paper stringlengths 9 16 | proof stringlengths 0 131k |
|---|---|
math/9910177 | CASE: Let MATH be a MATH - cobordism between MATH and MATH. Then MATH is a MATH - cobordism between MATH and MATH. By condition, MATH is simply connected, and MATH. We notice that there is a sequence of surgeries on the manifold MATH (relative to the boundary MATH) so that the resulting manifold is REF - connected (see an argument given in CITE). Let MATH be a trace of this surgery. Then its boundary is decomposed as MATH . We glue together the manifolds MATH and MATH: MATH . Then the boundary of MATH (as a MATH - manifold) is MATH . Now we use REF to ``push" a positive scalar curvature metric from MATH through MATH to MATH keeping it a product metric near the boundary. At this point a psc-metric MATH on MATH may be such that the product metric MATH is not of positive scalar curvature. We find MATH so that the product metric MATH has positive scalar curvature, and then we attach one more cylinder MATH with the metric MATH . We use metric MATH to fit together the metric already constructed on MATH with the metric on MATH. In particular, there is MATH so that the restriction of MATH on MATH has positive scalar curvature (since an isotopy of positive scalar curvature metrics implies concordance). By small perturbation, we can change MATH, so that it has positive scalar curvature and it is a product near the boundary. Then we do surgeries on the interior of MATH to make it REF - connected. Let MATH be the resulting manifold. In particular, MATH. Finally we use ``push" a positive scalar curvature metric from MATH to MATH through MATH keeping it a product metric near the singular stratum MATH. CASE: Let MATH be a simply connected MATH - manifold, with MATH, and MATH. By condition, the singular stratum MATH. Let MATH be a MATH - cobordism between MATH and MATH. In particular, we have MATH. Recall that MATH is embedded to the union MATH together with the colors MATH . By conditions, the manifolds MATH are simply connected, and MATH. As above, there is a surgery on MATH (relative to the boundary MATH) so that a resulting manifold is REF - connected. Let MATH be the trace of this surgery: MATH . We glue together the manifolds MATH to obtain a manifold MATH, where we identify MATH see REF . The resulting manifold MATH (after smoothing corners and extending metric according with the Surgery Theorem construction) is such that MATH is REF - connected cobordism between MATH and MATH. Thus we can ``push" a positive scalar curvature metric from MATH to MATH through the cobordism MATH. Thus we obtain a psc-metric MATH on MATH which is a product near boundary. In general, the product metric MATH on MATH is not of positive scalar curvature. Then we have to attach one more cylinder MATH to ``scale" the metric MATH to a positive scalar curvature metric MATH through an appropriate homotopy. CASE: Then we consider the manifolds MATH and MATH. Again, we perform surgeries on the interior of MATH, MATH to get REF - connected manifolds MATH and MATH. Let MATH, MATH be the traces of these surgeries: MATH . Now we attach the manifolds MATH and MATH to MATH by identifying MATH respectively. Let MATH be the resulting manifold (after an appropriate smoothing and extending a metric), see REF . Notice that MATH is still a MATH - cobordism between MATH and MATH. This procedure combined with an appropriate metric homotopy gives MATH together with a metric MATH on MATH, so that it is a product metric near the boundary, its restriction on MATH has positive scalar curvature, and its restriction on the manifolds MATH are psc-metrics MATH respectively (for some psc-metrics MATH, MATH, MATH). It remains to perform surgeries on the interior of MATH to get a REF - connected manifold, and finally push a psc-metric from MATH to MATH relative to the boundary. REF A proof of REF is similar. |
math/9910178 | We use the correspondence of REF: The set MATH is in bijection with the set of graded comodule morphisms MATH of degree MATH and the differential corresponds to the map MATH . This shows that we have a well-defined complex. The rest follows upon inspection of REF. |
math/9910178 | Define a differential on the graded module MATH by MATH . It is not hard to check that its square vanishes. On the other hand, MATH is endowed with the differential of REF. Now the morphism MATH defines an isomorphism onto a graded submodule and the differential given on MATH is induced by the one on MATH. |
math/9910178 | The claim follows from the equalities MATH . |
math/9910178 | Suppose that REF holds. The square MATH commutes in the homotopy category of complexes of MATH-modules thanks to REF. By assumption, the adjunction morphism MATH is a quasi-isomorphism. The right vertical arrow is the identity (since the MATH-module structure on MATH is induced from that of MATH). So if we apply the homology functor to the diagram, we see that REF holds. Suppose that REF holds. Consider the filtrations MATH and MATH . The morphism MATH, MATH, is compatible with the filtrations. The MATH-term of the spectral sequence associated with MATH is the NAME resolution of the graded MATH-module MATH. By our assumption, this module is unital and thus the NAME resolution is quasi-isomorphic to the module MATH and the map MATH induces a quasi-isomorphism. It follows that MATH induces an isomorphism in the MATH-terms of the spectral sequences. Since the filtrations are bounded below and exhaustive, the spectral sequences converge (by the classical convergence theorem CITE) and MATH is a quasi-isomorphism. |
math/9910187 | and are corollaries of NAME 's horizontal curvature equation. To prove notice that the curvature of MATH is positive if the curvature of its horizontal lift, MATH, is positive. The curvature of MATH is positive if its image, MATH, under MATH has positive curvature, proving . Let MATH be the projection onto the first factor. The curvature of MATH is also positive if its image under MATH is positively curved. If MATH, then this is the case, provided the image of MATH is nondegenerate, proving . |
math/9910187 | Just combine our description of MATH with the observation that if we square the expression in REF we get the reciprocal of the quantity in REF . |
math/9910187 | The proof of is a routine exercise in the definitions which we leave to the reader. Part gives us a commutative diagram MATH of Riemannian submersions from which readily follows. It follows from the diagram that if MATH is horizontal for MATH with respect to MATH, then MATH is horizontal for MATH with respect to MATH and MATH is horizontal for MATH with respect to MATH. This proves the ``only if" part of . The ``if" part of follows from the ``only if" part and the observation that MATH via a dimension counting argument. |
math/9910187 | Throughout the paper we will call the tangent spaces to the orbits of MATH and MATH, MATH and MATH. The orthogonal complement of MATH with respect to the biinvariant metric will be called MATH. According to REF, MATH is diffeomorphic to the pull back of the NAME fibration MATH via MATH, where MATH is the antipodal map and MATH is the NAME fibration that is given by right multiplication by MATH. Moreover, the metric induced on the pull back by the product of two unit MATH's is biinvariant. Through out the paper our computations will be based on perturbations of the biinvariant metric, MATH, induced by MATH, where MATH is the sphere of radius MATH. Observe that if MATH is a killing field on MATH whose length is MATH with respect to the unit metric, then the corresponding killing field on MATH with respect to either MATH or MATH has length MATH with respect to MATH. It follows from this that the quantity REF is constant when we do a NAME perturbation on MATH via either MATH or MATH. Thus the effect of these NAME perturbations is to scale MATH and MATH, and to preserve the splitting MATH and MATH. The amount of the scaling is MATH and converges to MATH as the scale, MATH, on the MATH - factor in MATH converges to MATH and converges to MATH as MATH. We will call the resulting scales on MATH and MATH, MATH and MATH, and call the resulting metric MATH. With this convention the biinvariant metric MATH is MATH. It follows that MATH is the restriction to MATH of the product metric MATH where MATH denotes the metric obtained from MATH by scaling the fibers of MATH by MATH. Since MATH and MATH are by symmetries of MATH in each column, they are by isometries on MATH and hence also on MATH. Let MATH denote the metric obtained from MATH via the NAME perturbation with MATH when the metric on the MATH - factor in MATH is MATH. An argument similar to the one above, using rows instead of columns, shows that MATH is by isometries on MATH. Since MATH commutes with MATH, it follows that MATH acts by isometries on MATH. Doing a NAME perturbation with MATH on MATH produces a metric MATH which can also be thought of as obtained from MATH via a single NAME perturbation with MATH. Since MATH acts by isometries on MATH and commutes with MATH, MATH acts by isometries with respect to MATH. But MATH can also be obtained by first perturbing with MATH and then perturbing with MATH. So repeating the argument of the proceeding paragraph shows that MATH acts by isometries with respect to MATH. |
math/9910187 | It is easy to see that MATH and MATH are free on MATH and hence that MATH and MATH are free on MATH, proving . It was observed in CITE that MATH is diffeomorphic to MATH and in CITE that MATH is the bundle of type MATH. Combining these facts proves . Part is a special case of REF. It follows from that if MATH is in the kernel of MATH, then MATH. On the other hand, MATH is the principal MATH - action for MATH. Combining these facts we see that the kernel has order MATH, and we observed above that MATH is in the kernel. |
math/9910187 | A proof of this for the case of the complex NAME fibration can be found in CITE. The proof for the quaternionic case is nearly identical. is an immediate consequence of the definition MATH. To prove we let a superscript MATH denote the horizontal component with respect to MATH and we compute MATH where for the last equality we have used the fact that MATH is invariant under right quaternionic multiplication. |
math/9910187 | If the projection of MATH onto MATH, MATH and MATH is degenerate, then by replacing the second vector with the appropriate linear combination of the vectors in REF we may assume that MATH has the form MATH where MATH and MATH are multiples of each other. By replacing the first vector in REF by the appropriate vector in MATH we can further assume that MATH, so MATH has the form MATH where MATH and MATH are multiples of each other. It follows that MATH where we have used the results of REF to conclude that many terms are MATH. and follow from REF . |
math/9910187 | If MATH where MATH and MATH, then MATH because the MATH - tensor of MATH is MATH and MATH is the kernel of the MATH - tensor of MATH. |
math/9910187 | The proof is just straight forward computations of inner products. Many of the required computations were done in REF. |
math/9910187 | We will combine REF with the following observations. CASE: Any plane in MATH has a nondegenerate projection (with respect to the splitting MATH) onto MATH. CASE: MATH. It follows from REF that if the projection of MATH onto MATH is nondegenerate, then so is the projection onto MATH (with respect to the splitting MATH). CASE: Any plane in MATH has a nondegenerate projection (with respect to the splitting MATH) onto MATH. CASE: MATH. It follows from REF that if the projection of MATH onto MATH is nondegenerate, then so is the projection of MATH onto MATH (with respect to the splitting MATH). |
