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math/9908135 | Because we have the short exact sequence MATH we can associate to any MATH bundle MATH over MATH the equivalence class of its lifting bundle gerbe. We want to show that this map is a bijection. It is enough to show that the characteristic class of a MATH bundle and the NAME class of its associated lifting bundle gerbe are the same. The characteristic class of a MATH bundle is the pull-back of the generator of MATH by any classifying map. Because MATH is a MATH the transgression for the fibering MATH is an isomorphism MATH. Fix a generator of MATH by defining it to be the negative of the transgression of the NAME class of MATH. A choice of classifying map defines a commutative diagram MATH . It follows that transgression maps of these two fiberings will commute with pull-backs and hence that the characteristic class of the bundle MATH is the negative of the transgression of the generator of two-dimensional cohomology in any of the fibres. The required result follows from REF. |
math/9908137 | Let MATH be a bounded operator on MATH which belongs to the commutant of the algebra of operators generated by the set MATH. Since MATH is dense in MATH and all the linear operators involved are continuous we can without loss of generality consider them as operating on MATH and satisfying MATH for MATH and for all MATH. Let MATH denote the projection of MATH onto MATH. Let MATH denote the restriction of MATH to MATH; then MATH is a bounded linear operator of MATH into MATH. It follows immediately that MATH is a bounded linear operator on MATH. Let MATH and suppose MATH belongs to MATH. If MATH we may use the isomorphic embedding MATH to identify MATH with an element of MATH so that MATH, and thus MATH . If MATH then use MATH to identify MATH with a subspace of MATH. Write MATH where MATH belongs to the identified subspace of MATH and MATH belongs to its orthogonal complement in MATH. Since all representations are unitary and for MATH we have MATH it follows that MATH . By REF, therefore MATH . Since this relation holds for all MATH and MATH it follows that MATH belongs to the commutant of the algebra of operators on MATH generated by the set MATH. NAME 's lemma for operator algebras (see, for example, CITE) implies that MATH, where MATH is a scalar depending on MATH and MATH is the identity operator on MATH. Now MATH is a map of inductive limit sets such that MATH, and it follows from the definition of an inductive limit map that MATH for sufficiently large MATH, MATH with MATH. Indeed, if MATH and MATH with MATH then MATH. For MATH we then have MATH . On the other hand, if MATH with MATH then for all MATH we have MATH . Since MATH, we must have MATH for all MATH. This implies that MATH where MATH is a constant and MATH is the identity on MATH. By the same NAME 's lemma quoted above the representation MATH on MATH must be irreducible. |
math/9908137 | As remarked above, the dual MATH-module MATH is irreducible (by REF ) with signature MATH, and isotypic components of different signatures are mutually orthogonal since their projections MATH are mutually orthogonal. Finally if a vector in MATH, which we may assume to belong to MATH for some MATH, is orthogonal to MATH for all MATH, it must therefore be orthogonal to MATH for all MATH, and hence must be the zero vector in MATH, and thus zero in MATH. A similar argument applies to the isotypic components MATH, and thus REF holds. Now fix MATH and MATH. Then the restriction of MATH to MATH decomposes into a (non-canonical) orthogonal direct sum of equivalent irreducible unitary representations of signature MATH. A representative of this representation may be obtained by applying REF to get the inductive limit MATH of the chain MATH; for example, when MATH, the representation MATH is given by REF on MATH. Considered as a MATH-module MATH decomposes into a (non-canonical) orthogonal direct sum of equivalent irreducible unitary representations of signature MATH. A representative of this representation may be obtained by applying REF to get the inductive limit MATH (note that although MATH is a stationary chain at MATH, the representations MATH depend on MATH even though they are all equivalent and belong to the class MATH); for example, when MATH the representation MATH is given by REF on MATH which is defined by REF . By an analogous argument we infer that the same conclusions hold for MATH, MATH, MATH, MATH. Now consider the decomposition of the restriction to MATH of the representation MATH of MATH. The multiplicity of MATH in MATH is the dimension of MATH, where MATH is the vector space of linear homomorphisms intertwining MATH and MATH. Since MATH and MATH are, by assumption, compact, this dimension is finite. If MATH is an element of MATH, where MATH (respectively, MATH) denotes the representation space of MATH (respectively, MATH), then since MATH and MATH it follows that we have an inductive chain of homomorphisms MATH. Let MATH (respectively, MATH) denote the inductive limit of MATH (respectively, MATH); then there exists a unique homomorphism MATH (see, for example, CITE, or CITE). Again by REF , MATH (respectively, MATH) is irreducible with signature MATH (respectively, MATH), and it is easy to show that MATH is an intertwining homomorphism. Conversely, all homomorphisms of inductive limits arise that way. Consequently, the chain MATH induces the inductive limit MATH. Obviously for sufficiently large MATH, MATH. By duality, we obtain in the same way the inductive limit MATH; actually this chain stabilizes for MATH sufficiently large. It follows from REF (see also the proof of REF) that MATH. |
math/9908137 | If MATH where MATH are integers such that MATH and MATH for all but a finite number of MATH, let MATH denote the total number of all nonzero entries MATH, MATH; then MATH can be realized as a subspace of the NAME - NAME - NAME space MATH. From REF it follows that MATH belongs to the isotypic component MATH of MATH, thus MATH is the inductive limit of the chain MATH. If MATH is an irreducible unitary representation of class MATH then by REF MATH where MATH (respectively, MATH) is the representation of MATH (respectively, MATH) dual to MATH(respectively, MATH). For sufficiently large MATH every MATH is the inductive limit of a chain MATH and for such a MATH . REF implies that MATH and this achieves the proof of REF . |
math/9908137 | To prove this corollary we apply REF to MATH and MATH, then apply REF to MATH and MATH, and finally apply REF to obtain the desired result at order MATH. The main difficulty resides with the definition of the identity representation on MATH, which we will construct below. For each MATH let MATH denote the identity representation of MATH on MATH. This means that if MATH occurs with multiplicity MATH in MATH then there exist MATH nonzero vectors MATH, MATH, such that MATH for all MATH. By construction each MATH is a polynomial function in MATH for some MATH. Thus MATH is a MATH-invariant polynomial in MATH. If MATH denotes the one-dimensional subspace spanned by MATH, then for sufficiently large MATH and for each fixed MATH we have a chain of irreducible unitary representations MATH. We can define the isomorphism MATH by MATH, MATH; then obviously MATH for all MATH. Also for all MATH, MATH, MATH with MATH we have MATH. Thus we can define the inductive limit representation MATH, where the action of MATH on MATH is defined as follows: Let MATH; then MATH for some MATH. If MATH for some MATH then MATH and MATH . Then it follows from REF that MATH is irreducible with signature MATH. The only problem with this approach is that the isomorphism embedding MATH is not the isomorphic embedding MATH. To circumvent this difficulty we define the inverse or projective limit of the family MATH where MATH denotes the subspace of all MATH-invariants in MATH, as follows: For each pair of indices MATH, MATH with MATH define a continuous homomorphism MATH such that CASE: MATH is the identity map on MATH, CASE: if MATH then MATH. Here we can take MATH as the truncation homomorphism, that is, MATH is defined on the generators MATH by MATH . The projective limit of the system MATH is then formally defined by MATH . Let MATH denote the projection of MATH onto MATH. Let MATH denote the representation of MATH on MATH; then MATH. Recall that if MATH denotes the subspace of all polynomial functions on MATH the MATH is dense in MATH. Let MATH denote the inductive limit of MATH; then clearly MATH is dense in MATH. Let MATH (respectively, MATH) denote the dual or adjoint space of MATH (respectively, MATH). Then since MATH is dense in MATH, MATH is dense in MATH. By the NAME representation theorem for NAME spaces, every element MATH is of the form MATH for some MATH, and the map MATH is an anti-linear (or conjugate-linear) isomorphism. Thus we can identify MATH with MATH and obtain the rigged NAME space as the triple MATH (see CITE for the definition of rigged NAME spaces). However, generally an element of MATH does not belong to MATH, but can still be considered as a linear functional (not necessarily continuous) on MATH, and furthermore, in this context the identity representation MATH will respect the isomorphic embedding MATH. |
math/9908138 | To begin, notice the long exact sequence of MATH implies that the function MATH is additive on graded MATH-modules MATH. For MATH, let MATH be the set of cones of codimension MATH. Consider the complex of MATH-modules MATH where the differentials are induced by the boundary maps from any cone MATH to its maximum proper subcones, with appropriate signs depending on the relative orientations of MATH and these subcones. It is straightforward to see that the cohomology of this complex occurs only at the MATH spot, and in fact equals MATH. Since MATH is additive, this means that we only need to show MATH for each cone MATH. So fix MATH. Let MATH be the set of generators of one-dimensional cones of MATH, and let MATH be the polynomial ring MATH. We claim that in the denominator of REF , we need to take the product only over MATH. Indeed, if MATH, then MATH acts trivially on MATH. Hence MATH is equal to MATH tensored with the NAME complex for MATH. This NAME complex gives rise to extra factors in the alternating sum of traces, which will cancel the factors MATH in the denominator. Hence we must now show that MATH . Let MATH satisfy REF , so that the series on the right of REF converges absolutely for all MATH for some MATH. We will prove the more general statement MATH for all finitely generated graded MATH-modules MATH. Indeed, in REF both sides are additive on finitely generated graded MATH-modules MATH, and coincide on MATH itself. Together with the existence of a free resolution, this implies REF, which finishes the proof of the lemma. |
