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math/9907034 | Let MATH be a curve in the moduli space MATH of deformations from the point MATH representing MATH and an arbitrary point MATH. Then there is a corresponding smooth map MATH such that MATH is the deformation of MATH. Now MATH is linearly equivalent to MATH if and only if MATH for each harmonic MATH-form MATH with integral cohomology class. Differentiating with respect to MATH at MATH this implies that MATH where MATH is the section of the normal bundle defining an infinitesimal variation of special Lagrangian submanifolds. This must hold for all harmonic MATH with integral cohomology class and so by linearity for all MATH. Now by NAME theory the map MATH defined by MATH is an isomorphism from harmonic REF-forms to harmonic REF-forms, so we can write MATH for a harmonic REF-form MATH. Now MATH but MATH is Lagrangian, so that MATH restricts to zero on MATH, thus REF becomes MATH . But now we know that MATH restricted to MATH is a harmonic form, and moreover any harmonic form is the derivative of a variation through special Lagrangian submanifolds. Thus if all deformations are linearly equivalent to MATH, MATH for all harmonic REF-forms MATH on MATH and all harmonic REF-forms MATH on MATH. Since the harmonic REF-forms run through each cohomology class this means that the cohomology class MATH of MATH on MATH vanishes, which is the theorem. The converse follows by differentiating the integral MATH with respect to MATH. |
math/9907036 | Put MATH. Since the maps MATH are injective and have disjoint ranges, there is actually one function MATH such that MATH . One now proves as in REF below (the tacit assumption there that MATH is MATH-quasi-invariant is not needed for this) that MATH where MATH . Note that the NAME operator MATH has the property MATH by the computation MATH . We observe that MATH is a positive function by REF and it is MATH-integrable, as can be seen by using REF on MATH. If MATH is a positive bounded function on MATH we have from REF : MATH . Putting MATH equal to characteristic functions, the MATH-quasi-invariance REF of MATH is immediate. |
math/9907036 | We will first verify that the relations REF - REF define a representation of MATH, and verify that its restriction to the abelian subalgebra MATH is the spectral representation. If MATH, we have MATH where MATH. (If it happens that MATH, the relation REF says that MATH is MATH-invariant, and MATH is then what is called a MATH-measure in CITE.) Applying REF to MATH we obtain MATH . Defining MATH by REF , we see immediately from the MATH term that the ranges of MATH are mutually orthogonal, and if MATH, then from REF : MATH so each MATH is an isometry, and hence MATH . Furthermore MATH and REF for MATH follows. If MATH with MATH, define MATH . One verifies from REF that MATH and MATH . Combining REF - REF with the relations MATH which are valid if MATH, we obtain MATH where MATH . This proves firstly that MATH and REF show that MATH is indeed a representation of the NAME relations. Secondly, REF shows that MATH maps onto the algebra of operators on MATH of the form MATH where MATH ranges over all continuous functions on the NAME set MATH. Thus the restriction of the representation MATH to MATH is indeed the spectral representation. To show the main part of REF , that is, the existence of the objects MATH, MATH, MATH, one does indeed start with a spectral measure MATH for the restriction of the representation to MATH. The spectrum of MATH is MATH, so this gives the decomposition REF , and the action of MATH on MATH is given by REF . If MATH, and MATH is the representative of MATH in MATH: MATH then MATH and the quasi-invariance of MATH under MATH follows. Thus one may define MATH by REF . Similarly, if MATH has support in MATH, one verifies that MATH . Thus the two representations of MATH given by MATH on MATH and MATH on MATH are unitarily equivalent. In particular, this means that MATH for MATH-almost all MATH, so the constancy of MATH almost everywhere over the orbits of MATH follows. But REF then implies that MATH is constant on MATH-orbits (actually the two forms of constancy are equivalent). Also it follows from the unitary equivalence REF that MATH is quasi-invariant under MATH and that MATH exists. See CITE or CITE for details on spectral multiplicity theory. Now, one may define a representation MATH of MATH on MATH by MATH . One checks that this is indeed a representation of MATH by the first part of the proof, and by the proof of REF it follows that MATH for all multi-indices MATH. Define an operator MATH by MATH . Using the NAME relations in a standard manner, one checks that MATH is a unitary operator, and MATH for MATH. Putting MATH we have by REF MATH and hence MATH commutes with the representatives on MATH of the algebra MATH. Hence MATH has a decomposition MATH where MATH is a measurable field on unitaries. It now follows from REF that MATH has the form REF . This ends the proof of REF . |
math/9907036 | Adopt the assumptions in REF and let MATH be an intertwiner between the two representations. In particular this means that MATH intertwines the two spectral representations of MATH on MATH and MATH, respectively, that is, MATH for all multi-indices MATH. But this is equivalent to MATH having the decomposition REF and MATH having the measurable decomposition MATH where MATH. We now compute, using REF , MATH and MATH . Using the intertwining REF we thus deduce that MATH . Conversely, if MATH satisfies REF , the intertwining follows from REF . This ends the proof of REF . |
math/9907036 | Note that the representation MATH in REF - REF is of the form MATH where MATH is the representation in REF corresponding to the scalar-valued case with MATH product measure and MATH. We then use Let MATH be a representation of MATH in a NAME space MATH and let MATH be a second NAME space. If MATH are operators in MATH, then MATH define a representation of MATH in MATH if and only if the MATH's are unitary. We have MATH . Hence MATH holds if and only if each MATH is isometric. We have MATH . But the projections MATH are mutually orthogonal. So MATH if and only if each MATH is coisometric. The result follows. Now we apply the lemma to MATH, MATH, and it remains to check that the vector state MATH yields the state MATH in REF . Let MATH for multi-indices MATH, MATH, MATH. Then MATH, and MATH where MATH, and where we used REF for the scalar-valued representation MATH in MATH and the observations from above on the trace of MATH. The term MATH is nonzero (and therefore MATH) if and only if MATH, that is, MATH . Since we are in the free group, this happens precisely when MATH and MATH. |
math/9907036 | From REF - REF , we have MATH where MATH, MATH are multi-indices, and MATH . Since MATH, the result follows. |
math/9907036 | We compute the action of MATH and MATH directly from the formulas given in REF . We have MATH proving REF , and MATH proving REF . The stated formulas for MATH and MATH, MATH, result from the following covariance principle: MATH, and, more generally, MATH. REF is a special case of REF . |