math/9910187 | Again the proof is just straight forward computations of inner products. In fact the computations that proved REF will suffice for all of the vectors except MATH and MATH. These are just multiples of the limits of MATH and MATH as MATH. |
math/9910187 | To prove split MATH into MATH where MATH and MATH. To be more concrete suppose that MATH is a multiple of MATH. Suppose MATH has the form MATH for some purely imaginary unit quaternion MATH. Notice that MATH and MATH if MATH. Combining REF , with REF we get that the curvature of MATH is positive except possibly if MATH. On the other hand, the projection of MATH onto MATH is nondegenerate unless MATH or MATH . So MATH is positively curved unless MATH or MATH is a multiple of the same convex combination of MATH as MATH is of MATH. To complete the proof of it remains to consider the case when MATH in which it is easy to see, using REF , that MATH is positively curved unless MATH. We saw in REF that at points where MATH, MATH . Using REF it is easy to see that the only planes in MATH whose projections onto both MATH and MATH are degenerate are those of the form REF . It follows that the only other possible MATH's are of the type described in REF , . Combining this with straightforward computations of MATH - tensors completes the proof of , and even supplies explicit solutions for MATH in terms of MATH and MATH. For example when MATH and MATH, MATH, and when MATH and MATH, MATH. |
math/9910187 | The main idea is that most planes are eliminated because their projection onto the orbits of MATH are nondegenerate. Combining REF with REF we see that it is actually enough to check that a projection is degenerate with respect to MATH. The tangent space to the orbit of MATH is spanned by the MATH - orthogonal basis MATH where MATH and MATH are purely imaginary, unit quaternions that satisfy MATH. Before we can find the projections we also need a formula for MATH along the orbits of MATH. At points of the form REF MATH is MATH . To find the projections we compute MATH . According to REF we should compute the inner products of MATH and MATH with the vectors in REF with respect to MATH. These are the same as the inner products of MATH and MATH with the vectors in REF with respect to MATH, where MATH. We will compute the latter inner products since the notation is simpler. We will only do this explicitly for MATH, since the computations for MATH are the same modulo obvious changes in notation. MATH where MATH stands for a nonzero, but irrelevant term. Combining the previous REF equations we get MATH . Combining REF we get MATH . MATH . Combining REF we get MATH . MATH . Combining REF we see that MATH . Combining REF , and REF we see that the projection of MATH onto MATH is nondegenerate unless MATH or MATH, proving the proposition. |
math/9910187 | Along the orbit of MATH the formula for MATH is MATH . So MATH . Similar computations show that MATH . Combining REF , and REF we see that the projection of MATH onto MATH is nondegenerate unless MATH or MATH, proving the proposition. |
math/9910187 | The point is that if MATH is any other convex combination of MATH, then the projection of the plane in REF onto MATH is nondegenerate. To see this we will need to compute the projection of MATH onto MATH. Although we will only need to know its values for MATH, we will do the computation for arbitrary MATH, since it may be of interest to some readers. Along the orbit of MATH the formula for MATH is MATH . So MATH . Similar computations show that MATH . Combining REF with REF we see that if MATH is any plane of the form of REF and not of the form of either REF or REF , then the projection of MATH onto MATH is nondegenerate and hence MATH is positively curved. |
math/9910187 | To prove we must check that the MATH - tensors of MATH and MATH vanish on these planes. These computations are rather long, but are quite straight forward so we leave them to the reader. Recall CITE that MATH denotes the circle in MATH that is fixed by the action induced from MATH via MATH. It is easy to see that MATH and MATH span a MATH - torus, MATH, in MATH whose tangent plane is horizontal with respect to MATH at every point. Since MATH is horizontal, MATH is a covering map. The order of the covering is at least two because MATH leaves MATH invariant under MATH. We show next that the order is exactly two. Let MATH and MATH be the geodesic circles that are generated by MATH and MATH. Then with respect to the join decomposition MATH, MATH is a radial circle and for MATH, MATH is an intrinsic geodesic in a copy of MATH. Using these observations we see that a point in MATH has only two preimages under MATH from which it follows that MATH is a two fold covering. The fields MATH and MATH are both parallel along MATH. Straightforward computation shows that the results of transporting them from a point MATH to MATH agrees with their images under the differential of the map induced on MATH by MATH via MATH. Therefore MATH is a REF - fold orientation preserving cover, and MATH is a torus rather than a NAME bottle. It follows from that MATH is flat and computations similar to those in the proof of show that MATH is totally geodesic. Combining part with REF , we see that MATH consists of the points with MATH or MATH or MATH. Using the join structure MATH we see that the union of the points with MATH, MATH and MATH is a MATH - sphere in MATH. (Keep in mind that MATH is MATH - ineffective on MATH.) For similar reasons the union of the points with MATH, MATH and MATH is another MATH - sphere. These two three spheres intersect along the common MATH - sphere MATH and their inverse images via MATH are of course diffeomorphic to MATH. Finally notice that every point in this inverse image has zero curvatures because MATH is the quotient map of MATH and MATH acts by isometries. |
math/9910187 | Translating REF on page REF into our classification scheme REF shows that the unit tangent bundle is the bundle of type MATH. We will show via direct computations (similar to those in CITE) that MATH is also the bundle of type MATH. As in CITE we define MATH by MATH and we define explicit bundle charts MATH by MATH and MATH . MATH and MATH are embeddings onto the open sets MATH and MATH . In fact their inverses are given by MATH and MATH . Thus MATH . So MATH is the bundle of type MATH and hence is the unit tangent bundle of MATH. |
math/9910187 | To compute the homology of MATH we decompose it as MATH where the gluing map MATH is given by MATH. From REF and the NAME - NAME theorem it follows that MATH is simply connected. The NAME - NAME sequence for the decomposition REF is MATH . Since MATH, MATH. Also since MATH, MATH. The next step is to compute MATH. Suppose MATH is a set of generators for MATH, and MATH generates MATH. Then using the fact that the map MATH given by MATH has degree MATH we can see that MATH ( To evaluate the first map represent MATH by MATH. ) Put another way, the matrix of MATH with respect to our sets of generators is MATH . From this point a routine algebraic computation shows that MATH . It follows from REF that MATH is injective if MATH, and hence we get using REF that MATH completing the proof of . Part is a corollary of . The known examples of simply connected MATH - manifolds with positive curvature are given in CITE, CITE and CITE. With the exception of MATH and the example, MATH, of NAME none of them are MATH - connected. Since MATH, MATH is not a homotopy sphere. It is not homotopy equivalent MATH since, according to REF, MATH is not a MATH - cohomology sphere. |
math/9910187 | From the long exact homotopy sequence of the fibration MATH we see that MATH and MATH. Thus MATH and MATH. To compute MATH and the cup products we appeal to the NAME sequence MATH of the fiber bundle MATH with integer coefficients. We already know that MATH. Moreover, it follows from the universal coefficient theorem that MATH since MATH has no torsion and MATH is a torsion group. Thus the NAME sequence shows that MATH and hence that MATH has the same integral cohomology groups as MATH . To see that MATH does not have the cohomology ring of MATH we look at the NAME sequence again: MATH . We therefore have MATH . This means that MATH maps the generator MATH for MATH to twice a generator MATH for MATH . If we let MATH we therefore have MATH . Therefore, it suffices to show that MATH in order to show that the cohomology ring of MATH is not that of MATH . We have that MATH is an isomorphism. Thus MATH must also be twice a generator for MATH . Therefore, if MATH we would have that MATH generates MATH . This, however, is impossible. Besides MATH and MATH, there are two examples given in CITE and CITE of simply connected MATH - manifolds with positive curvature. Neither of these has the cohomology modules of MATH so MATH does not have the homotopy type of any known example. |
math/9910191 | The first-named author would like to thank NAME for useful suggestions in regard to REF . He also thanks the Centre NAME en NAME at NAME University for allowing him to use their computing facilities. We are grateful to the referee for pointing out the reference to NAME, and for some interesting remarks concerning the product of two curves we considered. Finally we thank NAME for mentioning the formula for our cusp form in terms of eta-functions. |
math/9910191 | We first determine the torsion subgroup MATH. It is known that the specialization map MATH for any MATH is always injective even when the fiber is a bad fiber (see CITE); here MATH denotes the subgroup of MATH consisting of all points which specialize to smooth points in MATH. Moreover, the notation MATH is used for the group of smooth points in MATH, which is also the connected component of zero of the fiber over MATH in the NAME model of MATH. Considering the table of bad fibers above, one observes that MATH is a subgroup of both MATH and MATH. This implies that MATH is trivial. Since MATH is not the zero section, we conclude that it has infinite order. Also, since the NAME group is in fact a module over MATH, it follows that MATH and MATH are independent. The rank of the NAME group of MATH and the rank of the NAME group are related by the NAME formula, which says that MATH . From the calculation of the bad fibers and the fact that MATH one concludes that the rank of MATH is at most MATH, hence equal to MATH. It follows that the rank of MATH is REF and therefore MATH is a singular MATH surface. Next we compute the canonical height pairing for MATH. This is easily done using CITE, and one finds the matrix MATH . This shows again that MATH and MATH are linearly independent. We use the computation to conclude that MATH generates the NAME group. If the NAME group MATH would be strictly larger than the group generated by MATH and MATH, then MATH must contain an element whose canonical height is less than MATH. However, the a priori lower bound for the canonical height calculated by using the method of CITE is MATH in the present case. Hence MATH and MATH generate MATH. Let us denote by MATH the NAME lattice of MATH in the sense of CITE. The pairing on MATH is given by twice the canonical height pairing; that is, it is given by the matrix MATH . Hence, MATH is isomorphic to MATH, and we have MATH . Since by the NAME index theorem MATH is negative, one concludes that MATH. |
math/9910191 | Consider the family of planes passing through the line MATH. The intersection of MATH and a plane in this family is MATH and a plane cubic curve. This yields another elliptic fibration MATH. Setting MATH and using a standard algorithm we can convert the fiber at the generic point to the NAME form MATH where the transformation is given by MATH . This fibration has six fibers of type IV at MATH. The same fibration can be obtained by starting from the surface MATH given by MATH . Namely, with MATH, using the transformation MATH one finds exactly the same NAME equation. Since MATH is nonsingular and minimal, we conclude that MATH and MATH are isomorphic. A result of CITE shows that MATH corresponds to the matrix MATH in the NAME classification. This proves the assertion. |