math/9908138 | Without loss of generality, we may assume that MATH is simplicial. Indeed, we can freely subdivide MATH as long as we do not add any new one-dimensional cones. If we pick a generic collection of points on the one-dimensional faces of MATH, and construct the convex hull of these points for each cone, then we construct a simplicial refinement of MATH. It is more convenient now to equip MATH with a new MATH-grading that we will denote MATH. To define this grading, suppose that MATH lies in the cone MATH, and write MATH . Then MATH is defined to have component MATH for MATH, and component MATH otherwise. Obviously, this definition is independent of MATH, and it is easy to see that the old MATH grading is induced from this finer grading. Moreover, MATH is a graded MATH-module in the new sense for MATH. Suppose that MATH is a lattice point that violates the statement of the lemma. This means that MATH appears with a nonzero coefficient in REF, and that there exists MATH such that MATH . Using the notation MATH, REF translates to MATH . To calculate MATH, we can tensor MATH with the NAME complex and calculate the cohomology of the resulting complex. At the MATH-degree MATH we will have MATH . For each MATH the space MATH is one-dimensional, with MATH equal to MATH . Now the complex REF splits into a direct sum of subcomplexes according to MATH. If MATH appears with a nonzero coefficient in REF, then the NAME characteristic of one of these subcomplexes, say MATH, must be nonzero. We will show that if this MATH satisfies REF, then in fact MATH, which is a contradiction. Let MATH be the degree of MATH. Consider the set MATH of all MATH such that MATH . We claim that MATH has a unique maximal element with respect to inclusion. Indeed, suppose MATH, so that MATH . Then MATH which implies that MATH lies in the minimum cone containing MATH for all MATH, and that the corresponding coefficient of MATH is MATH. Hence MATH and MATH . Hence MATH. We will denote the maximum set for MATH by MATH. It is easy to describe the subsets MATH satisfying MATH . Indeed, REF happens if and only if there exists a cone MATH containing both MATH and MATH. Let us say that MATH is an allowed subset of MATH if this happens. Let MATH be the minimal cone containing MATH. We will now show that if MATH, then MATH is not a generator of MATH. Indeed, suppose MATH is a generator of MATH, and let MATH be an allowed subset. If MATH, then MATH is allowed, and if MATH, then MATH is allowed. Hence the set of all allowed subsets splits into pairs, and each pair contributes MATH to MATH. However, by REF. In addition, if any of the coordinates of MATH were at least MATH, then we could replace MATH with MATH, which contradicts its maximality. This means that MATH lies in the interior of the convex hull of all possible MATH, where MATH satisfies MATH . Therefore, MATH is in the interior of the convex hull of all MATH, which implies MATH for all MATH. Because MATH for all MATH, we conclude that MATH is nonnegative for all MATH, and is nonpositive for all other MATH. Let us now calculate the NAME characteristic of MATH. We have MATH . Here we have used MATH for every lattice point MATH. This is easy to verify by looking at the simplicial complex MATH induces on a small sphere about MATH. We can quotient MATH by the subspace generated by MATH, and can induce a fan there. Hence without loss of generality we may assume MATH. We now observe that the inner sum in REF is zero, unless no MATH with MATH lies in MATH. To see what this means, define MATH . Then MATH . Now define a subset MATH by MATH . We claim that the sum in REF is the NAME characteristic of MATH in NAME cohomology theory. Indeed, this cohomology could be calculated using the acyclic covering by the open stars MATH for MATH, and the cones MATH above are in one-to-one correspondence with intersections of MATH via MATH . Therefore, to show that MATH, it suffices to show that the space MATH is contractible. First of all, notice that MATH contains the entire open half-space MATH. Pick a point MATH inside this half-space. For every point MATH we will define a path MATH such that MATH, MATH as follows. Consider any cone MATH with MATH. There is a unique decomposition MATH such that MATH has nonzero coordinates only for MATH and MATH has nonzero coordinates only for MATH. Then define MATH . It is easy to see that these paths assemble together into a continuous map MATH providing a retraction of MATH onto MATH. Hence we have shown that MATH, which contradicts our assumption that MATH and MATH disobey the conditions of the lemma. This completes the proof. |
math/9908138 | The lattice MATH is a disjoint union of a finite number of regions, each characterized by the collection of signs of MATH (positive, zero, negative) for all MATH. Each such region is the interior of a rational polyhedral cone in some sublattice of MATH. It therefore suffices to show that each of the above convergence statements is true separately for MATH inside one such region MATH. Pick MATH such that MATH . Then for every MATH with MATH, we have MATH . For any MATH with MATH we just get a nonzero constant in the denominator, because we always assume that MATH are not integers. Finally, REF implies that for every generator MATH of the closure of MATH, and every MATH contributing to the numerator of MATH we get MATH . Since for MATH and MATH, we have MATH for each MATH, it follows that the terms of the MATH can be estimated by MATH, where MATH is the minimum of the linear functions (one for each MATH in the numerator of MATH) that are strictly positive on all generators of the closure of MATH. This implies the first statement. The second statement is treated similarly. |
math/9908138 | First consider the right-hand side of REF. Using MATH and applying REF, and the product formula for MATH CITE, the right-hand side becomes MATH . This is seen to be MATH which by NAME is MATH . Now we want to calculate MATH using the NAME complex associated to the covering of MATH by toric affine charts. Let MATH be a dummy multi-variable that keeps track of the torus action on MATH. For every cone MATH with generators MATH, let MATH be the corresponding affine chart. Let MATH be the graded dimension of MATH with respect to MATH. Let MATH be the dual basis to MATH. Then the space of sections of MATH on MATH, as a module over MATH, is generated by an element with grading MATH if MATH is one of the generators of MATH, and by an element with zero grading otherwise. Now it is easy to see that MATH where we expand everything as a power series around MATH. Here we have used the identity of power series in MATH whose proof can be found in CITE. Now all we need to prove the statement is to notice that, as power series in MATH, MATH . This follows easily from the description of NAME complex, see CITE for details. |
math/9908138 | The point is that sometimes when we desingularize and add extra MATH, we may get MATH, so REF cannot be directly applied. On the other hand, it is easy to see that in REF , the convergence is uniform in MATH. Hence MATH is continuous as a function of MATH, and the result follows. |
math/9908138 | This will follow from REF and the transformation properties of MATH. For any modular transformation MATH, we have MATH where MATH, and MATH depends on MATH but not on MATH or MATH CITE. Now use the notation of REF . We denote MATH by MATH and MATH by MATH. We have MATH where in the last step we used the elliptic property MATH . We can assume that the values MATH are chosen so that MATH for some generic MATH. This guarantees MATH is trivial in MATH. It is easy to see that the arguments of REF still work, provided MATH. Now we have MATH . |
math/9908138 | This can be shown by arguments similar to those used to prove REF , and so we omit the details. |
math/9908138 | In the smooth case this result follows immediately from REF and NAME 's formula. Let MATH. Then for each MATH we have MATH . Hence MATH is polynomial in MATH . Things are more complicated in the general case. Let MATH, where MATH is defined in REF . If MATH, then MATH . This allows us to write MATH as a power series in MATH and MATH, whose coefficients are polynomials in MATH and MATH. Now suppose MATH. Then MATH . Thus we can write MATH as MATH times a power series of MATH and MATH whose coefficients are polynomial of MATH and MATH. Since the limit as MATH exists, after we integrate we find no negative degrees of MATH. Finally, the statement about grading is proved by looking at the total degree in MATH and MATH and observing that at the end we put MATH. |
math/9908138 | From the definition of MATH and NAME 's formula, MATH . Let MATH . We want to show the residue of MATH at MATH is zero. But notice that MATH is a doubly-periodic function due to the condition on MATH. Therefore, the sum of its residues in a fundamental domain is zero. But MATH has a unique pole at MATH, which finishes the proof. |
math/9908138 | We need to show that for a fixed MATH all MATH can be expressed in terms of MATH. Denote the ring generated by MATH by MATH. Because of REF , for every set of nonzero residues MATH with MATH, we have MATH . If MATH, then this implies that for any four residues MATH with MATH, we have MATH . This implies MATH . We now go back to the original relation and see that if MATH, then MATH . Since MATH we can find two sets of MATH with different MATH, which shows that MATH. If MATH, we have MATH if MATH. Together with the symmetry MATH, this implies MATH if MATH. Therefore, MATH for MATH. It remains to consider MATH to show that all MATH lie in MATH. CASE: The only difference is that for MATH one can no longer use MATH. CASE: At MATH we can no longer find two sets of four MATH with different MATH. On the other hand at MATH we easily conclude that MATH, because of MATH. CASE: By symmetry, all we have is MATH for MATH. We notice that MATH is proportional to MATH of CITE, and is therefore a modular form. It is also known (see CITE) that the ring of modular forms for MATH is freely generated by an element of degree MATH and an element of degree MATH. It remains to observe that MATH is not proportional to MATH. |
math/9908138 | Consider the function MATH . This is easily seen to be a doubly-periodic function. Its only pole in the fundamental domain is at the origin, and the NAME expansion around it is MATH . On the other hand, consider the function MATH . The function MATH is even, so it has no residue at MATH. One easily sees that its NAME expansion is MATH where MATH is holomorphic. Thus, MATH is an even doubly-periodic holomorphic function, which implies that it is constant in MATH. As a result, the coefficients of its NAME expansion at MATH are zero. It remains to notice that all coefficients of MATH at MATH are polynomials in MATH. |
math/9908138 | Because of REF , we can express every element of MATH as a polynomial in MATH and MATH. The automorphic properties allow us to conclude that MATH is absent from these polynomial expressions. Together with REF , this implies that MATH is contained inside the ring generated by MATH if MATH, and by appropriate MATH for MATH. To prove the opposite inclusion, we first notice that MATH contains all MATH. Indeed, consider the degree function on the one-dimensional lattice MATH that equals MATH on MATH and MATH. There is only one possible complete fan, and the corresponding toric variety is MATH. By REF , we get MATH . Similarly, by using MATH we conclude that MATH, which helps us at levels less than MATH. To finish the proof, we must show that toric forms have the expected behavior at all cusps. Let MATH. Here is the argument for MATH. It is easy to see that MATH . We need to show that for any MATH, the function MATH is bounded in any neighborhood of MATH. Using the transformation properties of MATH, we can deduce that MATH . Using elliptic properties of MATH, we can reduce MATH to MATH where MATH and MATH. This introduces an extra summand that is irrelevant, since all we want to show is boundedness. We will closely examine the MATH-expansion of MATH. From the definition of MATH REF, we have MATH . Since MATH, the smallest (rational) power of MATH can occur at MATH if MATH or at MATH if MATH. In the first case, it is clear that the coefficient is nonzero. Since differentiation by MATH contributes a nonzero factor of MATH, the ratio in the formula above starts with MATH, and has nonnegative powers of MATH. In the second case, two terms that contribute to the smallest power of MATH have nonzero coefficients whose ratio is MATH. Hence there is no cancellation. As a result, the ratio again has only nonnegative powers of MATH. A similar argument works for all MATH, and so this completes the proof. |
math/9908138 | This follows from REF . |
math/9908138 | First we assume MATH and consider the operator MATH. Let MATH, and assume MATH is linear with respect to MATH. Using REF , we find MATH . Here MATH takes the value MATH if MATH is the residue MATH, and is MATH otherwise. We will compare this expression with MATH where the sum is taken over lattices MATH satisfying MATH and MATH. The dual of each MATH lies in MATH, and we have MATH . Here the sum over MATH in REF is taken over lattices with MATH and MATH. Now consider the contributions to the MATH-sum by different MATH. If MATH, then there is only one MATH for which MATH is always divisible by MATH: the set of all MATH with MATH. So for every MATH, each cone MATH contributes MATH . On the other hand, if MATH, then the sum REF is taken over all MATH. Letting MATH, we have for fixed MATH and MATH . Collecting these facts, we see MATH . This shows that MATH is a linear combination of toric forms. Now let MATH, and consider the operator MATH. To show that MATH is stable under MATH, we use a technique similar to the proof of REF . Let MATH be a parameter, and let MATH be a generic degree function with respect to MATH. By the argument above, we have MATH where the sum is taken over some collection of lattices MATH with MATH and MATH generic for all MATH. By REF, we have MATH . Let MATH be a complete set of residues for MATH. For any MATH, we write MATH for the function MATH. We have MATH . We have MATH . Using MATH this becomes MATH . REF imply MATH for some collection of lattices MATH and degree functions MATH. We can desingularize MATH separately with respect to each MATH. Then by REF equals MATH where each MATH is a nonsingular toric variety, and MATH is a MATH-family of rational functions in MATH and MATH. Since the action of MATH commutes with the MATH-limit, we see that MATH . Here MATH is a NAME series in MATH with coefficients that are polynomials in MATH and MATH (MATH), and by appropriate MATH for other MATH, as in REF . This limit exists and converges to a modular form. Hence MATH doesn't appear in the final expression, and by REF MATH . |
