATH the invariant of MATH with an even component labeled by MATH is the invariant of MATH with that component doubled and labeled by MATH and MATH respectively. Since MATH is degenerate we have MATH so in particular in any projection of this link any crossing involving MATH can have its parity switched (over to under or vice versa) without changing the invariant. A sequence of such changes can unlink and unknot the component labeled by MATH so the invariant of the doubled link is equal to the invariant of the original link time MATH where MATH is the framing of the component labeled by MATH . Now MATH is even and MATH so we get only the factor of MATH . But since MATH is trivial, we know MATH and since in a MATH-category quantum dimensions are positive, MATH . Of course MATH extends to a MATH action on MATH by linearity, which commutes with the duality map MATH . Again by linearity it is true that labeling a component by any MATH gives the same value to the invariant of an even link as labeling it by MATH so we can as well label components of an even link by equivalence classes MATH . Thus the pairing defined by MATH descends to a well-defined pairing on MATH . Now the tensor product MATH extends by linearity to an associative, distributive multiplication with identity on MATH (here we identify MATH with the vector MATH of MATH), thus making MATH an algebra. Since MATH the quotient MATH inherits the algebra structure. For each equivalence class MATH of a simple object MATH notice MATH is a nontrivial homomorphism from MATH to MATH . Supposing the pairing MATH is degenerate, then these MATH homomorphisms must be linearly dependent, and thus two must be equal. If MATH then MATH for all MATH . Now since MATH we have MATH for all MATH . In particular, since MATH (where MATH is the trivial object, which is the multiplicative identity) is nontrivial there is a minimal idempotent MATH such that MATH but MATH if MATH . Now MATH where MATH is the multiplicity of MATH in MATH . Thus MATH so there exists a MATH such that MATH and MATH . Of course if MATH then MATH or MATH so since MATH we conclude that there is a MATH such that MATH for all MATH . NAME proves that this property implies MATH is degenerate, so we reach a contradiction and conclude the pairing was nondegenerate. |
math/9910106 | CASE: It suffices to prove this for MATH with MATH . MATH . CASE: By the previous point, for any simple MATH so either MATH or for every MATH we have MATH . The second condition we have already noted is equivalent to the degeneracy of MATH so this only happens when MATH or MATH . In both cases the formula follows immediately. CASE: Using the previous point MATH . Of course MATH . |
math/9910106 | Notice first that MATH so MATH is unchanged by reversing the orientation of any component of MATH . For invariance under NAME move II as pictured in REF , notice by REF of the link invariant the value of MATH on the link on the right is MATH times that on the left, using REF . For invariance under NAME move MATH as pictured in REF , the argument is given pictorially in REF , where the first equality is by REF of the invariant, the second by REF , the third by the degeneracy of MATH and MATH the fourth by REF and the fifth by REF again. |
math/9910106 | The proof of REF shows that if MATH is any ribbon MATH-category and MATH is the quotient as in CITE of MATH by a cyclic group of invertible even degenerate objects, then the image of the expression MATH above under the functor from MATH to MATH is the same expression in the image category (the statement of the proposition discusses only the case when the quotient is by the full set of degenerate objects, but the argument does not use this fact in any way). Thus if the full set of even degenerate objects is a cyclic group of invertible elements and MATH is therefore modular, the formula MATH gives an invariant of MATH-framed three-manifolds as in CITE, and if not then MATH is spin-modular and MATH gives an invariant of compatibly framed three-manifolds as in the previous theorem. |
math/9910106 | By CITE, every closed subset of the NAME alcove (closed meaning that the truncated tensor product of two elements of the set is the sum of elements of the set) yields a semisimple ribbon MATH-subcategory of the standard ribbon category associated to the NAME alcove whose degenerate objects are invertible and by CITE form (except for cases mentioned) a cyclic group. In CITE and CITE it is determined when these closed subsets include odd degenerate objects, and thus whether the formula MATH gives an invariant of MATH-framed three-manifolds or compatibly framed three-manifolds. |