math/9910191 | Suppose that a polynomial MATH which is relatively prime to MATH divides MATH. Since MATH does not divide MATH, its cube MATH divides MATH. Thus, we may write MATH (MATH). Since MATH divides MATH, we write MATH. The polynomials MATH and MATH satisfy the equation MATH . The degree of the left hand side is MATH, but this equals MATH except in the case MATH. Since the degree of the right hand side is a multiple of REF, the degree of the left hand side cannot equal MATH. Thus, we have MATH. In other words MATH must be a constant and the only possibilities for this constant are MATH, which lead to the solutions MATH. |
math/9910191 | The function field of MATH over MATH is MATH (with the relations MATH and MATH). The function field of the quotient by MATH is the subfield of all functions invariant under MATH. Over MATH, these invariants are generated by MATH and MATH. The relation between these is MATH, so one finds precisely the function field of MATH over MATH. |
math/9910191 | Consider the surjection MATH given by MATH . This is in fact the quotient map for the group of automorphisms on MATH generated by MATH. This generator can be lifted to the automorphism MATH on MATH. Using REF above now finishes the proof. |
math/9910191 | By what is said above, it suffices to compute the action of a NAME element MATH on a set of generators of the NAME group. Using the elliptic fibration MATH, we study such generators. Firstly, there are the zero section and a fiber. On each of these MATH acts trivially, so they contribute MATH to our trace. Next we consider the sections MATH and MATH. Note that since MATH in the group law, it follows that MATH in case MATH. On the other hand, MATH is fixed under every element of MATH. So it follows that these two sections contribute MATH to the trace. It remains to compute the contributions from the irreducible components of the singular fibers, which do not meet the zero section. These fibers are over MATH, MATH and MATH. Over MATH, the fibre is of type MATH and all its components turn out to be rational. Hence they contribute MATH to our trace. Over MATH one finds fibers of type MATH. Of the components not meeting the zero section, one is always rational. With MATH a coordinate on this one, the other three meet it in points satisfying MATH. Hence two of them are interchanged by MATH and the third one is rational. This means that one obtains MATH from these fibers to the trace. To study the fiber over MATH, one changes coordinates using MATH and MATH. The elliptic fibration MATH is then given by MATH. Over MATH, which corresponds to MATH, one finds a fiber of type IV. The two components not meeting the zero section are interchanged by MATH. Hence they add MATH to the trace. Summing all contributions now proves the proposition. |
quant-ph/9910061 | The generator polynomial of the code MATH can be written as MATH . The dual code MATH has generator polynomial MATH, and its check polynomial is MATH. From MATH and REF we obtain MATH . From REF follows that MATH and thus MATH, in particular, MATH. Hence the number of terms in MATH is even. The degree of MATH is less than MATH, and therefore the degree of MATH is less than MATH, too. This implies that in the summation MATH no terms cancel each other, showing that the number of terms in MATH, denoted by MATH, is divisible by four. On the other hand, from REF , MATH can be written as MATH . Again from REF , we conclude that MATH, and thus the number of terms MATH of MATH is even. Hence for MATH all terms MATH in the summation REF cancel each other. For the remaining MATH terms, two terms cancel each other iff MATH. But then we have also MATH, so in total four terms are cancelled. Hence MATH for some integer MATH. We already know that MATH is divisible by four. Therefore MATH must also be divisible by four which implies that MATH is divisible by four since MATH is odd. From REF follows that MATH is a vector space basis of the code. The weight of each of these vectors is divisible by four. Being a weakly self-dual code, the inner product of any two codewords is zero, that is, the number of common ones is even. This implies that the weight of the sum of two codewords which are doubly-even is again divisible by four. (For the last implication see also CITE.) |
quant-ph/9910124 | Note first that the funtionals MATH and MATH are, as infima over continuous functions, upper semicontinuous. Together with the compactness of the set of admissible MATH this implies that the suprema MATH from REF are attained. In other words: optimal purifier MATH with MATH exist, and we can assume without loss of generality that they are fully symmetric (according to the discussion in REF). Hence we can apply REF and the decomposition REF to get in analogy to REF MATH and MATH . The last two Equations show that we have to optimize each component MATH of the purifier MATH independently. In the one qubit case this is very easy, because we can use REF to get MATH and MATH. Hence maximizing MATH is equivalent to maximizing MATH. But we have according to REF MATH which shows that MATH holds as stated. For the many qubit - test version the proof is slightly more difficult. However as in the MATH-case we can solve the optimization problem for each summand in REF separately. First of all this means that we can assume without loss of generality that MATH takes its values in MATH because the functional MATH which we have to maximize, depends only on this part of the operation. Full symmetry implies in addition that MATH is diagonal in occupation number basis (see REF ), because MATH commutes with each MATH (MATH, MATH) if MATH commutes with MATH. If MATH this means we have MATH where MATH is a positive operator with MATH. Inserting this into REF we see that MATH. Hence we have to maximize MATH. The first step is an upper bound which we get from the fact that MATH is a positive operator. Since MATH (another consequence of full symmetry) we have MATH . Multiplying this Equation with MATH and taking the trace we get MATH . However calculating MATH we see that this upper bound is achieved, in other words MATH maximizes MATH. If MATH holds we have to use slightly different arguments because the estimate REF is to weak in this case. However we can consider in REF the dual MATH instead of MATH and use then similar arguments. In fact for each covariant MATH the quantity MATH is, due to the same reasons as MATH diagonal in the occupation number basis and we get MATH where MATH is again a positive operator with MATH (MATH denotes again the occupation number basis) and MATH is a positive constant. Since MATH is unital we get from MATH the estimate MATH in the same way as REF . Calculating MATH shows again that the upper bound MATH is indeed achieved, however it is now not clear whether maximizing MATH is equivalent to maximizing MATH. Hence let us show first that MATH is necessary for MATH to be maximal. This follows basically from the fact that MATH is, up to a multiplicative constant, trace preserving. In fact we have MATH . This means especially that MATH holds, that is, decreasing MATH by MATH is equivalent to increasing MATH by the same MATH. Taking into account that MATH holds with MATH, we see that reducing MATH by MATH reduces MATH at least by MATH . Therefore MATH is necessary. The last question we have to answer, is how the rest term MATH has to be chosen, for MATH to be maximal. To this end let us consider the slightly modified fidelity MATH (which is in fact related to optimal cloning; see CITE and REF). It is in contrast to MATH maximized iff MATH. However the operation which maximizes MATH is obviously the optimal MATH cloner (up to normalization) which is according to CITE unique. This implies that MATH fixes MATH already. Together with the facts that MATH is necessary for MATH to be maximal and MATH is realized for MATH we conclude that MATH holds, which proves the assertion. |
quant-ph/9910124 | We substitute MATH in the sum, and get MATH where coefficients with MATH are defined to be zero. We can write the non-zero coefficients as MATH . Since MATH, for all MATH, the series for different values of MATH are all dominated by the geometric series, and we can go to the limit termwise, for every MATH separately. In this limit we have MATH for every MATH, and hence MATH. The limit series is again geometric, with quotient MATH and we get the result. |
cond-mat/9911406 | Consider the signed set MATH, MATH, where MATH is the empty set. The positive and negative sets are MATH . Let, MATH, then the sign of MATH is defined as MATH . For the MATH-walks of MATH we do not use the vertex weight of the MATH-walk as a whole, but rather define a ``product" weight, MATH in terms of its individual walks. In particular, MATH . We construct a sign reversing involution, MATH on MATH. Let MATH. The involution is an extension of the NAME involution and splits into three cases, CASE: No intersections (MATH). If none of the walks of MATH intersect, then MATH. CASE: NAME intersections (MATH). If any of the walks MATH has an edge of MATH in common with another walk, MATH then let MATH be the least integer for which a walk MATH shares an edge with another walk MATH. Of all the vertices adjacent to the edges in common with MATH and MATH choose the one with the smallest MATH coordinate and denote it by MATH, then MATH is defined as the MATH-walk obtained by exchanging MATH and MATH at MATH. CASE: Vertex only intersections (MATH). If any of the walks of MATH share vertices and none share edges, then let MATH be the least integer for which a walk MATH intersects another walk MATH. Of all the vertices in common with MATH and MATH choose the one with the smallest MATH coordinate and denote it by MATH, then MATH is defined as the MATH-walk obtained by exchanging MATH and MATH at MATH. Call the vertex, MATH at which the involution exchange takes place, the ``involution" vertex. The difference in the product weight of MATH and MATH is then MATH where MATH is the product weight of MATH with the contribution of the weight associated with MATH divided out. Since MATH is defined as the set of MATH-walks for which the involution vertex MATH exists and arises from the ``vertex only intersections" case of MATH, this means that all the MATH-walks in MATH have walks which only intersect at vertices - there are no shared edges. Define two MATH-walks to be related, MATH iff MATH can be obtained from MATH by the interchange of any number of osculations with switches (or visa versa). This relation is easily seen to be an equivalence relation and hence partitions MATH into disjoint subsets, MATH, MATH where MATH is some index set for the partitions. Define the canonical element, MATH of each partition, MATH, as the MATH-walk for which all the intersections are osculations. Let MATH be the set of vertices of MATH in common with at least two walks of MATH. Note, the cardinality of MATH is MATH. For the MATH-walks in MATH we have, MATH where MATH is the product weight of MATH with all the vertex weights associated with the vertices in MATH divided out. This follows since MATH allows for each vertex in MATH to be a switch (that is, weight MATH) or an osculation (that is, weight MATH). The sign, MATH is correctly obtained since it is just MATH to the number of occurrences of a switch that is, the number of factors of MATH. Thus we have the following: MATH as required. |
