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math/0102148 | It is obvious that MATH on MATH. We fix MATH and set MATH. Let MATH belong to an orthonormal basis for the orthogonal complement of MATH which we choose such that the submatrix MATH is diagonal. Assume that MATH and MATH, MATH correspond to the indices MATH and MATH, respectively. We use the NAME summation convention for MATH. The matrix MATH is positive, and thus testing with the vectors MATH gives MATH . We introduce the abbreviation MATH . Direct computations using REF give MATH and thus for MATH sufficiently large, MATH . Expanding the determinant and using that MATH is diagonal gives MATH . Hence the inequality for geometric and arithmetic means as well as REF show that for large values of MATH, MATH . |
math/0102148 | According to REF , the condition MATH on MATH follows if MATH . In view of REF , this can be arranged by choosing MATH sufficiently large. The property MATH now follows from the inequality MATH which holds for MATH sufficiently large. |
math/0102148 | The maximum principle applied to REF yields that MATH in MATH. Since MATH, it follows that MATH with MATH. Thus we obtain MATH and this finally gives REF . |
math/0102148 | We choose for fixed MATH an orthonormal basis as in REF , now such that the submatrix MATH is diagonal. We expand the determinant, MATH . Now we substitute in the estimates of REF , MATH . |
math/0102148 | We introduce functions MATH and MATH by MATH and define MATH by MATH . This MATH is zero on MATH, is close to a cone, and satisfies the regularity REF . Furthermore, MATH . We compute the NAME curvature of MATH, MATH . We conclude that in MATH . |
math/0102148 | We choose a smooth, strictly convex function MATH such that MATH and define MATH. Then MATH is a solution of the NAME problem MATH . We may assume that MATH in MATH because otherwise we solve a NAME problem (with small MATH) by using CITE with MATH as the subsolution, and take the solution to this NAME problem instead of MATH. Finally, we apply the continuity method and solve the NAME problems MATH for smooth strictly convex functions MATH, MATH. The solvability of this NAME problem is a consequence of the a priori estimates of CITE and the fact that MATH . The last inequality follows from the maximum principle because MATH for MATH. For MATH we obtain the function MATH we are looking for. |
math/0102149 | First, let's fix MATH. According to REF , we have MATH and this expression differs from MATH for at least one MATH, by the unitarity of MATH-matrices. Select such a MATH, and apply MATH to both sides of the equation. One gets that MATH differs from MATH, but this can only happen if MATH because MATH. Next, for MATH consider MATH . Applying MATH to both sides of the above equation we get from REF MATH . But the lhs. equals MATH according to REF if MATH. Equating both sides we arrive at MATH . As the lhs. is independent of MATH, we must have MATH and MATH as well as MATH . |
math/0102149 | Write MATH, with MATH coprime to MATH, and recall that MATH, where MATH is the mod MATH inverse of MATH. Consider MATH . Applying MATH to both sides of the above equation and taking into account REF , we get MATH . But form REF , the lhs. equals MATH so after rearranging, we get MATH . As the lhs. is independent of MATH . Of course, the point is that MATH is independent of MATH. If we introduce the diagonal matrix MATH then it is obvious that the order of MATH divides the denominator of MATH and the first equality in REF holds. The second follows simply from MATH. |
math/0102149 | This follows at once from MATH applied to MATH. |
math/0102149 | Let MATH, with MATH coprime to MATH. According to REF MATH . Applying MATH to both sides we get MATH that is, MATH by REF , that is, MATH . But this is equivalent to the assertion. |
math/0102149 | Take MATH equal to the order of MATH. Clearly, MATH divides the conductor. By REF of the Appendix, MATH . But MATH or in other words MATH because both MATH and MATH are diagonal, hence they commute. If MATH is even, then taking the MATH-th power of the last equation gives the result. If MATH is odd, then there exists MATH such that MATH, and taking the MATH-th power of the equation gives again REF . |
math/0102149 | We have just seen that it holds for MATH, and it also holds for MATH by REF . But MATH and MATH generate MATH, consequently the claim should hold for any MATH. |
math/0102149 | Let's write MATH. We know from CITE CITE that MATH where MATH is the exponentiated conformal weight of the primary MATH, that is, the matrix element MATH. By REF MATH that is, MATH or in other words MATH . Taking the MATH-th power of the last equation and using REF gives the result, because MATH and MATH commute. |
math/0102149 | MATH according to REF . |
math/0102149 | Let's apply MATH to both sides of the modular relation REF . One gets MATH . After rearranging and using REF we get the assertion. |
math/0102149 | Clearly, MATH divides the conductor, for the eigenvalues of MATH are roots of unity. But for MATH we have MATH and MATH according to REF , consequently all such MATH fixes MATH, that is, MATH. On the other hand, if MATH fixes MATH then MATH, that is, MATH. |
math/0102149 | MATH diagonal means that MATH is the identity permutation. If we apply MATH to MATH, we get MATH because MATH. But MATH, consequently MATH, proving the lemma. |
math/0102149 | That MATH divides MATH is obvious. If we consider the matrix MATH, then it will equal MATH times the identity matrix. But MATH is the exact order of MATH, consequently MATH is a primitive MATH-th root of unity. On the other hand, REF implies MATH for all MATH coprime to MATH, that is, MATH . In other words, the exponent of MATH should divide MATH, and this can only happen if MATH divides MATH. Next, let's consider MATH. By the above result, MATH is coprime to both MATH and MATH, consequently it is also coprime to MATH. If MATH denotes its inverse mod MATH, then we have MATH . According to REF , MATH . But if MATH is diagonal, then it equals MATH, so that MATH. Together with MATH, this yields MATH. Finally, let MATH denote the gcd of MATH and MATH, and consider MATH. It is straightforward that MATH is coprime to MATH, and its inverse modulo MATH is MATH. But MATH and once again this should equal MATH, consequently MATH divides REF. |
math/0102149 | If MATH denotes the central charge, then MATH. But MATH, so by the above MATH, proving the claim. |
math/0102149 | Let's denote by MATH the exponent of the symmetric group MATH of degree MATH. As the permutations MATH belong to MATH, it follows that MATH is the identity permutation, that is, MATH is diagonal. By REF this means that MATH, which in turn implies that MATH for all MATH, that is, the exponent of the group MATH divides MATH. Taking MATH to be the greatest integer satisfying this last condition proves the proposition. |
math/0102149 | According to REF MATH because MATH. But MATH so MATH . From REF MATH . Putting all this together, we get MATH which proves the claim according to REF . |
math/0102149 | If we take MATH and MATH in REF , then MATH and MATH. |
math/0102149 | If MATH, then MATH is the identity matrix, and REF implies the result. |
math/0102149 | If MATH, then REF reduces to MATH, whose only solution is MATH. |
math/0102149 | As a homomorphic image of MATH, MATH is generated by MATH and MATH, so it is enough to verify REF for these matrices. For MATH, the lhs. reads MATH while the right-hand side is MATH and these are clearly equal. For the case of MATH, the lhs. reads MATH according to REF , while the right-hand side is MATH . It is easy to check that MATH proving the claim. |
math/0102154 | This follows directly from classical work of NAME and the recognizability of the MATH - sphere, as follows. The first step is to find a complete system of MATH-spheres in MATH. A complete system is one which is guaranteed to contain enough spheres to completely decompose MATH into prime factors; the system may contain redundant spheres. The earliest algorithms to do this are due to NAME; more recent improvements can be found, for instance, in CITE. For each sphere MATH in the list, cut MATH along MATH. If the resulting manifold is connected, then MATH is a non-separating sphere, and so MATH is reducible. If MATH is disconnected, let MATH and MATH be the pieces. Each MATH has a single sphere boundary component; let MATH be the manifold obtained by capping off this boundary component with a MATH-ball. Now NAME and NAME 's MATH - sphere recognition algorithm may be applied to the MATH, and MATH is a reducing sphere if and only if neither MATH or MATH is MATH. |
math/0102154 | From the triangulation of MATH we obtain some presentation of MATH: MATH . We then consider the representation variety of homomorphisms MATH as defined for example, in CITE. Since MATH is finitely presented, this variety is constructible as the zero set of a finite set of polynomials with integer coefficients. Indeed, MATH can be thought of as the variety in MATH defined by MATH. The representation variety MATH is the subvariety of MATH obtained by adjoining the MATH polynomials coming from the matrix equations corresponding to the relations of MATH. For each pair MATH of generators of MATH, we construct the subvariety MATH consisting of those representations for which the equations MATH hold. In the notation introduced above, REF corresponds to the equations MATH, MATH, and MATH. Recall that a subgroup of MATH is said to be nonelementary if no element of MATH is fixed by then entire subgroup. Notice that if MATH is any representation so that MATH is nonelementary, then MATH is conjugate to a representation in MATH. Also observe that given some representation MATH of MATH which does not send MATH to the identity, there are at most four representations conjugate to MATH which satisfy REF . Indeed, if MATH and MATH both satisfy REF for some MATH, then MATH must be an eigenvector of MATH. Likewise, MATH must be an eigenvector of MATH. Up to scaling the columns of MATH, there are at most four ways to arrange this. Given one of the four choices, the requirements that the determinant of MATH be one and that MATH have a one in the upper right corner determine MATH up to multiplication by MATH. For each pair MATH, we would like to add the isolated points of MATH-to our list of candidate representations. Note that MATH-is given by a set of polynomials with integer coefficients. Let MATH