math/9908138 | As in REF , MATH . |
math/9908138 | By REF , we need only consider MATH, and for these levels, it suffices to verify the statement for MATH. We have MATH . After differentiating with respect to MATH and evaluating at MATH, we obtain MATH . Now differentiate REF with respect to MATH to obtain MATH . The right-hand side of REF becomes MATH where MATH is a constant. In REF we can take the first sum over MATH and the second over MATH, and absorb the correction in the constant. After simplifying, we find that REF becomes MATH where MATH is defined as in the proof of REF . Now it is not hard to show, using REF, that MATH . Convolving with roots of unity, we obtain MATH . Hence after subtracting a linear combination of MATH from MATH, we obtain a constant. Since this difference is modular, the constant must vanish, and the proof is complete. |
math/9908139 | Given MATH choose MATH and define MATH inductively by MATH for MATH. Then clearly MATH and MATH since MATH. Hence if MATH, that is, MATH and MATH then MATH . |
math/9908139 | We prove by induction that for each MATH the following statement holds: MATH . First observe that if MATH then MATH and MATH, which is by definition regular. Thus the first element in this reverse lexicographical ordering is MATH and REF holds trivially with MATH, where MATH and MATH . Now assume the statement holds for MATH. Let MATH denote the immediate successor to MATH, and consider the element MATH of the form MATH where the MATH-tuple MATH is yet to be determined. This implies that MATH . From REF it follows that MATH . Since MATH is right after MATH, the multi-indices MATH consist of MATH and MATH. Hence REF can be written as MATH . This implies that MATH . In REF we can solve for MATH provided that MATH is not zero. To insure this we now determine MATH as in REF by choosing the integer MATH such that MATH where MATH is the cardinality of MATH. Then MATH . It follows that MATH . Thus we have shown that the MATH-tuple MATH of positive integers can be chosen so that the coefficient of MATH in REF is a positive rational number, and the coefficients of MATH in the sum MATH are rational numbers. Obviously, MATH is a regular polynomial, so by dividing both sides of REF by the coefficient of MATH we can write MATH where MATH is regular and the constants MATH are rational. For MATH, that is, MATH . REF can be written as MATH . Set MATH and MATH, then it follows from REF that for all MATH, MATH where MATH is obviously regular and the coefficients MATH are obviously rational. Hence we have completed the induction. Now for the proof of the lemma in REF , choose MATH to be the last element of MATH. Then REF reads: ``For every MATH, where MATH is a regular element of MATH". This is exactly what the lemma affirms. |
math/9908139 | From our discussion pertaining to NAME 's discovery of the universal enveloping algebra of a NAME algebra, it follows that the quotient algebra of the polynomial algebra MATH modulo the equivalence relation MATH can be regarded as the universal enveloping algebra of the NAME algebra MATH generated by MATH. Define a map from MATH to MATH, the vector space of all regular polynomials in MATH, by assigning to each polynomial MATH in MATH the unique regular polynomial MATH equivalent to MATH as defined by REF . From the proof of REF it follows that this map is linear, and that it is surjective since the unique regular polynomial equivalent to a given regular polynomial is itself. The kernel of this homomorphism is, by the definition of the equivalence relation MATH, the vector space spanned by all trinomials of the form MATH for arbitrary MATH and MATH in MATH. Let MATH denote this kernel, then obviously MATH is a two-sided ideal of MATH. It follows from the first isomorphism theorem that MATH is isomorphic to MATH as vector spaces. From the remark following REF , it follows that the set MATH, forms a basis for MATH, and hence a basis for MATH via the isomorphism above. Note that we have shown following REF that MATH is isomorphic to MATH via the isomorphism MATH, therefore MATH is isomorphic to MATH. For the second part of the theorem, we remark that it follows from REF that each MATH is equipollent to a unique regular polynomial MATH for some MATH. Thus it suffices to consider the set MATH. We also remark that it suffices to show that the set MATH is linearly independent in MATH for all MATH, since MATH is a filtered algebra. From the proof of REF it follows that each MATH, is equivalent to a unique regular polynomial of the form MATH, where MATH is a regular polynomial of degree MATH. Thus if for some scalars MATH such that MATH is zero in MATH (that is, equivalent to MATH), then since the equivalence relation MATH is linear it follows that the regular polynomial MATH is equivalent to MATH. It follows from REF (or more precisely REF ) that MATH must be identically zero. Since MATH and MATH for all MATH, it follows that MATH and hence by the first part of the proof of the theorem, it follows that MATH for all MATH. This completes the proof of the theorem. |
math/9908149 | We now describe the construction of the constants in the statement of REF . Combining REF , there exists a compact neighborhood MATH of MATH such that: CASE: Every mapping MATH is NAME on MATH. Moreover, there exists a constant MATH such that for every MATH, the norm of the NAME constant of MATH is uniformly bounded by MATH, MATH as in REF . CASE: There exists MATH such that MATH we have MATH . We then define MATH as the minimum of MATH and the distance of MATH to the boundary of MATH. We may now conclude the proof of REF . Set MATH and MATH . In that case, we want to bound MATH . If MATH, one can bound MATH and MATH . By induction, MATH and hence MATH . The condition MATH guarantees that MATH. |
math/9908149 | Let MATH so that MATH . Choose MATH so that REF holds, that is, MATH, and such that MATH. This can be done for MATH for some constants MATH and MATH dependent of MATH. The total cost of computing MATH is MATH, where MATH is the arithmetic complexity of the algorithm and MATH is arbitrarily small. Since MATH is fixed, MATH is a constant and MATH the total cost is MATH for all MATH arbitrarily small. |
math/9908149 | The proof is divided in several steps. CASE: For any MATH, there is a full-measure set MATH such that MATH is well-defined, and a local diffeomorphism in MATH. Hence, there is a full-measure set MATH such that MATH is defined on MATH and for all MATH there is a neighborhood MATH of MATH such that MATH is a diffeomorphism MATH. CASE: Let MATH be the set of all MATH such that MATH is a diffeomorphism near MATH for all values of MATH that are large enough. Then MATH contains the set of complex polynomials without roots of the same modulus. Hence MATH has full measure. CASE: Let MATH. Then MATH has full measure. Moreover, let MATH. Then MATH is a local diffeomorphism with derivative of norm MATH in an open neighborhood MATH of every MATH, for all MATH. Indeed, if we write MATH, we can make the MATH norm of the second term arbitrarily small, namely less than MATH. We know that the norm of the derivative of MATH is precisely REF, hence the bound MATH . In the particular case MATH we can set MATH. CASE: Since MATH pointwise in the MATH topology and for MATH almost everywhere, we can assume that MATH contains an open ball MATH of center MATH, where MATH is a local diffeomorphism with derivative bounded by MATH. CASE: Since MATH is compact, the union MATH has a finite sub-cover MATH, and we set MATH. Then we set MATH, and we obtain that for any MATH, MATH is a local diffeomorphism in MATH, with derivative of norm bounded by MATH. |
math/9908149 | Our iteration MATH can be written in terms of the following real operations: MATH, MATH, MATH, MATH, MATH, multiplication by MATH, by MATH, absolute value, MATH, MATH. The set of inputs such that a `log of zero' or an `absolute value of zero' occurs has zero measure. Therefore, for almost every MATH, MATH is computed by a composition of analytic functions. Also, for almost every MATH, none of the output values is zero. Therefore, for every intermediate quantity MATH, the derivative of any output MATH with respect to MATH is finite (say MATH). Therefore, a relative perturbation of MATH in MATH leads to a perturbation of size MATH in MATH. Thus, we set MATH. By continuity, a perturbation of MATH in each intermediate value MATH leads to a perturbation smaller than MATH, for input MATH in a certain neighborhood of MATH. |
math/9908149 | Let MATH be the set of all MATH such that MATH is well-defined. Recall that MATH and that the operator MATH is not defined for MATH. Therefore, MATH is open and has full measure. Let MATH and MATH be a small connected neighborhood containing MATH. Let MATH, then by taking MATH small enough and MATH large enough, we can guarantee that MATH and MATH are well-defined. Since MATH is connected, all the branching outcomes in the computation of MATH and MATH are the same, hence MATH restricted to MATH is a composition of locally analytic functions. We can assume without loss of generality that all derivatives are bounded, hence there is a constant MATH such that MATH for MATH small enough, but still independent of the choice of MATH. |
math/9908149 | We start with the easy case and assume that MATH. Set MATH. Then, MATH. Now, MATH and hence MATH. For the general case, we will use MATH-invariance of MATH and MATH. Let MATH be a convenient unitary matrix: MATH . Set MATH and MATH. The choice of MATH has the particularity that MATH. We can compute MATH in terms of MATH: MATH . Using the easy case, MATH . By MATH-invariance, MATH . |
math/9908149 | MATH . |
math/9908149 | Let MATH and order the MATH's such that MATH . Define MATH. Then MATH . Hence, MATH . Hence, MATH . |
math/9908149 | We set MATH, where MATH is the constant such that the volume of a MATH neighborhood of MATH is less than MATH. With probability larger than MATH, MATH . We now use the fact that, for any MATH, we have MATH. We set MATH and using MATH in the previous formula, we obtain: MATH . An extra MATH iterations ensures that MATH provided that MATH . We claim that we have in this case that the sum MATH where MATH . Indeed, assume we have reordered the roots so that MATH. It then follows that MATH . Now, for MATH fixed, let MATH . Note that MATH, and MATH. Also, MATH . In general, every multi-index MATH may be obtained by starting from MATH and increasing one of the indices, in such a way not two indices are equal. Then MATH is the set of multi-indices obtained after MATH steps. Therefore, we may bound MATH to get MATH under the assumption that MATH. This concludes the proof of REF . |
math/9908150 | Write MATH . In the sum above, MATH for any choice of MATH, MATH except MATH. Since there are MATH other terms, we obtain that: MATH . |
math/9908150 | Write MATH where MATH ranges over the MATH such that MATH and MATH, , MATH. Of course, MATH ranges over all the other terms. We can rewrite MATH as: MATH . Hence, MATH . The terms in MATH are all smaller than MATH. Since there are MATH of them, MATH . Adding those two bounds, we obtain indeed: MATH . |
math/9908150 | By using REF with MATH, we obtain: MATH . Using the same lemma with MATH, we get: MATH . Subtracting the two previous expressions and dividing by MATH we get: MATH . This shows the first part of the Lemma. By using REF , we can also bound: MATH where MATH is as in REF . We can now estimate REF minus REF , altogether divided by MATH: MATH . We can also estimate REF minus REF , altogether divided by MATH: MATH . |
math/9908150 | CASE: Assume that MATH, MATH are successive elements of MATH. Then, REF implies: MATH . For the evaluation of MATH, we have to distinguish two cases: If MATH, then MATH . If MATH, let MATH be such that MATH and MATH are successive elements of MATH. Recall that MATH by hypothesis. Using REF , we get: MATH . In any case, MATH . We use the hypothesis MATH to deduce that MATH, and: MATH REF : Using REF , we have: MATH . Subtracting, we obtain: MATH . |