cs/9911002 | Let MATH be the minimal automaton of MATH. For MATH, MATH, we introduce the following series of MATH . If MATH, MATH, then we have the following relations MATH . To check relation MATH, one has to compute MATH. Notice that MATH iff MATH . Use REF and treat the case MATH separately. For relations REF, if MATH belongs to MATH then MATH and MATH . In MATH, one observes that MATH. Relation MATH is immediate. Therefore the submodule MATH of MATH finitely generated by the series MATH's, MATH's, MATH's, MATH's, MATH's is stable for the operation MATH, MATH. By associativity of the operation MATH, this module is stable. By CITE, the series of MATH are recognizable. To conclude the proof, notice that MATH . Indeed, if MATH then MATH. |
cs/9911002 | A state of MATH is a derivative of MATH of the form MATH . Since MATH, then MATH. We consider the morphism MATH defined by MATH if MATH for some MATH. We can verify the properties of MATH using the definition of the minimal automaton CITE, CASE: MATH for some MATH, MATH, MATH. So MATH and MATH. CASE: MATH and MATH. CASE: A state MATH belongs to MATH if MATH therefore MATH and MATH. |
cs/9911002 | The condition is sufficient. The support of a recognizable series belonging to MATH is a regular language CITE. The condition is necessary. By REF , one has a morphism MATH where MATH (respectively, MATH) is the minimal automaton of MATH (respectively, MATH). We proceed as in the proof of REF . Let MATH be the set of states of MATH; for MATH, MATH, we introduce the following series MATH . We conclude as in REF . |
cs/9911002 | Assume MATH. Consider the congruence of the semiring MATH defined by MATH. We denote by MATH the finite semiring MATH and by MATH the canonical morphism MATH. The characteristic series of MATH, MATH, is recognizable (see CITE). So MATH is rational (see CITE). Since MATH is finite and MATH is rational, the set MATH is a regular language (see CITE). If MATH and MATH, then consider the series MATH and the set MATH. |
cs/9911002 | This is a direct consequence of REF . |
cs/9911002 | Let MATH. We have by definition of MATH, MATH . It is sufficient to show that if MATH then MATH . Since MATH is strictly increasing, MATH. Since the characteristic sequence of MATH is ultimately periodic, there exists MATH such that MATH. Then MATH. There exists MATH such that MATH. Now, assume that MATH. Therefore MATH and MATH . So we have MATH which is a contradiction and MATH. |
cs/9911002 | We proceed by induction on the degree of MATH. If MATH is a polynomial of degree one then one has MATH with MATH and MATH. Assume that the result holds for polynomials of degree MATH. If MATH is a polynomial of degree MATH, then there exists a polynomial MATH of degree MATH such that MATH, MATH. Therefore MATH and MATH. We can conclude by induction on MATH because MATH. |
cs/9911002 | One can build a polynomial MATH of degree MATH such that MATH and for all MATH, MATH. Indeed, let MATH. The conditions on MATH gives the following triangular system MATH . This polynomial MATH has some useful properties. We have the polynomial identity MATH for MATH. Then it holds for MATH if we extend the definition of MATH to MATH. By REF , MATH. One shows by induction on MATH that MATH (respectively, MATH) is an integer since MATH (respectively, since MATH by REF ). Let MATH, notice that MATH . Indeed, an integer MATH has a representation of length MATH if MATH and MATH . Notice that MATH is a translation of the set MATH of the first words of each length. Therefore MATH is MATH-recognizable, see CITE. Let MATH. Our aim is to show that MATH is not MATH-recognizable. For MATH large enough, we first show that MATH . The first inequality is obvious. In view of REF , to satisfy the second inequality, one must check whether MATH . We can write MATH as MATH with MATH and MATH being a polynomial of degree not exceeding MATH. Then, MATH . The coefficient of MATH is MATH. So, there exists MATH such that for all MATH, this polynomial expression of degree MATH is strictly positive and MATH. If MATH is sufficiently large, we show that MATH . Let MATH. In view of REF , one has to verify that MATH . By definition of MATH and by REF , one has MATH . Therefore it is sufficient to check whether MATH, which occurs if and only if MATH . To verify that this inequality holds, remember that MATH and for MATH, MATH. Thus one studies the quotient MATH when MATH, MATH . So there exists MATH such that for all MATH, MATH. Assume that MATH is regular then the set MATH is a finite union of arithmetic progressions. We may apply REF ; indeed, the function MATH is strictly increasing in MATH and there exist MATH and MATH (simply written MATH) such that MATH. Let MATH be such that MATH. By REF , there exists MATH (simply written MATH) such that for all MATH and for all MATH, MATH . Let MATH. In view of REF , one has MATH . Since MATH must be positive for all MATH, the coefficient of the greatest power of MATH, MATH, must be strictly positive. This coefficient is MATH and we have the condition MATH . Notice that the coefficient vanishes only if MATH. By hypothesis, this case is excluded (notice that MATH). But MATH must be negative for all MATH. The coefficient of the greatest power of MATH is also MATH and must be strictly negative. Then we have simultaneously the condition MATH which leads to a contradiction. |
cs/9911002 | In the proof of REF , we introduced a polynomial MATH such that MATH. In view of REF , we have to find an integer MATH such that for MATH large enough MATH . The coefficient of MATH vanishes in REF . The coefficient of MATH in REF is MATH with MATH. It is strictly increasing with MATH and equals zero for MATH . The same coefficient in REF is MATH. It is strictly decreasing with MATH and equals zero for MATH. If MATH and MATH are not integers then there exists MATH such that the coefficients of terms of maximal degree are both strictly positive. Otherwise, one has to consider the integer case MATH or MATH (it is obvious that any other MATH leads to a strictly negative expression for REF or REF ). Moreover, if MATH (respectively, MATH) then REF (respectively, REF ) is satisfied for MATH large enough. Notice that for MATH the coefficient of MATH in REF with MATH is the opposite of the coefficient of MATH in REF with MATH since MATH. Notice also that the independent term in REF for MATH is MATH. In REF for MATH this term is MATH. Thus we can write REF with MATH as MATH and REF with MATH as MATH . If there exists MATH such that MATH then let MATH. If MATH (respectively, MATH) then one takes MATH (respectively, MATH). Now, assume that MATH for MATH. If MATH then one takes MATH. Otherwise, MATH is a striclty positive integer (remember the properties of MATH obtained in the proof of REF ). Therefore MATH and one takes MATH. |
cs/9911002 | By REF , there exists a subset MATH of MATH such that MATH is MATH-recognizable and MATH is not. Assume that the graph of the addition MATH is regular. Let MATH be the canonical homomorphism defined by MATH. It is clear that the set MATH is regular. Therefore MATH is regular. Thus MATH is also regular, a contradiction. |
cs/9911002 | It is obvious that there exists a word MATH which is MATH-tiered (see CITE), MATH. Let MATH. As shown in CITE, there exists a constant MATH such that the number of words of length MATH is greater than MATH for any integer MATH. |
cs/9911002 | CASE: There exist MATH and a constant MATH such that for all MATH, MATH. If one replaces MATH by a bigger constant then the latter inequality holds for all MATH. For MATH sufficiently large, there exists a constant MATH such that MATH . CASE: With the sequence MATH of REF , one has MATH . Since MATH, then MATH is a linear function of MATH and for MATH large enough, there exists a constant MATH such that MATH . |
cs/9911002 | Assume that for all MATH, MATH only for a finite number of integers MATH. In other words, MATH. By successive applications of NAME 'stheorem, there exist complex numbers MATH and a subsequence MATH such that MATH . Since MATH, then MATH. For MATH, one gets in the same manner MATH . Therefore one has MATH . This equality leads to a contradiction since the NAME determinant does not vanish. |
cs/9911002 | The sequence MATH satisfies a recurrence relation. Therefore, if MATH is a root of multiplicity MATH of the characteristic polynomial of MATH then one can write MATH where MATH is a polynomial of degree less than MATH. Moreover MATH is MATH; in other words, we have a constant MATH such that MATH . This latter inequality has important consequences. CASE: We first show that MATH implies MATH. Otherwise, let MATH and MATH the maximal degree of polynomials MATH corresponding to the different roots of modulus MATH. So we can write MATH . In the last expression, MATH is made up of two sorts of terms, namely MATH where MATH . So MATH if MATH. Therefore, by REF , there exists an infinite sequence of integers such that MATH . For MATH large enough, MATH and MATH occurs infinitely often which contradicts REF . CASE: In the same way, one can verify that if MATH then the degree of the corresponding polynomial MATH cannot exceed MATH. CASE: If we are interested in the behaviour of MATH when MATH, then in REF , we simply focus on the terms of the form MATH for MATH such that MATH. Indeed, any other term in MATH provides MATH with a term which converges to zero (all these terms are included in MATH). So, if we assume that MATH has a multiplicty MATH and if MATH,MATH, MATH are the other roots of modulus one with MATH, MATH and MATH; then one can write MATH with MATH and MATH . Therefore, it is easy to see that MATH . Moreover, we see that MATH has, necessary, a multiplicity MATH; otherwise, MATH, which is a contradiction with REF . |
cs/9911002 | For MATH, we have MATH . To avoid any misunderstanding, MATH is the sequence associated to the language MATH of the numeration MATH and MATH is related to MATH. So, MATH and MATH . We take MATH, a MATH-recognizable set. We have, for MATH sufficiently large, MATH . Indeed, MATH. By REF , there exists MATH such that MATH. So MATH. On the other hand, MATH has an exponential dominant term. Then MATH. For all MATH sufficiently large, there exists a unique MATH such that MATH . Then MATH with MATH and MATH. Notice that as a function of MATH, MATH is increasing and not bounded. To show that MATH if MATH. Assume that MATH is bounded, divide all members of REF by MATH. Let MATH and obtain a contradiction. Suppose that MATH is accepted by an automaton with MATH states. There exist MATH, MATH and MATH such that MATH with MATH and MATH. Then using the pumping lemma, we obtain a contradiction. |
cs/9911002 | We consider words of length at least MATH. Indeed, there is only a finite number of words of length less than MATH and they can be treated separately. Let MATH be a word of MATH of length MATH with MATH. We compute MATH applications of REF on MATH and we obtain MATH . Recall that the notation MATH is written in place of MATH. We will denote by MATH the sum of the last three terms. For all MATH, MATH and MATH, let us define MATH and MATH . With these notations, we can rewrite MATH as MATH . Therefore, using REF , we have MATH . It is obvious that the MATH's take their values in a finite set MATH. Therefore sums of MATH elements of MATH also take their values in a finite set, say MATH. Notice that the MATH's (respectively, the MATH's) are completely determined by the letter MATH (respectively, MATH) and the state MATH reached after the lecture of the first letters of MATH (respectively, the state MATH). Therefore, we extend the notation MATH to a meaningful one: MATH with MATH, MATH and MATH. We are now able to build a finite letter-to-letter MATH-tape automaton MATH over MATH with MATH some finite alphabet. The formula expressing MATH can be interpreted in the following way. The reading of MATH, MATH, provides the decomposition of MATH with MATH; MATH; MATH; MATH. The reading of MATH gives a coefficient MATH for MATH. The other MATH coefficients can be viewed as ``remainders". Roughly speaking, if we have already read the word MATH and if we are reading MATH, then we have to consider the state MATH. (Therefore it seems natural to mimic MATH in MATH.) The coefficients MATH are nothing else but MATH. Thereby we can give a precise definition of MATH. The set of states is MATH where MATH does not belong to MATH and is the unique final state of MATH. The copies of MATH will be used to store the ``remainders". The start state is MATH. The transition relation MATH is defined as follows. If MATH, MATH, MATH . These transitions compute an output MATH from MATH. The alphabet MATH is finite since MATH is finite. But we have still to read the last MATH letters of MATH. For each state MATH, MATH is finite (recall that MATH are the words accepted from MATH). So, for each state MATH and each word MATH, we construct an edge from MATH to MATH labelled by MATH. (This kind of edge can naturally be split in MATH elementary edges using MATH new states.) Indeed, notice that MATH is a constant which only depends on the state MATH reached (the first component in MATH) and the remainding word MATH. |
