paper stringlengths 9 16 | proof stringlengths 0 131k |
|---|---|
math/0008202 | We have already observed that MATH and thus MATH. In particular, by REF there exists MATH with MATH. CASE: Let MATH. If MATH, then REF yields MATH and the result follows from REF . To show that this is actually the only possible case, assume on the contrary that MATH. Note that MATH is the canonical linear series by RE... |
math/0008206 | CASE: Otherwise there would be sequences MATH and MATH REF such that MATH for all MATH; this implies MATH which shows that MATH has an accumulation point MATH with MATH; but then MATH is also an accumulation point of MATH and therefore an element of MATH - a contradiction. CASE: See CITE, proof of REF . Assume that REF... |
math/0008206 | For any MATH there exists an open conic neighborhood MATH such that MATH such that REF holds with this set of constants on MATH. The sets MATH are open in MATH and form a covering of the compact set MATH. Thus there exist MATH such that MATH . Consequently, MATH is contained in the union of the MATH (MATH). Now set MAT... |
math/0008206 | Choose representatives MATH, MATH with compact support; with the short hand notation MATH we have to estimate MATH in a suitable conic neighborhood of any point MATH in the complement of the right hand side of REF . Let MATH be an open cone containing MATH such that MATH. By REF there exist open cones MATH REF such tha... |
math/0008206 | Let MATH . Then for any MATH with MATH and support sufficiently close to MATH we have MATH (MATH). Also, by REF , since MATH is not contained in MATH there exist open conic neighborhoods MATH of MATH in MATH such that MATH and MATH. Thus, by REF , if the support of MATH is close enough to MATH we also have MATH . Since... |
math/0008206 | This is a straightforward adaptation of the proof of the corresponding distributional result (see for example, REF ). |
math/0008208 | By integration by parts, we obtain that MATH . Using REF we get MATH which proves REF. We now observe that MATH is a solution of the linear ODE MATH . Since MATH and MATH, we have MATH. We also see that MATH whenever MATH and MATH whenever MATH. Since MATH belongs to the interval MATH, MATH must remain in this interval... |
math/0008208 | We have MATH . Applying REF to the last expression, we obtain REF. |
math/0008208 | Let MATH . For MATH, we define the partition times MATH by the relation MATH which is possible since MATH is continuous and decreasing. This definition implies in particular that MATH and, therefore, MATH. Bounding the integral in REF by MATH, we obtain MATH . We have MATH and MATH . Since MATH, this implies MATH and t... |
math/0008208 | M. enumREF CASE: The difference MATH satisfies MATH with MATH-a.s. Now, MATH which implies MATH for all MATH. CASE: Let us assume that MATH. Then we have for all MATH and thus by REF, MATH . The integral on the right-hand side can be estimated by REF, yielding MATH . Therefore, MATH which proves REF because MATH. CASE:... |
math/0008208 | By REF , MATH where MATH, which implies MATH where we have used the relation MATH. |
math/0008208 | By integration by parts, we obtain that MATH which implies that MATH . By our hypothesis on MATH, the first term in brackets is positive. |
math/0008208 | M. enumREF CASE: Let MATH and let MATH be any partition of the interval MATH. We define the events MATH . Let MATH be a deterministic upper bound on MATH, valid on MATH. Then we have by the NAME property MATH . CASE: To define the partition, we set MATH for some MATH to be chosen later, and MATH . Since MATH, we have M... |
math/0008208 | The upper bounds are easy to obtain. For MATH we have, using MATH, MATH . For MATH, the hypothesis MATH implies MATH . For MATH, a similar estimate is obtained by splitting the integrals for MATH and MATH. For MATH, we have MATH . To obtain the lower bound, we first consider the interval MATH, where we use the estimate... |
math/0008208 | Let MATH be a partition of the interval MATH. By REF , the probability in REF is bounded by MATH, where MATH . If MATH, we define the partition by MATH . Estimating MATH as in the proof of REF , we obtain MATH . Therefore, REF holds with MATH. For MATH, we define the partition separately in two different regions. Let M... |