math/9908150 | Let MATH. It is easy to check that MATH, hence MATH. So we can bound: MATH . Therefore, we are in REF , with MATH. Correctness of REF can be proved now by induction. At step MATH, the list MATH contains MATH, MATH where MATH are all successive elements of MATH and MATH, MATH, MATH are not in MATH. (Possibly, we can have MATH). The induction hypothesis is true at REF , with MATH, and MATH. At each step, there are two possibilities: CASE: MATH. In that REF few of the MATH may be discarded; but REF prevents the algorithm from discarding elements of MATH. CASE: MATH. In that case, REF guarantees that all the MATH will be discarded. Hence, the induction hypothesis is true at step MATH. At step MATH, the last point MATH is added to MATH. Since MATH, MATH. A note on the running time: although the usual complexity of a convex hull algorithm is MATH for MATH points in the plane, the complexity is smaller when those points are `ordered' like ours: MATH. (Compare with REF). REF has a running time of MATH operations (including a fixed number of transcendental operations). Indeed, each point is added to the list MATH precisely one time. It can be discarded only once, so the interior `while' loop is executed at most MATH times in one execution of the algorithm. |
math/9908153 | For MATH we have MATH, and hence REF implies MATH . Thus REF follows from REF Locally on MATH the morphism MATH is a projection of the form MATH, and thus MATH. Hence REF holds. REF is already known (see CITE). Then REF follows from REF and the injectivity of MATH. |
math/9908153 | Let MATH and set MATH . By the definition of MATH we have MATH and hence we obtain MATH by REF . By the definition of MATH we have MATH and by the definition of MATH we have MATH . By the argument similar to CITE (see also CITE) we have MATH . In particular, we have MATH . Thus we obtain MATH by REF . Hence MATH. By REF we have MATH and MATH for any MATH. The proof is complete. |
math/9908153 | Let MATH and MATH. Since MATH is a MATH-bundle, we have MATH, and hence MATH . Thus REF follows from REF. REF follows from the fact that MATH is an inductive limit of projective morphisms and hence MATH commutes with the duality functor MATH. REF is proved similarly to CITE, and we omit the details (see also CITE). Then REF follows from REF and surjectivity of MATH. |
math/9908153 | Let MATH and set MATH . By the definition of MATH we have MATH. Hence we obtain MATH by REF . By the definition of MATH we have MATH and by the definition of MATH we have MATH . Moreover, by the argument similar to CITE and CITE we have MATH . In particular, we have MATH . Thus we obtain MATH by REF . Hence MATH. By REF we have MATH and MATH for any MATH. The proof is complete. |
math/9908157 | The proof is by induction on MATH. If MATH we can take MATH and take MATH to be a minimal extension of MATH satisfying REF ; then REF holds because MATH. If MATH, without loss of generality we can assume that MATH. Let MATH . As in the case MATH, there exists MATH containing MATH such that REF hold for MATH for MATH. Then apply the inductive hypothesis to MATH, MATH, and the MATH REF to obtain MATH for MATH and a minimal MATH. |
math/9908157 | Since MATH is isomorphic to MATH, MATH is isomorphic to MATH and hence not isomorphic to MATH because its cardinality is different. We remark also that MATH is not slender, but MATH is slender since it is a NAME group - see CITE. |
math/9908157 | The proof is by contradiction. Suppose that MATH such that for some MATH and some MATH . Let MATH be the set of all MATH such that MATH does not force MATH; then MATH is a stationary subset of MATH. For each MATH choose MATH such that MATH. We can assume that each MATH satisfies: CASE: MATH; MATH; for each MATH, MATH is a function in MATH and not just a name; MATH (MATH) is independent of MATH; if MATH, MATH implies MATH and MATH REF is MATH and independent of MATH. Moreover, if MATH and MATH, then MATH. When we say that ``MATH occurs in MATH" we mean that MATH is non-empty, or MATH occurs in MATH for some MATH or MATH, MATH, or MATH for some MATH. Without loss of generality we can assume (passing to a subset of MATH) that REF there exists MATH such that for all MATH, MATH and every ordinal MATH which occurs in MATH is less than MATH; MATH forms a MATH-system, whose root we denote MATH (that is, MATH for all MATH in MATH); MATH REF and MATH REF are independent of MATH. Moreover, for every MATH, MATH forms a MATH-system and for all MATH in MATH, MATH and MATH agree on MATH, so MATH. Also, MATH and MATH are independent of MATH. Finally, MATH (=MATH), MATH (= MATH) and MATH are independent of MATH for each MATH. Let MATH denote the ``heart" of MATH; that is, MATH and for all MATH, MATH (= MATH, say) for MATH; and MATH. The conditions in REF insure that if MATH are members of MATH such that every ordinal which occurs in MATH is MATH, then MATH and MATH are almost compatible; however, there may be problems in determining a value for MATH for MATH (independent of MATH); it is because of these that the following argument is necessary. We can assume that MATH and that for all MATH. Choose MATH such that MATH for all MATH. Let MATH . To obtain a contradiction, it suffices to prove that MATH forces: CASE: MATH is a group of cardinality MATH . This is a contradiction since MATH. If MATH does not force (MATH), then there is a finite subset MATH of MATH and a condition MATH such that MATH forces REF MATH has cardinality MATH. (Note that it follows from REF that MATH is generated by MATH.) We can assume that if MATH occurs in MATH, then MATH. Let MATH be the subset of MATH composed of all elements of the form MATH where MATH. Let MATH; so MATH has MATH elements; list them as MATH. Now choose elements MATH of MATH listed in increasing order and such that the smallest, MATH, is larger than MATH and such that every ordinal which occurs in MATH is MATH. Moreover, we can choose them so that for any MATH, the ``common part" of MATH and MATH is MATH; that is, MATH and for all MATH, MATH. (So, in particular, MATH.) Choose new ordinals MATH for MATH such that MATH . Moreover, we make the choice so that for all MATH, MATH is larger than any ordinal MATH which occurs in any MATH. There is a condition MATH which extends MATH and each MATH REF such that MATH forces for all MATH: MATH . This MATH will force versions of REF for the MATH. Also choose MATH to force values for MATH for any MATH so that REF hold for MATH for any MATH and any MATH. We claim that REF There is a subset MATH of MATH of size at least MATH and a condition MATH which extends MATH and MATH for every MATH and satisfies MATH . Assuming REF , let us deduce a contradiction, which will prove that MATH forces (MATH). Work in a generic extension MATH such that MATH. For MATH in MATH we have MATH because MATH for some MATH and, letting MATH, MATH so since MATH is pure-closed, MATH, and hence MATH. Therefore MATH has cardinality at most MATH which is a contradiction of the choice of MATH. In order to prove REF we define inductively, for MATH, a condition MATH (where MATH is as in the enumeration of MATH for MATH and MATH) such that MATH and for MATH, MATH. We also define a subset MATH of MATH of size at least MATH such that for each MATH, MATH. (So in the end we let MATH and MATH.) To begin, let MATH and let MATH be any common extension of MATH and the MATH. (There is no problem finding such an extension.) Suppose now that MATH and MATH have been defined for some MATH. Choose MATH in MATH such that MATH decides for all MATH the value of MATH for all MATH and all MATH. For each MATH fix MATH such that MATH extends MATH and satisfies MATH, MATH, and MATH for all MATH. (This is possible by the proof of REF since we only need to find solutions to REF modulo MATH-since MATH for MATH.) Define an equivalence relation MATH on MATH by: MATH iff MATH is a function. By choice of MATH and MATH, there is an equivalence class, MATH, of size at least MATH. For MATH we can define a common extension MATH of MATH and the MATH and let MATH. This completes the inductive construction. |
math/9908157 | If not, then since MATH is MATH-free, it contains a free pure subgroup of countably infinite rank. Let MATH be a basis of such a subgroup. For any MATH, MATH since MATH does not divide MATH mod MATH (or even mod MATH). Therefore MATH is an infinite subset of MATH, which contradicts REF . |
math/9908157 | We give an absolute construction of MATH using the fact that a torsion-free group is MATH-free if and only if every finite rank subgroup is finitely-generated (compare CITE), that is, if and only if the pure closure of every finitely-generated subgroup is finitely-generated. For any group MATH, let MATH be the sum of all finite rank subgroups MATH of MATH which are not free but are such that every subgroup of MATH of smaller rank is free; it is easy to see that for such MATH, MATH and hence MATH. Moreover, the definition of MATH is absolute. Now define MATH by induction: MATH, MATH-and for limit ordinals MATH, MATH. It follows by induction that MATH for all MATH. We claim that MATH; it suffices to prove that MATH is MATH-free. If not, then there is a finite rank subgroup of MATH which is not finitely-generated. We can choose one, MATH, of minimal rank, so all of its subgroups of smaller rank are free; say MATH is the pure closure of MATH; but then for some MATH, the pure closure of MATH is not free, but still has the property that every subgroup of smaller rank is free; hence MATH, which is a contradiction. |
math/9908157 | Suppose MATH, MATH and MATH are as in the hypotheses. First we claim that MATH belongs to MATH. Indeed we can define MATH in MATH as follows: MATH if for some MATH, MATH belongs to the subgroup generated by MATH and MATH; and otherwise MATH for some fixed MATH. In fact, the second case never occurs because MATH is defined in MATH. Next we claim that MATH belongs to MATH. The proof is similar in principle, using the inductive construction of MATH given by the proof of REF . But then by REF , MATH is determined by only finitely many more values, so also MATH belongs to MATH. |
math/9908157 | If MATH is defined to be the pure subgroup of MATH generated by MATH and MATH is such that MATH, then MATH induces an isomorphism of MATH with cl MATH. Hence MATH cl MATH is finite rank free; therefore, arguing as in REF , MATH belongs to MATH. |
math/9908164 | REF has already been established. Differentiating it and skew-symmetrising yields MATH, where MATH denotes the curvature of MATH on MATH. Since MATH is NAME, MATH where (for any MATH-form MATH and vector fields MATH) MATH. REF now follows by taking a trace. |
math/9908164 | Since MATH is closed and MATH is skew, MATH. The usual formulae for the NAME tensor of MATH CITE yield the first result by direct calculation. Next observe that MATH and that MATH. Also, by CITE, MATH and MATH, which leads to the following formula for MATH: MATH . Applying the star operator gives the second formula. |
math/9908164 | Differentiating MATH gives MATH where MATH. This is skew and so MATH is a divergence-free conformal vector field. Also MATH (since MATH is orthogonal to MATH and MATH) so MATH is twist-free. Finally, to show that MATH preserves the NAME connection, it suffices to show that the NAME derivative MATH of the NAME derivative on MATH vanishes. Now MATH is orthogonal to MATH and MATH, so MATH, and MATH since MATH is divergence-free. |
math/9908164 | The curvature of MATH is: MATH . Now since MATH is a conformal vector field preserving MATH, MATH. Also MATH, so MATH. Contracting with MATH and using the fact that MATH and MATH (that is, MATH) gives MATH where MATH and MATH denote the components of MATH parallel and orthogonal to MATH. Substituting this into the formula for MATH gives, for MATH orthogonal to MATH, MATH . This vanishes for all MATH because MATH and MATH is orthogonal to MATH, so MATH. |
math/9908164 | Let MATH be the vector field dual to MATH with respect to MATH, where MATH is the trivialisation of MATH determined by MATH. Now by REF , MATH (here MATH is viewed as a skew endomorphism). Since MATH, MATH. Now if MATH is orthogonal to MATH then MATH and MATH for some MATH-form MATH. Hence MATH is a divergence-free twist-free conformal vector field, and it preserves the NAME connection since MATH, by definition. |