cs/9911002 | Let the regular language MATH be the graph of the function MATH defined in REF . We denote by MATH and MATH the canonical homomorphisms of projection. Let MATH . If MATH is MATH-recognizable then MATH is regular and MATH. So MATH is MATH-recognizable since MATH is regular. Conversely, if MATH is regular then MATH is also regular. For each MATH, MATH can take more than one value but only one is in MATH. So the set MATH is regular and equal to MATH. |
cs/9911002 | It is well known that for such a system MATH the normalization MATH is computable by finite letter-to-letter MATH-tape automaton for any alphabet MATH (see CITE). So by the previous corollary, MATH is MATH-recognizable if and only if MATH is MATH-recognizable. Another well-known fact related to NAME numeration systems is that a subset MATH is MATH-recognizable if and only if it is definable in the structure MATH (see CITE). In particular, multiplication by a constant MATH is definable in MATH. So MATH is definable in the structure if and only if MATH is definable. |
cs/9911003 | The only property of a tree decomposition that does not follow immediately is the requirement that each edge connect two vertices contained in the label of some node. Since this is true of MATH and MATH, any induced subgraph edge MATH must have MATH for some MATH, but MATH may not be a descendant of MATH. However, if not, MATH belongs to both MATH and (by assumption) MATH where MATH is a descendant of MATH. Therefore, by contiguity, MATH, and similarly MATH, so in this case MATH still both belong to the label of at least one node in the subtree. |
cs/9911003 | We perform dynamic programming in tree MATH. Let a partial isomorph at a node MATH of the tree be an isomorphism between an induced subgraph MATH of the pattern MATH and the induced subgraph of MATH associated with the subtree rooted at MATH. We let MATH be formed by adding two additional vertices MATH, MATH to the subgraph of MATH induced by vertex set MATH. We connect each of the two additional vertices to all vertices in MATH, and each of the two additional vertices also is given a self-loop. Then from any partial isomorph at MATH we can derive a graph homomorphism from all of MATH to MATH, which is one-to-one on vertices in MATH, maps the rest of MATH to MATH, and maps MATH to MATH. Let a partial isomorph boundary be such a map; REF illustrates a partial isomorph and the corresponding boundary. Since a partial isomorph boundary consists of a map from a set of at most MATH objects to a set of at most MATH objects, there are at most MATH possible partial isomorph boundaries for a given node. Suppose that node MATH has children MATH and MATH . We say that two partial isomorph boundaries MATH and MATH are consistent if the following conditions all hold: CASE: For each vertex MATH, if MATH or MATH, then MATH. CASE: For each vertex MATH, if MATH then MATH. CASE: At least one vertex MATH has MATH. We say that two partial isomorph boundaries MATH and MATH form a compatible triple with MATH if the following conditions both hold: CASE: MATH and MATH are both consistent with MATH. CASE: For each MATH with MATH, exactly one of MATH and MATH is equal to MATH. For each partial isomorph boundary MATH, let MATH be the number of partial isomorphs which give rise to that boundary, and include a vertex of MATH. Let MATH be the number of partial isomorphs which give rise to that boundary, and do not include a vertex of MATH. These values can be computed in a bottom-up fashion as follows: CASE: If there is no MATH for which MATH, then all partial isomorphs having boundary MATH involve only vertices in MATH, and can be enumerated by brute force in time MATH. CASE: Otherwise, we initialize MATH and MATH to zero. Then, for each partial boundary MATH that is consistent with MATH, and such that there is no MATH with MATH and MATH, we increment MATH by MATH and increment MATH by MATH. Finally, for each compatible triple MATH we increment MATH by MATH and increment MATH by MATH. The total time for testing all triples for compatibility and performing the above computation is MATH. At the root node of the tree, we compute the number of isomorphs involving MATH simply by summing the values MATH over all partial isomorph boundaries for which MATH for all MATH. |
cs/9911003 | We first follow the above dynamic programming procedure, to compute the values MATH and MATH for each partial isomorph boundary. We then compute top-down in the tree the set of pairs MATH where MATH is a partial isomorph boundary and MATH is either MATH or MATH, such that the value MATH contributes to the final count of subgraph isomorphs. These pairs can be identified as the ones such that MATH was included in the computation of some pair higher in the tree that has been previously identified as contributing to the total, and that caused a nonzero increment in this computation. Finally, we compute bottom-up again, listing for each contributing pair MATH the partial subgraph isomorphs counted in the value MATH. This step can be performed by mimicking the initial computation of MATH described in the previous lemma, restricted to the boundaries known to contribute to the overall total, replacing each increment by a concatenation of lists, and replacing each multiplication with the construction of partial isomorphs from a Cartesian product of two previously-computed lists. The number of steps for this computation is proportional to the number of steps in the previous algorithm, together with the added time for each combination of a pair of partial isomorphs. Each such combination can be charged to a subgraph isomorph included in the output, and each output isomorph is formed by a binary tree of combinations that takes MATH time to perform, so the total added time is MATH. |
cs/9911003 | Without loss of generality (by adding edges if necessary) we can assume MATH is embedded in the plane with all faces triangles (including the outer face). Form a tree with one node per triangle, and an edge connecting any two nodes whenever the corresponding triangles share an edge that is not in MATH REF . Label each node with the set of vertices on the paths connecting each corner of the triangle to the root of the tree. Then each edge's endpoints are part of some label set (namely, the sets of the two triangles containing the edge), and the labels containing any vertex form a contiguous subtree (namely, the path of triangles connecting the two triangles containing the edge from the vertex to its parent, and any other triangles enclosed in the embedding by this path). Therefore, this gives us a tree decomposition of MATH. The number of nodes in any label set is at most MATH, so the width of the decomposition is at most MATH. |
cs/9911003 | We choose an arbitrary starting vertex MATH, and let MATH consist of the vertices at distance MATH from MATH. We then let MATH be the graph induced by the vertex set MATH, as shown in REF . Clearly, the sets MATH form a partition of the vertices of MATH, and each vertex is in at most MATH subgraphs MATH. Then for MATH, MATH consists of the vertices at distance at most MATH from MATH, so by applying REF to its breadth first spanning tree we can find a tree decomposition with width at most MATH. To show that each MATH with MATH has low tree-width, form an auxiliary graph MATH from MATH by collapsing into a single supervertex all the vertices at distance less than MATH from MATH, and deleting all the vertices with distance at least MATH. MATH is a minor of the planar graph MATH and is therefore also planar. Then similarly collapsing a breadth first spanning tree of MATH gives a spanning tree of MATH with depth at most MATH, so MATH has a tree decomposition with width at most MATH, in which each node of the decomposition includes the collapsed supervertex in its label. MATH is formed by deleting this supervertex from MATH, so we can form a tree decomposition of MATH with width at most MATH by removing the supervertex from the decomposition of MATH. Next, we need to show that any connected subgraph MATH of MATH with MATH is contained in MATH, where MATH is the smallest value such that MATH. But MATH is formed from MATH simply by removing the sets MATH where MATH or MATH. No MATH with MATH can contain a vertex of MATH, or else MATH would have been smaller. And no MATH with MATH can contain a vertex MATH of MATH, or else we could find a path of length at most MATH from MATH to MATH by concatenating a path in MATH from some vertex MATH to MATH (which has length at most MATH) with the breadth first tree path from MATH to MATH (which has length MATH), contradicting the placement of MATH in MATH. Therefore, none of the vertices that were deleted from MATH can belong to MATH, so MATH remains a subgraph of MATH. Finally, the condition that MATH can not be a subgraph for MATH where MATH is clearly true, since no such MATH can include any vertex of MATH. The time bound is dominated by the time to perform the tree decompositions on the graphs MATH, which by REF is MATH. |
cs/9911003 | The algorithm consists of the following steps: CASE: Apply the method of REF to find a partition of the vertices into sets MATH associated with graphs MATH having low width tree decompositions. CASE: For each MATH, count or list the subgraph isomorphs of MATH in MATH that involve at least one vertex of MATH, using the algorithm of REF or REF respectively. CASE: Sum all the counts or concatenate the lists, to get a count or list of the isomorphs in MATH. By REF , each isomorph of MATH in MATH occurs in exactly one way as an isomorph in MATH that involves at least one vertex of MATH, so the algorithm produces the correct total count or list. The time for the first step is MATH, and the time for the last step is MATH, both dominated by the time for the second step which is MATH, where the MATH factor arises by plugging the MATH treewidth bound of REF into the analysis in REF . This can be simplified to MATH. |