math/0008208 | Assume first that MATH and let MATH. Then we have MATH by assumption. By REF, the difference MATH satisfies MATH . We consider the first exit time MATH . For all MATH in the set MATH and MATH, we have by the hypotheses on MATH and MATH together with REF MATH . Therefore, REF yields MATH . The integral is bounded by MAT... |
math/0008208 | Using integration by parts, we have MATH . The upper bound follows immediately, and the lower bound is obtained by bounding the exponential in the last integral by MATH. |
math/0008208 | M. enumREF CASE: For MATH, we introduce a partition MATH of the interval MATH, which will be chosen later, and for each MATH, we define a linear approximation MATH by MATH where MATH. Assume that MATH for all MATH. Then by REF MATH for MATH, provided the partition is chosen in such a way that for all MATH . CASE: If MA... |
math/0008208 | M. enumREF CASE: Let MATH. By assumption, MATH is non-negative for MATH. The difference MATH satisfies the equation MATH with MATH. Since MATH for MATH, MATH follows for all such MATH and, therefore, NAME 's lemma yields MATH . This shows MATH for those MATH. Now assume MATH and MATH. Then, REF implies that MATH for al... |
math/0008208 | M. enumREF CASE: Since by symmetry, MATH on MATH, we have by the strong NAME property MATH . We now observe that MATH which implies MATH . CASE: Next, we use that MATH is a Gaussian random variable with mean MATH and variance MATH . By REF , MATH and we thus have MATH which proves REF, using REF. CASE: In order to comp... |
math/0008208 | M. enumREF CASE: We first introduce some notations. Let MATH and define MATH. We may assume that MATH for all MATH (otherwise we replace MATH by its maximum with MATH). For MATH we define the quantities MATH . CASE: Let us first consider the case MATH. Recall that MATH. By REF and the strong NAME property, we have the ... |
math/0008208 | M. enumREF CASE: Whenever MATH, we have MATH which shows that MATH can never become larger than MATH. Similarly, whenever MATH, we get MATH provided MATH, which shows that MATH can never become smaller than MATH. This completes the proof of REF. CASE: We now introduce the difference MATH. Using NAME 's formula, one imm... |
math/0008208 | M. enumREF CASE: By integration by parts, we find MATH . The relation MATH together with REF yields MATH . The second term in brackets gives a contribution of order MATH. In order to estimate the contribution of the first term, we perform the change of variables MATH, thereby obtaining MATH where MATH and MATH. The las... |
math/0008208 | Let MATH and define a partition MATH of MATH by MATH . Since MATH, we obtain MATH for all MATH. Now we can proceed as in the proof of REF . |
math/0008208 | Assume first that MATH. We introduce the difference MATH, set MATH, and define the first exit time MATH . On MATH, we get by the estimate REF on MATH, REF MATH for all MATH, which leads to a contradiction for MATH. We conclude that MATH and thus MATH for all MATH in MATH, which proves REF. The inclusion REF is a straig... |
math/0008208 | Let MATH denote the left-hand side of REF. For any MATH, we have MATH where MATH is an (exponential) martingale, satisfying MATH, which implies by NAME 's submartingale inequality, that MATH . This gives us MATH and we obtain the result by optimizing REF over MATH. |
math/0008208 | First note that for all MATH, MATH which implies, by integration by parts, MATH where we have used MATH and MATH. This proves the assertion in the case MATH. In the case MATH, we have MATH where we have used that MATH holds for all MATH. This proves the assertion for MATH. |
math/0008209 | For each chord diagram, a subgraph consisting of its chords is a REF-factor. The chords of all MATH-diagrams constitute the complete graph MATH, the chords of each single MATH-diagram being again a REF-factor of MATH. The action of MATH on MATH induces an action on the set MATH of all REF-factors of MATH. The orbits of... |
math/0008209 | Each permutation MATH has cycle type MATH, MATH. If MATH has a cycle of length MATH the product in REF equals zero, otherwise it reduces to a single term MATH since MATH. The double sum in REF can then be replaced with a single sum over all MATH. The group MATH contains MATH permutations of cycle type MATH. Denoting by... |