be the ideal in MATH generated by these polynomials. There exist algorithms (see, for instance REF ) to decompose this ideal into prime ideals. REF gives an algorithm to determine the dimension of such an ideal. Each isolated point in MATH-is part of the variety determined by some zero - dimensional ideal MATH. Let MATH be some such solution. Each coordinate MATH is a root of the monic polynomial MATH which generates the elimination ideal MATH. An algorithm to find this polynomial is given in REF (see also REF). Possibilities for MATH are determined by finding the irreducible factors of MATH and isolating intervals or rectangles for the various roots. There is now a finite list of possible values for each MATH. Not every combination of choices necessarily corresponds to a representation, so we must check the original relations in MATH by matrix computations over MATH, for each of the finite number of combinations of choices. Those MATH - tuples of choices which satisfy the relations are the candidate representations coming from MATH. Since there are finitely many pairs MATH, the process described ends up with a finite list of representations. Suppose that MATH is hyperbolic. Then there is a discrete faithful representation MATH of MATH into MATH which gives the hyperbolic structure on MATH. This representation lifts to MATH as proved in CITE. We abuse notation slightly by continuing to refer to the lifted representation as MATH. Note that the image MATH cannot be cyclic, and that every nontrivial element is loxodromic. Therefore, there must be some pair of generators MATH so that MATH and MATH are loxodromics with distinct axes. As MATH is discrete, MATH is nonelementary (see REF ). Therefore MATH is conjugate to a representation MATH in MATH. It is a well - known consequence of NAME Rigidity that MATH is unique up to conjugacy and orientation, and that nearby representations are necessarily conjugate, so MATH is an isolated point in MATH. The discrete faithful representation MATH must therefore appear in our list of candidate representations. |
math/0102154 | For MATH a two - by - two complex matrix, define: MATH . The norm MATH gives rise to a metric on the space of two - by - two complex matrices defined by MATH. The restriction of this metric to MATH gives the same topology as the ordinary one (see CITE for details). Therefore the hypothesis that MATH is indiscrete is equivalent to the existence of nontrivial elements MATH of MATH so that MATH is arbitrarily small. It is shown in CITE that if MATH acts on the upper half space model of hyperbolic space in the usual way, and MATH is the point MATH, then MATH, where MATH is the hyperbolic metric. A simple calculation shows that if there are nontrivial matrices MATH with MATH arbitrarily small, then there must also be matrices with MATH (equivalently MATH) arbitrarily small. Since the only elements of MATH which fix MATH (and thus have MATH) are elliptics whose axes pass through MATH, this means there must be nontrivial matrices in MATH with MATH arbitrarily small but positive. We claim that we can find such matrices which do not commute. Suppose we cannot. Then all matrices in MATH which move MATH a sufficiently small distance commute, and thus share a common axis, which we'll call MATH. Let MATH be the subgroup of MATH which sends MATH to itself. Rotation angle and translation length are continuous functions on MATH, so we can find elements of MATH which both translate MATH an arbitrarily small distance along itself and rotate MATH about MATH by an arbitrarily small angle. By hypothesis, MATH is nonempty. Let MATH. The subgroup MATH of MATH preserves the geodesic MATH and so its elements do not commute with those of MATH. Since rotation angle and translation length are invariants of conjugacy class, there are elements of MATH with arbitrarily small such. There are therefore elements of MATH which move MATH as small a distance as we like. This contradicts our assumption that all matrices in MATH which move MATH a sufficiently small distance commute. Finally let MATH be an arbitrary point in MATH. If MATH is an element of MATH which takes MATH to MATH we may apply the above argument to MATH. |
math/0102154 | Suppose MATH. If MATH then MATH has MATH as eigenvectors. If MATH then MATH has eigenvectors MATH and MATH, unless MATH in which case MATH has MATH as its sole eigenvector. Of course any multiple of an eigenvector is an eigenvector, but we have chosen specific ones precisely to remove that ambiguity. Since we can do arithmetic and square roots in MATH and compare elements of MATH, it is possible to decide whether or not two vectors of the given form are the same. |
math/0102154 | A matrix in MATH acts elliptically or parabolically on hyperbolic space if and only if its trace is in the interval MATH. For each MATH we can find the algebraic number MATH. As discussed in REF, there are algorithms to find the imaginary part of MATH, and to test equalities and inequalities, so it can be determined whether or not MATH. If MATH, we may then check whether or not MATH. |
math/0102154 | For each MATH, we compute MATH (which involves just multiplication, addition, and a square root). For each pair of matrices MATH, MATH of MATH so that MATH and MATH we check whether or not MATH is the zero matrix. |
math/0102154 | The idea is to detect the failure of REF - REF listed at the beginning of this section. By REF , one of these conditions must fail if MATH is not discrete and faithful with torsion - free image. We start with REF . Let MATH from the presentation REF. The representation MATH is reducible if and only if all the elements of MATH share a common eigenvector. This condition is checkable by REF . If the representation is irreducible, we begin systematically listing elements of the free group MATH in such a way that any element is eventually listed. For each finite list produced this way, we look for evidence that REF - REF have failed for MATH. Let MATH be the finite list at some the listing process. By performing matrix multiplications over MATH, we can determine MATH for any MATH. Let MATH. If REF fails, then there will eventually be elements of MATH which are not equivalent to the identity in MATH, but which are in the kernel. If MATH for some MATH, then we apply the solution of the word problem in MATH to determine whether or not MATH is the identity element in MATH. Thus if REF fails, we will eventually discover it. Note that this is the first use of the assumption that MATH has solvable word problem. If REF fails, then eventually MATH will contain nontrivial elliptics or parabolics. By REF , we can detect this. If REF fails, then there are arbitrarily small noncommuting matrices. In particular, MATH will eventually contain noncommuting matrices MATH and MATH with MATH and MATH. By REF , we can detect such matrices. If MATH is discrete and faithful with torsion - free image, then none of REF - REF can fail, and so this procedure will continue endlessly. Otherwise, one of the four conditions will eventually be discovered to fail and the procedure will stop. |
math/0102154 | This follows from our ability to compute traces and from REF . |
math/0102154 | Given three spheres in Euclidean space, it is not hard to write down a set of inequalities in terms of their centers and radii which are satisfied if and only if the spheres intersect in a pair of distinct points. Moreover, these inequalities involve only arithmetic and square roots, and so may be checked algorithmically if the centers and radii are given by algebraic numbers. If the spheres intersect in a pair of points, then the coordinates of these points can be explicitly written down (again using only arithmetic and square roots) in terms of the centers and radii of the spheres. Therefore, for each triple of isometric spheres of elements of MATH it is possible to decide if they intersect in MATH, and if so to write down a triple of algebraic numbers which are the coordinates of their intersection. These triples are potential vertices of MATH. Let MATH be the set of potential vertices of MATH. Since every vertex of MATH is contained in at least three distinct isometric spheres, every vertex of MATH is in MATH. However MATH may contain vertices not in MATH. For each MATH, MATH, we check that MATH is outside or on the sphere MATH. If MATH is inside MATH, MATH is outside MATH, and so it is discarded. After discarding these vertices, we continue to refer to the smaller set of vertices as MATH. After all sphere - vertex pairs have been checked, this set is equal to the zero - skeleton of MATH. Now we find the edges of MATH by looking for pairs of spheres whose intersection contains two vertices. For each pair of vertices MATH in MATH, we can construct the list of matrices MATH in MATH so that MATH and MATH are both on MATH. There is an edge connecting MATH to MATH in MATH if and only if this list contains at least two elements. Let MATH be the set of unordered pairs of vertices contained in at least two isometric spheres. Clearly every edge of MATH is represented by some element of MATH. Conversely, suppose MATH for some MATH,MATH. Since MATH is convex, the geodesic segment between MATH and MATH is contained in MATH. Since MATH is convex, this segment is in fact contained in the boundary of MATH, and is therefore part of MATH. Thus we can construct the MATH - skeleton of MATH. It remains only to find the MATH - cells of MATH. For each MATH, the set MATH can be constructed. There is a MATH - cell of MATH formed from a part of MATH if and only if the set MATH contains at least three vertices and these vertices are linked up by edges in MATH to form a circuit. These conditions can be checked on MATH for each MATH, so we can produce a list of faces MATH. This completes the combinatorial data needed to construct MATH. |
math/0102154 | To check that MATH is connected, it suffices to check that any pair of vertices in MATH can be connected by a sequence of edges. The NAME characteristic of MATH is equal to MATH. |
math/0102154 | We first check, for each MATH, that we also have MATH. Since faces are uniquely determined by their vertices, it then suffices to check that MATH sends the vertices of MATH to those of MATH. Since we have kept track of the coordinates of the vertices, and can explicitly compute the action of MATH on points in MATH with algebraic coordinates, this is straightforward. |