cs/9911003 | Let MATH denote the number of isomorphs of MATH in MATH. Rather than counting the isomorphs of a single pattern, we count the isomorphs of all planar graphs having at most MATH vertices. There are only MATH such graphs CITE, so this factor does not change the overall form of our time bound. We order these graphs by the number of connected components, so that when we are processing a particular graph MATH we can assume we already know the values of MATH for every MATH with fewer components. Our algorithm them performs the following steps on each graph MATH: CASE: If MATH is connected, compute MATH using the algorithm of REF . CASE: Otherwise, let MATH be the disjoint union of two subgraphs MATH and MATH, and let MATH, where the sum is over all graphs MATH with fewer components than MATH, and MATH denotes the number of different ways MATH can be formed as the union of MATH and MATH. The product MATH counts the number of ways of mapping MATH into MATH such that both MATH and MATH are isomorphically mapped but their instances may overlap. The term MATH corrects for these overlaps by subtracting the number of overlapped occurrences of each possible type. The coefficients MATH may be computed by brute force enumeration of all possible ways of marking a vertex of MATH as coming from MATH, MATH, or both, combined with a planar graph isomorphism algorithm, in time MATH. Therefore, the overall time taken in the second step of the algorithm is MATH, independent of MATH, and the total time is dominated by the first step, in which we apply REF to MATH connected graphs, taking time MATH. |
cs/9911003 | The map MATH can be defined by specifying which (if any) vertex of MATH maps to each vertex in MATH (using MATH bits per vertex of MATH to specify this information) and also specifying which of the remaining vertices in MATH map to the vertex MATH in MATH and which ones map to the vertex MATH. However, it is not possible for the boundary to come from an actual subgraph isomorphism unless each connected component of MATH is mapped consistently either to MATH or to MATH, since any path from MATH to MATH must pass through a vertex of MATH. So, to finish specifying the boundary, we need only add this single bit of information per component of MATH, and by definition there are at most MATH such components. |
cs/9911003 | For Hamiltonian graphs and bounded degree graphs this is straightforward. For REF-connected graphs, assume without loss of generality that no edge can be added to MATH connecting two vertices in MATH; then each component of MATH must occupy a distinct face in the planar embedding of MATH induced by the unique embedding of MATH. |
cs/9911003 | The proof consists simply of plugging the improved analysis of REF into the algorithm of REF . |
cs/9911003 | For each vertex MATH, we count the number of cycles of length MATH in the neighbors of MATH. The sum of the sizes of all neighborhoods in MATH is MATH. Each neighborhood has treewidth at most REF by REF . Any partial isomorphism of a MATH-cycle in a node MATH of this decomposition can only consist of a single path of at most MATH vertices, which starts and ends at some two of the at most three vertices in MATH and may or may not involve the third vertex; therefore we need only keep track of MATH different partial isomorph boundaries at each node. A careful analysis of the steps in the algorithm of REF then shows that the most expensive step (finding compatible triples) can be performed in time MATH per node, giving an overall running time of MATH. |
cs/9911003 | We test for each integer MATH whether there is a cycle (or nonfacial cycle) of length MATH. To test if there is a nonfacial cycle, we count the total number of cycles of length MATH in the graph and subtract the number of faces of length MATH. The total time for this procedure is MATH. For the nonfacial cycle problem, once the length of the cycle is known, we can find a single such cycle by performing our subgraph isomorph listing algorithm, stopping once MATH cycles are generated, where MATH is the number of faces of the given length. By radix sorting the list of cycles (in lexicographic order by their sequences of vertex indices) we can then test in linear time which of the generated cycles are nonfacial. |
cs/9911003 | We describe how to test for the existence of a separating cycle of length MATH; the shortest such cycle can then be found by a sequential search similar to the computation of the girth. We first modify the construction of the graphs MATH in REF , by including in MATH not only all the vertices in layers MATH through MATH but also a single supervertex for each connected component of the graph induced by the vertices in layers MATH, MATH, and a supervertex for the (single) connected component of the vertices in layers MATH, MATH. Then a cycle that uses only vertices in layers MATH through MATH (and does not use any of the supervertices) is separating in the modified MATH if and only if the corresponding cycle is separating in MATH. Note that the added supervertices only add one level to the breadth first search tree of MATH and hence the tree-width is still MATH. Then we need merely modify the dynamic program of REF , to use a definition of a partial isomorph boundary that, in addition to the map MATH, specifies which of the remaining unmapped vertices of MATH are in each of the two subsets of MATH separated by the cycle, enforcing the requirement that no vertex in one subset be adjacent to a vertex in the other subset. This modification multiplies the number of boundaries by MATH, but this increase is swamped by the MATH terms from our previous analysis. |
cs/9911003 | We simply to test subgraph isomorphism for all possible planar graphs on MATH vertices, and return the subgraph isomorph with the most edges. |
cs/9911003 | Let MATH be a connected component of MATH, let MATH be formed from MATH by contracting MATH into a supervertex, and let MATH be the set of faces and vertices adjacent to the contracted supervertex. Then (since it is just the neighborhood of a vertex) MATH has the structure of a cycle in MATH, and separates MATH from MATH. If MATH is minimal, then it must consist of exactly the original vertices in MATH. The converse is immediate, since no edge in the embedding of MATH can cross a face or vertex in MATH. |
cs/9911003 | We can assume without loss of generality that MATH is two-connected, so the graph MATH described in REF has no multiple adjacencies. We form MATH as above and find the shortest cycle in MATH that separates two original vertices. As with the shortest separating cycle problem REF this can be done by a slight modification to our dynamic programming method that decorates the dynamic programming states with MATH bits of additional information regarding the separated vertices. Since any planar graph has a vertex of degree at most five by NAME 's formula, the shortest cycle in MATH must have length at most ten, so the algorithm takes time MATH. |
cs/9911003 | Assume without loss of generality that MATH is two-edge-connected. Embed MATH, and form a graph MATH by subdividing each edge of MATH and connecting the resulting subdivision points to new vertices in each adjacent face; this construction is illustrated in the right of REF . Then MATH is a planar graph with MATH original vertices, MATH edge-vertices on each edge of MATH, and MATH face-vertices in each face of MATH, so its total complexity is MATH. By the assumption of two-edge-connectivity, MATH is simple. One can use an argument similar to the one in REF , in which we delete a cutset from MATH, contract a connected component, and examine the neighborhood in MATH of the contracted supervertex, to show that MATH is MATH-edge-connected iff there is no cycle of fewer than MATH edge- and face-vertices in MATH, which separates two original vertices. As before, the degree bound on a planar graph imposes a limit of ten on the length of the shortest such cycle, and as before this cycle can be found by a minor modification to our dynamic programming algorithm. |
cs/9911003 | Without loss of generality, the multiplicity of any edge is at most MATH, as higher multiplicities can not improve the overall connectivity. After the edge subdivision step in the construction of MATH, the resulting graph is a simple planar graph with MATH vertices, after which we can proceed as in the remainder of REF . |
cs/9911003 | As in REF , we compute the distances of each vertex from some arbitrary starting vertex MATH. We then let MATH consist of those vertices with distance at least MATH and at most MATH from MATH, and we let MATH be the graph induced by the set of vertices with distances at least MATH and at most MATH from MATH. The proof that these graphs have treewidth MATH and that each MATH contains the MATH-neighborhood of MATH is essentially the same as that of REF . |
cs/9911003 | By performing the decomposition of REF we can assume without loss of generality that we have a tree decomposition MATH for MATH of width MATH. As with any tree, we can find a node MATH the removal of which disconnects MATH into subtrees of size at most MATH. Our primary data structure consists of the distances MATH from each vertex MATH to each vertex MATH, together with a recursively constructed data structure in each subtree. To answer a query pair MATH where the two vertices belong to different subtrees of MATH, we can simply try each of the MATH values MATH where MATH ranges over all the members of MATH. To answer a query where the two vertices belong to the same subtree, we can use the recursively defined structure in that subtree. It remains to show how we quickly determine which node MATH is eventually used to answer each query. To do this, define the level of a node to be the the recursive subdivision process at which the node was chosen, and define the superior of a node MATH to be the node chosen at the next earlier level in the subtree containing MATH. The links from a node to its superior define a tree structure MATH different from the original decomposition tree MATH. Further, define the home node of a vertex MATH to be the node with the earliest level with MATH. Note that, because of the requirements that the labels containing MATH are contiguous, the home node is uniquely defined. Then, the node to be used in answering a query pair MATH is simply the least common ancestor in MATH of the home nodes of MATH and MATH. |
cs/9911003 | As above, we can assume MATH has a tree decomposition of width MATH, which we assume has the form of a rooted binary tree. Define levels in this tree and home nodes of vertices as above, except that we terminate the recursive subdivision process when we reach subtrees with fewer than MATH nodes (which we call small subtrees). If the node labels containing a vertex MATH belong only to nodes in a small subtree MATH, then MATH does not have a home node, instead we say that MATH is MATH's home subtree. Define a pair of nodes in MATH to be related if there is no node between them with an earlier level than both. Then, each node MATH is related to MATH nodes at each level: one node at an earlier level than MATH, and at most one node in each later level in each of the at most three subtrees formed by removing MATH from MATH. Therefore, there are MATH pairs of related nodes. Our data structure then consists of the matrix of distances from vertices in MATH to vertices in MATH, for each pair MATH of related nodes. The space for this data structure is MATH. It can either be built as a subset of the data structure of REF , in time MATH, or bottom-up (using hierarchical clustering techniques of CITE to construct the level structure in MATH, and then computing each distance matrix from two previously-computed distance matrices in time MATH) in total time MATH; we omit the details. To answer a query, we form chains of related pairs connecting the home nodes (or small subtrees) of the query vertices to their common ancestor in MATH. The levels of the nodes in these two chains becomes earlier at each step towards the common ancestor, so the total number of pairs in the chain is MATH. We then build a graph, in which we include an edge between each pair of vertices in the labels of a related pair of nodes, labeled with the distance stored in the matrix for that pair. We also include in that graph the edges of MATH belonging to the small subtrees containing the query vertices, if they belong to small subtrees. The query can then be answered by finding a shortest path in this graph, which has MATH vertices and MATH edges. |