math/9910106 | As in the previous subsection let MATH be the formal vector space spanned by isomorphism classes of simple objects in MATH and let MATH and MATH be the corresponding vector spaces for MATH and MATH . Of course the truncated tensor product gives an algebra homomorphism MATH which by point REF in the definition above is onto. It is shown in CITE that the link invariant with a component labeled by MATH is the product of the link invariants with components labeled by MATH and MATH respectively. From this we see that MATH has the property that MATH because MATH has this property in MATH . Thus MATH the last equality being a consequence of the behavior of the invariant under MATH . Thus MATH and MATH are nonzero multiples of each other, since MATH and MATH are all nonzero. Since MATH is easily seen to be unchanged if we replace MATH by a nonzero multiple, the result follows from the invariant's behavior under MATH . |
math/9910106 | Given MATH let MATH be MATH if all MATH in MATH satisfy MATH . If MATH contains any elements with MATH then those with MATH form an index two subgroup, and since MATH is cyclic, MATH too must have an index two subgroup whose intersection with MATH has only objects with MATH . In that case let MATH be the corresponding index two subgroup of MATH . Then MATH is generated by MATH and any element of MATH with MATH . Thus in particular MATH is generated by MATH and elements of MATH . So whatever MATH is MATH and MATH generate the same set MATH generated by MATH and MATH . The previous proposition shows MATH and MATH generate MATH and that MATH consists of even degenerate objects for MATH so MATH is the product of MATH and MATH . |
math/9910106 | Recall from CITE MATH if and only if MATH for all MATH since MATH is cyclic, this is equivalent to MATH for MATH a generator. So MATH is equivalent to MATH for all MATH and for some generator MATH which is to say MATH for some generator MATH or equivalently for all MATH . Now it is shown in CITE that MATH so MATH with MATH is always degenerate for MATH if it is in MATH thus we need only check the additional assertions in REF . Again from CITE, MATH so this is one if and only if MATH is even. Finally, notice the set of MATH such that MATH is even forms a subgroup. If it is proper it has index two, because the product of two elements not in this subgroup is in this subgroup. |
math/9910106 | Since MATH it follows MATH . We claim there is a MATH such that MATH is nontrivial and hence is equal to the unique nontrivial element of MATH . Specifically, MATH is the obstruction to lifting a MATH bundle to a MATH bundle, MATH being its double cover, which necessarily exists because of the condition on MATH . To see that MATH is nontrivial, consider MATH the four-manifold constructed by the surgery process of REF from the MATH-framed unknot. More precisely, it is formed by attaching a single MATH-handle to a MATH-handle along the framed unknot and then attaching a MATH-handle to the resulting MATH boundary. We construct a MATH-bundle over MATH as follows. Attach the trivial bundle over the MATH-handle to the trivial bundle over the MATH-handle via an overlap map on the boundary which is homotopic to an element of MATH not in the index two subgroup. Extend this bundle over the MATH-handle. Of course this bundle does not lift to MATH so the image MATH of MATH in MATH is nontrivial. The image of MATH is MATH which is nontrivial because the intersection pairing given by the MATH identity linking matrix is nondegenerate. On four-manifolds with even forms (which includes all spin four-manifolds) MATH so the integral of every class of MATH against such four-manifolds is even. |
math/9910106 | NAME and NAME 's techniques, in conjunction with the previous proposition, predict a spin TQFT associated to the NAME group MATH with half-integer levels, where MATH is a simply-connected group and MATH is a subgroup of the center containing an index two subgroup. Now integer levels of MATH correspond as levels of MATH to integer multiples of MATH being the least integer such that MATH is an even integer for all MATH . So half-integer levels correspond to odd multiples of MATH . Thus NAME and NAME predict a spin TQFT for MATH at level MATH if MATH is an integer for all MATH (because it is a half-integer multiple of an even number) but at least one of these numbers is odd (otherwise MATH is a multiple of MATH for all MATH and MATH would not be the least such meeting the defining condition). |