math/0102154 | We use a combination of exact and numerical computations to check whether the conditions of REF are satisfied at each edge. Specifically, let MATH be some edge of MATH. Adjacent to MATH are precisely two faces, MATH and MATH. Let MATH be one of the two faces. Let MATH. By hypothesis, MATH is an edge of MATH. The edge MATH is adjacent to two faces, one of which is MATH. The other face is MATH for some MATH. We inductively define MATH to be MATH and MATH to be the face adjacent to MATH across MATH. Let MATH be the first integer for which MATH, the edge we started with, in which case MATH and MATH. If MATH satisfies the edge condition, then, in particular, MATH. If the exact calculation shows that this is the case, then the angle sum is some integer multiple of MATH. Suppose MATH and MATH are two spheres whose intersection contains an edge. Then their dihedral angle is given by MATH, if MATH is the sphere of radius MATH centered at MATH. In the case where MATH and the coordinates of MATH are given by algebraic numbers (or any numbers which can be determined to any desired accuracy) it is not hard to show that the angle can then be determined numerically to any desired accuracy. We can therefore do a calculation of the angle sum around an edge with tolerance less than, say MATH. Together with the exact calculation, this tells us whether the edge condition is satisfied. As there are a finite number of edges in the complex MATH, it will take a finite amount of time to check them all. |
math/0102154 | For each generator MATH of MATH, we wish to check whether MATH is in MATH. Let MATH be the generators of MATH, which are also the face pairings of MATH. Each face of MATH is part of the isometric sphere MATH for some MATH. Note that MATH is in the interior of MATH. Since hyperbolic space is tiled by copies of MATH, MATH is in MATH for some MATH. But the following are clearly equivalent: CASE: MATH CASE: MATH CASE: MATH is outside or on MATH for all MATH. From our earlier remarks, this last statement is clearly checkable. If MATH then MATH whenever the statement is true. We can systematically list all possible words in the free group MATH and for each word MATH in the list check to see whether REF above holds for MATH. We must eventually find such a MATH, since the copies of MATH tile hyperbolic space. When this MATH is found, test whether MATH. If so, then MATH, and we go on to the next generator. If every generator has image in MATH, then MATH. Otherwise, MATH is a proper subgroup of MATH. |
math/0102154 | We can now think of MATH as an epimorphism of abstract groups MATH where the MATH are words coming from the edge cycles. Recall that we started with some finite set MATH of words in the free group MATH. The MATH correspond to the matrices in MATH which pair the faces of the polyhedron MATH, and generate MATH. To check whether or not MATH is faithful, we attempt to construct an inverse map. For each MATH pick some MATH in the finite set MATH. The MATH thus chosen determine a homomorphism from the free group MATH to MATH. We can check whether this map factors through MATH by using the word problem algorithm to check that each MATH is sent to the identity element of MATH. If the map does not factor, then MATH does not have an inverse and therefore must have nontrivial kernel. Otherwise, we have constructed a homomorphism MATH from MATH to MATH-and we have MATH by construction. To determine whether MATH the procedure checks that for each generator MATH, again using the word problem algorithm. If any MATH, then MATH implies that MATH has kernel. Otherwise, MATH is an isomorphism onto MATH. |
math/0102154 | Let MATH from the presentation REF. We begin systematically listing elements of the free group MATH in such a way that an element and its inverse are always added at the same time. At each stage of the listing, we have some finite list MATH of words and a list MATH of matrices in MATH. From this data we attempt to build a fundamental domain. Taken together, REF give an algorithm which either constructs or fails to construct a fundamental domain for the action of MATH, given some finite MATH which is closed under inverses. This algorithm is outlined in REF . If this algorithm finds a fundamental domain for the action of MATH, then it follows that MATH acts as a discrete fixed point free group of isometries of hyperbolic space. According to REF , there is then an algorithm to determine whether MATH is faithful. Let MATH be a discrete faithful representation with torsion - free image MATH. We must show that the algorithm we have described eventually discovers a fundamental domain for the action of MATH. Note first that any subset of MATH satisfies condition MATH. For suppose that MATH and MATH are hyperbolics in MATH which have the same isometric sphere MATH. Then MATH fixes MATH, and so must be trivial or elliptic. But MATH contains no elliptics, so we must have MATH. The intersection of the exteriors of all the isometric spheres of elements of MATH is the NAME domain for MATH centered at MATH. REF implies that the quotient of hyperbolic space by MATH is homeomorphic to MATH, so it is a closed REF - manifold. In particular, MATH is geometrically finite, so the NAME domain centered at MATH has finitely many sides. Each of these sides is part of the isometric sphere of a unique loxodromic element of MATH. There is therefore a finite subset of MATH containing all the matrices whose isometric spheres form faces of the NAME domain. Eventually our list MATH will contain all these matrices, and we suppose for the remainder of the proof that it does. But now the complex MATH constructed in REF is precisely the boundary of the NAME domain centered at MATH, which is homeomorphic to a sphere. Face pairings are given by the isometries whose isometric spheres form the faces, and the algorithm of REF will verify this. The NAME domain satisfies the hypotheses of the NAME Polyhedron Theorem, as will be verified by the algorithm of REF. The face identifications of the NAME domain for an action generate that action, so the algorithm of REF will correctly determine that MATH. Thus if MATH is large enough, the algorithm of REF will eventually produce a fundamental domain for the action of MATH. |
math/0102154 | By REF MATH admits a hyperbolic structure if and only if CASE: MATH is irreducible, and CASE: there is an injective homomorphism MATH so that MATH acts freely and cocompactly on MATH. By REF we may suppose that MATH is irreducible. REF gives a way to construct a finite list MATH of representations of MATH into MATH, with the property that if a representation satisfying REF exists, then it is in MATH. REF together give a way to decide whether a particular representation from MATH satisfies REF or not. The manifold MATH admits a hyperbolic structure if and only if some representation in MATH satisfies the condition. |
math/0102154 | Note that by REF , it is sufficient to find a triangulated hyperbolic manifold which is homeomorphic to MATH. If it is known that MATH is hyperbolic, it follows that the fundamental group is automatic. There are algorithms to find the automatic structure once one is known to exist CITE. This automatic structure immediately provides a procedure for solving the word problem. We can then use the algorithms of REF to find a particular representation of MATH which is discrete and faithful, and a fundamental domain for the action of MATH on hyperbolic space through that representation. The fundamental polyhedron with face - identifications give a cell - structure for MATH which is easily turned into a triangulation. |
math/0102157 | Necessity. Let the collections MATH and MATH of inputs and states of MATH satisfy the assumptions of REF . Apply induction on MATH. Suppose that for MATH, where MATH, we have MATH. Then, by virtue of REF, MATH . This implies MATH for MATH. Therefore, the latter holds for an arbitrary MATH. Since for any MATH we have MATH (see, for example, CITE), for any MATH the series MATH converges absolutely. Now, by virtue of REF and from REF we get for any MATH that is equivalent to REF, and we have proved the necessity of REF . The necessity of REF is established analogously, by rewriting REF for MATH and using REF. Sufficiency. Let us set for arbitrary MATH . Clearly, the collections MATH and MATH of inputs and states of MATH satisfy the assumptions of REF . Then we can write down for them REF, with MATH, as follows: MATH or equivalently, MATH . By using system REF , we obtain MATH . Since MATH are arbitrary, and the operators MATH and MATH are bounded and selfadjoint, the latter implies REF. Analogously, REF implies REF. Now, for arbitrary MATH, and MATH set MATH . Clearly, the collections MATH and MATH of inputs and states of MATH satisfy the assumptions of REF . Then we can write down for them REF, with MATH, as follows: MATH or equivalently, MATH . By using system REF , we obtain MATH that is equivalent to MATH . By REF established formerly, we obtain: MATH . One can substitute MATH instead of MATH, and MATH instead of MATH, and obtain MATH . Since MATH can be taken arbitrary, REF follows. Analogously, REF implies REF. The proof is complete. |
math/0102157 | Set MATH where this supremum is taken over all MATH-tuples of commuting contractions MATH on a common separable NAME space MATH. If MATH then MATH, and by REF the assertion of REF follows with MATH. Suppose that MATH. Then MATH. By REF, there exist separable NAME spaces MATH and holomorphic MATH-valued functions MATH on MATH such that MATH . Setting MATH for MATH, we obtain MATH: MATH . Since MATH, again by REF, there exist separable NAME spaces MATH and holomorphic MATH-valued functions MATH on MATH such that MATH . Setting MATH for MATH, we obtain MATH: MATH . Set MATH, and according to this orthogonal decomposition define MATH by MATH and MATH for MATH. By subtracting REF from REF for each MATH and MATH, we obtain REF, that completes the proof. |
math/0102157 | By REF, MATH can be realized as the transfer function of some multiparametric linear system MATH, that is, MATH in some neighbourhood MATH of MATH in MATH. If MATH is a MATH-valued function holomorphic on some neighbourhood of MATH (here MATH is some input multisequence of MATH, which has the support in MATH) then (see CITE) one can write down the so-called MATH-transform of MATH: MATH with holomorphic functions MATH on some neighbourhood MATH of MATH in MATH. We can consider without a loss of generality that MATH. Set MATH, and let MATH be arbitrary. Then by using REF , and equalities REF twice, for MATH and for MATH, we have for all MATH: MATH . Setting MATH we obtain REF. |