math/9907036 | This is immediate from REF . |
math/9907036 | We check that MATH for all MATH, and MATH. We may assume that MATH for MATH (MATH the free semigroup on the generators MATH). The inner product is MATH and MATH, since there is no solution MATH to the equation MATH. |
math/9907036 | If MATH is a KMS state at value MATH, then MATH . If MATH, MATH, this says MATH . But MATH and hence MATH is a solution of REF . But this equation has solutions MATH if and only if all MATH are nonzero, and all have the same sign; and, in that case, the solution MATH is unique. For definiteness, assume that all MATH are positive, and then the solution MATH of REF is also positive. Because of the NAME relations, the element MATH is either MATH (if MATH), MATH, or of the form MATH or MATH for some MATH. But from REF , we have MATH and thus MATH for all nonempty strings MATH. Hence it follows from REF again that MATH which is REF . But this expression does indeed define a state by REF . The case that all MATH are negative is treated similarly, so this proves REF . |
math/9907036 | REF was established in REF . CASE: Assume that REF does not hold. Then there exist MATH with MATH, MATH. Put MATH . Then MATH is an isometry in MATH which is not unitary. Hence MATH cannot be an NAME. This also establishes the final remark in REF . CASE: We may assume that all MATH are positive. We have noted that MATH is the closure of the linear span of MATH with MATH. If MATH, we define MATH and we set MATH. Now, if MATH, MATH are in MATH then either the product MATH is zero, or we have MATH and the product is MATH, or we have MATH and the product is MATH. In the latter two cases MATH, and, in the former case, MATH. Thus in general, MATH . Thus if we define MATH then MATH is a MATH-algebra, and MATH is finite-dimensional since MATH for MATH. Since any MATH has a grade, it follows that MATH is dense in MATH. Thus MATH is an NAME, and REF is proved. |
math/9907036 | Consider arbitrary multi-indices MATH, MATH, MATH, and MATH built from MATH. The product MATH is nonzero only if either MATH is of the form MATH, or MATH is of the form MATH. If each of the factors in REF is in MATH, then MATH and MATH. Recall that the grade of the first factor is MATH, and that of the second is MATH. If the two factors have different grades, and if the product is nonzero, then MATH or MATH. In the first case, the product is MATH, and in the second it is MATH. In either case, if MATH or MATH, the product of the two elements from MATH will be nonzero with MATH, see REF , so MATH will not then be a set of matrix units, that is, REF will not hold. This proves the ``only if" part of REF . Conversely, if there exists a MATH such that MATH for all MATH, then the case MATH with MATH is excluded since MATH. For if MATH, then MATH, and MATH. The same argument also excludes MATH with MATH. It follows that REF will be satisfied for MATH with MATH as an index set. |
math/9907036 | Let us use the shorthand notation MATH . It follows from the computations in the proof of REF that, given two projections MATH, MATH, then MATH, MATH are either mutually orthogonal, or one is contained in the other; and the latter case occurs in, and only in, the following two cases: MATH . Using this, it follows easily from case-by-case considerations that the projections in the family MATH are mutually orthogonal. For example, the projections MATH, MATH are mutually orthogonal by REF , and if MATH and MATH with MATH, MATH, then both REF are excluded, so MATH; and similarly, if MATH and MATH are in MATH, then MATH implies MATH. It remains to show that the projections in these two families add up to MATH. If not, there would exist a multi-index MATH such that MATH is orthogonal to all projections in the two families. If then MATH, we could find a MATH such that MATH or MATH, but since MATH, this is impossible. If MATH, then MATH, which is impossible. If MATH, write MATH. If there exists a MATH such that MATH, then MATH, which is impossible; and, if not, there is a MATH with MATH and MATH. But then MATH, and MATH, so this is equally impossible. Thus the projections in the two families add up to MATH, and REF is proved. |
math/9907036 | Referring to REF , define MATH and MATH for MATH; for greater MATH's, MATH becomes the empty set. MATH may also be the empty set for some MATH, but we will prove in a moment that if the greatest common divisor of MATH is MATH, this only happens for finitely many pairs MATH. Now, define MATH as the linear span of elements MATH with MATH, MATH, with the convention that MATH is empty if MATH and MATH, MATH for MATH. It follows from REF that each MATH is a full MATH matrix algebra, and that the units of MATH are orthogonal and add up to MATH as MATH runs over MATH for fixed MATH. Put MATH . If MATH, then MATH and MATH and hence MATH is partially embedded in MATH with multiplicity equal to the number of MATH's such that MATH. We also have MATH for MATH, and thus MATH is embedded into MATH with multiplicity MATH for MATH. It follows that MATH is indeed an increasing sequence of finite-dimensional subalgebras, and in particular MATH contains all monomials MATH of grade MATH. Thus MATH is dense in MATH, reestablishing that MATH is an NAME, and the description of the embedding MATH proves REF . The remaining statements in REF will be proved after REF , below. |
math/9907036 | Since MATH have greatest common divisor MATH, there are MATH such that MATH and hence there are MATH such that MATH . Now, if ad absurdum MATH is infinite we may find arbitrarily large MATH, but then MATH, MATH are not contained in MATH; thus MATH, MATH, MATH are not in MATH, etc., and thus we can find arbitrarily long sequences of the form MATH not in MATH. But as MATH contains MATH, this is impossible. Thus MATH is finite. |
math/9907036 | Since any node in the NAME diagram is connected to a node of the form MATH further down, and MATH is connected to all nodes MATH where MATH, it follows that all nodes in a row will be connected to all nodes in some row further down, which means that MATH has strictly positive matrix elements for some MATH. Therefore MATH is simple CITE, and MATH has a unique trace state CITE, CITE. This ends the proof of REF . |
math/9907036 | By REF - REF the NAME - NAME eigenvalue MATH and the normalized left NAME - NAME eigenvector MATH are the same for MATH and MATH. But REF states that MATH and MATH are the same subgroup of MATH, and by REF the positive cones are the same. By REF , MATH is represented by the same element of the two cases, and thus the complete invariants REF are the same. Thus MATH and MATH are isomorphic MATH-algebras. |
math/9907036 | In this case MATH so MATH and the result follows from REF . |