math/9910106 | We shall confirm invariance of the second quantity, that of the first follows. Invariance under orientation reversal is clear. According to NAME and NAME an invariant of a link with a characteristic sublink is an invariant of the spin manifolds if it is unchanged by the following move: add a MATH framed unknot to the link (possibly linking with other components), apply a positive or negative full twist to the disk it bounds (so as to change the linking matrix of the link) and add it to the characteristic sublink if and only if the sum of its linking numbers with the existing characteristic sublink is even. Notice first that as argued earlier the formula MATH applied to a link gives the same answer as the same formula interpreted in the quotient of MATH by MATH so we may assume that MATH is modular and that MATH is spin modular. Let us call the unique odd degenerate object in MATH . Observe first that if MATH is a simple object of MATH then MATH . To see this, note that since MATH is the trivial object and MATH we have that MATH . But supposing MATH then if MATH is an other simple object in MATH but not in MATH then MATH so that MATH and we conclude MATH . But this is certainly true for MATH so MATH is degenerate for MATH . This is a contradiction so MATH . Consider the result of NAME and NAME 's move, and suppose first that the unknot is to be added to the characteristic sublink, because its linking number with the old sublink is even. The invariant MATH is a sum over labelings of the components of MATH with those in MATH labeled by elements of MATH and those not in MATH labeled by elements of MATH . Choose such a labeling, and consider the invariant of this labeled link with a particular label MATH on the new unknot. The condition on the linking number means that the new unknot surrounds a collection of strands an even number of which have labels not in MATH and therefore the tensor product of all these labels is a sum of labels in MATH (here we use REF of the ribbon category). If MATH is in MATH then MATH is a distinct label with MATH and MATH for MATH and we see that labeling the new unknot by MATH versus MATH contributes the same amount with opposite sign to the computation of the total invariant. Thus labeling the new unknot by MATH would give a total invariant of MATH so that in the computation of MATH we might as well replace the label MATH on the new unknot with MATH . The same argument applies to MATH and MATH so we see that in this case the invariance of MATH under this move is equivalent to the invariance of the ordinary manifold invariant of MATH under this move. Now suppose that the new unknot is not to be added to the characteristic link, because its linking number with the characteristic link is odd. The same argument as above shows the truncated tensor product of the labels of the strands going through the new unknot is a sum of elements of MATH . If MATH is a label for the new unknot which is not in MATH and we compare the effect on the invariant of replacing MATH by MATH we see that MATH but now an unknot labeled by MATH surrounding a sum of labels not in MATH contributes MATH so again MATH and MATH contribute opposite amounts to the sum (in this case they may not be distinct, but then MATH contributes zero) and thus labeling the unknot by MATH results in a total invariant of zero. Once again we may as well replace the label of MATH with MATH and again the results follows from the invariance of the standard invariant under the move. |
math/9910106 | If we present the compatibly framed manifold by an even link MATH then notice that the empty link is a characteristic sublink, and MATH is the NAME presentation of this MATH-framed spin three-manifold. Of course MATH as defined in the second subsection of this section, and thus the invariants are equal as long as the normalizations are equal, that is if MATH with MATH the NAME link labeled by MATH . But of course the NAME link represents MATH with the standard spin structure and MATH-framing, so MATH which means MATH . |
math/9910106 | For invariance under choice of basepoint, it suffices to check the quantity is unchanged if one basepoint is moved past a crossing or a critical point. This fact is generally true for decorated projections and does not rely on special properties of the MATH-matrix. To see that it is unchanged when a basepoint is moved past a crossing or any decorating NAME algebra element, consider for a given state and component, the product MATH where MATH is the last decoration encountered in the traversal about the component and MATH is the product of the rest of the decorating elements with appropriate powers of MATH . The MATH represents the integer rotation number of the point decorated by MATH which because of the winding number condition on the projection is MATH if the direction of traversal agrees with the orientation and MATH if it disagrees. In either case the defining property of the quantum character gives MATH so the computed quantity is the same as for the case where the base point is just