math/0102157 | Let, as in the proof of REF , MATH where this supremum is taken over all MATH-tuples of commuting contractions MATH on a common separable NAME space MATH, and MATH be the MATH-tuple of operators REF corresponding to a given multiparametric linear system MATH. If MATH then MATH, and by REF, system MATH has a conservative dilation MATH, that is, a MATH-conservative one with MATH. Suppose now that MATH. Applying REF to MATH, we have the existence of separable NAME spaces MATH with canonical symmetries MATH, and holomorphic MATH-valued functions MATH on MATH such that REF holds. Let us define these spaces MATH, operators MATH, and functions MATH exactly as in the proof of REF , that is, MATH, MATH so that REF hold. Set MATH, MATH. Then MATH. It follows from REF that MATH . In particular, MATH is a bounded and boundedly invertible operator, and MATH is a closed lineal, that is, a subspace in MATH. Analogously, it follows from REF that MATH . In particular, MATH is a bounded and boundedly invertible operator, and MATH is a closed lineal, that is, a subspace in MATH. It follows from REF that MATH . Taking into account REF, rewrite REF as MATH and by virtue of REF, we get MATH: MATH . Taking into account REF, rewrite REF as MATH and by virtue of REF, we get MATH: MATH . Now, adding REF to REF, we get MATH: MATH and subtracting REF from REF, we get MATH: MATH . It follows from REF that there exists unique unitary operator MATH such that MATH: MATH . It follows from REF that MATH that is, MATH is a MATH-semiunitary operator. Moreover, MATH is a closed lineal, that is, a subspace in MATH. In addition, from REF one can see that MATH: MATH hence MATH . Now let us show that the subspace MATH in MATH is a NAME space with respect to the metric MATH induced by the canonical symmetry MATH. By REF, in order that MATH is a canonical symmetry on MATH, it is necessary and sufficient that any MATH has a MATH-orthogonal projection onto MATH, that is, a vector MATH such that MATH. For an arbitrary MATH set MATH. Since, due to REF, MATH we get MATH. For an arbitrary MATH we have: MATH . Thus, MATH is a desired MATH-orthogonal projection of MATH onto MATH, and we have proved that MATH is a canonical symmetry on MATH (that is, MATH is a NAME space with respect to the metric MATH). Further, MATH . Define on the space MATH the canonical symmetry MATH. Due to REF, the operator MATH, defined by REF, can be considered as a bounded and boundedly invertible on its domain MATH-isometric operator from MATH to MATH, whose domain and range are given by MATH . By REF, there exist a separable NAME space MATH, a canonical symmetry MATH, and a MATH-unitary operator MATH, with MATH such that MATH is an extension of MATH, that is, MATH. Since for any MATH we have MATH we get for any MATH: MATH . Set for all MATH, where MATH denotes the NAME symbol, MATH . Clearly, MATH is a MATH-unitary operator, where MATH for any MATH is a MATH-unitary operator, and MATH is a MATH-unitary operator, where MATH . Therefore, for any MATH the operator MATH is MATH-unitary. Consider the following partitioning of MATH: MATH . Then MATH is a multiparametric MATH-conservative scattering system. Since, by virtue of REF, MATH we have for all MATH: MATH . Therefore, in some neighbourhood MATH of MATH the resolvent MATH is well-defined and holomorphic, and we have in this neighbourhood: MATH . Thus, MATH is a MATH-conservative scattering system realization of the linear operator-valued function MATH. From REF we obtain: MATH . As it was shown in the proof of REF (see also Subsection V. REF), the latter means that the system MATH which is determined by the system operators MATH coinciding with the system operators MATH of MATH, however with another block partitioning of these operators, is a dilation of MATH. Defining a canonical symmetry of MATH: MATH we obtain that MATH is a desired MATH-conservative scattering system dilation of a given system MATH. The proof is complete. |
math/0102157 | Let MATH be a given MATH-valued function holomorphic on a neighbourhood MATH of MATH in MATH and vanishing at MATH. Then, by REF, there exists a realization MATH of this function, that is, MATH in some neighbourhood MATH of MATH. By REF of this paper, there exists a multiparametric MATH-conservative scattering system MATH, with a canonical symmetry MATH, which is a dilation of MATH. By REF, the transfer functions of a multiparametric linear system of the form REF and of its dilation coincide in some neighbourhood of MATH in MATH. Hence, MATH in some neighbourhood MATH of MATH. Thus, the system MATH is a desired MATH-conservative scattering system realization of MATH. The proof is complete. |
math/0102161 | We calculate MATH. Differentiation produces MATH . In order to obtain MATH and MATH, we differentiate problem MATH with respect to MATH: for all MATH, MATH . Denoting MATH, we are thus led to the initial value problem MATH which can be solved by variation of constants. The expressions we want to calculate are evaluations of MATH and MATH at MATH. We find MATH . Thus MATH if, and only if, MATH, which may happen only if MATH. So, if MATH, there exists MATH such that MATH and the result now follows from the Implicit Function Theorem. Suppose now MATH. Since MATH is an isolated root of MATH and we must have MATH, it follows that MATH. But, in this case, denoting MATH, we see that MATH solves MATH and MATH if and only if MATH. Again, MATH is a manifold. |
math/0102161 | For each MATH and MATH fixed, we consider the straight line MATH. We will show that MATH always intercepts each manifold MATH once and transversally. Uniqueness and transversality follows from MATH . Smoothness (and local smooth invertibility, for a fixed MATH) in MATH, in turn, follows by setting MATH in the formula above. Suppose that MATH: in this case, MATH is increasing, MATH and MATH. Here MATH. Clearly, the range MATH of MATH is an open set. From standard oscillation theory, the solutions of the three problems below (with initial position and velocity at REF equal respectively to REF) have increasing arguments, MATH . We will see that the asymptotics of the argument at both ends of MATH will be given by the argument of the solutions of the leftmost and rightmost problems. For a fixed value of the parameter MATH, let MATH be the solution of the problem MATH . We denote MATH . It follows immediately that MATH . We first study the behavior of MATH when MATH. Fix MATH. It is easy to prove (see CITE) that the argument function satisfies the differential equation MATH . Then MATH . Thus MATH, for MATH, MATH and MATH . Since MATH is continuous and uniformly bounded in MATH, we can define the integrating factor MATH. Multiplication by this factor and integration produces MATH . If MATH, we choose MATH and have MATH, yielding MATH for all MATH. The estimate MATH where MATH, shows that MATH when MATH. This shows that MATH converges increasingly to MATH, for MATH, if MATH. If MATH, we have MATH in the interval MATH if MATH is big enough. Defining MATH, we obtain MATH when MATH. Consequently, the integral in REF is negative, if MATH is sufficiently big. We conclude that MATH. Since MATH, and we again have that MATH increases to MATH. In a similar fashion, we obtain that MATH converges decreasingly to MATH, when MATH. Thus, the line MATH intersects the same manifolds MATH irrespective of MATH: the values of MATH for which MATH trespasses MATH lie strictly between MATH and MATH. The case MATH is similar. |
math/0102162 | We apply REF below to obtain a NAME cobordism MATH . Then Legendrian surgery corresponds to attaching a REF-handle along MATH in a NAME (respectively, symplectic) manner, which yields a NAME (respectively, symplectic) cobordism from MATH to MATH. (See CITE.) |
math/0102162 | If MATH, then the manifold MATH is simply the connected sum of several copies of MATH. There is a unique tight contact structure MATH on MATH, and it is NAME fillable. The uniqueness of MATH on MATH follows from the unique connect sum decomposition theorem of CITE and the uniqueness on MATH due to CITE. Assume MATH consists of a single positive NAME twist along MATH. Then the manifold MATH is obtained from MATH by a NAME surgery along MATH with surgery coefficient one less than the framing induced on MATH by the fiber. But we can also make MATH a Legendrian curve in MATH so that the framings given by the contact structure and the fibers agree. (In other words, the twisting number of MATH relative to MATH is zero.) This is made possible by applying the Legendrian Realization Principle. Note that to apply the Legendrian Realization Principle, a fold may be necessary (for details see CITE). Thus MATH is obtained from MATH by a Legendrian surgery and hence is NAME fillable. Now, if MATH is the product of MATH positive NAME twists, we perform MATH . Legendrian surgeries on different leaves, in order. |
math/0102162 | If MATH is NAME fillable, then we are done by REF . Therefore, let MATH be a contact structure which is not NAME fillable. By REF , there exists an open book decomposition for MATH which is adapted to MATH. Let MATH be the binding, MATH the fibering of the complement, MATH the fiber, and MATH the monodromy map. Since MATH is not NAME fillable, any product decomposition of MATH into NAME twists must contain some negative NAME twists. We view each NAME twist as being done on a separate fiber. On a fiber just after one on which a negative NAME twist was done along MATH, we can take a parallel copy of MATH and perform a positive NAME twist, which is tantamount to a Legendrian surgery. If a compensatory positive NAME twisted is added whenever there is a negative NAME twist, then we will have a new monodromy map MATH with only positive NAME twists. Of course MATH will define a different manifold MATH and a different contact structure MATH . However, since the difference in between the monodromy for MATH and for MATH is just several positive NAME twists, we can get from MATH to MATH by a sequence of Legendrian surgeries. Thus we have a NAME cobordism from MATH to MATH . |