math/0008209 | Dividing out the first term in REF gives MATH . We begin with establishing an upper bound for MATH using NAME 's formula in the following form MATH where MATH satisfies MATH (see CITE). We have MATH for MATH, MATH, MATH. Then we find MATH for MATH odd. Using the following estimate MATH (see CITE) for the binomal coeffi... |
math/0008209 | As MATH it follows from REF that MATH where MATH represents the contribution of those permutations of MATH which are not in MATH. Such permutations have cycle type either MATH or MATH and there are MATH permutations of each type in MATH. The product in REF equals MATH for permutations of cycle type MATH and MATH for pe... |
math/0008209 | From REF we get MATH . Clearly MATH for MATH. NAME 's formula shows that MATH and hence MATH. But MATH which completes the proof. |
math/0008210 | From CITE, it suffices to show that the NAME DGAs corresponding to MATH and MATH have nonisomorphic graded homology algebras. The NAME DGA associated to MATH is the free associative unital algebra MATH over MATH generated by the crossings of MATH, which have been labelled in REF as MATH. Of these generators, we calcula... |
math/0008213 | We let MATH have the natural pulled back metric MATH with respect to which MATH becomes a Riemannian submersion with totally geodesic fibers REF . For any MATH we denote with MATH its horizontal lift on MATH. Let MATH, MATH, MATH be the unit Killing vector fields which give the usual MATH-Sasakian structure of MATH (na... |
math/0008213 | Recall that MATH is a connection iff for any MATH and any vertical MATH, the brackets MATH are horizontal. As any horizontal field is of the form MATH, we have MATH hence MATH is horizontal iff MATH and MATH are horizontal. We can take MATH. The above two brackets are surely vertical, thus they will be horizontal iff t... |
math/0008213 | A standard complex structure on MATH and MATH is obtained by applying REF to the highest vertical arrows in the diagram: MATH where REF is applied to zero level sets MATH and MATH to obtain their induced NAME structures on them. The same diagram tells that MATH and MATH are bundles in NAME surfaces MATH over the comple... |
math/0008214 | From the upper semicontinuity it follows that for each MATH there is a MATH such that for any MATH . Now applying these inclusions and the assumption of the theorem, we obtain the following: for any MATH . Consider the open cover of MATH given by MATH . From NAME 's Lemma CITE it follows that there exists a locally fin... |
math/0008214 | We know that MATH . Let MATH be the projections. Then there is MATH such that MATH . Assume that MATH and suppose MATH where MATH . Let MATH . Suppose MATH . For MATH let MATH . Then MATH . We have an element of MATH . Consider MATH . Thus MATH is a fixed point of MATH . |
math/0008214 | Assume that there is a convex neighborhood MATH of MATH such that MATH and for each MATH there is MATH such that MATH . Now, for each MATH let MATH and let MATH . Then the set MATH is nonempty by assumption, convex since MATH is locally convex, compact as the intersection of a compact set and a closed set. Also MATH is... |
math/0008214 | MATH is a linear operator, so MATH. Now MATH implies MATH . |
math/0008214 | The ``if" part is obvious. Next, if MATH is a compact linear operator, then for any bounded MATH there is a compact MATH such that MATH . Now since MATH is a finite dimensional subspace of a normed space, it is topologically complemented, that is, MATH is homeomorphic to MATH . Then MATH is compact in MATH and MATH. |
math/0008214 | Let MATH be a bounded subset of MATH . Then by REF , MATH where MATH is compact. Therefore MATH . It is easy to show that there is a bounded set MATH in MATH such that MATH . Let MATH . Then MATH is compact as MATH is finite dimensional. In particular, MATH is closed, so by CITE, MATH is u.s.c. Therefore MATH is compac... |
math/0008214 | First, MATH is closed under linear combinations. Indeed, for MATH we have MATH . Second, if MATH both commute with MATH then MATH commutes with MATH . Indeed MATH . |
math/0008214 | Let MATH then MATH for some MATH . Suppose MATH and let MATH . Then MATH . |
math/0008214 | Let MATH . Then MATH so MATH . |
math/0008214 | Let MATH . Then MATH . In particular, there is some MATH such that MATH . Suppose MATH . Now we use REF , as follows: MATH . Hence MATH, so MATH . |