below instead of just above the decorated point. To see that it is unchanged when the base point is moved past a critical point, consider a base point at which the orientation of the component is upwards, which is to be moved past a minimum of the height function configured so that the orientation rotates clockwise around it (the other cases all work similarly). Viewing the rest of the link as a fragment with the fragment basepoint just above the basepoint for the component in the link and decorated by MATH the value of the decorated link projection before moving the basepoint is MATH and after moving it is MATH . To see that the invariant is independent of the choice of projection, notice by REF we may check that it does not change under regular isotopy and the height function moves. The first regular isotopy move reduces to the equation MATH which is a restatement of REF . The second isotopy move is exactly the NAME REF . The height function moves are immediate from the definition. |
math/9910106 | We have already argued that this invariant is unchanged by orientation reversal of any component. Invariance under spin NAME move II follows from the normalization of MATH and REF of the list at the end of the previous subsection (the invariant of the NAME link is MATH). Before proving invariance under move MATH, consider the fragment in REF . By repeated application of REF we see that the sum over all states of the values assigned to the fragment is MATH where the MATH entries of the tensor product label the components from left to right at the bottom of the fragment. Similarly, using REF , we see that the same fragment with the opposite parity is associated to MATH the mirror image of this fragment is associated to MATH and the mirror image with opposite parity by MATH . With this in hand we see that the fragment on the left-hand side of REF corresponds (with the same convention of decorating the strands from left to right at the bottom of the picture) to MATH the last line by REF . Of course, replacing each MATH in the above computation by MATH and every MATH with MATH we see that the right-hand side equals the same thing, and thus the two fragments are interchangeable in the calculation of the invariant. |
math/9910107 | Since pseudo-finite fields are NAME fields in the terminology of CITE, the result is a special case of REF. |
math/9910107 | Take MATH a formula in prenex normal form, which is logically equivalent to MATH. Since, for every pseudo-finite field MATH containing MATH, MATH, the result follows from REF and the observation made at the end of REF. |
math/9910107 | This follows from REF applied to the sentences MATH and MATH in REF , respectively. |
math/9910107 | We first assume that MATH is true in every pseudo-finite field MATH containing MATH. If, for every MATH in MATH, there would exist a closed point MATH in MATH such that MATH is false in MATH, a suitable ultraproduct of the fields MATH would yield a pseudo-finite field containing MATH in which MATH is false, since the ultraproduct construction commutes in the present case with the NAME symbol. Conversely, suppose that there exists MATH in MATH such that, for every closed point MATH in MATH, the formula MATH is true in MATH. By the quantifier elimination REF , there exists a NAME formula MATH, over a suitable MATH, with no free variables and no quantifiers, such that MATH holds in every pseudo-finite field MATH containing MATH. It follows from the first part of the proof that, maybe only after replacing MATH by a multiple, MATH holds also in MATH for every closed point MATH in MATH. Thus we may suppose that MATH has no quantifiers. Let MATH be the NAME stratification over MATH belonging to MATH. Because MATH has no free variables, MATH consists of only one cover MATH, with MATH a field which is NAME over MATH. Assume now that there exists a pseudo-finite field MATH containing MATH in which MATH is false. Let MATH be a topological generator of the absolute NAME group of MATH, and denote by MATH the restriction of MATH to MATH. Then MATH where MATH denotes the subgroup of MATH generated by MATH. Since we suppose that MATH is false in MATH, we have MATH. By NAME 's Theorem for MATH (see, for example, CITE), there exists a closed point MATH of MATH such that MATH . Thus MATH, whence MATH is false in MATH, which contradicts our assumption that MATH is true in MATH. |