math/9908166 | First, let us consider the coefficients MATH for non-prime MATH. So, let MATH with prime MATH. Since MATH, we have: MATH . Taking the coefficient of MATH in both sides of the above identity, we get MATH where MATH is a polynomial with integer coefficients (and zero constant term). Hence, we can write MATH . Therefore, the coefficients MATH, MATH can be excluded from the set of generators for the MATH-module MATH. Now, if MATH is still not prime, we repeat the above procedure until we arrive at a set of generators consisting only of coefficients MATH with prime MATH. Now, what we need to show is that this set of generators can still be reduced to the set REF . Note, that for any (prime) generator MATH of the cyclic group MATH one can take the coefficient MATH as a generator of MATH in the dimension MATH (that is, in MATH). Indeed, let MATH be any prime. Then MATH. Hence, MATH . Taking the coefficient of MATH in both sides of the above identity, we get MATH. Hence, MATH. Since MATH is a generator of MATH, the element MATH is invertible in MATH. So, MATH with MATH. Thus, for any prime MATH the coefficient MATH is a multiple of MATH, and that is why it can be excluded from the set of generators for MATH. Now, consider the coefficient system MATH introduced in the theorem. (That is, MATH is the coefficient of MATH in the series MATH if MATH is not divisible by MATH, and is the coefficient of MATH in the series MATH if MATH is divisible by MATH.) By induction, we may suppose that this coefficient system is a set of generators for MATH in all dimensions up to MATH. Hence, for any MATH and MATH one has MATH with MATH. We are going to prove that MATH can be also decomposed in such a way. It follows from the above argument that we can consider only prime MATH. First, suppose that MATH is not divisible by MATH. Hence, MATH, where MATH is a generator of MATH. Let MATH be any prime. Taking the coefficient of MATH in both sides of REF , we obtain MATH . Here we expressed coefficients MATH, MATH, MATH, as a linear combinations of generators MATH, that is, MATH. Therefore, MATH . Since MATH is a generator of MATH and MATH is not divisible by MATH, we deduce that MATH is invertible in MATH. Thus, it follows from REF that MATH is a linear combination of MATH and MATH with coefficients from MATH. Now, suppose that MATH is divisible by MATH, that is, MATH. Before we proceed further, let us make some preliminary remarks. It is well known (NAME, NAME), that the complex cobordism coefficient ring MATH is a polynomial ring: MATH, MATH. The ring MATH is the coefficient ring of the (universal) formal group law of geometric cobordisms (compare CITE, CITE). This formal group law has a logarithm series with coefficients in MATH, namely MATH, compare CITE. Hence, the coefficient ring of the logarithm is MATH, where MATH. It is well known that this ring is the maximal subring of MATH on which all cohomological characteristic numbers take integer values. One can choose generator sets MATH, MATH for the rings MATH, MATH such that the inclusion MATH is as follows: MATH . Let MATH be the set of elements of degree MATH in the ring MATH . Then MATH consists of elements in MATH that are decomposable into the product of two non-trivial factors. The map MATH takes the coefficients MATH of the series MATH to the element of the form MATH (compare CITE). Therefore, the coefficients MATH can be taken as multiplicative generators of MATH in dimensions MATH. In other dimensions MATH we have MATH, that is, MATH is divisible by MATH in MATH. Now, let us return to the proof of REF . Let us rewrite the identity REF substituting MATH for MATH: MATH . Taking the coefficient of MATH in both sides, we get MATH where MATH denotes the coefficient of MATH. Let us write again the coefficients MATH for MATH as linear combinations of generators MATH. Since MATH for MATH, the last identity can be rewritten as MATH for some MATH. The last two summands in the above formula can be rewritten as MATH, where MATH, MATH. The coefficients MATH are multiplicative generators of MATH in the dimensions MATH. Since MATH is a polynomial ring, one has MATH, that is, MATH is divisible by MATH in MATH. Let MATH with MATH. Then REF gives MATH where MATH, MATH. Since MATH is divisible by MATH, it follows that MATH is divisible by MATH (for MATH). Hence, the whole above identity is divisible by MATH. Dividing it by MATH and mentioning that MATH is invertible in MATH, we obtain that MATH is decomposable as MATH with MATH. Thus, setting MATH we get a decomposition of type REF for MATH (note that MATH), which completes the proof of REF . |
math/9908166 | Consider the set of generators for MATH constructed in REF . If MATH is divisible by MATH and MATH, the elements MATH are divisible by MATH, that is, lie in MATH. All other MATH do not belong to MATH. |
math/9908166 | CASE: Necessity. Let MATH. Note that the set of generators for the MATH-module MATH described in REF has the following property: each of its elements MATH is a multiplicative generator of MATH in dimension MATH. However, this set of generators for MATH has no elements in dimensions MATH such that MATH is divisible by MATH and MATH. So, we add any generators MATH in these missing dimensions to get the whole set of multiplicative generators for MATH. Now we have MATH. Since MATH, it follows from the NAME - NAME theorem that all the MATH-characteristic numbers MATH, MATH, lie in MATH. If MATH is not divisible by MATH, then there are no partitions MATH divisible by MATH. Now, let MATH. One can write MATH as a homogeneous polynomial of degree MATH in MATH: MATH where MATH for MATH. It follows from the description of MATH given in REF that MATH if and only if the coefficients MATH in the decomposition REF are zero modulo MATH for all non MATH-adic and divisible by MATH partitions MATH. Consider the NAME - NAME character MATH in cobordisms CITE: MATH . Here MATH is the first cobordism NAME class of the universal line bundle, MATH is the same NAME class in cohomologies, and the coefficients MATH are from MATH. Then for any MATH holds MATH where MATH for MATH. The coefficient ring MATH of the NAME - NAME character coincides with MATH (compare CITE). Hence, MATH with invertible MATH. Now, let us write MATH as a homogeneous polynomial in MATH. Since all MATH have integer cohomological characteristic numbers, to prove the necessity of the theorem it suffices to show that the coefficient of MATH in the decomposition of MATH is zero modulo MATH if the partition MATH is divisible by MATH. This coefficient is the homological characteristic number MATH (see REF ), which can be decomposed as follows (see REF ): MATH where MATH means that MATH refines MATH. This coefficient is divisible by MATH. Indeed, if the partition MATH is divisible by MATH and non MATH-adic, then MATH is zero modulo MATH, since MATH (see above). If there are some summands of the form MATH in the partition MATH, then MATH, that is, MATH is divisible by MATH. Anyway, the whole sum in REF is divisible by MATH. The necessity of the theorem is proved. CASE: Sufficiency. Since all the MATH-characteristic numbers of MATH are in MATH, it follows from the NAME - NAME theorem CITE that MATH. Besides, suppose that the characteristic numbers MATH are zero modulo MATH for all divisible by MATH partitions MATH, MATH . Consider again the constructed above generator set MATH for MATH. In order to prove that MATH one needs to show that for every divisible by MATH and non MATH-adic partition MATH the coefficient MATH in decomposition REF is zero modulo MATH. Let MATH be such a partition. We can rewrite identity REF as follows: MATH . One can assume by induction that if a partition MATH such that MATH, MATH, MATH, is non MATH-adic, then the coefficient MATH is divisible by MATH. If the partition MATH is not non MATH-adic (that is, there some summands of the form MATH), then MATH is divisible by MATH. Anyway, the second summand in the right hand side of REF is zero modulo MATH. The left hand side of REF is zero modulo MATH by assumption. Since MATH is non MATH-adic, we have MATH with invertible MATH. So, MATH is not divisible by MATH. Thus, it follows from REF that MATH is zero modulo MATH. |
quant-ph/9908033 | Let MATH and MATH. Because the eigenvalue MATH is real and MATH, we have the equality MATH. That is MATH, or MATH is orthogonal to MATH. |
quant-ph/9908033 | Suppose that MATH and MATH are bounded and defined everywhere so that any algebraic expression involving them is defined everywhere. Moreover assume that MATH is invariant under MATH and that MATH for arbitrary positive integer MATH. By assumption all the operators involved are linear; this implies that MATH is it self linear. By the linearity of MATH, we have MATH. The assumed invariance of MATH under MATH and the commutativity of MATH and MATH in MATH yield MATH. Then by induction we get the expression MATH for all MATH and for every positive integer MATH. Taking the norm of both sides of REF gives the inequality MATH . Now let us restrict REF in MATH. Dividing both sides of REF by MATH and taking the supremum in MATH, MATH where MATH is the restricted norm in MATH. The restriction above in MATH is necessary to enable us to have the equality in REF . Cancelling the common factors in REF finally leads to the inequality MATH which is impossible for arbitrary integer MATH because MATH, MATH and MATH are bounded, that is, have finite norms. If MATH is nilpotent of index MATH, then REF implies that MATH for all MATH, which is also impossible for MATH. Thus MATH and MATH can not be simultenously bounded and defined everywhere. The same conclusion can be arrived by interchanging the roles of MATH and MATH. |
quant-ph/9908033 | CASE: The canonical pair MATH and MATH, and the domain MATH constitute the commutator MATH where MATH is the identity in MATH. Assuming that MATH and MATH have eigenvectors in MATH, their eigenspaces are obviously invariant under MATH. Thus by REF the eigenvectors do not belong to MATH. CASE: If MATH is invariant under either MATH or MATH, then it follows immediately from REF that MATH and MATH can not be both bounded because MATH is an invariance commutator, and it has the well defined inverse MATH, and MATH obviously commutes with any operator which leaves MATH invariant. |
quant-ph/9908033 | CASE: Consider the following sequence of bounded, everywhere defined operators, MATH, for MATH, for all MATH. Since MATH is invariant under MATH, we have MATH for all MATH. The invariance of MATH under MATH and the boundedness of MATH ensure that MATH and MATH; therefore MATH is defined in the entire MATH. The linearity of MATH leads to MATH for all MATH and for every positive integer MATH. Using this result we get MATH . To establish commutation relation REF we need to get the limiting form of REF . Note that MATH and MATH are strong NAME sequences converging to MATH and MATH, respectively. Also the sequence MATH is strongly NAME. This follows from REF , MATH. Because the two terms in the right hand side of the equation are themselves strong NAME sequences, then by a MATH argument MATH is itself a strong NAME sequence converging to some MATH. Since the restriction of MATH in MATH is closed, and since MATH for all MATH and is strongly convergent, then MATH and MATH. Thus in the limit MATH for all MATH, MATH, which is what we have sought to prove. CASE: Let MATH and MATH. For the given MATH, we have, by the first half of the theorem (on using the conjugate relation of REF ), MATH . Because MATH is invariant under MATH by assumption, we have MATH . Since MATH and MATH are arbitrary and MATH is dense, that is, orthogonal to MATH, we finally have MATH for all MATH. |
quant-ph/9908033 | CASE: Let MATH be an arbitrary element of MATH. Then MATH because MATH. Then, by induction, for every MATH and for every positive integer MATH, MATH. That is MATH for all positive integer MATH. Now for every MATH, MATH; MATH is unique because MATH is unitary. This implies that for every MATH there exists a unique MATH such that MATH. This further implies that MATH. Using the same argument we used above, we find that MATH for all positive integer MATH. Thus MATH for all integer MATH. REF constitutes the commutator MATH in MATH, where MATH. MATH belongs to MATH by REF requires that MATH REF map MATH into its orthogonal complement. Thus MATH . That is the MATH's are orthonormal. Since MATH, REF implies that MATH. This and the orthonormality of the MATH's mean that for all integer MATH, MATH is an eigenvector of MATH with the eigenvalue MATH. CASE: If MATH has a multiplicity of MATH, there are MATH linearly independent eigevectors corresponding to the eigenvalue MATH. The rest of the eigenvalues will have the same multiplicity because MATH is a bijection. CASE: By REF some of the eigenvalues of MATH are given by MATH, MATH. This implies that the eigenvalues of MATH extend from negative to positive infinity. The countability of the eigenvalues follows from the separability of the NAME space. CASE: Let MATH but there are no MATH and MATH such that MATH. By REF MATH is an eigenvector of MATH with the eigenvalue MATH, which is a contradiction. Thus MATH consists only of the eigenvalue differences of MATH. |