math/9907036 | We already noted above that REF is CITE. But MATH follows from CITE. The implication MATH follows by noting that the stabilization of the dual actions of MATH, MATH is outer conjugate to MATH, MATH by NAME duality. The only remaining nontrivial implication is REF ; as noted in CITE, the relations REF imply that MATH and MATH are similar, and thus have the same characteristic polynomial. But by REF , the characteristic polynomial determines MATH and thus MATH uniquely. Thus REF . |
math/9907036 | For MATH, let MATH. Then MATH which proves the lemma. |
math/9907036 | This was verified in REF . |
math/9907036 | We have MATH showing that MATH if and only if MATH. But MATH is divisible by MATH if and only if MATH is a root. |
math/9907036 | The representation MATH is unique since MATH is irreducible. To see that MATH is invariant under MATH, use REF directly, or substitute MATH into REF as follows: If MATH satisfies MATH, then MATH where we used the identity MATH which in turn is equivalent to REF . |
math/9907036 | We leave the modifications needed to cope with the extremal cases MATH to the reader, and consider the generic situation MATH. We use REF in calculating MATH in the basis defined from REF . Define MATH . Then MATH is a basis for MATH by REF . Furthermore, MATH . Since MATH it follows from REF that the remainder is zero (this accounts for REF ), and MATH which accounts for the upper left column in REF via REF . Since for MATH the rest of the left half of the matrix REF is accounted for. For the rest of the entries in REF for MATH, pick the monomials MATH as a basis for the remainder terms in the Euclidean representation of MATH. Using again REF , we get MATH which accounts for the MATH'st column in REF . For MATH such that MATH we have, using REF : MATH and that accounts for the remaining columns in REF . |
math/9907036 | Use the standard rules for computing determinants on REF , and use REF . |
math/9907036 | This proposition is essentially also true in the more general situation where MATH is a primitive nonsingular matrix. Using the standard basis for MATH, we saw in REF that MATH . But MATH is in MATH if and only if (using REF ): MATH that is, if and only if MATH, that is, MATH . Using the basis REF , this is REF . More specifically, we saw in REF that MATH takes the block form MATH relative to the decomposition MATH . The submatrices MATH and MATH are both invertible in dimensions MATH and MATH, respectively. Moreover REF shows that each of the submatrices MATH and MATH has a form which is similar to that of MATH itself. The MATH matrix MATH was also computed in REF . For MATH, we therefore have the formula MATH and, similarly, MATH . |
math/9907036 | CASE: We have already commented that MATH is the rank of the group MATH, so MATH is an isomorphism invariant. CASE: If MATH, let again MATH denote the set of prime factors of MATH, with the convention MATH. If MATH and MATH are isomorphic, it follows from REF by taking the determinant on both sides that MATH where we used REF . Hence MATH so MATH . Thus in particular MATH is an isomorphism invariant, as claimed. As the exact sequence MATH is uniquely determined by the dimension group MATH, the group MATH is an isomorphism invariant. But if MATH denotes the restriction of MATH to MATH, then MATH identifies with the upper left-hand part of the matrix REF . But MATH by REF and hence MATH by REF . It follows from REF that MATH is an isomorphism invariant and thus MATH is so. Thus REF is proved. Furthermore, if MATH is another nonsingular primitive incidence matrix defining the same dimension group as MATH, it follows from REF that MATH and thus MATH and MATH are related as MATH and MATH in REF , except that the MATH, MATH now are just (necessarily nonsingular) integer matrices, without any positivity. (See an elaboration of this in the following paragraph.) But positivity did not play any role in the first part of the present proof, and hence MATH . But MATH and MATH, so MATH is an isomorphism invariant, which shows REF . By CITE, the groups MATH and MATH order isomorphic. Let MATH be the corresponding order isomorphism. It follows from REF that MATH restricts to an isomorphism of MATH onto MATH. We have shown in REF that MATH is constructed from MATH the same way MATH is gotten from MATH as an inductive limit. Now apply REF to both MATH and MATH. Then the argument from REF yields MATH where MATH, MATH are natural numbers, and the matrices MATH and MATH are MATH over MATH. The argument which yields MATH as the inductive limit MATH in REF also yields the following associated isomorphism: MATH . This follows by general category theory from the commutativity of REF and exactness of the vertical short exact sequences of this diagram. Let us elaborate on this: Use induction, and REF for MATH, starting with the obvious isomorphism MATH given by MATH and arriving at MATH . Since MATH for a suitable matrix MATH by REF , we get MATH with the isomorphism induced by MATH of REF . It is an isomorphism, for if MATH then MATH since MATH is nonsingular. So then MATH by REF . This proves REF . By REF , we get MATH and so MATH by REF . To see this, we must also check that the defining homomorphism REF does indeed pass to the respective inductive limit groups MATH and MATH. But note that MATH and the right-hand side is mapped into MATH under MATH from REF . So we have the commutative diagram MATH of homomorphisms of abelian groups. As a result, there is an induced homomorphism of the respective inductive limit groups MATH where MATH for short. REF shows that MATH where we used REF in the last step. Hence, by the homomorphism theorem, we have MATH which is REF . Let MATH and MATH be as in REF with associated matrices MATH and MATH, and suppose the MATH-algebras MATH and MATH are isomorphic. The corresponding order isomorphism MATH therefore induces isomorphisms MATH and MATH . It follows further that MATH then induces an isomorphism MATH . This makes sense since we have already concluded that MATH and MATH. (Recall that MATH.) Now the argument after REF applies to MATH and MATH, the same way as we used it to get identity of the sets of primes for MATH and MATH. Using finally MATH we conclude that MATH which is the claim. The final statement of REF is clear from the proof in the special case that MATH has the form REF . The only thing that separates the general case from the special one is the special form of MATH, MATH, MATH in REF , and thus the formulae MATH, MATH and MATH. |
math/9907036 | The result follows from the equality MATH in REF , and the natural isomorphism MATH coming from REF above. See also REF - REF below. |
math/9907036 | If for example MATH is rational then MATH is integral since it is a solution of a monic polynomial. If MATH and MATH define isomorphic NAME, then it follows from CITE that MATH, that is, MATH is rational and thus integral, and MATH, MATH are products of the same primes, MATH. But in this case MATH, and by REF MATH. |