math/9910107 | First consider the case where MATH. Let MATH be the set of all subfields of MATH which are of finite type over the prime field of MATH and which contain the constants in MATH and MATH, and assume there exists for each MATH in MATH a pseudo-finite field MATH containing MATH in which MATH is not equivalent to MATH. Choose an ultrafilter MATH on MATH containing MATH, for every MATH in MATH. Let MATH be the ultraproduct of all the fields MATH with respect to MATH. Clearly MATH is a pseudo-finite field in which MATH is not equivalent to MATH. Moreover MATH is imbedded in MATH by the map MATH, where MATH if MATH and MATH if MATH. This contradicts MATH. The proof for MATH is just the same, considering now the set MATH of all subfields of MATH which are of finite type over the prime field of MATH and contain the constants in MATH, MATH and MATH, where MATH is a formula satisfying the conditions in the definition of MATH and replacing everywhere ``MATH is not equivalent to MATH in the pseudo-finite field MATH" by ``MATH is not the graph of a bijection between MATH and MATH". |
math/9910107 | This is REF. When MATH is abelian it is REF where the hypothesis MATH abelian was not used seriously in the proof. |
math/9910107 | REF amd REF follow from REF applied to the morphism MATH and MATH, respectively. Similarly, REF follows from a dual form of REF which is as follows (notation being as in CITE) : for MATH a morphism of finite groups, MATH a character of MATH and MATH a MATH-variety, we have MATH. The proof is just similar to the one of REF using the projection formula REF and the fact that the functor MATH of CITE commutes with MATH. Another way to prove REF would be to remark that when MATH is projective and smooth it is just a consequence of REF together with elementary theory of representations of finite groups and then to deduce the result for arbitrary MATH by additivity of MATH. |
math/9910107 | By a classical result of NAME, MATH is a MATH-linear combination of characters of the form MATH with MATH a cyclic subgroup of MATH. It follows from REF that, for every MATH-variety MATH, MATH where in the middle term MATH is viewed as a MATH-variety, whence MATH belongs to MATH. |
math/9910107 | Follows directly from the additivity of MATH and REF . |
math/9910107 | The first statement is clear and the second follows from REF , since MATH. |
math/9910107 | Follows directly from the next lemma. |
math/9910107 | By its very definition, MATH is equal to MATH. As in the proof of REF we may write MATH as a MATH-linear combination MATH with MATH in MATH and MATH running over the set of cyclic subgroups of MATH, and it follows from REF that the left hand side of REF is equal to MATH . The right hand side of REF is equal to MATH with MATH the NAME automorphism at MATH (up to conjugation) and hence may be rewritten as MATH . Now it follows from NAME 's trace formula together with standard constructibility and base change theorems for MATH-adic cohomology that there exists a non zero element MATH in MATH such that, for every closed point MATH of MATH, REF is equal to REF . |
math/9910107 | After refining MATH and MATH one may assume MATH and MATH for MATH. Hence it is sufficient to prove the following: let MATH be an integral and normal MATH-scheme of finite type and let MATH and MATH be NAME covers with NAME groups respectively MATH and MATH provided with a family of subgroups MATH (respectively, MATH) of MATH (respectively, MATH) which is stable by conjugation under elements in MATH (respectively, MATH); assume there exists a non zero element MATH in MATH such that for every closed point MATH in MATH and every closed point MATH of MATH, MATH if and only MATH, then MATH . But this follows from the more general REF . |
math/9910107 | Replacing MATH and MATH by multiples, we may assume that MATH and MATH are MATH-linear combinations of irreducible characters of MATH and MATH, respectively. By REF , maybe after replacing MATH by MATH for some non zero MATH in MATH, we can assume MATH and MATH by going to a suitable commun NAME cover of MATH and MATH. By NAME 's Theorem, we have then MATH, and the result follows. |
math/9910107 | It is enough to prove that MATH. Hence, by additivity of MATH, one may assume that the NAME stratification MATH consists of a single colored NAME cover MATH, with MATH a normal variety and MATH non empty. Denote by MATH the support of MATH and by MATH the restriction of the projection MATH to MATH. Write the NAME stratification MATH as MATH . It follows from NAME 's Theorem and the NAME estimate CITE, that the morphism MATH is generically finite and that MATH. Hence, maybe after performing a finite partition of MATH into locally closed normal subschemes, we may assume that, for every MATH in MATH, MATH is mapped by MATH onto MATH and that the restriction of MATH to MATH is a finite étale morphism. Furthermore, maybe after replacing MATH by MATH for a suitable non zero MATH, we may, by REF , replace the MATH's by suitable NAME covers. It follows that, maybe after performing another finite partition of MATH