cs/9911003 | We begin by performing a breadth first search from an arbitrary vertex. This will produce a tree of height at most MATH, so by REF we can find a tree decomposition of width MATH, which as usual we can assume has the form of a rooted binary tree. We first perform a bottom-up sweep of this tree to compute for every node MATH the distances between every pair of vertices in MATH, in the graph associated with the subtree rooted at MATH. These MATH distances can be found by combining the distance matrices of the two children of MATH, in time MATH, so this phase takes time MATH. We then sweep the decomposition top down, computing for every node MATH the distances between every pair of vertices in MATH, in the whole graph MATH. The first pass correctly computed these distances at the root of the tree, and at any other node MATH the distances can be computed by combining the distance matrices of the parent of MATH (previously computed in the top-down sweep) and its two children (computed in the bottom-up sweep), again in time MATH per node. We finally sweep through the tree decomposition bottom up, keeping at each node MATH a subset MATH of the vertices seen so far in the subtree rooted at MATH, together with the distances from each member of MATH to each member of MATH. When we process a node MATH, we perform the following steps: CASE: Let the set MATH for node MATH consist of the union of the corresponding sets MATH and MATH for its children MATH and MATH, together with MATH. CASE: Compute the distances from each member of MATH to each member of MATH, by combining the previously computed distances to MATH or MATH with the distances within MATH. CASE: Compute the distance between each pair of nodes MATH where MATH and MATH, by testing the distances through all MATH possible intermediate nodes in MATH. CASE: Radix sort the members of MATH according to the lexicographic ordering of their MATH-tuples of distances to MATH, and eliminate all but one member for each distinct tuple. The value returned as the diameter is then the maximum of the distances from MATH to MATH computed in the second step, and the distances of pairs MATH comuted in the third REF any eliminated member MATH of a tuple belongs to a diametral pair MATH, where MATH is not in the subgraph associated with MATH, then the uneliminated member MATH with the same tuple would have the same distance to MATH, and would form another diametral pair. Therefore, the algorithm above will correctly find and report a diametral pair. The number of distinct MATH-tuples of integers in the range from MATH to MATH is MATH, hence this gives a bound on the size of each set MATH. The time to compute distances between pairs MATH is MATH times the square of this quantity, which is still MATH. |
cs/9911014 | Clearly, MATH satisfiability is in NP (and thus NAME), since every satisfiable formula is satisfiable on a linear frame with MATH worlds, where MATH is the modal depth of MATH. This immediately gives the following NP algorithm for MATH satisfiability: Guess a linear frame of size MATH, and for every world in the frame, guess a valuation on the propositional variables that occur in MATH. Accept if and only if the guessed model satisfies MATH. It is easy to see that the following polynomial-time algorithm decides poor man's satisfiability with respect to MATH. Let MATH, where the MATH-s are literals. MATH is MATH satisfiable if and only if CASE: MATH is satisfiable (that is, for all MATH and MATH, MATH), and CASE: MATH, (that is, MATH does not contain conjuncts of the form MATH, in which case the formula is satisfied in a world with no successors), or CASE: MATH is MATH satisfiable (the world has exactly one successor). |
cs/9911014 | Because the frame is finite, both satisfiability problems are in NP. Thus it suffices to show that poor man's satisfiability with respect to is NAME. Since we are working with a fragment of propositional modal logic, it is extremely tempting to try to reduce from an NAME propositional satisfiability problem. However, because poor man's logics contain only a fragment of propositional logic, these logics don't behave like propositional logic at all. Because of this, propositional satisfiability problems are not the best choice of problems to reduce from. In fact, they are particularly confusing. It turns out that it is much easier to reduce a partitioning problem to our poor man's satisfiability problem. We will reduce from the following well-known NAME problem. GRAPH REF-COLORABILITY: Given an undirected graph MATH, can you color every vertex of the graph using only three colors in such a way that vertices connected by an edge have different colors? Suppose MATH where MATH. We introduce a propositional variable MATH for every edge MATH. The three leaves of will correspond to the three colors. To ensure that adjacent vertices in the graph end up in different leaves, we will make sure that the smaller endpoint of MATH satisfies MATH and that the larger endpoint of MATH satisfies MATH. The requirements for vertex MATH are given by the following formula: MATH . Define MATH. MATH is clearly computable in polynomial time. To show that MATH is indeed a reduction from GRAPH REF-COLORABILITY to poor man's satisfiability with respect to , first note that it is easy to see that for every set MATH, the following holds: MATH is satisfiable if and only if no two vertices in MATH are connected by an edge. It follows that MATH is satisfiable on if and only if there exist sets of vertices MATH, and MATH such that MATH and MATH is satisfiable for MATH. This holds if and only if there exist sets of vertices MATH, and MATH such that MATH and no two vertices in MATH are adjacent for MATH, which is the case if and only if MATH is REF-colorable. (We obtain a coloring by coloring each vertex MATH by the smallest MATH such that MATH.) |
cs/9911014 | The proof that satisfiability with respect to MATH is NAME is very close to the proof that MATH satisfiability is CITE and therefore omitted. For the poor man's satisfiability problem, note that a simplified version of Ladner's PSPACE upper bound construction for MATH can be used to show the following. Let MATH, where the MATH-s are literals. MATH is MATH satisfiable if and only if CASE: MATH is satisfiable, CASE: for all MATH, MATH is MATH satisfiable, and CASE: MATH is MATH satisfiable. (only relevant when MATH.) Note that this algorithm takes exponential time and polynomial space. Of course, we already know that poor man's satisfiability with respect to MATH is in PSPACE, since satisfiability with respect to MATH is in PSPACE. How can this PSPACE algorithm help to prove that poor man's satisfiability with respect to MATH is in REF Something really surprising happens here. We will prove that for every poor man's formula MATH, MATH is MATH satisfiable if and only if (the conjunction of) every pair of (not necessary different) conjuncts of MATH is MATH satisfiable. Using dynamic programming, we can compute all pairs of subformulas of MATH that are MATH satisfiable in polynomial time. This proves the theorem. It remains to show that for every poor man's formula MATH, MATH is MATH satisfiable if and only if every pair of conjuncts of MATH is MATH satisfiable. We will prove this claim by induction on MATH, the modal depth of MATH. In the proof, we will write ``satisfiable" for ``satisfiable with respect to MATH." If MATH, MATH is a conjunction of literals. In that case MATH is not satisfiable if and only if there exist MATH and MATH such that MATH. This immediately implies our claim. For the induction step, suppose MATH (where the MATH-s are literals), MATH, and suppose that our claim holds for all formulas of modal depth MATH. Suppose for a contradiction that MATH is not satisfiable, though every pair of conjuncts of MATH is satisfiable. Then, by the NAME construction given above, we are in one of the following three cases: CASE: MATH is not satisfiable, CASE: for some MATH, MATH is not satisfiable, or CASE: MATH is not satisfiable. By induction, it follows immediately that we are in one of the following four cases: CASE: There exist MATH such that MATH is not satisfiable, CASE: there exist MATH such that MATH is not satisfiable, CASE: there exist MATH such that MATH is not satisfiable, or CASE: there exists a MATH such that MATH is not satisfiable. If we are in REF is not satisfiable. In REF is not satisfiable. In REF is not satisfiable. So in each case we have found a pair of conjuncts of MATH that is not satisfiable, which contradicts the assumption. |
cs/9911014 | Satisfiability with respect to MATH is NAME by pretty much the same proof as the NAME proof for MATH CITE. To show that the poor man's version remains NAME, first note that a formula is MATH satisfiable if and only if it is satisfiable in the root of a binary tree. CITE showed that the set of true quantified REFCNF formulas is NAME. Using padding, it is immediate that the following variation of this set is also NAME. QUANTIFIED REFSAT: Given a quantified Boolean formula MATH, where MATH is a propositional formula over MATH in REFCNF (that is, a formula in conjunctive normal form with exactly REF literals per clause), is the formula true? We will reduce QUANTIFIED REFSAT to poor man's satisfiability with respect to binary trees. To simulate the quantifiers, we need to go back to the formula that forces models to be of exponential size. MATH . MATH is clearly satisfiable in the root of a binary tree and if MATH is satisfied in the root of a binary tree, the worlds of depth MATH form a complete binary tree of depth MATH and every assignment to MATH occurs exactly once in a world at depth MATH. We will call the worlds at depth MATH the assignment-worlds. The assignment-worlds in a subtree rooted at a world at distance MATH from the root are constant with respect to the value of MATH. It follows that MATH QUANTIFIED REFSAT if and only if MATH is satisfiable with respect to binary trees. This proves that satisfiability for MATH is NAME, but it does not prove that the poor man's version is NAME. Recall that MATH is in REFCNF and thus not a poor man's formula. Below, we will show how to label all assignment-worlds where MATH does not hold by MATH (for MATH). It then suffices to add the conjunct MATH to obtain a reduction. How can we label all assignment-worlds where MATH does not hold by MATH? Let MATH be such that MATH, where each MATH is the disjunction of exactly REF literals: MATH. We assume without loss of generality that MATH is even and that each MATH contains REF different propositional variables. For every MATH, we will label all assignment-worlds where MATH does not hold by MATH. Since MATH, this implies that we have to label all assignment-worlds where MATH holds by MATH. In general, this cannot be done in poor man's logic, but in this special case we are able to do it, because the relevant part of the model is completely fixed by MATH. As a warm-up, first consider how you would label all assignment-worlds where MATH holds by MATH. This is easy; add the conjunct MATH . NAME can label all assignment-worlds where MATH holds as follows: MATH . This can easily be generalized to a labeling for MATH: MATH . Note that we can write the previous formula in the following suggestive way: MATH . In general, suppose you want to label all assignment-worlds where MATH hold by MATH, where MATH, MATH, and MATH are literals. Suppose that MATH, MATH, and MATH's propositional variables are MATH, MATH, and MATH, respectively. Also suppose that MATH. The labeling formula MATH is defined as follows. MATH . If MATH is satisfied in the root of a complete binary tree, then there exist at least MATH worlds at depth MATH such that MATH holds. If MATH is satisfied in the root of a binary tree, then the worlds of depth MATH form a complete binary tree and there are exactly MATH assignment-worlds such that MATH holds. It follows that if MATH is satisfied in the root of a binary tree, then MATH is satisfied in the root if and only if MATH holds in every assignment-world where MATH holds. Thus, the following function MATH is a reduction from QUANTIFIED REFSAT to poor man's satisfiability with respect to MATH. MATH . |