math/9907036 | In general, if MATH is a lattice in MATH and if MATH is a MATH matrix such that MATH, then MATH is a finite abelian group of order MATH. See, for example, CITE. Applying this to REF for each MATH, we get that MATH has order MATH. Note from REF that MATH . A further calculation shows that the usual matrix multiplication, MATH, induces an isomorphism of abelian groups MATH for each MATH; so, in proving cyclicity, it is enough to deal with MATH where the assertion amounts to the There is an isomorphism MATH given by MATH . Since MATH has order MATH it is enough to show that the lattice element MATH is in MATH if and only if the number MATH is divisible by MATH. Hence, we must show that, if MATH, then the equation MATH is solvable in MATH if and only if MATH. But MATH is a solution in MATH; in fact, the explicit formula is given by REF or REF as follows: MATH . This proves that MATH if and only if MATH as claimed. As mentioned, the claim proves REF . |
math/9907036 | Follows directly from REF , and REF . See in particular REF - REF . |
math/9907036 | Immediate from REF - REF . |
math/9907036 | The details are contained in the last part of the proof of REF . |
math/9907036 | As the proofs of REF . |
math/9907036 | If MATH is a polynomial in MATH, we may assume that all the coefficients of MATH are in MATH by reducing modulo MATH. Let MATH be the leading term in MATH. If MATH, there is nothing to do. If MATH add an integer multiple of MATH to MATH such that the leading coefficient is contained in MATH. Then do the same thing for the second leading term in the new polynomial, etc. It is clear that this procedure determines the coefficients MATH uniquely. In the special case that MATH, the expansion REF reduces to MATH and hence MATH in that case. |
math/9907036 | Let MATH, MATH, and MATH. Then MATH and in MATH: MATH . This proves the relations REF . To prove that the relations REF actually characterize MATH we use the polynomial representation REF - REF . There MATH is characterized as the additive group MATH modulo the linear combinations of the elements MATH for MATH, where MATH is given by REF as MATH . Thus MATH is characterized as the abelian group generated by elements MATH with the relations MATH for MATH. But then the abelian group defined by the relations REF above is isomorphic to this polynomial group through the map MATH for MATH. This proves that the abelian group defined by REF is isomorphic to MATH, and furthermore, an isomorphism between the groups is given by MATH . Using REF , we thus see that REF is valid. Since MATH identifies with the free abelian group generated by MATH in the above picture, the remaining statement about MATH is immediate. |
math/9907036 | The proof is similar to the proof of REF . We use the fact that if MATH represents a step in the algorithm, and if MATH, MATH and MATH are the corresponding characteristic polynomials, then MATH. As described in REF , the argument is by recursion: Suppose MATH is a triangular representation as in REF . Then REF yields divisibility for the respective characteristic polynomials MATH . If this first reduction decreases the rank, then REF shows that MATH could not be irreducible. At the first step in the reduction, REF show that the NAME - NAME eigenvalue MATH is a root of MATH. We must show that, if MATH factors nontrivially, that is, MATH, with MATH, and say MATH irreducible, then the process may continue. Since the matrices MATH and MATH have the same form as MATH at the outset, we would get MATH again with the properties from the proof of REF . Let MATH. Then MATH may be represented, via REF , as multiplication by MATH on MATH. Let MATH denote the following induced linear mapping (quotient by ideals): MATH, for MATH. It is well defined and injective due to the assumptions made on MATH. Since MATH is represented as multiplication by MATH in MATH, the range of MATH is then a nontrivial invariant subspace (over MATH) for MATH, and we arrive at the triangular form REF . The argument from the proof of REF shows that the entries of REF must have the same standard form as described in the previous step. Hence the process may continue until at some step, MATH say, MATH is irreducible. |
math/9907036 | The respective quotient groups have the following generators: MATH and a use of REF yields: CASE: MATH, MATH, MATH, and CASE: MATH, MATH, MATH . Hence MATH, and MATH. |
math/9907036 | It is enough to show that MATH and MATH represent different elements of MATH. This can be done by recursion, and use of the relations REF - REF . Alternatively, it can be checked directly by the argument from the proof of REF below that the two MATH-elements MATH and MATH are different in MATH. Both arguments are essentially based on the MATH-number, even though MATH. In the present case, MATH, which is good enough. The present argument is essentially a ``baby" version of the argument in the rest of this chapter. We sketch the details. It is a proof by contradiction. Suppose MATH were an isomorphism of the ordered MATH-groups, say MATH, which made them the same element of MATH. Since MATH, there is a MATH such that MATH. We then claim that MATH for all MATH. The argument for this is based on REF - REF for the generators: Let MATH. To verify that MATH, we must check that MATH. This holds since MATH. Hence MATH, and therefore MATH as claimed. These maps may be represented with matrices MATH as follows: MATH with the consistency conditions MATH . Thus MATH, and MATH (see REF ). This defines the sequence MATH as a sequence of linear endomorphisms of MATH, and so each MATH is represented by a matrix in MATH. That turns out to be very restrictive. It is not satisfied for MATH. In fact, even MATH has a non-integral entry. (The matrices MATH play the role of the intertwiners MATH in REF , with MATH, MATH, etc., but in the reasoning here positivity does not play a role.) Let MATH and MATH be the normalized NAME - NAME eigenvectors: MATH . Hence MATH by the NAME - NAME theorem. But MATH since MATH preserves the normalized trace; see REF . Hence, by taking MATH large, MATH REF will be arbitrarily close to MATH, where we used the numbers from the last column in REF . If MATH denote the matrix entries of MATH, then MATH, and MATH. But they are integers, so there is a MATH such that MATH for all MATH. Since MATH does not divide MATH, this is a contradiction. We have proved that the MATH-elements are different as claimed. |
math/9907036 | If MATH, then MATH, and hence MATH has the form MATH where MATH for MATH, and MATH. But MATH can contain at most MATH elements of the form MATH where MATH, and since this number is finite and MATH is an increasing union, it follows that there is a MATH such that all these elements are contained in MATH. But as MATH, the implication MATH follows. Next, choose MATH such that REF holds. We prove by induction with respect to MATH that REF holds. MATH is REF , so assume REF holds for MATH, and assume that MATH and MATH . By the induction hypothesis, we then have MATH . But applying MATH to both sides, we have MATH . Applying the case MATH, one obtains MATH and applying MATH to both sides MATH and this proves REF . |
math/9907036 | Since MATH where MATH is a matrix with integer coefficients, it is clear from REF that MATH . But it follows from REF that MATH for MATH and hence the converse inclusion is valid. |