math/0008214 | For each MATH is MATH-invariant. Indeed, suppose MATH and MATH . Then MATH for some MATH . Therefore MATH and MATH . Suppose now that MATH is not dense in MATH for some MATH . Then we can assume that MATH because otherwise MATH is MATH-invariant. Then MATH is the desired subspace. First, MATH is a linear subspace of MA... |
math/0008214 | Let MATH . Then MATH and MATH is a semialgebra. To check the rest of the conditions of the theorem, observe that since MATH we have MATH for all MATH . Therefore MATH for all MATH so that MATH for any polynomial MATH without constant term-MATH . Thus for all MATH . Therefore by the theorem there is a nontrivial closed ... |
math/0008214 | By REF MATH is a semialgebra, so we need only to consider the following. CASE: Suppose MATH and MATH . Then MATH because MATH is a linear subspace. Thus MATH is closed under nonzero linear combinations. CASE: Suppose MATH commutes with MATH . Now since MATH we have MATH . Hence MATH is closed under compositions. |
math/0008214 | The set MATH is closed because MATH is u.s.c. |
math/0008214 | By REF MATH is a semialgebra. Therefore by REF there is a nontrivial closed subspace MATH that is either REF MATH-invariant or REF finite-dimensional and MATH-weakly-invariant. In case of REF MATH is MATH-invariant because MATH . Consider REF . If MATH then MATH is a linear relation with nonempty values. Then by REF , ... |
math/0008214 | Apply the above theorem to the linear relation MATH . |
math/0008215 | As noted in CITE, this follows from REF together with the main result of CITE. |
math/0008215 | Since MATH has rank greater than one, its NAME graph MATH is a bushy tree, meaning that each point of MATH is within some fixed distance MATH of some vertex MATH such that MATH has at least REF unbounded components. We can thus apply the following result, which is REF, to the metric fibration MATH. Let MATH, MATH be me... |
math/0008215 | Recall that we have a MATH-equivariant MATH-bundle MATH, carrying a MATH-equivariant piecewise Riemannian metric whose restriction to each fiber is isometric to MATH, with MATH acting cocompactly. Each of the spaces MATH is embedded in MATH as the inverse image of MATH. We now define a MATH-equivariant MATH-bundle MATH... |
math/0008215 | The horizontal foliation of MATH is an example of a uniform foliation, which means that any two leaves have finite NAME distance. Any map between two horizontal leaves which moves points a bounded distance in MATH is a quasi-isometry between those leaves. It follows that there is a canonical coarse equivalence class of... |
math/0008215 | We clearly have inclusions MATH and so REF implies REF implies REF . To prove REF implies REF , suppose that MATH is periodic in MATH, that is, MATH is the axis of some pseudo-Anosov element MATH. Choose a point MATH, and consider the sequence MATH, MATH. Since MATH acts cocompactly on MATH, there is a sequence MATH su... |
math/0008215 | A nontrivial isometry MATH of MATH must map some vertical geodesic to a different vertical geodesic. Since any two vertical geodesics in MATH have infinite NAME distance, MATH is an infinite distance from the identity, so that MATH is injective (injectivity of MATH therefore is true regardless of periodicity). We now p... |
math/0008215 | By REF , and the fact that the projection MATH induces an isometry between the space of horizontal leaves of MATH with the NAME metric and the quotient tree MATH, it follows that any quasi-isometry MATH induces a quasi-isometry of MATH and so MATH induces a homeomorphism MATH of the NAME set MATH. Suppose that MATH is ... |
math/0008215 | To justify this computation, first we show MATH. Consider a quasi-isometry MATH, regarded as an element of MATH; we must show that MATH. By REF the quasi-isometry MATH coarsely respects the horizontal foliation of MATH and so MATH lies over a quasi-isometry of MATH, also denoted MATH. From REF it follows that MATH indu... |
math/0008215 | Suppose MATH is a commensuration of MATH such that MATH equals the identity in MATH. Then there is a bounded function MATH so that for all MATH we have MATH . Since MATH is bounded, the cardinality MATH is finite. Plugging the above equation into MATH gives MATH . Note that this is true for all MATH, and the right hand... |