quant-ph/9908033 | In MATH the sequence of operators MATH converges strongly to MATH, that is, MATH for all MATH, for some MATH, the MATH's being analytic vectors of the unbounded operator MATH. Also because MATH is invariant under MATH and the restriction of MATH in MATH is closed, MATH. Now the invariance of MATH under MATH and MATH, and the boundedness of MATH ensure that MATH and MATH, which implies that MATH is defined in the entire MATH. Since MATH and MATH are a canonical pair and MATH is linear, it can be shown that MATH for all MATH. Following the same steps employed in REF , we get MATH for all MATH. Because MATH by hypothesis is invariant under MATH, the sequence MATH converges strongly to MATH. Also since MATH is bounded, thus continuous, the sequence MATH likewise converges strongly to MATH. Thus in the limit REF reduces to MATH for all MATH and MATH. For MATH, let MATH be an integer such that MATH. Then for all MATH because MATH. Thus for all MATH, MATH . Thus REF holds for all MATH. Now REF is the commutator MATH where MATH is the restriction of MATH in MATH. Because MATH for all MATH for every MATH, MATH is an invariance commutator in MATH. Since MATH is invertible, by REF , MATH and MATH can not be both bounded and defined everywhere. But they are. Therefore in order to maintain the equality in REF MATH must be the zero vector - and no other else. Thus MATH is the trivial subspace. |
quant-ph/9908079 | Let MATH, MATH, and denote the group action of MATH by MATH. By means of this action to each MATH a vector field MATH on MATH is associated, whose flow we denote as MATH. Its pushforward with MATH acting on a function MATH on MATH is MATH . If MATH is a Hamiltonian vector field, then we have, furthermore, MATH because the group action is Hamiltonian. For MATH in the center of MATH we have MATH and these equations imply MATH. The generating function MATH thus has to fulfill MATH . Here MATH is a linear map from the NAME algebra of MATH, which we identify using the momentum map with its isomorphic NAME algebra of generating functions of the group action on MATH, to MATH. This map is in fact a REF - cocycle in the cohomology of this NAME algebra: MATH implies MATH which leads to MATH. This observation already proves our first assertion: If MATH is semisimple, we have MATH and MATH; MATH vanishes for any MATH. This means that each of the generating functions, and therefore any generated function, is invariant with respect to the action of MATH. But the center of MATH acts nontrivially, because the group action of MATH is effective, and not any phase space function, which is in general not invariant, can be generated. For groups with MATH (nonperfect groups, compare the remark following this proof) the above argument cannot be used. However, if MATH is finite, there is for each MATH a MATH with MATH. Due to MATH (This follows from REF and MATH for all MATH, which in turn is a consequence of MATH for any constant function MATH on the phase space.) we again have MATH for any MATH and MATH. |
quant-ph/9908079 | According to the invariance of MATH with respect to MATH, the action of Diff-MATH does not lead out of the subbundle MATH. For dim-MATH each of the components of MATH is invariant only with respect to orientation preserving diffeomorphisms (and we thus restrict to MATH in this case). The momentum map of the action has been provided already above, furthermore, and its effectiveness on MATH is obvious. Transitivity on MATH (MATH for dim-MATH) follows as MATH REF acts fiber transitively on MATH (that is, it acts transitively on the space of fibers), while on the fiber of MATH REF over the origin MATH of some particular local coordinate system of MATH the vector fields MATH act transitively. |
quant-ph/9908079 | Let MATH be a finite - dimensional subalgebra of MATH with at least three generators. Without any restriction these generators can be assumed to be of the form MATH with MATH, MATH, MATH and MATH. Otherwise they can be brought into this form by appropriate linear combinations. Let MATH, MATH, be two of the generators and MATH as defined above. Then we can reveal the following conditions for MATH to be finite - dimensional: CASE: MATH, and analogously MATH: Otherwise MATH would contain a contribution of MATH with nonzero coefficient. By induction, repeated commutators would contain contributions from MATH with arbitrary MATH. Therefore, the subalgebra could not be finite - dimensional. CASE: MATH, and analogously MATH: Otherwise by appropriate linear combinations we could trade the generators for two new generators not fulfilling REF . We conclude that all generators are of the form MATH, that is, there are only three linearly independent generators MATH. This proves that a finite - dimensional subalgebra is at most three - dimensional. We can now determine the form of these subalgebras MATH: From the commutation relations of the MATH it follows that MATH has to be proportional to MATH in order for MATH to be closed under commutation. This can only be the case if there is a MATH such that MATH. Analogously there must be a MATH such that MATH. Now we must have MATH because otherwise MATH would not be closed. The only three - dimensional subalgebras of MATH are, therefore, given by MATH for MATH, which are easily seen to be isomorphic to MATH. |
quant-ph/9908079 | A two - dimensional NAME algebra generated by MATH and MATH can, without restriction, be assumed to be of the form MATH or MATH, respectively. In close analogy to the proof of the preceding lemma, one may show that the former case implies MATH (in contradiction to the linear independence of MATH and MATH) and that the latter case is possible only if MATH and MATH for a MATH. |
quant-ph/9908079 | As abelian subalgebras of MATH are at most one - dimensional (MATH has rank one), any two - dimensional subalgebra may be spanned by generators MATH and MATH satisfying MATH. In the complexified NAME algebra MATH they span a NAME subalgebra, which is a maximally solvable subalgebra and unique up to conjugation. Thus we know that for any two - dimensional subalgebra of MATH there is, in the fundamental representation of the algebra, a complex two - by - two matrix MATH of unit determinant such that MATH and MATH (and, up to a sign, MATH is unique). Reality of the matrices MATH and MATH implies that MATH is either real or purely imaginary. In the former case, MATH and the conjugation is compatible with the reality condition leading from MATH to MATH. In the latter case, MATH where MATH. Conjugation with the imaginary piece MATH maps MATH into MATH. The assertion of the lemma then follows upon the isomorphism REF and exponentiation to group level. |
quant-ph/9908079 | According to REF the finite - dimensional subgroups of MATH which are generated by elements of MATH can be at most three - dimensional because the finite-dimensional subalgebras of MATH, which is the complexification of MATH, are at most three-dimensional. The three-dimensional of these subgroups are isomorphic to MATH - fold covering groups of MATH spanned by MATH, MATH, and MATH. All the two - dimensional subgroups are subgroups of these three - dimensional ones, moreover. Finally, there are the one - dimensional subgroups of MATH which are generated by exponentiation of an arbitrary element of MATH. We now investigate the action of these subgroups when lifted to MATH. One - dimensional groups cannot have orbits filling all of the two - dimensional half - cylinder. So they cannot act transitively. According to REF , all two - dimensional subgroups are in one of the two conjugacy classes, representatives of which are generated by the vector fields MATH and MATH. Their lifts MATH and MATH to MATH fix the fiber over MATH and therefore the groups cannot act transitively. (The other two - dimensional subgroups, being conjugate to one of these two groups, can just as less act transitively.) The only candidates with transitively acting lift are now the covering groups of MATH. That they act indeed transitively can be seen from the following consideration: The lift of the action of a MATH - fold covering group of MATH is generated by the two vector fields given in the previous paragraph together with the vector field MATH. The former two act transitively in some fibers and the latter one acts fiber transitively. Thus their joint action is transitive on MATH. |
cond-mat/9909233 | For each time MATH, define Boolean variables MATH for each particle MATH and MATH for each site MATH. Then the effect of a move MATH is simply that of a comparator gate with inputs MATH and MATH, and outputs MATH and MATH: MATH . This converts the list to a comparator circuit of size MATH and width MATH, where MATH is the number of particles and MATH is the total number of sites named in the list. The outputs MATH and MATH give us the set of occupied sites and active particles at time MATH. |
cond-mat/9909233 | We will use sites of the cluster to store truth values, with occupied and unoccupied sites representing true and false wires respectively. However, the same site will represent two different wires at different times. Our basic tool is the walk shown in REF , in which a particle comes from the origin and moves down a horizontal conduit. It steps off the conduit to visit site MATH, continues to MATH if MATH is already occupied, and continues to a previously unoccupied site MATH if MATH is already occupied. If MATH and MATH are times before and after this walk, the effect on MATH, MATH and MATH is as follows: MATH . Thus if the old values of MATH and MATH are the inputs to a comparator gate, the new values of MATH and MATH are its outputs. If a comparator circuit has MATH gates, it has at most MATH wires, which need at most MATH sites to represent their inputs and outputs. If we place these sites contiguously along a row adjacent to the conduit the particles use, and if the origin is at one end of this conduit, each walk takes at most MATH steps, and the total time for MATH such walks is MATH. |
cond-mat/9909233 | Using the MATH processors, we keep MATH particles active at any given time, all moving in parallel. Whenever one or more reaches an unoccupied site, the particle on the processor with the lowest index is deposited there, the other particles remain active, and that processor starts a new particle at the origin. As we showed above, this will give us the same probability distribution of clusters as if we added particles one at a time. Since each processor adds MATH particles on average, the mean time for each processor to complete its task is MATH where MATH. However, the running time of the algorithm is the time it takes the last processor to finish, which is at most MATH plus the length of the last particle's walk. Since these times are distributed with an exponential tail MATH for large MATH, and since the average maximum of MATH things distributed with probability MATH is MATH where MATH is NAME 's constant, the last processor finishes in average time MATH as promised. |
cond-mat/9909233 | In one dimension, the lattice sites are the integers and the cluster is a line segment MATH. Initially, MATH. Each particle's walk is a mapping that increases either MATH or MATH by one, by adding a particle at the left or right end of the cluster. Thus, the history of the cluster can be represented by a directed path starting at the origin in one quadrant of a two-dimensional square lattice where the MATH and MATH coordinates represent MATH and MATH respectively. For each particle, we have a list of sites it will visit. The first step in the algorithm is to convert this list to a mapping on the square lattice, that is, a table of entries MATH where MATH is either MATH or MATH, depending on whether MATH or MATH first appears in that particle's list. Since MATH and MATH are bounded by MATH, this table has length MATH. We can do this conversion in MATH parallel time with MATH processors where MATH is the length of the particle's walk. If neither MATH nor MATH appears in a particle's list then MATH and that particle is not incorporated into the cluster. We then calculate the composition of all these maps by composing the maps of adjacent pairs of particles, then composing these pairs, and so on. This takes MATH parallel time and can be done by MATH processors, one for each entry in the map. The final state of the cluster is this composed map applied to the initial state MATH. |