math/9907036 | Assume first that MATH is given, and define the matrix MATH by MATH . If MATH, then MATH for some MATH, and hence MATH, that is, MATH. But then MATH and as MATH has no torsion, MATH which shows REF . But then REF follows from MATH. Furthermore, as MATH is finitely generated, it follows that MATH for some MATH, that is, MATH which shows REF . REF follows likewise from MATH for some MATH. REF follows from MATH . Conversely, if MATH is given with REF - REF , one deduces that MATH defined by REF has the properties: MATH so MATH is an order-automorphism from REF , that is, MATH . The last statements in REF are straightforward from MATH. |
math/9907036 | If MATH is a morphism, there is a MATH such that MATH . Then MATH . Thus, if MATH, then MATH for some MATH and all MATH, and hence MATH. |
math/9907036 | By REF in the form REF , it follows that there exists a MATH such that MATH for MATH, when MATH is the matrix associated to the homomorphism MATH by REF . We have MATH . Since MATH for all MATH, it follows from NAME - NAME theory that there exists a constant MATH such that MATH . But the first component of MATH is MATH by REF and MATH for all MATH, and since all components of MATH are integer by REF and MATH, it follows that MATH in an integer. But it follows from REF that MATH where we used REF . It follows that MATH so MATH . |
math/9907036 | In this case MATH. |
math/9907036 | First apply REF below to deduce MATH and MATH. But then we may apply REF with MATH and MATH. |
math/9907036 | If MATH, the matrix MATH in REF takes the form (now MATH): MATH where MATH . Note in passing that MATH is the right NAME - NAME eigenvector of MATH by REF . Using REF one computes that MATH . Iterating, one computes that MATH where MATH and the degree of the polynomial MATH is: MATH for MATH. Using this and the transformation matrix MATH and the definition of MATH together with REF one has to show that for any MATH with MATH and any MATH that MATH (The last equivalence is trivial.) This can be done by brute force, looking at highest order terms in MATH. We do omit the painful details, however, since the result can also be proved by another method described below. |
math/9907036 | From the relations REF , it follows that MATH . Inserting these relations in the relations REF for MATH in REF gives MATH . We know already from REF that MATH for MATH. Assume by induction that MATH for all MATH. It follows from REF that MATH . This shows REF , and the remaining statements in REF are obvious. |
math/9907036 | Use the notation of REF and define MATH . It follows from REF that MATH . The conclusion in REF is MATH . But MATH so this conclusion is equivalent to MATH . But the last statement is obvious from REF . This proves REF . |
math/9907036 | Let MATH be the MATH-theory morphism defined by the morphism MATH. Since MATH we have MATH and since MATH and thus MATH there is a MATH such that MATH . Apply MATH to both sides MATH . But since the morphism is assumed to be unital, we have MATH by uniqueness of the trace, and hence MATH . This proves REF . Since REF means MATH, REF follows. Finally, if the homomorphism MATH is an isomorphism and REF holds, it follows by reverting the proof that MATH, and thus MATH has a multiplicative inverse in MATH. But multiplicative invertible elements of MATH have the form on the right-hand side of REF . |
math/9907036 | Since MATH, it is clear that MATH . Conversely, if MATH, then MATH for some MATH since MATH . But as MATH, it follows that MATH for some MATH. But MATH has MATH and MATH, so MATH. Thus MATH . But as MATH and the first component of MATH is MATH, it follows that MATH. Thus MATH for a MATH, so MATH . This together with REF proves the lemma. |
math/9907036 | If there exists an isomorphism, then MATH and hence it follows from REF that MATH. The rest is straightforward from REF . |
math/9907036 | It suffices to show that MATH . For this, let MATH. Then MATH for a MATH for all MATH. Using REF we have MATH so MATH . Thus MATH . But MATH, being the NAME - NAME eigenvalue of the primitive matrix MATH, is strictly larger in absolute value than any other eigenvalue, and since MATH is a matrix with integer matrix elements, it follows that MATH . But since the first component of MATH is REF by REF , it follows that MATH so MATH and this proves REF and thereby REF MATH . |
math/9907036 | This follows from REF . |
math/9907036 | Once one notes that REF , MATH means that MATH for MATH, this is clear from the discussion preceding the proposition. |
math/9907036 | CASE: If the sequence splits, let MATH be a section, and put MATH. Then MATH . CASE: If REF holds, we can define a section MATH by MATH and since MATH, and MATH, MATH extends uniquely to MATH. |
math/9907036 | Since always MATH, the first statement follows from REF . The last statement follows from REF : If MATH and MATH, then MATH and if MATH is a section for REF , then MATH. Thus, there is a MATH such that MATH. But then MATH and it follows that MATH has the multiplicative inverse MATH in MATH, so MATH has the form REF . |
math/9907036 | The left inclusion follows from MATH. Next note that MATH is a MATH-invariant sublattice of MATH. Note that if MATH, then MATH, so we may assume MATH, and hence MATH if and only if MATH . But then by iteration MATH for MATH, and expanding those polynomials we see that MATH . Thus MATH . But MATH is also a MATH-invariant sublattice of MATH, and thus a free abelian group, and the restriction of MATH to MATH is clearly surjective. Since MATH is injective, it follows that MATH is invertible and thus MATH. But the characteristic polynomial of MATH is a factor of the characteristic polynomial of MATH, and since the constant term of the former polynomial is MATH, it is a factor of MATH. It follows that MATH, which means MATH. |
math/9907036 | We know from REF that REF is equivalent to CASE: MATH. We now argue that REF is equivalent to CASE: MATH. But since both MATH and MATH are the free abelian groups generated by MATH points in MATH, and there is an element of MATH transforming these MATH-tuples into each other, it follows that there is a natural number MATH such that MATH . Thus the equivalence of REF follows by linearity. Next, put MATH . Clearly, MATH is a subgroup of MATH containing MATH. We now show MATH . But this follows from MATH . But since REF , it follows from REF that MATH. But if MATH, then MATH for all MATH by REF . Thus MATH contains a subgroup isomorphic to MATH. |