into locally closed normal subschemes, we may assume that the morphisms MATH obtained by composition with MATH are all NAME covers and that they are all equal to a same NAME cover MATH. Moreover we may assume that the cover MATH coincides with the cover MATH. By assumption there exists a non zero element MATH of MATH such that, for every closed point MATH of MATH, the map MATH induces a bijection between MATH and MATH. Hence, for such a MATH and such a MATH, a point MATH belongs to MATH if and only if there exists MATH in MATH and MATH in MATH such that MATH belongs to MATH. Furthermore, since a given point MATH in MATH should be the image under MATH of a unique point in MATH, one deduces from NAME 's Theorem, that, for MATH in a subset of closed points of MATH of NAME density REF, the element MATH in MATH and the class of MATH in MATH should both be unique. Since, for such a MATH, MATH it follows, again by NAME 's Theorem, that MATH . By REF , one deduces now that MATH. |
math/9910107 | It is enough to prove the following: Let MATH be a NAME cover of MATH-varieties with NAME group MATH, MATH a connected component of of MATH, MATH the decomposition subgroup of MATH at the generic point of MATH, MATH a character of MATH, then MATH. Since MATH, this is equivalent to the equality MATH which follows from REF . |
math/9910107 | With the preceding notations take a NAME stratification MATH of MATH, such that, for every closed point MATH of MATH, MATH. By REF , MATH for every pseudo-finite field MATH containing MATH. Hence, MATH for every pseudo-finite field MATH containing MATH. Let MATH with MATH the normalisation of MATH in MATH. Again by REF we get that for some non zero element MATH of MATH, MATH, for every closed point MATH of MATH, with MATH the stratification obtained by base change. Hence, by REF , MATH and the result follows by REF . |
math/9910107 | By REF we may assume that MATH is of finite type over MATH, so we may apply REF to get a certain formula MATH and then, by REF , we may replace the formula MATH by a NAME stratification and the result follows from REF . |
math/9910107 | To prove REF first observe that if MATH and MATH are formulas with free variables MATH, and MATH and MATH with MATH and MATH . NAME stratifications of MATH with disjoint support then MATH. Hence we have MATH, MATH and MATH, whence the statement follows. REF are proven by taking respectively the product and the complement (in the obvious sense) of the corresponding NAME stratifications, while REF is just evident. |
math/9910107 | Follows directly from REF . |
math/9910107 | Indeed, this follows from REF . |
math/9910107 | This follows from REF together with the observation at the begining of REF. |
math/9910107 | Indeed, it follows from REF , since, by a classical result of CITE, MATH has quantifier elimination in the language MATH. |
math/9910107 | Let us first prove the equivalence of REF . By REF , there exists a MATH-sentence MATH without quantifiers over the valued field sort and the value sort such that MATH is equivalent in MATH to MATH, for every henselian MATH-extension MATH of MATH. Hence there exists MATH in MATH such that, for every closed point MATH in MATH, the sentence MATH is equivalent to MATH in MATH. Indeed, if this would not be the case, a suitable ultraproduct of the fields MATH would yield a henselian MATH-extension MATH of MATH in which MATH would not be equivalent to MATH. Hence we may assume MATH is a sentence (that is, is without free variables) and has quantifiers only over the residue field sort, in which case the result follows from REF . The proof of the equivalence of REF is completely similar. |
math/9910107 | One may assume MATH and MATH is associated to a formula MATH on MATH. Then MATH is associated to the formula MATH and it follows from REF that MATH is a definable subassignement of MATH associated to a special formula on MATH - without quantifiers in the valued field sort - say MATH. One can formally write MATH and then, by expanding the variables MATH into series in the formulas of type REF appearing in MATH, one obtains a infinite set of conditions in the variables MATH. If one substitutes in these conditions MATH whenever MATH, then only a finite number of conditions which involve only coefficients MATH with MATH remain. In this way one obtains from the formula MATH a formula MATH with coefficients in MATH and free variables MATH, MATH and MATH is the definable subassignement of MATH associated to MATH. |