cond-mat/9909233 | We can convert MATH random bits into the coordinates of a MATH-dimensional random walk of MATH steps in MATH parallel time, since the MATH'th coordinate is the sum of the first MATH moves. We then add particles one at a time, by letting our list of moves be a concatenation of walks, one for each particle. Note that the particles will not actually take these walks; they will only take them as long as they are active, that is, until they reach an unoccupied site. Since in time MATH a particle will reach the boundary of a MATH-dimensional sphere with MATH sites, the probability of the MATH'th particle still being active after MATH steps has an exponential tail of the form MATH, and the probability of some particle still being active at the end of its walk is at most MATH times this. Setting this equal to MATH tells us that we can ensure with probability MATH that no particles are left active at time MATH by giving the MATH'th particle a walk of length MATH . Using the construction of REF gives a comparator circuit of depth MATH and width MATH where MATH is the total number of sites named in the walks. We then use NAME and NAME 's simplification algorithm to evaluate this circuit. In the worst case where every walk heads away in a different direction from the origin as fast as it can, MATH is proportional to MATH, and the simplification algorithm runs in time MATH, no better than explicit simulation. However, MATH is almost always significantly less than MATH, making this circuit narrower than it is deep. In particular, since the probability of a particle being at a site a distance MATH from the origin after MATH steps is roughly MATH, a crude union bound shows that the probability of any particle reaching any site MATH from the origin in MATH steps is at most MATH . Setting this equal to MATH tells us that with probability MATH, all the particles are confined to a ball of radius MATH which is in the crossover regime for multiple random walkers studied in CITE. The volume of this ball is MATH and the simplification algorithm works in time MATH plus smaller corrections. The two sources of possible error - failing to have all the particles' walks terminate, or having some walker exceed the radius in REF - both have probability MATH. By rescaling these to MATH, we can keep the total probability of error below MATH. |
cond-mat/9909233 | First we generate, in advance, the paths of all MATH walkers; this can be done in parallel time MATH as in the algorithm of REF. We then grow the cluster in a series of shells. At each step we take the current cluster MATH and determine, in parallel, what site outside MATH each active particle hits first, which is where it will stick if no other particle gets there first. We then look at the set of particles at each sticking point, attach the one with the lowest index, deactivate it, and keep the other particles active. We repeat this with the new cluster, and continue until there are no active particles left. In the early stages, the cluster will be diamond-shaped, since almost every site at its perimeter becomes occupied at each step. Later on, it approaches its final shape which is roughly spherical, and every site on the perimeter has a roughly equal probability of becoming occupied. Thus each step adds a shell of constant thickness, and the algorithm will grow a cluster of size MATH in MATH steps. Finding the first sticking point of a walk of length MATH can be done in MATH parallel time with MATH processors, so doing this for all MATH particles takes MATH processors. Finding the particle with the lowest index at each sticking point can be done in MATH time with just MATH processors. Therefore, the total running time is MATH, and the number of processors we need is MATH, which is polynomial in MATH. |
cs/9909004 | Let MATH denote a tour of MATH in MATH. Such a smooth path must intersect MATH. Translate MATH along the bisector of the angle defined by the tangents of MATH at MATH and MATH. Now, let MATH denote the furthest position of MATH tangent to MATH and let MATH be some tangent point (see REF ). Suppose that MATH and MATH coincide on a small interval around MATH. In this interval, the curvature of MATH is the same as the curvature of MATH which is MATH. Now, suppose that MATH is strictly below MATH just after MATH. Notice that such a tangent point always exists if MATH does not coincide with MATH. It follows from REF of the appendix that the curvature of MATH is strictly greater than the curvature of MATH which is MATH. |
cs/9909004 | Perform a tree traversal on MATH. Each time a vertex MATH is visited, such that MATH, there exists a circle of radius REF/MATH tangent to the boundary of MATH, and centered on the edge joining MATH and MATH. This circle can be computed easily once the edges of MATH defining the edge joining MATH to MATH are known. Then, the subtree of MATH rooted at MATH is pruned and the traversal continues from MATH. In this way, all the MATH circles with radius MATH inscribed in MATH are found in order of their appearance on MATH. Hence, the maximal path MATH corresponding to the convex hull of the circles can be obtained easily by joining two consecutive circles by their common supporting segment. The MATH time complexity of the algorithm follows from the fact that the number of vertices visited during the transversal of MATH is in MATH. |
cs/9909004 | Perform a tree traversal on MATH. The traversal can be oriented such that the children of any node are visited in counterclockwise order. An arc is produced each time a vertex MATH is visited from its parent MATH. This arc is less than a semicircle, centered at MATH, and tangent to the two edges of MATH whose bisector contains the edge MATH of MATH. Finally, a degenerated arc is produced if MATH is a leaf of MATH. To see that the arcs are produced in the right order, observe that the tree traversal can be performed by moving a point MATH continuously along the edges of MATH. Let MATH be the orthogonal projection of MATH on the edge of MATH belonging to the NAME cell on the right-hand side of MATH with respect to the direction of the traversal. Since MATH corresponds to the medial axis of a convex polygon, MATH moves continuously around the boundary of MATH in counterclockwise direction. Now, consider the arc computed while MATH traverses the edge MATH of MATH. By construction, the first endpoint of this arc corresponds to MATH when MATH coincides with MATH. Thus, the arcs are produced during the traversal of MATH such that the first endpoints of the arcs appear in counterclockwise order on the boundary of MATH. The MATH time and space complexities of the algorithm follow from the fact that MATH has at most MATH vertices. |
cs/9909004 | Let MATH be a normal parametrization of MATH such that MATH. Let MATH be the angle made by the tangent to MATH at the point MATH with the MATH-axis. The functions MATH, MATH and MATH are related as follows: MATH and MATH. Since MATH is convex and lies above the MATH-axis for MATH, there is an interval MATH on which MATH is continuous and strictly increasing. The function MATH is defined similarly and has the same properties. Hence, there is an interval MATH on which, either MATH, or MATH, or MATH. Suppose that MATH. By definition, MATH and MATH. Since MATH is continuous, there is a value MATH such that MATH. Furthermore, MATH. Thus, the point MATH is below the point MATH which is impossible. The case MATH is even simpler. Hence, MATH. This implies that MATH . Thus, the average curvature of MATH is greater than MATH. |
cs/9909019 | For REF , suppose that MATH is a hypercube and consider any two elements MATH, MATH in MATH such that MATH. (Note that the first component of these tuples must be the same if MATH is a hypercube.) Then for all MATH in MATH, MATH. Therefore, by definition of the relations MATH, for all MATH in MATH we have MATH, that is MATH. For REF , suppose that MATH is full and let MATH and MATH be points in MATH. Since MATH is full there exists MATH such that MATH is in MATH. Clearly MATH. Thus, there is a path from MATH to MATH of length two. For REF , suppose that MATH is full and consider any MATH. Since MATH is full there exists MATH such that MATH. By REF , the world MATH is in MATH and by construction for each MATH, we have MATH. |
cs/9909019 | We first show REF . Let MATH be a MATH frame. Take MATH to be the set of tuples MATH where MATH. We show that MATH is a full system and such that MATH. MATH. To see that MATH is full, let MATH. We show that there exists MATH such that MATH. Since MATH is a MATH frame, there exists a world MATH such that MATH for each MATH. Thus MATH. This shows that MATH is full. Write MATH. To show that MATH, define the mapping MATH by MATH. It is clear that MATH is a bijection. Moreover, MATH . Thus, MATH is a frame isomorphism, establishing MATH. This completes the proof of REF . For REF , let MATH be a MATH frame. Define MATH. Clearly MATH is a hypercube. Write MATH. We show that MATH. Consider an element MATH of MATH. Since MATH is a D frame, there exists MATH such that MATH for each MATH. Moreover, because MATH is an I frame this MATH is unique. Define the mapping MATH by taking MATH to be the unique MATH such that MATH for each MATH. The mapping MATH is surjective because for each MATH we have MATH. Moreover MATH is injective because if MATH then for each MATH we have that MATH is in both MATH and MATH. Thus, these equivalence classes must be the identical, and hence the tuples MATH and MATH are identical. It remains to show that MATH has the homomorphism property. For this, note that by construction, for each MATH we have MATH. Thus if MATH then MATH, hence MATH. Conversely, suppose MATH and MATH and MATH. By definition of MATH, MATH and MATH. Since MATH it follows that MATH. Thus, MATH. Thus, MATH is a frame isomorphism, establishing MATH. This completes the proof of REF . |
cs/9909019 | Suppose the opposite and assume there is a formula MATH that corresponds to REF. Consider the frame MATH in REF . The frame MATH is an I frame, so MATH. Consider now the frame MATH and a function MATH such that MATH maps points in MATH according to the names in the Figure. It is easy to see that MATH is a p-morphism from MATH to MATH. Since p-morphisms preserve validity on frames REF we have that MATH. But MATH is not an I frame and we have a contradiction. |
cs/9909019 | Left to right. Let MATH be a model based on MATH such that MATH. Since MATH, then MATH. Analogously, suppose MATH. Since MATH we have MATH. Right to left. Suppose MATH and for all MATH we have MATH. Take a valuation MATH such that MATH if and only if MATH. Since MATH and MATH, we have MATH and so MATH. But since MATH, it must be that MATH. |
cs/9909019 | Consider the set MATH of the transpositions of MATH, that is, functions MATH where MATH, such that if MATH then MATH, if MATH then MATH, and MATH otherwise. We have MATH. But by set theory (CITE page REF for example) MATH, and so MATH. So, by induction, we have MATH. Call MATH the bijection MATH, and define MATH. To prove the lemma holds we consider two cases: MATH and MATH. For MATH, assume any MATH, and any MATH. Take any MATH for MATH. Then MATH is a transposition of MATH. So, there exists a MATH such that MATH. So MATH. For MATH, assume again any MATH, and any MATH. Consider the transposition MATH; we have MATH. But MATH for some MATH. So MATH. |
cs/9909019 | Consider a set MATH such that MATH and MATH have the same cardinality, and let MATH be a bijection from MATH to MATH. Then there is a function MATH, satisfying the property expressed by REF . Define now a function MATH, such that MATH, otherwise MATH, where MATH is any element in MATH. Define the function MATH by MATH. We claim MATH has the property required. For, let MATH and take any MATH and MATH. Then MATH, and so by REF there exist MATH, such that MATH. Define MATH for MATH. We then have MATH since MATH. |
cs/9909019 | Let MATH be an infinite set with cardinality at least as great as the cardinality of any MATH. This can be constructed by taking the union of these sets MATH or by considering the set of the natural numbers MATH if all the sets MATH are finite. For each MATH, let MATH be the function promised by REF . Define MATH by MATH. It is immediate that this function has the required property. |