math/0102162 | Let MATH be two overtwisted contact manifolds. It is a well-known fact in REF-manifold topology that we can find a link MATH in MATH such that a certain integer NAME surgery on MATH will yield MATH . Thus we can construct a topological cobordism MATH from MATH to MATH by attaching REF-handles with the appropriate framing to MATH . Moreover, one can adapt the proof of REF to show that we may assume that MATH has an almost complex structure with complex tangencies MATH on MATH . We now apply REF : Let MATH be a compact, almost complex (real) REF-manifold with boundary MATH. Assume MATH is MATH-concave, MATH is integrable near MATH, and the corresponding contact structure MATH is overtwisted. If the cobordism MATH from MATH to MATH consists of only REF-handle attachments, then there exists a deformation of MATH (rel MATH) to an integrable complex structure MATH on MATH. Using this theorem, we obtain a NAME structure on MATH for which the complex tangencies on MATH are MATH and on MATH are some contact structure MATH homotopy equivalent to MATH . Now, we are done if MATH is overtwisted, since overtwisted contact structures are classified by their REF-plane field homotopy type CITE. But we can easily ensure that the contact structure on MATH is overtwisted by adding some extra NAME twists to MATH that are disjoint from the regions where REF-handles are attached. |
math/0102162 | Given MATH, take a Legendrian curve MATH and its standard neighborhood MATH. Choose a framing as in REF so that the slope of the dividing set of MATH is MATH. Now, identify slopes MATH with their respective ``angles", MATH. In order to distinguish the different amounts of ``wrapping around", we will choose a lift MATH instead. There exists an exhaustion of MATH by concentric MATH, where the angles of the dividing curves on the tori monotonically increase over the interval MATH as the MATH move towards the core. Now, let MATH be the overtwisted REF-manifold obtained by performing a full NAME twist along MATH. This replaces MATH by the solid torus MATH, where the angles of the dividing curves of an exhaustion by tori monotonically increase over the interval MATH. We claim that a full NAME twist MATH is the inverse process of a sequence of Legendrian surgeries along the same core. In fact, take a Legendrian curve MATH in MATH in the same isotopy class as MATH, whose standard neighborhood MATH has an exhausting set of tori which spans the interval MATH. After Legendrian surgery, the new MATH ``rotates" in the interval MATH. Repeated application (total of REF times) of Legendrian surgery will get us back to MATH. Note, however, that the intermediate manifolds are not necessarily diffeomorphic to MATH. |
math/0102162 | Given any overtwisted contact structure MATH, we know by REF that there is a NAME cobordism MATH from MATH to MATH . Let MATH be any closed symplectic REF-manifold. Use NAME 's theorem to excise a small standard ball around a point in MATH and obtain a manifold MATH with concave boundary MATH . We then obtain a concave filling of MATH by gluing MATH to MATH. It is clear that there are infinitely many choices for MATH that will yield infinitely many different concave fillings for MATH. |
math/0102162 | Let MATH be NAME filled by MATH . According to REF, there is a symplectic embedding of MATH into a compact NAME minimal surface MATH of general type. If we take MATH, then MATH will be a concave symplectic filling of MATH . A slight modification of the above argument will produce infinitely many concave fillings. Specifically, in a small standard REF-ball MATH, there exist a right-handed Legendrian trefoil knot with MATH and a linking Legendrian unknot with MATH. If we add REF-handles to MATH along these Legendrian knots, we obtain a new NAME manifold MATH. Embed MATH in a compact NAME surface MATH and remove MATH to obtain a concave symplectic filling MATH of MATH . In the layer MATH in MATH there exists a symplectically embedded torus MATH (see CITE). Let MATH be the elliptic surface obtained by taking the normal sum CITE of MATH copies of the rational elliptic surface along regular fibers. Then consider the symplectic manifold MATH, obtained by taking the normal sum of MATH along MATH and MATH along a regular fiber. These concave fillings of MATH are not related by blowing up and down, since if they were then the compact manifolds MATH obtained from MATH by normal summing with MATH would also be so related. However, this is not the case, as MATH and MATH is unchanged by blowing up and down. |
math/0102164 | It is sufficent to verify that the right hand side of the Fay's formula has zero MATH-periods. Writing MATH, where MATH is a regular bidifferential on MATH holomorphic with respect to MATH and anti-holomorphic with respect to MATH, we get by NAME theorem that the periods of MATH with respect to the variable MATH are the same as periods of MATH. Next, it follows from REF and the NAME theorem that MATH . Here MATH is a pull-back of the bidifferential MATH by the map MATH, defined by MATH. These properties uniquely characterize MATH as a kernel of the NAME projection operator MATH onto the subspace of harmonic MATH-forms on MATH, so that MATH . Therefore MATH and the MATH-periods of the right hand side of the Fay's formula are indeed MATH. |
math/0102164 | It is a straightforward application (compare CITE) of the NAME summation formula MATH where MATH is a function of the NAME class and MATH is its NAME transform, where MATH is the standard Euclidean inner product in MATH. Let MATH be the basis for the lattice MATH dual to a symplectic basis MATH in MATH, so that for MATH . Using MATH and the formulas MATH where MATH, the instanton part MATH can be represented as the following theta-series MATH . Here MATH is the following MATH matrix MATH and MATH . We apply the NAME summation formula to the function MATH . The inverse matrix MATH is readily computed MATH (this is the place where the extra term MATH in the topological action is crucial, compare CITE), so that MATH . The Gaussian integral for MATH is computed explicitly MATH so that MATH . Using the modular transformation formula for the NAME theta-function completes the proof. |
math/0102164 | Directly follows from the definition of normalized reduced correlations functions and the arguments used above. Note that when MATH all reduced correlation functions vanish. |
math/0102164 | As in CITE, we start with the following NAME problem. Find functions MATH and MATH that are holomorphic in the domains MATH and MATH respectfully, are smooth in the corresponding closed domains and on their common boundary MATH satisfy MATH . With the normalization MATH this problem has a unique solution given by the NAME integral MATH where MATH for the MATH sign and MATH for the MATH sign. Now let MATH be a deformation of MATH satisfying conditions in REF . By the calculus formula, MATH so that on MATH . Since in some neighborhood of MATH we see that if MATH for all MATH then MATH in MATH. This implies that on MATH and MATH admits holomorphic continuation as a MATH-form on MATH. Since MATH and MATH we get MATH, so that MATH is also regular at MATH. Since, by definition, REF-form MATH is pure imaginary on MATH, by NAME reflection principle it can be analytically continued to a holomorphic MATH-form on MATH. Since MATH has genus REF we conclude that MATH. In particular, MATH on MATH, which by REF is equivalent to the condition that vector field MATH corresponding to the deformation MATH is tangential to MATH. This proves REF . For the proof of REF , set MATH. We have for MATH so that the MATH-form MATH admits holomorphic continuation to the domain MATH and is regular at MATH. Similarly, for MATH the MATH-form MATH is holomorphic on MATH with a simple pole at MATH with residue MATH. As before, we conclude that REF-form MATH admits a meromorphic continuation to MATH with only poles at MATH and MATH. For MATH they are simple poles with residues MATH and MATH, so that using the global coordinate MATH on MATH we get MATH . In particular, MATH on MATH and MATH on MATH. For MATH we get MATH . Similarly, setting MATH we get for MATH . From here we conclude that for MATH the MATH-forms MATH are meromorphic on MATH with the only pole at MATH and MATH . Similarly, REF-forms MATH are meromorphic on MATH with the only pole at MATH and MATH . Using characteristic property of the NAME polynomials MATH we get that on MATH and on MATH . The proof of REF is now straightforward. By definition of the harmonic moments MATH, the calculus formula and the NAME theorem we get MATH so that REF-forms MATH correspond to the vector fields MATH on MATH. In particular, the same formula shows that corresponding differentials MATH - MATH-forms on MATH, are given by REF-forms MATH on MATH and satisfy MATH . |
math/0102164 | It is another standard computation. Consider the Gaussian integral MATH and make the change of variables MATH, where MATH is uniquely determined by the condition that MATH does not contain linear terms in MATH and by the normalization MATH. Using the NAME theorem we get MATH where MATH. The function MATH is smooth, has compact support and MATH so that MATH . Since MATH we finally obtain MATH . Multiplying by the regularization factor and passing to the limit MATH we see that MATH is well-defined and is given by the formula above. |
math/0102164 | Indeed, the energy MATH of a NAME measure MATH on MATH is defined by (see, for example, CITE) MATH . In our case, due to the presence of a delta-measure, we formally have MATH. However, with the above regularization MATH. |
math/0102164 | Immediately follows from the definition of the metric MATH in REF. |
math/0102165 | The justification can be made with one inequality. Note that the inequality below is not derived by NAME, but by the fact that all the terms are positive and the left hand side is just the diagonal of the product of the sums on the right. MATH . |
math/0102165 | To see that MATH embeds in MATH we look at what the NAME transform does on MATH. Let us define MATH . Now let MATH so MATH . It is easy to see from here that the NAME transform is an isometry between these two spaces. Further, since the MATH norm is bounded by the MATH norm, we see that MATH for MATH. |
math/0102165 | Consider the function MATH. Clearly MATH. Now REF from CITE states that a positive real-valued function is in MATH iff MATH. However computation shows that MATH for the above MATH. These are square summable but not summable. |
math/0102165 | Let MATH in MATH and consider MATH. MATH . |