math/0008215 | There is a NAME subgroup MATH of finite index; it follows that MATH has finite index in MATH. Choose a splitting MATH, consider the action of MATH on MATH by automorphisms. Choose a finite surface cover MATH with corresponding subgroup MATH, and consider the orbit of MATH under the action of MATH. This orbit consists o... |
math/0008216 | If MATH or MATH then MATH and REF follows easily. Hence assume MATH. We may also assume MATH. If MATH then from symmetry and convexity we have MATH and MATH, and REF follows easily. Hence we assume MATH. If MATH is above MATH, let MATH be the reflection of MATH across MATH. Then MATH and MATH, and it follows easily fro... |
math/0008216 | By REF , MATH has the ratio weak mixing property. Let MATH. Then provided MATH is large, MATH . Thus we need only consider paths inside MATH. Let MATH to be specified and consider MATH. For such MATH, there exist MATH and paths MATH in MATH occuring at separation MATH. Now MATH so using REF , provided MATH is large, MA... |
math/0008216 | We may assume MATH and MATH, with MATH to be specified. Let MATH and MATH, where MATH are to be specified and MATH denotes the integer part. Provided MATH is large (depending on MATH), we have MATH. Let MATH and let MATH . By REF , for some MATH, MATH . Note this probability is for the infinite-volume limit. For each d... |
math/0008217 | First suppose MATH has the ratio weak mixing property. For MATH, let MATH and MATH. Define events MATH . Then by REF , MATH . Let MATH and MATH be as in REF . Then for some MATH depending on MATH, provided MATH is chosen large enough MATH . Therefore MATH . Suppose MATH, and consider MATH with MATH and MATH for which M... |
math/0008217 | Let MATH and MATH, with MATH to be specified. Let MATH be the MATH-hull skeleton of MATH. For each MATH let MATH be a dual site with MATH and MATH. By REF , provided MATH is large enough we have MATH . Further, MATH (with MATH.) Therefore using the FKG property, REF , MATH . |
math/0008217 | Suppose MATH contains a MATH-bottleneck MATH. We have two disjoint paths from MATH to MATH: MATH and MATH (traversed backwards.) Each of these may intersect MATH a number of times. Accordingly, MATH contains a finite sequence of sites MATH such that the segment MATH of MATH between MATH and MATH satisfies MATH for all ... |
math/0008217 | We may assume MATH contains a clean MATH-bottleneck MATH, for otherwise REF is immediate from REF . We have MATH . Let MATH denote the union of MATH and all (MATH)-small offspring of MATH, and let MATH . Then MATH and MATH . Note that the set MATH of (MATH)-large offspring of MATH can be divided into two disjoint class... |
math/0008217 | From the definition of MATH we may assume MATH. It follows easily from REF that for some MATH, MATH so by REF it suffices to consider MATH . Suppose MATH. Fix MATH to be specified, let MATH and suppose MATH. By REF , MATH . Let MATH. Let MATH. For each MATH there is a segment MATH entirely outside MATH, with MATH and M... |
math/0008217 | We will refer to the requirement MATH as the size condition, and to all other assumptions of the Proposition collectively as the basic assumptions. We first prove REF . We proceed by induction on MATH, using REF for MATH. Fix MATH and define MATH where MATH . If MATH is large enough then, from REF and the lattice natur... |
math/0008217 | We may assume MATH has at least one bottleneck. If MATH is a primary bottleneck in MATH, and the MATH-large offspring of MATH are MATH, then MATH and therefore from REF , MATH can be surrounded by a (non-lattice) loop of MATH-length at most MATH . Since MATH minimizes the MATH-length over all such loops, it follows tha... |
math/0008217 | Let MATH and MATH and MATH . Then using REF , MATH . The events MATH are empty unless MATH (compare REF ); if MATH, and hence MATH, is large enough, this implies MATH, so we may restrict the sums in REF to such MATH. Presuming MATH is large enough, MATH is strictly positive for all MATH. For MATH we apply REF to get MA... |
math/0008217 | From the definition of MATH and REF , for any MATH, if MATH is sufficiently large, MATH . If we take MATH sufficiently large, this and REF prove that REF holds with conditional probability approaching REF as MATH. Next, from the quadratic nature of the NAME variational minimum (see CITE, CITE), for any MATH, if MATH is... |