math/9907036 | Setting MATH in REF we obtain MATH and hence MATH . Next setting MATH in REF we obtain similarly MATH so MATH . We conclude that MATH . Conversely, if MATH it follows from REF that MATH has the form MATH where the pair MATH has the form MATH for suitable MATH, MATH. But writing MATH and choosing MATH large enough, it follows that MATH and using the same reasoning on MATH, the inclusion MATH follows. Applying REF to MATH, it follows that MATH is positive if and only if MATH. We have MATH if and only if MATH, and hence REF . But if REF is not fulfilled, then MATH or MATH, and choosing MATH (or MATH) in REF we see that MATH contains elements that are not in MATH. Thus REF . Next, define MATH for MATH, MATH. Then MATH and MATH so MATH is spanned over MATH by MATH and the elements MATH . It remains to prove the last statement in the proposition. So assume MATH and define MATH by REF as the product of the primes in this set. It follows that the matrix elements in the left column of MATH all are divisible by MATH and hence all matrix elements of MATH are divisible by MATH, that is, MATH and hence, applying MATH to both sides, MATH . It follows that MATH . Conversely, MATH is spanned over MATH by MATH and the elements REF for MATH. But MATH . But since MATH, we have MATH, so the vector MATH has integral components. It follows that the elements REF are contained in MATH. Thus MATH . Now, REF finally establish REF . (The last argument was also used in CITE.) |
math/9907036 | As already remarked, REF is a special case of REF and CITE. As for REF shows that MATH as unordered groups, with positive cones determined by MATH . But then the map MATH defines an isomorphism of ordered groups MATH for MATH. Thus MATH, MATH, are both isomorphic to MATH, and REF follows. |
math/9907036 | This follows from the discussion before the Proposition. |
math/9907036 | Writing MATH in the form MATH note that then REF becomes MATH and the eigenvector MATH may be computed by directly solving MATH: MATH . Since MATH, with MATH, and MATH, we get MATH where we used the fact that MATH . We claimed in REF - REF that MATH, and this now follows. The scaling REF is immediate from this. |
math/9907036 | If MATH, then recursion in REF yields MATH for MATH. Since MATH and MATH take the same values on MATH and MATH, the identity (in each case), MATH shows that then also MATH, and therefore, by REF , MATH. The converse is clear. |
math/9907036 | Clearly MATH and MATH, so MATH, and hence MATH . Since MATH is the smallest MATH-invariant subgroup of MATH, in order to show the converse inclusion it suffices to show that MATH . But as MATH, it suffices to show that MATH . Then it suffices to show that the right column vector in MATH in REF is in MATH. But using MATH and REF we see that this column vector has the form MATH, where MATH. This proves REF and thus REF is proved. |
math/9907036 | This is an immediate consequence of REF in the proof of REF . |
math/9907036 | Let MATH. We will also prove that MATH by establishing the following relations between the right-hand sets of REF : MATH . The first inclusion to the left is immediate, and since the vectors MATH are all in MATH by REF , the second inclusion follows. But if REF MATH, then MATH where MATH, and hence REF MATH. |
math/9907036 | By REF we have MATH so MATH (see REF for the definition of MATH). But MATH must map MATH onto MATH, from which the assertion follows. |
math/9907036 | We have MATH and MATH by REF . From REF , it follows that MATH . |
math/9907036 | Using REF we have MATH and a similar expression is valid for MATH. Combining this with REF we find an element MATH with a multiplicative inverse such that MATH . Now, we may find two disjoint subsets of MATH, say MATH and MATH, such that MATH where MATH, MATH. The relation above may be written MATH . Since the primes MATH are all distinct, and all of them are factors of both MATH and MATH, and thus none of them are factors of the polynomials MATH above, it follows that MATH. Thus MATH. But this means MATH and MATH . |
math/9907036 | The first statement follows from the formula MATH in the beginning of the proof of REF , as well as from the corollary itself, and the fact that MATH is an isomorphism invariant REF . For the second statement, note that REF implies MATH and the expression in parentheses is positive. We have assumed that MATH is an integer multiple of MATH, and if this multiple is MATH we will show the contradiction MATH . This will prove the lemma. Since MATH by REF will follow if we can show that MATH or MATH . But as MATH is an integer multiple of MATH, this says MATH which is obvious when MATH. |
math/9907036 | Put MATH. Since MATH, this sum is really direct, and it follows from REF that MATH. Conversely, if MATH, define MATH and write MATH. But MATH since MATH, and hence MATH. Thus MATH and MATH. Since MATH is positive if and only if MATH the last statement is clear. |
math/9907036 | Let MATH be the matrix in REF implementing the isomorphism. By REF , MATH, and by the proof of REF , MATH and thus MATH . This shows that MATH for all MATH. Furthermore we must have MATH where MATH and MATH. Now, REF follows from REF . For MATH to be onto, MATH must be surjective and hence MATH. REF is equivalent to MATH for all MATH and since MATH, this is REF . Finally noting that REF is a transcription of REF : MATH . But since MATH, this is equal to MATH which is equivalent to REF . For the converse statement, one has to verify that if MATH is defined by REF , then MATH satisfies the conditions in REF , but this follows by the same computations as above. (Note that as MATH, REF are automatic. To show MATH, note first that MATH, and next, since MATH is onto, there is for any MATH a MATH with MATH, but then MATH, thus MATH is surjective. It is clearly injective.) |
math/9907036 | Note first that by the beginning of the proof of REF , MATH has the form MATH where MATH is a positive integer. But it follows from REF that MATH . But on the other hand MATH by REF , and since MATH is mutually prime with MATH, REF follows. The remaining statements in REF are obvious. |
math/9907036 | We know that REF - REF are necessary and sufficient. But with MATH given, one may use REF to define MATH, MATH and since MATH is relatively prime to both MATH and MATH, we see that REF is necessary and sufficient for REF by putting MATH. Finally if MATH is defined as above we have MATH so the last condition in REF is fulfilled. Finally, REF follows by putting MATH in REF . |
math/9907036 | The last statement follows from REF . |
math/9907036 | The necessity of the conditions was established in REF . The sufficiency follows from REF as argued before the theorem. |
math/9907036 | This follows from the remarks before the theorem, REF , and the formula MATH . |
math/9907036 | By CITE, there exists for any MATH a matrix MATH such that the first column of MATH is MATH: this means MATH. But if MATH, this means MATH and transitivity follows. |
math/9907036 | (Due to NAME and NAME.) If REF holds, then MATH . If, conversely, REF holds, first choose matrices MATH with MATH . This is possible by REF . It follows from REF , below, that there exists a matrix MATH such that the first column in MATH is MATH and the first row in MATH is MATH. For this, we note that MATH and MATH, and the first component of MATH is MATH the first component of MATH. Now put MATH . Then MATH and MATH . |