math/0102165 | To see this we apply a standard technique when using the NAME transform. Applying the NAME transform to the frame operator of the system MATH we get: MATH . This implies that the system MATH has a finite upper frame bound iff MATH. In view of this and the proposition above all we need to show is that MATH bounded. This follows easily from the fact that MATH . |
math/0102165 | First we point out that because we are considering the periodic extension of the NAME transform in MATH we really only need MATH in place of MATH. Given MATH we have MATH. The result follows from the above inequality. |
math/0102165 | As usual, to highlight the essential components of the argument we prove the case MATH. Let us point out that MATH need not be in MATH even if MATH is, since MATH. By the theorem above it is enough to show that MATH has a finite upper frame bound less than MATH. In view of REF it is enough to show MATH. MATH . Where the fact that MATH follows from REF . In this case we get frame bounds MATH and MATH. |
math/0102165 | CASE: This follows directly from the theorem above since MATH is a subspace of MATH. CASE: For any MATH consider MATH. Then MATH and hence MATH. Since MATH is a modular frame for MATH, MATH . So we are done by the previous theorem. |
math/0102166 | We show the moduli space MATH as the minimal blow-up of the NAME group MATH modulo the action of MATH inside the translation subgroup. Note that there are MATH simplices in MATH, each with a particular ordering given by the hyperplanes. It follows from REF that the iterated blow-ups of MATH do indeed give a tiling by cyclohedra MATH. The affine braid arrangement is simply a way of keeping track of all the gluings defined by the twisting operation in a global setting. Adjacent polytopes can be accessed by crossing a hyperplane REF or an intersection of hyperplanes, thereby permuting their labelings. Indeed, our notation of labeled polygons with twisting of diagonals mimics the information found in the hyperplane arrangement. |
math/0102166 | We show that the blow-up of a minimal element MATH results in MATH. From the combinatorics of the affine braid arrangement, it is not hard to see that each MATH corresponds to a choice of MATH elements from the set MATH. Choose an arbitrary minimal cell and assign it such a choice, say MATH. We view this as a centrally symmetric MATH-gon having a diagonal partitioning it into a pair of MATH symmetrically labeled sides MATH, with the sides of the central polygon using the remaining labels. Note the correspondence of this dissected MATH-gon to the product MATH. As this minimal cell is blown-up, we see MATH different ways in which the labels can be arranged on the non-central polygons. However, since twisting is allowed along the diagonals, we get MATH different labelings, each corresponding to a MATH. Indeed, this is exactly how one gets MATH, with the associahedra glued as defined above. Therefore, a fixed labeling of the central polygon gives MATH, while allowing all possible labelings gives the result. |
math/0102168 | Let's define four regions as follows: CASE: MATH, CASE: MATH, CASE: MATH, CASE: MATH. For every subset MATH, define MATH . Then MATH . Now, MATH . So MATH and there are exactly MATH more REF's of MATH than REF's of MATH in region MATH. This finishes the proof. |
math/0102168 | First we prove the case of MATH. Note that MATH since MATH. Let MATH. Choose MATH as large as possible such that MATH and MATH (see REF). Such a MATH must exist since MATH. Now choose MATH as small as possible such that MATH and MATH for all MATH with MATH. Then there exists a MATH, MATH such that MATH. By construction, MATH . That is (in the notation of REF), MATH, MATH and MATH. So by this lemma, there exists a MATH such that MATH and MATH. Then MATH, so by REF , MATH. This proves our claim. To prove the case of MATH, it is easiest to use dual rank and difference functions: MATH . One can check that MATH if and only if MATH and then argue as above using this new rank function. (Note, to define MATH using MATH, we have to modify our fixed flag MATH and REF .) |
math/0102168 | The first equivalence follows from REF by comparing MATH and MATH. The second follows from the first by noting that MATH whenever MATH. |
math/0102168 | Pick some MATH. Now, MATH . We know by REF that MATH. Applying REF with MATH and MATH, we see that MATH is also REF. This proves REF follows immediately from REF and the definition of MATH by comparing MATH and MATH. REF follows from the first two parts along with REF (stated below). |
math/0102168 | Notice that MATH for MATH and MATH for MATH. By REF, MATH if and only if MATH. Since MATH, MATH. Combining this with our first observation implies that MATH. |
math/0102168 | First we show that MATH - that is, MATH for all MATH. It is clear from the definition that MATH. So this amounts to showing that one of the two dotted arrows in REF (corresponding to relation under the NAME order) exists. First suppose MATH - hence MATH. We wish to show that MATH. Letting MATH and MATH, this reduces to showing that MATH. Now MATH implies that MATH. So MATH if MATH. But this last inequality holds by choice of MATH. Now suppose that MATH. In all of the cases in REF except A.iii and B.iii, MATH. Since MATH and MATH agree at all indices except MATH, we can use REF to conclude that MATH. So MATH for all cases except possibly A.iii and B.iii. In Case A.iii, we know that both MATH, MATH - that is, MATH. Since MATH and MATH, we can therefore invoke REF to conclude that MATH. CASE:iii is similar. It is clear from REF that MATH can share at most one index with MATH. As we have already shown the inclusion MATH, we conclude that MATH. Now we show that MATH is an injection. Suppose MATH, MATH. No matter which case of REF we are in, we see that MATH or MATH. In other words, the index MATH doesn't share with MATH must be an index of MATH. In particular, if MATH such that MATH, the index they don't share with MATH must be the same. It is then easy to check by inspection of REF that MATH implies that MATH. Since MATH whenever MATH and MATH don't share any indices, we conclude that MATH is, in fact, an injection. |
math/0102168 | The last statement follows immediately from the decomposition of MATH. So, taking advantage of REF, we need only show that if MATH, then MATH. First consider the case where MATH. Then MATH. As MATH, we can apply REF to conclude that MATH. Finally, since MATH, MATH. So MATH in this case. Now, as in the proof of injectivity, we can restrict to the cases where MATH. We group into cases according to MATH, MATH and MATH. Cases MATH have MATH; Cases MATH have MATH. In each case we show that MATH. Cases I-VI are the only ones possible since MATH cannot have MATH as hypothesized. |
math/0102168 | By REF, MATH. The result follows from the identity MATH. |
math/0102168 | Suppose MATH. We will show that MATH. First, consider the case where MATH. From the definition of MATH, we see that MATH. So MATH. This implies that MATH and therefore MATH. Now we suppose MATH. So MATH are determined such that MATH implies MATH. Let MATH. Note that: CASE: By hypothesis, MATH. CASE: If MATH then at most one of MATH; not two. Hence, the cases below are the only ones we need consider. CASE: MATH. Then MATH and MATH. By REF, MATH. So MATH and, as we are in Case D.ii of REF , MATH. Similarly, MATH and, as we are in Case C.i of REF , MATH. CASE: MATH. The argument is parallel to that in the previous case. CASE: MATH. Here MATH. (Note that MATH is not considered since MATH.) Suppose MATH. From REF , this implies that MATH . Then, from REF , we see that MATH. The argument is analogous if we instead assume MATH. |
math/0102168 | We only prove REF as the proof for REF is entirely analogous. Diagrams for MATH, MATH and MATH are given in REF. We see that MATH implies MATH and MATH implies MATH. So, by REF, MATH. Equivalently, MATH and MATH. So, (Case C.ii of REF ) MATH and (Case D.i of REF ) MATH. |
math/0102168 | The proof of Patch Incompatibility is clear. To prove Link Incompatibility, it suffices to consider Cases A and B of REF . |
math/0102168 | For any point MATH, MATH. Similarly for the pair MATH and MATH. Now, MATH implies that MATH and MATH on MATH. This implies that MATH and MATH. |
math/0102168 | First consider the case where both MATH. Suppose there is a point MATH in region A of REF. Choose such a MATH as small as possible. Then we see that MATH and MATH are patch incompatible reflections for MATH and MATH. By REF , MATH is then singular. This contradicts the fact that MATH is an msp for MATH. Now suppose region A of REF is empty - this is shown in REF. Then MATH and MATH are incompatible reflections for MATH and MATH. Since MATH, this contradicts the fact that MATH is an msp for MATH. We now argue the case of MATH. (The arguments for MATH are parallel.) Clearly MATH and MATH. There are four possibilities with regard to the position of MATH. CASE: MATH. We are in Case A.iii of the definition of MATH. Hence, MATH, which is a contradiction. So this case cannot occur. CASE: MATH, MATH. This case cannot occur as it violates MATH. CASE: MATH, MATH. This case is depicted in REF. Suppose MATH. Then MATH and MATH are patch incompatible for MATH. This contradicts the fact that MATH is an msp for MATH. If MATH, then we can argue as in REF to obtain our contradiction. CASE: MATH, MATH. See REF. This is analogous to the previous case. |
math/0102168 | Suppose that MATH and MATH. We will obtain a contradiction. Let MATH. Choose MATH as large as possible such that MATH. Note that MATH and MATH. Since MATH is an msp, we can invoke REF to find a MATH. Suppose MATH. Since MATH, MATH is shaded so MATH. Hence MATH and MATH are incompatible for MATH and MATH. This contradicts MATH. Otherwise, MATH overlaps both MATH and MATH, so, by REF, we are in one of the following two scenarios. CASE: MATH. By choice of MATH, MATH. (Note that MATH.) So either MATH, MATH or MATH, MATH (the latter case is shown in REF.). In either case, we can apply the Ell REF to conclude that MATH. This contradicts the choice of MATH. CASE: MATH. Since MATH, for MATH to be an element of MATH, we need MATH, MATH and MATH for some point in each of regions A and B in REF. But then MATH and MATH are link incompatible for MATH. Furthermore, by having chosen MATH as large as possible, we ensure that MATH. This contradicts MATH. |