math/0008217 | Let MATH be the non-maximal final MATH-descendants of MATH, and MATH. Then MATH while (compare REF ) MATH . The lemma follows easily. |
math/0008217 | The proof is partly a modification of that of REF , so we use the notation of that proof. The basic idea is that a large inward deviation of MATH from MATH for some MATH reduces the factor MATH in REF . First observe that the proof of REF actually shows that MATH . Let MATH be the site in MATH most distant from MATH. L... |
math/0008217 | Let MATH and MATH, where MATH is from REF . Let MATH denote the NAME shape of area MATH centered at the center of mass of MATH. From translation invariance, REF we have (recalling MATH) MATH . With this fact, we can repeat the argument of REF , but excluding reference to MATH, to obtain (using again MATH) MATH so that ... |
math/0008217 | Let MATH. Since MATH is a decreasing event, we have MATH so it is enough to show MATH . Let MATH . Fix MATH and let MATH with MATH and MATH, where MATH is to be specified. Then MATH and MATH provided MATH and MATH are small and MATH is large enough. Let MATH denote the outermost open dual circuit surrounding MATH in MA... |
math/0008217 | We begin with REF . The proof of REF is valid for MATH through REF (see REF ), which gives that for MATH sufficiently large, MATH . If MATH, we can complete the proof as we did in REF . Thus suppose MATH . Since MATH, provided we choose MATH small enough the set MATH satisfies MATH. Let MATH denote the event that MATH ... |
math/0008217 | We begin by obtaining an analog of REF , by mimicking its proof. We omit some details becuase of the similarity. First fix MATH and let MATH be a constant to be specified. Our induction hypothesis is that for every MATH and enclosure event MATH, we have MATH and MATH and MATH . We need only consider parameters satisfyi... |
math/0008217 | For REF we need only observe that REF is valid for MATH in place of MATH and only MATH considered. After these same changes, REF follows from REF with the last inequality omitted. |
math/0008217 | As mentioned above, we omit some details. Suppose MATH and MATH for some MATH, via open dual paths. If MATH is large, one can trivially dispose of the case in which MATH, so we hence forth tacitly consider only connections occuring inside MATH; in particular this means MATH. There are then two cases: either there is a ... |
math/0008217 | Fix MATH and let MATH, with MATH from REF . Let MATH denote the bond boundary of MATH in MATH, and define MATH analogously. For MATH let MATH denote the event that all dual bonds in MATH are open, MATH and MATH are open and all other bonds in MATH are closed; define MATH analogously for MATH. Define the event MATH (com... |
math/0008217 | Fix MATH large and let MATH. Let MATH denote the vertical coordinate of the point where MATH meets the positive vertical axis. Let MATH and MATH, with MATH as in the proof of REF . Let MATH and let MATH be the MATH-hull skeleton of MATH. It is an easy exercise in geometry to see that the natural half-slabs MATH, are di... |
math/0008220 | NAME MATH with open sets around each asymptotic height function MATH, such that the entropy for normalized height functions in these sets is strictly less than MATH unless MATH. By compactness, only finitely many of the sets are needed to cover MATH; we include the one corresponding to MATH. Then REF implies that for M... |
math/0008220 | We begin with the first of the two properties. Recall that NAME functions are differentiable almost everywhere (NAME 's theorem) CITE. For any point MATH at which MATH is differentiable, we have MATH if MATH is sufficiently small, say MATH with MATH (where MATH depends on MATH). If MATH lies within an equilateral trian... |
math/0008220 | Because MATH is concave, MATH where MATH and MATH . We have MATH (as one can see by computing the average by integrating over cross sections and applying the fundamental theorem of calculus), and hence MATH since MATH is a continuous function of MATH. Now combining REF with MATH yields the desired result. |