cs/9911014 | The reduction from poor man's satisfiability with respect to MATH is obvious. It suffices to use the number restriction MATH to make sure that every world in the relevant part of the model has at most two successors. Let MATH be the modal depth of MATH. All worlds that are of importance to the satisfiability of MATH are at most MATH steps away from the root. The reduction is as follows: MATH . |
cs/9911014 | It is easy to see that these five sets are exactly the minimal complete bases for modal logic. All these satisfiability problems are polynomial-time equivalent, and their NAME follows immediately from Ladner CITE. |
cs/9911014 | First note that all these cases are clearly NAME, because they contain propositional satisfiability. It remains to show that MATH satisfiability is in NP if MATH, MATH, or MATH. The first case is exactly propositional satisfiability and thus in NP. For the second case, a MATH formula is satisfiable if and only if substituting MATH for every outermost MATH subformula gives a propositionally satisfiable formula. For the last case, the following NP algorithm decides MATH satisfiability. Given a formula MATH, guess a valuation on all the subformulas of MATH and accept if and only if this valuation makes MATH true, the valuation is propositionally consistent, and for all MATH that are set to true, MATH is satisfiable. It is crucial that we do not have to verify anything if MATH is set to false in the valuation, because MATH can only occur positively. |
cs/9911014 | As mentioned in the introduction, the MATH case follows from CITE. It is easy to see that MATH satisfiability with respect to any class of frames MATH is polynomial-time reducible to MATH satisfiability with respect to the same class of frames: Introduce two new variables MATH and MATH to simulate MATH and MATH, respectively, and define reduction MATH as follows. MATH . This completes the proof of REF , that is, poor man's logic with or without constants. It remains to show that MATH satisfiability is coNP-complete. It follows from careful inspection of the proof of CITE that satisfiability for variable-free MATH formulas is coNP-hard. Let MATH be a MATH formula. It is easy to see that MATH is satisfiable if and only if the MATH formula MATH is satisfiable. |
cs/9911014 | CASE: Every MATH formula can in polynomial time be transformed into an equivalent MATH formula, by moving the negations inward. Since all operators are unary, every MATH formula is of the form MATH, where MATH is a literal or a constant. It is easy to see that the unsatisfiable formulas of this form are exactly the formulas of the form MATH. CASE: MATH formulas are of the form MATH, where the MATH-s are literals or constants, and the MATH-s and MATH-s are MATH formulas. If MATH or if one of the MATH-s is a literal or MATH, this formula is satisfiable. Otherwise, the formula is satisfiable if and only if one of the MATH-s is satisfiable. This gives a recursive polynomial-time algorithm. CASE: MATH formulas are of the form MATH, where the MATH-s are literals or constants. This formula is satisfiable if and only if MATH is satisfiable. CASE: MATH formulas are of the form: MATH, where the MATH-s are literals or constants. This formula is satisfiable if and only if MATH is satisfiable and for every MATH, MATH is satisfiable. This is gives a recursive polynomial-time algorithm. CASE: A MATH formula is satisfiable if and only if substituting MATH for every propositional variable and for every outermost MATH subformula makes the formula true. CASE: Satisfiability for MATH formulas can be recursively computed as follows. Replace every outermost MATH subformula by MATH if MATH is satisfiable and by MATH if MATH is not satisfiable and replace all propositional variables by MATH. MATH is satisfiable if and only if the resulting propositional sentence evaluates to true. CASE: Every MATH formula is satisfiable. |
cs/9911014 | We will use the following theorem. Satisfiability for modal formulas with one propositional variable is NAME. We will reduce satisfiability for modal formulas with one propositional variable to satisfiability for modal formulas with zero propositional variables. The reduction has the same flavor as the proof of REF from CITE, in that we will encode the truth of the propositional variable by the frame. Let MATH be a modal formula with one propositional variable. We assume that MATH is built from the sole propositional variable MATH, and operations MATH. The main idea of the reduction is the following. Suppose MATH is satisfiable. Then MATH is satisfiable on a model MATH in which every path has length MATH. We extend MATH in such a way that the assignment to MATH is encoded in the frame. Define MATH as follows. CASE: MATH. CASE: MATH. CASE: MATH is irrelevant, since there are no propositional variables. Thus, MATH contains all the information of MATH, and there is a maximal path of length MATH from a world MATH in MATH if and only if MATH. We will simulate MATH by REF. Suppose MATH for an arbitrary subformula MATH of MATH. In the reduction, we need to make sure that we do not force MATH to be true on the new world MATH. So, we will enforce that MATH holds in all successor worlds that do not have a maximal path of length MATH. Let MATH, where MATH is defined inductively as follows. CASE: MATH CASE: MATH CASE: MATH CASE: MATH . It is straightforward to show that for all MATH and all formulas MATH with MATH as only propositional variable and such that MATH, MATH if and only if MATH. This implies that if MATH is satisfiable, then MATH is satisfiable. For the converse, suppose that MATH is satisfiable. Let MATH be an acyclic model and MATH such that MATH. Define MATH as follows. CASE: MATH. CASE: MATH. CASE: MATH. Again, it is easy to show that for all MATH and all formulas MATH with MATH as only propositional variable and such that MATH, MATH if and only if MATH. It follows that MATH is satisfiable if and only if MATH is satisfiable. This proves the theorem, since MATH is clearly computable in polynomial time. |
cs/9911014 | First suppose that MATH. If REF does not apply, then MATH or MATH. In the first case, REF applies. For the second case, either REF applies or MATH, in which REF applies. Next suppose that MATH and MATH. If REF does not apply, then MATH is not a subset of MATH, MATH, or MATH. This implies that MATH contains MATH or MATH. In the first case, REF applies or REF applies. In the second case, REF applies or REF applies. Finally, suppose that MATH and MATH. That is, we are in a fragment of positive modal logic (see for example CITE). If REF does not apply, then MATH, since MATH. Also, MATH, since MATH. If REF does not apply, then MATH is not a subset of MATH or of MATH. It follows that MATH contains MATH and REF or REF applies. |
gr-qc/9911032 | Point REF is a well known consequence of the uniqueness result for constant mean curvature hypersurfaces of CITE, compare CITE. Point REF is essentially CITE. For the proof, note that as MATH is compact we can construct a MATH invariant time function on MATH by averaging any global time function MATH on MATH with respect to the MATH action, compare the proof of CITE. The level sets of the averaged time function are NAME surfaces and are invariant under the action of MATH. The result follows. The following argument for point REF is due to NAME. Let MATH and assume for a contradiction MATH is nonempty. Choose MATH and a time function MATH on MATH so that MATH and MATH. Let MATH denote the intersection of the past of MATH with the future of MATH and let MATH. The set MATH is closed and by global hyperbolicity MATH is compact and hence MATH and there is a MATH with MATH. If MATH is nonzero and null, then using the equation MATH which holds since MATH is Killing, gives a null curve of points in MATH where MATH is null. Following this into the past, shows that there is a MATH with MATH, which gives a contradiction. In case MATH, the linearization of MATH acts by isometries on MATH, and as the sphere of null directions in MATH is two dimensional it leaves a null direction fixed. Using the exponential map shows that the action of MATH near MATH leaves a null geodesic invariant, along which MATH must be zero or null. This leads to a contradiction as above. |
gr-qc/9911070 | Let us consider REF . The solution is given by MATH . Substituting MATH, MATH on MATH is asymptotically MATH . This means that if we take MATH large enough, MATH on MATH for any MATH and hence MATH on each MATH . REF null segment between the BEH and MATH by REF . |
gr-qc/9911070 | Let us consider each MATH null segment: MATH, MATH. In the former null segment, MATH, as already discussed before. In the latter null segment, MATH because MATH by REF and the previous lemma. This indicates that MATH on MATH by REF . |
hep-th/9911188 | If MATH then MATH, therefore it is verified that: MATH. The reciprocal is also right, if MATH then MATH, therefore it is verified that: MATH. |
hep-th/9911188 | It follows from the NAME Lemma CITE(p REF), CITE(p REF). In this case MATH is closed, MATH, then it must be exact, MATH. However, the vector field MATH is not univocally defined, being possible a transformation as: MATH, where MATH is a scalar field. It is verified that: MATH, due to: MATH. To fix this scalar it is imposed an additional equation as the NAME gauge: MATH. This restriction becomes: MATH . |
hep-th/9911188 | From the double duality it is concluded that if a p-form is null also is null its dual, so from MATH it is concluded that MATH is closed. Based on the NAME Lemma it must be exact, this is: MATH, therefore exist MATH. This is not univocally defined, it is verified that: MATH, where MATH. |
hep-th/9911188 | Firstly if MATH or MATH the theorem is immediately proved by choosing null any of the two auxiliary fields. The other case is MATH with non-null sources. The auxiliary fields MATH and MATH are chosen according with REF . From the previous Proposition it is proved that there are solution for both fields, however these solutions are not univocally defined. Therefore these solutions can verify that: MATH however it is obtained that MATH. Due to MATH, it can be introduced a new tensor: MATH, consequently verifying that: MATH. |
hep-th/9911188 | From the previous conditions must be: MATH and also MATH and MATH. However it is verified that from: MATH, it must be: MATH. Also from: MATH, and due that MATH, it must be: MATH. Therefore in the general case the MATH class is non-reducible to MATH and reciprocally because both are lineally independent. |
hep-th/9911188 | It is based on the properties of the MATH and MATH tensors CITE(p REF). |
hep-th/9911188 | The first is obtained directly from the previous Proposition, the second is obtained by using the intermediate step: MATH. The two last expressions are contraction of the previous. |
hep-th/9911188 | It is deduced directly applying REF . |
hep-th/9911188 | This result is based on REF , each tensor is expressed as: MATH . The last tensor can be simplified by using: MATH. Based on REF , it is verified that: MATH and MATH. It is obtained that: MATH. |
math-ph/9911013 | Set MATH and MATH. We have that MATH since MATH. Therefore, MATH which completes the proof. |
math-ph/9911013 | With MATH and MATH and MATH . Therefore, MATH and MATH . The lemma follows from REF . |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.