math/0102168 | We can visualize the situation as in REF. Since MATH, there is necessarily a point MATH in region A for which MATH. Suppose MATH. Then there is a point MATH in region B such that MATH. Then we can apply REF (with MATH, MATH) and REF to conclude that there is a point MATH of MATH in region MATH (see REF). If we choose MATH to be as low as possible in our diagram, then MATH (see REF). But then MATH and MATH are patch incompatible for MATH. This contradicts MATH. Therefore MATH and we can shade the entire region B. To shade the lower left corner, apply the preceding argument to MATH and MATH. |
math/0102168 | We only prove that MATH must avoid the configuration in REF. The proof of the other case is parallel. First suppose that MATH (see REF). Then by the Cross REF , MATH. But this contradicts REF as MATH. We get a similar contradiction if MATH. So we can henceforce assume that our configuration is actually as in REF - that is, MATH, MATH. Since MATH is an msp, we must have some reflection MATH. Suppose MATH . In the case of REF , MATH and MATH are incompatible for MATH. This contradicts MATH. (Similarly for REF .) So these intersections must be non-empty. It is clear from REF that for these intersections to be non-empty, we need MATH to be patch (rather than link) incompatible with each of MATH, MATH and MATH. By REF and because the intersections in REF must be non-empty, it is readily seen that we require MATH and MATH. Hence, there are only four possible ways in which MATH may overlap MATH, MATH and MATH. These are shown in REF. CASE: MATH, MATH. If MATH then MATH and MATH are patch incompatible for MATH. This contradicts MATH. Otherwise, we can apply the Cross REF to MATH and MATH to conclude that MATH. This contradicts REF. CASE: MATH, MATH. The argument is parallel to the previous case. CASE: MATH, MATH. If MATH or MATH then we can apply the Cross REF to MATH and MATH to conclude that MATH. This contradicts REF. The only alternative is that MATH on regions C and D. But then MATH and MATH are patch incompatible for MATH. This contradicts MATH. CASE: MATH, MATH. Here MATH and MATH are patch incompatible for MATH. This contradicts MATH. |
math/0102168 | Since MATH is an msp for MATH, there exists some MATH. Clearly, if MATH then MATH and MATH are (patch or link) incompatible reflections for MATH. This would contradict MATH. So, to ensure that REF does not hold, we need MATH in region A of REF and MATH in region B. Here we are including the possibilities that MATH or MATH. Note that (as is shown in REF) REF requires that MATH and MATH. Clearly if MATH then MATH and vice versa. Hence, by symmetry, we can treat only the cases where MATH. These two cases are illustrated in REF . For both cases, we can apply the Cross REF to MATH and MATH to conclude that MATH. Then MATH, which contradicts REF. |
math/0102168 | By passing to inverses, it is enough to prove that MATH is REF-avoiding. So choose MATH with MATH such that MATH. CASE: Assume MATH. We have the situation of REF. By definition of MATH, there exists a MATH with MATH. NAME assume that MATH as all cases where MATH are analogous to one of the cases we cover by transposing over the anitdiagonal. Clearly MATH. Since MATH is an msp, there exists a MATH. We claim that MATH . Suppose the intersection is empty. Then MATH and MATH are (patch or link) incompatible for MATH. This contradicts MATH. So we may assume that this intersection is non-empty. There are three cases according to whether MATH and MATH are CASE: link incompatible CASE: patch incompatible with MATH. CASE: patch incompatible with MATH. We only describe the arguments explicitly in the case of link incompatibility - the arguments are similar in the latter two cases. We will argue only MATH as the case of MATH is analogous. By NAME REF, there are three possibilities for the relative positions of MATH and MATH. They are displayed in REF. CASE: MATH (that is, MATH). We have that MATH and MATH are link incompatible for MATH. This contradicts MATH. CASE: MATH (that is, MATH). The argument is the same as in the previous case. CASE: MATH (that is, MATH). If MATH, then MATH and MATH are link incompatible for MATH. This contradicts MATH. Otherwise, we can apply the Cross REF to MATH and MATH to conclude that MATH. Then, as in REF, MATH and MATH are link incompatible for MATH. This contradicts MATH. CASE: Assume MATH. Since MATH, we can find MATH such that MATH. If CASE: MATH is in region A or A' of REF or CASE: MATH is in region C or C' of REF or CASE: MATH is in region E or E' of REF, then we can reduce to the previous case (of MATH) or we violate REF. So we must have CASE: MATH CASE: MATH CASE: MATH . As the argument of MATH is analogous, we will assume MATH. We now argue the four cases we have left according to whether MATH and whether MATH. CASE: MATH, MATH. See REF. Note that we can reduce to REF if MATH. So we assume MATH. We have already shown in REF that this configuration contradicts MATH. CASE: MATH, MATH. Since MATH is an msp, there exists some MATH. As MATH, one can see in REF that MATH. Hence, MATH and MATH are (patch or link) incompatible for MATH. CASE: MATH, MATH. See REF. We have already shown in REF that this case contradicts MATH. CASE: MATH, MATH. See REF. Note that if MATH, then in this case we can reduce to REF (with MATH). So we assume MATH. We have already shown in REF that this case contradicts MATH. This completes the proof that MATH is REF- and REF-avoiding. |
math/0102168 | Suppose we have MATH with MATH and MATH. We will obtain a contradiction. By REF, no points of MATH may occur in regions I or II of REF. Since MATH, there exist MATH such that MATH. As MATH is an msp, there also exists some MATH. Using REF, it is easy to check that if MATH then MATH and MATH are incompatible for MATH. This contradicts MATH. So, in particular, MATH. A typical (allowable) pair of positions for MATH and MATH is shown in REF. As we have now constructed MATH with MATH, we will assume that MATH and MATH were chosen initially such that MATH. Note that by the above construction, we can assume that MATH. Suppose MATH. Then MATH and (MATH or MATH) are link incompatible for MATH. This contradicts MATH. So our diagram looks like that pictured in REF and we have MATH. Therefore, we can find a point in each of the regions U V such that MATH. Choose the point in region U to be as low as possible. Choose the point in region V to be as far right as possible. Such points are shown in REF. Apply REF to the rectangle determined by these two points with MATH and MATH. This, along with REF, implies that there is another point MATH in either region P or Q. Without loss of generality, assume it is in region P. By having chosen the point in region U as low as possible, we find that MATH (see REF). Hence, MATH and MATH are link incompatible for MATH. This contradicts MATH. |
math/0102168 | For brevity in the following, we use the convention that MATH and MATH. CASE: If MATH or MATH, then by REF, MATH. This contradicts REF : We split this proof into proving the following facts: CASE: MATH. CASE: MATH for all MATH and MATH. CASE: MATH. CASE: MATH is as in the statement of REF . We now prove these claims. CASE: MATH. Assume MATH. We will obtain a contradiction. By REF, we can find MATH such that MATH (see REF). Choose MATH as large, MATH as small as possible subject to this restriction. If MATH, then an application of the Cross REF would offer the desired contradiction. So assume that this is not the case (that is, assume MATH on region R of REF). Since MATH is an msp, by REF , we can find some MATH. Recall that we chose MATH as large as possible such that MATH. It follows then that MATH. Similarly, our choice of MATH as small as possible such that MATH, in conjunction with the Cross REF and Ell REF , implies that MATH. Suppose MATH. This is depicted in REF. We see that MATH and MATH are patch incompatible for MATH. This contradicts MATH. So we may assume MATH as in REF. Suppose that MATH on region A. Then MATH and MATH are patch incompatible for MATH. This contradicts MATH. So there is at least one point in region A for which MATH has value REF. Now we can apply the Cross REF to the patch incompatible pair MATH to conclude that MATH on regions B and C. We display this knowledge in REF. Now suppose that there is a point in region D for which MATH. Then MATH and MATH are patch incompatible reflections for MATH. This contradicts MATH. Since, by construction, MATH, the only possibility left is that MATH (as in REF). We can now apply the Cross REF to MATH and MATH to conclude that MATH on region F. Hence MATH as claimed. The proof that MATH is entirely analogous when one uses MATH from REF . CASE: MATH for all MATH and MATH. By the previous step, we know that we can shade regions I and II in REF. For every MATH, by the definition of MATH, we can shade the corresponding regions U and V, respectively. This completes the claim. CASE: MATH. Suppose, on the contrary, that MATH for some point on region A in REF. Since MATH is non-decreasing as we move down or left in region A, we can assume that MATH for MATH. But then there must be some MATH with MATH and MATH with either MATH or MATH and MATH (see REF). Note that MATH by choice of MATH. If MATH then MATH and MATH are patch incompatible for MATH (see REF). This contradicts MATH. So we can assume MATH. Since MATH, and MATH is non-decreasing in region A as move down or left, we can now assume that MATH. But then MATH and MATH are patching incompatible for MATH (see REF). This contradicts MATH. So MATH as claimed. CASE: By the previous step, there is at most one point of MATH in region A. But as MATH, MATH and MATH, this fixes all the remaining points and we see that MATH as claimed. This is displayed in REF. |