cs/9909019 | Let MATH be a frame with MATH relations on its support set MATH. Write MATH for the relation MATH. Since each of the MATH is an equivalence relation, so is MATH. Since the set of worlds MATH of the frame MATH is non-empty, it can be viewed as the union of the equivalence classes of the relation MATH, which we call clusters. Write MATH for the set of clusters of MATH. Consider the infinite set MATH and a function MATH as described in REF , and define the frame MATH as follows: CASE: MATH, CASE: MATH if MATH and there exist worlds MATH and MATH such that MATH. We can prove that: CASE: The frame MATH is EDI. CASE: MATH is clearly an equivalence frame. CASE: We prove MATH satisfies REF. Write MATH for MATH. Suppose MATH. Then for all MATH we have that MATH, and there exist MATH and MATH such that MATH. Since MATH and MATH are equivalence classes of MATH, it follows from the latter that MATH, and consequently that MATH. Thus, MATH. CASE: We prove MATH satisfies property D. Consider MATH tuples MATH in MATH. For each MATH let MATH be a world in cluster MATH. Since MATH has property MATH, there exists a world MATH such that MATH for each MATH. Let MATH be the cluster containing MATH. Then, by construction, for each MATH we have MATH. CASE: The function MATH is a p-morphism from MATH to MATH. That the function MATH is surjective follows from REF . Next, we show that MATH is a frame homomorphism. Consider two tuples MATH, MATH in MATH such that MATH. Then there exists MATH and MATH such that MATH. By REF , we have MATH and MATH. Since MATH is an equivalence relation, it follows that MATH. To show the backward simulation property, consider a tuple MATH, and assume MATH for some world MATH of MATH. Let MATH be the cluster containing MATH. By REF , there exist MATH for MATH such that if MATH, then MATH. Since MATH by REF , it is immediate that MATH. |
cs/9909019 | Suppose the opposite and assume there is a formula MATH that corresponds to n-directedness. Consider two disjoint frames, MATH and MATH, where MATH, such that both MATH and MATH are n-directed. the frame MATH. Since by REF and MATH, it follows that MATH. This is because satisfaction of a formula of MATH at a world MATH depends only on worlds connected to MATH REF . But, then MATH is valid on a frame which This is the opposite of what we assumed at the beginning. |
cs/9909019 | Suppose that MATH is weakly directed and connected. Let MATH be any MATH worlds in MATH. We show that there exists a world MATH such that MATH for each MATH. Since MATH is connected, the worlds MATH are in the same connected component, and hence all connected to some world MATH. Since MATH is an equivalence frame the relations MATH are symmetric, so we may assume that for each MATH there exists a path directed from MATH to MATH. We now claim that none of these paths need to be any longer than one step, for if so, we can reduce their length. For, suppose without loss of generality that the path from MATH to MATH involves more than one step. Write this path as MATH and write the remaining paths as MATH to MATH. Using weak directedness (and an ordering of the worlds MATH such that MATH occurs in position MATH), we obtain a world MATH such that MATH and for each MATH we have MATH for some MATH. By symmetry of the relations, we obtain paths from MATH to the worlds MATH. For MATH these paths are of the form MATH and have the same length as the path connecting MATH to MATH. For MATH we have the path MATH, which can be shortened to MATH by transitivity of MATH. This argument establishes that there exists a world MATH such that for each MATH we have MATH for some MATH. Since MATH is weakly directed, it follows that there exists a world MATH such that MATH for each MATH. |
cs/9909019 | It is immediate from the definition that the class of weakly directed equivalence frames contains the ED frames and is closed under disjoint unions and isomorphism. To show that it is the smallest such class, we show that any weakly directed equivalence frame is isomorphic to a disjoint union of directed equivalence frames. For let MATH be weakly directed, and let MATH be a subset of the set of worlds of MATH containing exactly one world from each connected component of MATH. For each MATH let MATH denote the connected component of MATH containing MATH. By REF , each MATH is directed. It is then possible to show that MATH is isomorphic to the disjoint union of the frames MATH as MATH ranges over MATH. |
cs/9909019 | Since every ED frame is EWD, every formula valid on the EWD frames is valid on the ED frames. Conversely, suppose that MATH is not valid on some EWD frame MATH. Then there exists a valuation MATH and a world MATH such that MATH, where MATH. Let MATH be the connected component of MATH containing MATH and MATH the corresponding frame. Then MATH is a directed equivalence model, and by REF we have MATH. Consequently, MATH is not valid on the ED frame MATH. |
cs/9909019 | We first show that if MATH is a WD frame then MATH. For, suppose that MATH is an interpretation of MATH and MATH a world of MATH such that MATH. Then for each MATH there exists a world MATH such that MATH for some MATH and MATH. Since MATH is weakly directed there exists a world MATH such that MATH for each MATH. By REF , we have MATH. Since MATH, it follows that MATH. This establishes MATH. Conversely, suppose MATH. We show that MATH is weakly directed. Let MATH be worlds of MATH such that for each MATH there exists MATH such that MATH. We need to show that there exists a world MATH such that MATH for each MATH. To achieve this, let MATH be MATH distinct propositions and define the interpretation MATH by MATH if and only if MATH, for each MATH. (The interpretation MATH may be defined arbitrarily on all other propositions.) Note that we have MATH. Since MATH, and each formula MATH is MATH-local, it follows that MATH. In particular, there exists a world MATH such that MATH, hence MATH. But, by definition of MATH, this means that MATH for each MATH, as required. |
cs/9909019 | Soundness follows from what was proved in the first part of REF and the fact that all axioms and rules of MATH are sound for equivalence frames CITE. To prove completeness we use the canonical model technique. It is easy to show that the frame MATH of the canonical model for Section is reflexive, symmetric, and transitive with respect to the MATH relations. We prove it is also WD. Suppose that MATH are worlds of MATH such that MATH, for MATH. Consider the set MATH . We show that MATH is Section -consistent. It then follows by the maximal extension theorem that there is a maximal Section -consistent extension MATH, which satisfies MATH by construction. This will establish that the frame is WD. To show MATH is Section -consistent, we assume it is not Section -consistent and obtain a contradiction. It follows from the assumption that for each MATH there are formulae MATH, with MATH for each MATH, such that MATH . Let us now call MATH. Note that by MATH reasoning, we have MATH. It follows that MATH. (For else, MATH, hence MATH, contradicting consistency of MATH.) By propositional logic we obtain MATH. Thus, MATH. Now the formulae MATH are MATH-local, so using WD it follows that MATH. By MATH reasoning we get MATH. But by MATH reasoning and the fact that MATH this leads to the conclusion that MATH is inconsistent. This is the contradiction promised. |
cs/9909019 | If MATH then WD is MATH (with MATH-local). WD to NAME: Put MATH and MATH (note that these are REF-local and REF-local respectively). WD now becomes MATH . Now we drop the disjuncts MATH and MATH (this strengthens the antecedent and hence weakens the whole formula) to obtain as a consequence MATH which can be simply rearranged to obtain MATH as required. NAME to WD: From MATH we want to obtain MATH in the case that the MATH are MATH-local. Since the MATH are MATH-local, we have MATH. Assume MATH which, on distribution, is MATH . From each of these disjuncts, we will derive either MATH or MATH, thus proving WD. The derivations are as follows: CASE: From MATH, apply NAME 's axiom together with uniform substitution to the second term to obtain MATH. Use the MATH axioms MATH and MATH to obtain MATH. From this we deduce MATH and from the axiom T: MATH and substitution we obtain MATH. CASE: From MATH: the first conjunct gives MATH, then MATH, then MATH by MATH axioms. The second conjunct gives MATH, so putting them together we have MATH, from which we obtain MATH as a consequence, and hence MATH. CASE: From MATH, we obtain MATH by applying NAME to the first term. This now implies MATH, which in turn implies MATH. CASE: From MATH: this case is similar to the first one. |
cs/9909019 | All we need to prove is that the relations MATH are suitable. CASE: Consider worlds MATH and world MATH such that MATH and MATH. We need to prove that MATH, that is, that for all formulae MATH such that MATH-we have MATH if and only if MATH. We prove it from left to right; the other direction is similar. Note that MATH if and only if MATH because MATH is an equivalence model; but MATH and so MATH. But MATH and MATH, so MATH, which is what we wanted to prove. CASE: Consider worlds MATH such that MATH. This means that for all MATH, we have MATH if and only if MATH. Since MATH is an equivalence model it follows that MATH. |
cs/9909019 | We prove that MATH is an ED frame. The relations MATH are clearly equivalence relations. All it remains to show is that MATH is directed. To do that, consider any MATH. Since MATH is directed, there exists MATH such that MATH for MATH. But each MATH is suitable and so, by a consequence of REF of suitability we have that MATH, for MATH. Therefore the frame MATH is directed. |
cs/9909019 | Suppose MATH. Since by the proof of REF the logic MATH is canonical, the canonical model MATH for MATH is an equivalence model, it is weakly-directed and there is a point MATH, such that MATH. Consider the model MATH generated by MATH. By REF , we have MATH. The model MATH is clearly an equivalence model and, since it is connected it is also directed, by REF . Consider now the filtration MATH of MATH through MATH according to REF ; by REF , MATH is an equivalence directed model and it is finite by construction because MATH is a finite set. But MATH is a filtration, and by REF , MATH, which is what we needed to prove. The bound on the size of MATH follows from the observation above. |
cs/9909019 | By REF , to check that MATH is valid, it suffices suffices to check that MATH has no countermodel with at most MATH worlds. |
cs/9909019 | Note that MATH implies that MATH and MATH have the same length. It is immediate from the comments above that MATH for MATH and MATH. It therefore suffices to show that MATH is a trace. We do this by induction on the length of the trace MATH. The base case is straightforward. If MATH is a trace of length one, then it consists of an initial state MATH. Similarly, MATH consists of an initial state MATH. It is immediate from the assumption that MATH that MATH is a trace. Assume that the result has been established for traces of length MATH, and consider traces MATH and MATH of length MATH with MATH. Write MATH where MATH is the final state of MATH and MATH is the next-to-final state of MATH, and similarly write MATH. By the induction hypothesis, MATH is a trace indistinguishable to agent MATH from MATH, and indistinguishable to all other agents from MATH. Let MATH be a joint action enabled at MATH such that MATH, and similarly, let MATH be a joint action enabled at MATH such that MATH. Note that because state transitions record the joint external action component of a joint action in the resulting state, and because MATH, we have MATH. Write MATH for the common joint external action of these states and joint actions. Then we may also write MATH and MATH. To show that MATH is a trace we show that the joint action MATH is enabled at MATH and satisfies MATH. To show that MATH is enabled at MATH we show that each of its components is enabled at MATH. In the case of agents MATH, we need to show that the action MATH of agent MATH is enabled at MATH. This follows, using REF , from the fact that MATH is enabled for agent MATH at MATH, and from the fact that MATH. For agent MATH, we need to show that the action MATH is enabled at MATH. This follows, again using REF , from the fact that MATH is enabled for agent MATH at MATH, and from the fact that MATH. It therefore remains to show that MATH. Note first that MATH. Thus, the states MATH and MATH record the same joint external action. We show that they also have the same private state for each agent. In case of agents MATH, we have MATH . In case of agent MATH, we have MATH . This completes the proof. |
cs/9909019 | CASE: We prove that MATH is connected to MATH if and only if MATH. Left to right is immediate from REF . Right to left follows from REF For each agent MATH, let MATH be any trace of MATH with MATH. To show that the connected component is a hypercube, we prove that there is a MATH such that MATH. In fact, define MATH. By REF , MATH is a trace of MATH, with MATH for all MATH and MATH. It is immediate that all traces MATH of MATH with MATH are connected, and that the component containing MATH is isomorphic to the hypercube MATH. |
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