math/9910107 | The proof of REF using ultraproducts may be directly adapted to the present situation. Indeed, we may assume MATH is a closed subvariety of MATH, and it is enough to prove that if MATH, MATH, are weakly stable definable subassignements of MATH such that, for every finite subset MATH of MATH, MATH is not the empty subassignement, that is, for some field MATH, MATH is not empty, then MATH is not the empty subassignement. Since every weakly stable definable subassignement of MATH may be defined by an infinite conjunction of formulas in the language of rings with coefficients in MATH involving each only a finite number of coefficients of the power series MATH, MATH, compare the proof of REF , it follows that MATH is not empty for MATH the ultraproduct of the fields MATH with respect to a suitable ultrafilter. |
math/9910107 | Clear. |
math/9910107 | The first assertion follows directly from REF . The second assertion is a consequence of REF , together with the fact that MATH if MATH. |
math/9910107 | One reduces first to the case when MATH is affine irreducible with a closed immersion MATH, and then, by REF , we may assume MATH is associated to a special formula MATH on MATH which is obtained by repeated application of conjunction and negation from formulas of the form REF . Choose a nonzero regular function MATH on MATH which vanishes on the singular locus of MATH and let MATH denote the product of MATH and all the polynomials MATH, assumed to be non zero, occuring in REF . Now set MATH and define MATH as the locus of MATH. The definable subassignements MATH defined by MATH are stable by REF and the statement on dimension follows from REF. We still have to check that the denominators of MATH in MATH are bounded. For that it is enough to know that one can bound uniformly the degree of the coverings in the NAME stratifications associated by quantifier elimination REF to MATH. But consider the NAME stratifications associated by REF to the formulas MATH occuring in REF and let MATH the maximum of the degrees of the coverings appearing in these NAME stratifications. The NAME stratifications associated to MATH may be expressed in terms of the former ones, and the integer MATH is still a bound for the degrees of the coverings. For definable partitions, the construction of the MATH's is done in exactly in the same way. |
math/9910107 | The proof is just the same as the proof of Definition-REF, if one replace REF by REF . |
math/9910107 | Again the proof is essentially the same as the proof of REF, if one replace REF by REF . |
math/9910107 | Follows from REF similarly as the proof of REF. |
math/9910107 | The proof of REF may be directly adapted to carry over to the present situation. |
math/9910107 | Let us first prove REF . By REF one may use a resolution of singularities and assume that MATH is smooth. Also we may assume MATH is affine. By Pas's Theorem, REF , and REF , there exists a NAME function MATH such that the series REF is equal to MATH . By REF one reduces as in CITE to the case where the family MATH has bounded representation. Furthermore, one may assume MATH in the bounded representation of the family MATH by special formulas obtained by repeated application of conjunction and negation from formulas of the form REF . We denote by MATH the product of all the polynomials MATH, assumed to be non zero, occuring in these formulas of the form REF and we consider an embedded resolution of singularities MATH of the locus of MATH in MATH, with exceptional locus contained in MATH. The variety MATH admits a covering by affine open subsets MATH on which there exist regular functions MATH inducing an étale map MATH such that on MATH each MATH is a monomial in MATH multiplied by a regular function with no zeros on MATH. One may assume furthermore that the variables MATH appearing in at least one of these monomials are exactly MATH, MATH, , MATH. Now for MATH a definable subassignement of MATH and MATH in MATH, we denote by MATH the definable subassignement of MATH defined by MATH . It now follows from REF and the fact that MATH is bounded on MATH, that, uniformly in MATH, MATH, MATH is a finite MATH-linear combination of terms of the form MATH where MATH is a condition defining a NAME subset of MATH, MATH is a linear form with coefficients in MATH, and MATH is a definable subassignement of MATH, for MATH as above. Since MATH is bounded on MATH, the sum in REF is finite. Let us denote by MATH the definable subassignement of MATH defined as MATH with MATH the subvariety of MATH defined as the locus of points MATH in MATH such that MATH when MATH and MATH when MATH. But, by REF, with MATH, we have MATH hence we can rewrite, uniformly in MATH, MATH, MATH as a finite MATH-linear combination of terms of the form MATH with MATH, MATH and MATH as above. As in CITE, one may now conclude the proof by using REF. The proof of REF is similar except for the fact we have to replace MATH by MATH and the finite sums are replaced by infinite sums which converge in MATH. In particular, MATH is still a finite MATH-linear combination of terms of the form REF , but the number of terms in the series REF may now be infinite. |