math/9907036 | (Due to NAME and NAME.) We will use the fact that row, respectively column, operations on a matrix can be effectuated by multiplying from the left, respectively right, by matrices in MATH. For example, interchanging the first two rows in MATH corresponds to multiplying MATH from the left by MATH, and adding MATH times row MATH to row MATH corresponds to left-multiplying by MATH. The corresponding column operations follow by taking the transpose. If now MATH is a matrix of the form MATH let MATH, and choose MATH such that MATH. Let MATH be a matrix in MATH with first column MATH (it exists by CITE). Now multiply by MATH from the left to obtain MATH where the remaining elements MATH on the first row are linear combinations of MATH, and thus multiples of MATH. By subtracting integer multiples of the second column from the remaining columns, one finally finds a matrix MATH such that MATH . Putting MATH and transposing all these operations, one finds a MATH such that MATH where MATH. Thus, if we can prove REF for this kind of matrices, the general REF follows by multiplying from the left and right by the inverses MATH, MATH. This reduces the proof of REF to the case MATH and MATH where still MATH. But to this end one can use the matrix MATH . The determinant is MATH, but as MATH, this can be made equal to MATH by choosing the integers MATH and MATH by the Euclidean algorithm. This ends the proof of REF , and thus of REF . Note that if MATH the proof above does not work: One must choose an integer MATH such that MATH when MATH, MATH, MATH are given with MATH, and this is clearly impossible in general. |
math/9907036 | (Due to NAME and NAME.) If MATH, MATH are matrices of the form REF with MATH and MATH, then MATH and MATH by REF , and then MATH by REF . This proves REF . The converse implication follows in the cases MATH by REF and the discussion around REF , so we may assume MATH from now on. Assuming REF it follows from REF that MATH and MATH are isomorphic if and only if there exists a matrix MATH such that MATH where we now define MATH and MATH and MATH . Now, one checks that MATH and MATH and thus MATH . Now we cannot apply REF directly on MATH, MATH, MATH, MATH, for two reasons: we do not have MATH, and REF is only fulfilled modulo MATH. But let us remedy this by modifying MATH, MATH as follows: First add an integer multiple of MATH to MATH to obtain a new MATH, called MATH, such that MATH. This is possible since the last component of MATH is MATH by REF . Now modify the new MATH to MATH by adding integer multiples of the vector MATH to MATH until the second-to-last component contains none of the prime factors in MATH. This is possible since MATH is relatively prime to MATH and MATH. Then MATH since MATH is orthogonal to MATH and MATH. Finally, modify MATH to MATH by adding multiples of MATH until the second-to-last component is relatively prime to the first MATH components. Then MATH and hence MATH . But since MATH we may now apply REF to find a MATH such that MATH . But since MATH and MATH it follows that MATH . Thus REF are fulfilled and REF is proved. |
math/9907036 | The equivalence REF is REF , and REF is trivial. Thus it suffices to show that REF , so assume REF . Then one establishes MATH and MATH exactly as in REF , and it remains to establish MATH. For this one notes that REF remains true in the context of nonunital isomorphism with the exception that the condition MATH is just removed, and condition MATH is replaced by MATH where MATH is a positive scalar. But since MATH induces an automorphism on the range MATH of the trace, it follows that MATH is an invertible element of the ring MATH. Now REF - REF do not involve MATH and are still valid, and then REF is valid with the same proof. Now the equation in the proof of REF becomes MATH so REF is still valid. The proof that MATH is now exactly as in the proof of REF . |
math/9907041 | This theorem goes back to NAME, for a proof see CITE. |
math/9907041 | This follows (among other things) from REF on the change of length of curves under earthquake deformations. See also CITE. |
math/9907041 | This was also known to NAME, for a proof see CITE. |
math/9907041 | See REF . |
math/9907041 | See REF . |
math/9907041 | There are no geodesic bigons in MATH. |
math/9907041 | Exercise. |
math/9907041 | Let MATH be the MATH-punctured sphere in question. Pick a simple loop MATH, separating MATH into two thrice-punctured spheres, MATH, and MATH (see REF ). Let MATH be a simple geodesic. There are a finite number of such which do not intersect the separating curve MATH (either MATH or MATH, depending on whether or not one counts boundary curves). The curve MATH could be homotopic to MATH, but if not, it must intersect MATH transversely in MATH points. Let MATH be such a curve. Note that MATH consists of MATH geodesic segments, having all of their endpoints on MATH. Up to homotopy, there is only one way to thus place MATH segments in MATH, by REF . See REF . The intersection of MATH with MATH looks similar. Note that the length of MATH . Consider the inverse operation: Given two diagrams which look like REF we can glue them together with a rational twist MATH. The integer part MATH corresponds to twisting MATH times around MATH. The fractional part correspond to the change of the identification map: A diagram (with an orientation) has a canonical labelling (shown in REF ). Gluing with no twist corresponds to attaching the strand labelled MATH to one labelled MATH, and so on. Twisting by MATH corresponds to gluing the strand labelled MATH to the strand labelled MATH, if MATH, to MATH if MATH, MATH and so on. It is not hard to see that the twist MATH leads to a connected curve if and only if MATH and MATH are relatively prime. By REF , the length of MATH twisted by MATH is bounded above by MATH, while the number of twists not exceeding MATH leading to connected curves is MATH where MATH denotes the NAME totient function. By the previous observations, the length of curves obtained thereby is bounded above by MATH, so to obtain curves of length not exceeding MATH, we must take MATH, thus, for a fixed MATH we have MATH curves. Since MATH could be anything up to MATH (the constant MATH depending on the metric on MATH), we see that MATH where we use MATH to denote the number of simple geodesics on a surface MATH of length not exceeding MATH. The last inequality in REF is the consequence of REF below (the argument is standard in number theory; we give it here for completeness). |
math/9907041 | First, note that MATH . This follows, for example, from the observation that the number of elements of order MATH in the cyclic group of order MATH is equal to MATH. From REF , we have, by NAME inversion, that MATH . Using REF we have MATH . Changing the order of summation, we see that MATH where MATH denotes the integer part of MATH. Since MATH we have the estimate MATH where MATH . Note that MATH where MATH is the NAME MATH function, while MATH . Putting all these estimates together, we get the conclusion of the lemma. |