math/0102168 | Again for purposes of brevity, we'll assume throughout this proof that MATH, MATH and MATH. We now prove a series of claims elucidating the structure of MATH. As the chain of reasoning has several steps, we summarize them here before beginning. CASE: There exist MATH such that MATH. CASE: MATH and MATH are link incompatible. CASE: Such MATH and MATH exist for any MATH. CASE: MATH for any triple MATH. CASE: Conclude that we can shade the diagram as in REF. CASE: MATH in REF. CASE: REF holds. CASE: MATH is as claimed in REF . CASE: REF holds. CASE: There exist MATH such that MATH. Suppose there is no such triple of indices. Then by the definition of MATH, for given MATH there exist MATH REF such that MATH. By the Ell REF , along with the assumptions that MATH is an msp and that such triples do not exist, we can find MATH and MATH such that MATH (illustrated in REF). But then MATH and MATH are patch incompatible for MATH. This contradicts MATH. So we must be able to find a triple as claimed. CASE: If MATH then MATH. Suppose, on the contrary, that MATH. Then MATH on either all of region A or all of region B in REF. Assume that MATH on region A. Now, since MATH is an msp, by REF there exists MATH. We now consider the two possibilities for the relative positions of MATH and MATH. Suppose that MATH and MATH are link incompatible - that is, we have REF . For MATH to be link incompatible with MATH, we need MATH (as depicted in REF) since we are assuming MATH. Additionally, as MATH, MATH must have value REF for at least one point on each of regions C and D. Thus MATH and MATH are link incompatible for MATH. This contradicts MATH. On the other hand, MATH and MATH may be patch incompatible. Then there are four possibilities for the relative positions of MATH, MATH, MATH and MATH depending on whether MATH and whether MATH (see REF). In each situation, MATH and MATH are patch incompatible for MATH. This contradicts MATH. We have obtained a contradiction for every scenario in which MATH on region A. Arguing similarly if MATH on region B, we conclude that MATH. CASE: Given MATH, there exist MATH such that MATH. By REF , there exist MATH such that MATH. If MATH then we are done - so assume not. We can at least find a MATH with MATH. Without loss of generality, assume MATH for some MATH. We split into cases according to whether MATH, MATH or MATH. These are depicted in REF. Note that by the previous step, MATH, so MATH. In addition, if MATH, then MATH as desired. So in the following arguments (and REF), we assume MATH and derive a contradiction. Assume MATH. If MATH, then MATH are link incompatible for MATH. This contradicts MATH. Otherwise, by the Cross REF , MATH. Then MATH are link incompatible for MATH. This contradicts MATH. If MATH, then MATH and we get a contradiction as above. CASE: For every MATH, we have MATH. Suppose MATH. By the definition of MATH and the fact that MATH, we know that there exists MATH such that MATH. Now we can apply the previous step to obtain a MATH such that MATH. Note that by REF , MATH and MATH are link incompatible. So our situation is as depicted as in REF. Using the logic of the previous step, we see that MATH contradicts MATH. Hence MATH as desired. The argument for showing MATH is analogous. CASE: We can shade our diagram as in REF. This follows immediately from the previous four steps. CASE: MATH in REF. We start by showing that if MATH and MATH, then MATH implies MATH. By REF , MATH for some MATH (see REF). Suppose MATH - that is, MATH on region A. Then MATH and MATH are link incompatible for MATH. This contradicts MATH. A similar argument can be used to show that if MATH and MATH, then MATH implies MATH. The claim of MATH then follows by inspection from these two facts. CASE: REF holds, namely MATH. Recall that in REF we showed that MATH for any MATH. This implies that we can shade MATH as in REF. By REF , MATH and MATH are link incompatible. This implies that MATH. Similarly, MATH. It then follows from our explicit description of MATH that the values of MATH for MATH are as shown in REF. Arguing with MATH (see REF) and region B, we see that MATH for MATH is as shown in the same figure. But this means that MATH and MATH. This can only happen if MATH. CASE: MATH is as stated in REF tells us that we can conclude that MATH on all shaded areas of REF. Therefore, MATH for MATH. A similar argument to that in REF shows that MATH for MATH. So we need only investigate the values of MATH for MATH. To do this, assume that MATH for MATH for some MATH with MATH. Then, as in REF, we see that MATH on region B. Therefore, MATH and MATH are link incompatible for MATH. This contradicts MATH. Hence MATH for all MATH with MATH. So MATH as desired. CASE: REF holds. We need to show that if MATH then MATH. So assume MATH. By REF , MATH and MATH are patch incompatible reflections for MATH (see REF). This contradicts MATH. The argument showing that MATH is analogous. This completes the proof of REF. |
math/0102168 | REF tells us that REF is necessary. REF tell us that REF are necessary. So all we need to show is sufficiency. Let MATH be a reflection such that MATH. As MATH is injective, to calculate MATH from MATH we need only count how many reflections in MATH are not in the image of MATH. Note by REF that MATH and MATH. Consider first the case shown in REF of two decreasing sequences for MATH. Note that MATH . Hence, MATH . Since MATH, REF is negative. So by REF, MATH is a singular point of MATH. To prove that it is a maximal singular point, we consider some MATH and let MATH. Then, viewing REF, it is easily seen that MATH. Since MATH, by REF , MATH is a smooth point of MATH. Since MATH was chosen as an arbitrary cover of MATH, MATH is an msp for MATH. Now we prove the case shown in REF of three decreasing sequences for MATH. Note that MATH . Hence, MATH . Since MATH and MATH, REF is negative. So by REF, MATH is a singular point of MATH. To prove it is an msp for MATH, as above we consider some MATH and let MATH. We have MATH. Viewing REF, it is clear that MATH. If MATH, then MATH. If MATH, then by REF, we have that MATH, and MATH. So, in either case, MATH is a smooth point of MATH. So in both cases, MATH is an msp of MATH as claimed. |
math/0102168 | We only give the argument for singular points of the type REF (that is, those described in REF). The argument for singular points of type REF and MATH is analogous. We start by proving that MATH. To do this, fix some MATH (of type REF). We will choose indices MATH and MATH as described in the definition of MATH, and show that REF is satisfied for our choice of indices. So, using the notation of REF, let MATH correspond to a type REF pattern in MATH. Then set MATH . Now, recall from REF that MATH if and only if MATH is everywhere non-negative. But then REF follows from REF along with the observation that MATH in each of these diagrams. Now we need to show that any MATH satisfying REF for some MATH and some set of indices is a singular point of MATH. Since MATH by hypothesis, we see that MATH. Combining this with REF and the fact that MATH, we conclude that MATH . By hypothesis, we also know that MATH. Finally, recall that MATH . One possible configuration of the points of REF is shown in REF. Note that REF implies that MATH on region MATH. Hence, MATH are patch incompatible for MATH. If MATH, then we can conclude that MATH is a singular point of MATH by REF . Otherwise, consider REF. Pick the index MATH as small as possible such that MATH. Then MATH are patch incompatible for MATH. But then MATH is a singular point MATH as MATH by construction. Carrying out the analogous arguments for REF and MATH type singularities (using link incompatible reflections) completes the proof of Conjecture REF. |
math/0102168 | Fix MATH and pick some MATH. We know by REF that MATH. By REF , this implies that if MATH for some MATH, then MATH. With these facts, the result then follows easily by induction on MATH using REF . (Note that our base case of MATH is trivial.) |
math/0102168 | We apply induction on MATH. The case of MATH can be checked from REF . We assume then, without loss of generality, that MATH. Also, for brevity we will often write MATH and MATH in place of MATH and MATH, respectively. Consider MATH. The pairs MATH and MATH and MATH are shown in REF. We claim that MATH . The second equality follows from REF and the induction hypothesis. To obtain the first equality, we notice that MATH. Then, since MATH, REF follows from REF . Substituting this information into REF , we obtain MATH . Now we need to investigate the possible terms in the sum of REF . We will first determine which MATH with MATH can contribute to REF . By REF , if MATH then MATH. To have MATH while MATH, we need MATH. Since MATH is the only msp for MATH, this tells us that MATH. But by REF , MATH, so this implies that we must have MATH by the degree bound. It is then easy to check from REF that MATH if and only if MATH. We now split into two cases depending on the relative values of MATH and MATH. CASE: MATH. Since MATH, we know by the previous paragraph that the only MATH that can contribute to the sum in REF are those with MATH. Furthermore, since we are summing only over MATH for which MATH, the only possibility is MATH. In this case, by induction, along with REF , MATH. So MATH . CASE: MATH. MATH will contribute as in the previous case. However, from our discussion above, MATH will also contribute. For MATH, we have MATH and MATH. Plugging this term into REF , along with the term coming from MATH, we get MATH . |
math/0102168 | We apply induction on MATH. The case of MATH is easy to check using REF . For brevity, we abbreviate MATH and MATH by MATH and MATH, respectively. Let MATH. Now, as seen in REF, MATH is REF- and REF-avoiding, hence smooth CITE. Therefore, MATH for all MATH. Clearly MATH. Thus the first two terms of REF together contribute MATH. We now show that the sum in REF is empty. Since MATH for all MATH, MATH implies that MATH. But, as seen from REF, no MATH satisfying this length condition can satisfy the additional constraint of MATH. This proves the theorem. |
math/0102168 | We apply induction on MATH. The case of MATH is covered by REF. So we assume MATH. For clarity, we abbreviate MATH and MATH by MATH and MATH, respectively. In REF, we depict the pairs MATH and MATH and MATH. We claim that the first two terms in REF contribute MATH. First consider the pair MATH. Since MATH, by the induction hypothesis, REF , we see that MATH. Now consider the pair MATH. Since MATH and MATH, it follows that MATH. But since MATH, we get that MATH also. Plugging this information into REF , we can write MATH . Now we check which MATH will appear in the sum in REF . First note that MATH is the unique msp for MATH. By induction, MATH. By REF , MATH. Hence, the only MATH such that MATH and MATH is maximized is MATH. However, MATH, so MATH does not appear in the sum. So the only possible terms in the sum are those with MATH. From REF, we see that MATH is the only MATH satisfying both this length condition and MATH. Using REF and the induction hypothesis, one can check that MATH. Hence, the sum in REF contributes MATH. Simplifying, we see that MATH as claimed. |
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