math/0008220 | Let MATH be any asymptotic height function, and consider the neighborhood MATH of MATH consisting of all asymptotic height functions within MATH of MATH. We need to show that given MATH, if MATH is sufficiently small, then for all MATH, MATH. Let MATH. It follows from REF that if MATH is small enough, then the piecewis... |
math/0008220 | For existence, we can use a compactness argument, since MATH is compact. Because the local entropy integrand is bounded, MATH is bounded above, and we can choose a sequence MATH such that MATH approaches the least upper bound as MATH. By compactness, there is a subsequence that converges, and by upper semicontinuity, t... |
math/0008220 | Let MATH be the largest height function on MATH whose normalization is less than or equal to MATH, that is, the lattice sup of all height functions below MATH. (Technically, we must pick a lattice point and restrict our attention to height functions that give it height MATH modulo MATH. This ensures that all our height... |
math/0008220 | Start at any lattice point and work outwards, labelling the other points with the lengths of the shortest increasing paths to them. It is easy to prove by induction that on the square at sup norm distance MATH from the starting point, on two opposite sides the lengths alternate between MATH and MATH, and on the other t... |
math/0008220 | Consider a large torus, with edge weights MATH chosen to give tilt MATH and satisfying MATH (see Subsection REF), and view MATH as being contained in the torus. We will look at random free tilings of MATH generated according to the probability distribution on weighted tilings of the torus. If we fix the height of one p... |
math/0008220 | Without loss of generality, suppose that MATH and MATH are positive. Consider any vertical line through MATH on which the MATH-coordinate is integral. If the segment that lies within MATH has length MATH, then for every tiling of MATH, the difference between the number of north-going dominos bisected by the line and th... |
math/0008220 | Let MATH be the tilt of the plane. If MATH, then the conclusion follows from REF . Thus, we can assume that the tilt satisfies MATH. Suppose MATH is another boundary height function on MATH, which agrees with the same plane as MATH, to within MATH. We need to show that extensions of MATH and MATH have nearly the same e... |
math/0008220 | We know from REF that the entropy is independent of the precise boundary conditions, but we still need to prove that it equals MATH. To do so, we will compare with a MATH torus that has edge weights MATH satisfying MATH and yielding tilt MATH. (We can suppose that MATH, since otherwise the result follows from REF .) Th... |
math/0008220 | It is not hard to check that equilateral triangles satisfy the hypotheses of REF , and REF . Thus, we just need to deal with the case of a equilateral triangle whose boundary heights are within MATH of being planar. To prove that the entropy of the triangle is at least what we expect, tile the triangle with smaller squ... |
math/0008220 | Notice that the set of tilings whose normalized height functions are within MATH of MATH is non-empty, by REF . Call the set of such tilings MATH. Fix MATH. Choose MATH small enough that we can apply REF to the piecewise linear approximation MATH to MATH derived from a MATH-mesh (with approximation tolerance MATH). The... |
math/0008220 | Recall that MATH by hypothesis. Suppose MATH. We have MATH. For MATH write MATH. Then MATH . Thus as MATH runs counterclockwise around MATH, MATH runs clockwise around the ellipse MATH. Since MATH has no critical points in the unit disk, it is injective on the unit disk, mapping it to the exterior of MATH (with MATH ma... |
math/0008220 | From REF , for positive reals MATH we have MATH for all MATH. Combined with REF , this gives MATH . This implies that MATH (assuming these limits exist). Here note that the MATH for which MATH may also depend on MATH. Let MATH denote the integral in the statement of the theorem. We show that MATH converges to MATH. Let... |
math/0008220 | We must show MATH . For this proof only, let MATH and MATH be square roots of MATH, respectively, with signs chosen so that MATH (compare REF ). We can then factor REF as MATH and the second quotient is MATH. That is, we are left to show that MATH . Separating the real and imaginary parts of REF we have MATH . Solving ... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.