math/9907041 | As before, take the sphere MATH and cut it into a sphere MATH with MATH boundary components and a pair of pants MATH. We will, again, count the simple geodesics MATH which intersect MATH times, then sum over the possible values of MATH. The intersection of such a curve with MATH was already studied in the proof of REF . It remains to analyse MATH. This is a collection of MATH disjoint segments having all of their endpoints on MATH, and we will analyse this inductively. We will prove the following Let MATH be a sphere with MATH boundary components. The number of MATH component multicurves on MATH of length bounded above by MATH is bounded below by a constant times MATH. Consider a sphere with MATH boundary components, MATH . We cut off a MATH-punctured sphere MATH. The intersection of MATH with this sphere is a collection of segments, MATH of which go from MATH to MATH, while MATH go from MATH to itself (see REF ). There is another combinatorial possibility, shown in REF , but as we are only interested in a lower bound, we ignore it. The intersection of MATH with MATH has MATH connected components. Thus, we have the inequality MATH where the integral is in the sense of NAME, and MATH is the number of MATH-component multicurves of length bounded above by MATH on a sphere with MATH boundary components beginning and ending on a fixed component. MATH, then MATH integrating by parts. Note that each term in the sum comes from the intersection of MATH with MATH having MATH components. In each case, if that intersection has length MATH, we can twist the length of that intersection is MATH, and we can twist a number of times around MATH to bring the length up to MATH. This number is proportional to MATH, (up to a constant, of order of length of MATH). For example, for MATH, MATH . For MATH, MATH by virtue of REF . An easy inductive argument shows that MATH . To complete the proof of REF , use REF , and essentially repeat the counting argument in the proof of that Lemma verbatim. |
math/9907045 | For the convenience of the reader, we first briefly recall from CITE the relevant facts about NAME 's resolution. Let MATH be monomials in MATH. NAME 's free resolution of the ideal MATH has the form MATH where the free modules and maps are defined as follows. Let MATH be the free module on basis elements MATH, where MATH is a subset of length MATH of MATH. Set MATH . For each pair MATH such that MATH has MATH elements and MATH has MATH elements, let MATH and suppose that MATH. Define MATH . Then we define MATH by sending MATH to MATH. Clearly NAME 's resolution does not, in general, have the same length as the minimal free resolution of MATH since the length is equal to the number of generators of MATH and, for example, MATH has rank REF. Now, a key observation is given in CITE at the end of REF (page REF): the exact same construction can be used in a much more general setting to produce at least a complex. In particular, in our situation, replacing the monomial ideal MATH by the ideal MATH , we see that we at least have a complex. Furthermore, since MATH is also a monomial it can be lifted as above, and one can immediately check that we have MATH . Note that we do not necessarily have the same kind of lifting for the MATH; it is a lifting, but the products of MATH do not necessarily begin with MATH, as we indicated above (see REF ). However, this does not matter. We conclude that we have a complex MATH where the MATH restrict to the MATH. It remains to check that this complex is in fact a resolution for MATH. By the NAME exactness criterion, we have to show that REF MATH for all MATH and REF the ideal of MATH-minors of MATH contains a regular sequence of length MATH, or is equal to MATH. Both of these follow immediately from the fact that the restriction of the MATH is MATH for each MATH, and we know that the restriction is NAME 's resolution and hence satisfies these two properties. Finally, an entry of one of the MATH is REF if and only if the corresponding lifted matrix has a REF in the same position. Hence also the minimal free resolutions agree, as claimed. |
math/9907045 | MATH . |
math/9907045 | Suppose that the codimension of MATH is MATH. Clearly MATH is the intersection in MATH of MATH with the codimension MATH linear space defined by MATH. Hence MATH. So we only have to prove MATH. Since MATH is a monomial ideal, all associated primes are of the form MATH. By hypothesis, then, there exist MATH such that every element of MATH is in the ideal MATH. By the construction of MATH it is clear that every element of MATH is in the ideal MATH. Hence MATH as claimed. |
math/9907045 | REF is immediate from REF . For REF , MATH follows from REF (the converse is immediate). REF is obvious. REF follows from REF and a computation of the NAME polynomial. To prove REF we use induction on MATH. The case MATH follows from REF . Let MATH REF denote the ideal in MATH which lifts MATH to that ring via our construction. Note MATH; in this case we continue to refer to the ideal as MATH. By induction we may assume that MATH is a regular sequence on MATH. Then by REF we see that setting MATH, MATH, MATH and MATH, all the hypotheses of REF are satisfied, and hence MATH is a lifting of MATH to MATH. The result follows immediately. |
math/9907045 | We adopt the following notation: MATH . Let MATH and put MATH. Let MATH be the NAME resolution of MATH. From the proof of REF we know that MATH has a resolution MATH such that MATH. In other words, we have a short exact sequence of complexes MATH . Dualizing we obtain the short exact sequences of complexes MATH since MATH induces MATH REF gives MATH . Thus we get the short exact sequence of complexes MATH . From the exact sequence MATH and using REF and twisting by MATH we get the induced long exact homology sequence MATH . The multiplication map MATH is injective for all MATH. The claim immediately proves the assertion of this proposition since it allows us to break the homology sequence into short exact sequences of the form we want. In order to show the claim we consider the commutative diagram MATH where the center column comes from REF . Denote by MATH the multiplication on the left-hand side. Then the snake lemma provides a homomorphism MATH. Since MATH we have MATH. Thus MATH is induced from the embedding MATH, that is, MATH is injective. Therefore the snake lemma shows that MATH is injective. Since MATH is a submodule of MATH, our claim is proved. Now let MATH. In the argument above, we have not used the specific lifting from MATH to MATH, but only the fact that a resolution of MATH lifts to a resolution of MATH. Hence our assertion follows by induction on MATH. |
math/9907045 | We saw in REF that MATH is NAME if and only if MATH is NAME. Assume, then, that MATH is not NAME. Suppose MATH has codimension MATH; note that MATH. It is enough to show that MATH does not have finite length for some MATH in the range MATH. We have by assumption that MATH is non-zero for some MATH in the range MATH. Hence the result follows immediately from REF . |
math/9907045 | This is REF . |
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