math/9910107 | For MATH in MATH, denote by MATH the ``characteristic polynomial" MATH of MATH on MATH (since we are dealing with virtual representations MATH is a rational function). Assume MATH is in MATH for every MATH in MATH. By the part of the NAME conjectures proven by CITE, we have then MATH for MATH in a set of NAME density REF. Since a virtual MATH-adic representation of MATH is determined by the corresponding characteristic polynomials, we deduce from NAME 's Theorem that MATH. The result follows. |
math/9910107 | We shall prove directly the more general second statement. The proof will proceed by comparison with the proof of REF . We shall make here no notational difference between MATH and MATH and between MATH and MATH. We set MATH. In CITE we first used a resolution of singularities of MATH to reduce to the smooth case, and then considered an embedded resolution of a divisor MATH on some affine open subsets of MATH. By REF, there exists a non zero multiple MATH of MATH such that these resolutions extend over MATH to resolutions with good reduction mod MATH, for every closed point MATH in MATH, in the sense of CITE. Here MATH denotes the maximal ideal at MATH. Now, by the local calculations of MATH-adic integrals on resolutions with good reduction in CITE, we deduce that, for every closed point MATH in MATH, MATH is a finite MATH-linear combination of terms of the form MATH where MATH, MATH and MATH are the same as in REF , and furthermore the coefficients of the terms REF in MATH are the same as the coefficients of the terms REF in MATH. Indeed, we may assume MATH is affine and, by REF , we may also assume, maybe after replacing MATH by some non zero multiple, that the subassignements MATH are defined by repeated application of conjunction and negation from formulas of the form MATH where MATH are regular functions, MATH is a polynomial with coefficients in MATH and degree MATH, MATH is in MATH, and MATH is a MATH-formula in MATH free variables. Furthermore we may assume there exist regular functions MATH on MATH inducing an étale map MATH such that on MATH each MATH is a monomial in MATH multiplied by a regular function with no zeros on MATH, and that the variables MATH appearing in at least one of these monomials are exactly MATH, MATH, , MATH. With the notations from REF, we deduce from the local calculations in REF that MATH is a MATH-linear combination of terms of the form MATH with MATH a condition defining a NAME subset of MATH, MATH a linear form with coefficients in MATH, and MATH a definable subassignement of MATH. Here MATH and MATH depend only on the monomials appearing in the functions MATH, the coefficients of the polynomials MATH and the integers MATH, and MATH depends furthermore on MATH. In the good reduction case, local calculation of MATH-adic integrals as performed in CITE just provides the same expression for MATH as a MATH-linear combination of terms of the form MATH with the same coefficients in MATH as for MATH, and for each term the same functions MATH and MATH and the same definable subassignement MATH as in REF . The result now follows, since, by REF , MATH for every closed point MATH in MATH, for a suitable non zero multiple MATH of MATH. |
math/9910107 | The rationality of MATH follows from REF and the fact that it satisfies the required specialization property follows from REF just follow from REF . |
math/9910107 | One reduces to the case where MATH is a closed subvariety of MATH. Then the result follows from REF , since MATH and the subassignements MATH are stable definable subassignements of MATH depending on the parameter MATH. Here MATH and MATH denote the truncation morphisms associated to MATH and MATH, respectively. |
math/9910107 | We may assume MATH is a closed subvariety of MATH defined by MATH, MATH, and that MATH is defined by MATH, MATH, MATH, MATH, with MATH in MATH. We may also assume that MATH is a multiple of the discriminant of MATH. Consider the following formula MATH in MATH in the MATH free variables MATH and MATH over the valued field sort, MATH . For every closed point MATH in MATH, the equality MATH holds, as an direct calculation shows, compare REF. Similarly, the equality MATH follows from the very definitions, hence the result is a direct consequence of REF . |
math/9910107 | For every integer MATH such that MATH, we set MATH, with MATH and MATH. Similarly, we consider the formula MATH. So we have MATH and MATH, and the result follows directly from the next lemma. |
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