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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 REF , and hence that MATH for any MATH-order MATH. Recall that MATH by REF . Now, REF together with MATH leads to the following two cases. CASE: MATH . Here, MATH by REF . Thus MATH. Now, MATH is a MATH-order and hence MATH, a contradiction. CASE: MATH . Here, MATH is a MATH-order (as MATH) and so MATH cannot belong to MATH, a contradiction. CASE: MATH . As above, if we show that MATH, the result will follow from REF . If MATH, REF together with MATH yields MATH. Now, REF implies that MATH for every MATH-order MATH. On the other hand, as MATH, MATH, a contradiction. |
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 does not hold; for each MATH choose conic neighborhoods MATH REF in MATH with the following property: MATH the projection MATH to MATH has distance less than MATH to the compact set MATH. By assumption we can choose MATH such that MATH. In particular, MATH is a nonempty closed cone in MATH and therefore also MATH with MATH is contained in MATH. Now we choose MATH such that MATH and set MATH . By REF of the current lemma MATH is a closed cone in MATH; it intersects MATH only in MATH, so we can apply REF : for all MATH we have for some MATH . Sending MATH we conclude that MATH. By construction the summands of MATH are bounded: MATH (MATH). If MATH would tend to MATH then so would MATH. But since MATH are normalized this would imply that both MATH REF tend to zero yielding the contradiction MATH. Therefore the norms of (suitable subsequences of) MATH are bounded away from zero and above. There are subsequences MATH REF such that MATH; then MATH and therefore MATH - a contradiction. |
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 MATH to finish the proof. |
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 that MATH. Further, we set MATH. We claim that MATH is rapidly decreasing in MATH. To show this we write MATH and estimate the summands individually. Substituting MATH, MATH takes the form MATH . By REF MATH: MATH and MATH: MATH . Thus by NAME 's inequality we obtain (for MATH small and MATH): MATH . This last integral is convergent for MATH, so MATH is rapidly decreasing in MATH. MATH . We abbreviate the domain of integration by MATH. For MATH, MATH is tempered in MATH and MATH is rapidly decreasing in MATH. (For later use we note here that since MATH, the same decrease properties for MATH and MATH in fact hold on all of MATH. The following estimates thus remain valid upon replacing MATH by MATH.) Hence MATH . Supposing MATH and setting MATH this equals MATH . Since MATH, by REF we have MATH . We now split the domain of integration in REF into the parts MATH and MATH. Then by REF MATH . Also, by REF , MATH for MATH. It follows that MATH and hence also MATH is rapidly decreasing in MATH. Turning now to MATH, we first note that MATH is tempered on the domain of integration. Moreover, since MATH it follows that MATH is rapidly decreasing in MATH in said domain (again by REF ). Thus the same reasoning as in the case of MATH (compare the above remark) shows that MATH is rapidly decreasing in MATH as well, which completes the proof. |
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, MATH . Thus by REF , MATH. Again from REF the claim follows. |
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 for all MATH. |
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 the result follows from REF . |
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: Let us now assume that MATH. Then we have MATH for all MATH as in REF. For MATH, we have MATH by assumption. We consider the first exit time MATH and the event MATH . If MATH, then for all MATH, we have MATH, and thus by REF, MATH . However, by the definition of MATH, we have MATH, which contradicts REF for MATH. Therefore MATH, which implies that for almost all MATH, we have MATH for all MATH, and hence MATH for these MATH, which proves REF. |
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 MATH, and using NAME 's formula, we find for all MATH and all MATH where MATH is the upper bound on MATH, see REF. In order to estimate MATH, we introduce linear approximations MATH for MATH, defined by MATH where MATH is a Brownian motion with MATH which is independent of MATH. If MATH, we have for all MATH where MATH. This shows that on MATH, MATH . CASE: We are now ready to estimate MATH. REF shows that on MATH, MATH where MATH denotes the conditional variance of MATH, given MATH. As in REF, MATH . It follows that MATH . Inserting this into REF, we obtain for each MATH on MATH the estimate MATH . Note that for any MATH, there exist MATH and MATH such that MATH for all MATH and all MATH. Since MATH is an obvious bound, we obtain from REF MATH . Choosing MATH so that MATH holds, yields almost the optimal exponent, and we obtain MATH . |
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, valid for all MATH, which yields MATH . For MATH, we have MATH for all MATH, and thus MATH where we used the relation MATH in the last step. By the same relation, we obtain MATH . Finally, assume that MATH for all MATH, and recall that MATH is the solution of the initial value problem MATH . Since MATH, MATH for all positive MATH. For negative MATH, MATH is positive whenever the function MATH is negative. We have MATH and MATH . Since MATH whenever MATH, MATH can never become positive. This implies MATH. |
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 MATH . The partition times are defined via MATH . In the first case, we immediately obtain the bound REF. In the second case, estimating MATH in the usual way shows that MATH . Finally, let us note that, for MATH, MATH which concludes the proof of the proposition. |
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 MATH, which can be estimated by REF once again. Thereby, we obtain MATH which leads to a contradiction for MATH. We conclude that MATH, and thus MATH for MATH-almost all MATH. This shows that MATH and thus MATH for all these MATH and all MATH, which proves REF. The proof of the inclusion REF is straightforward, using the same estimates. |
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 MATH, then we have MATH where the variance MATH can be estimated by REF . We thus have by the NAME property MATH . CASE: We now choose the MATH in such a way that MATH is approximately constant. Given MATH, let MATH (Observe that MATH.) Choosing MATH as the smallest integer satisfying MATH we define the partition by the relations MATH . Then we have MATH which proves REF. CASE: It remains to show that REF is satisfied. Since MATH the condition reduces to MATH which is satisfied whenever REF is satisfied. |
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 all MATH, which shows the first inequality in REF. CASE: MATH being distributed according to a normal law, we have MATH where the variance MATH can be estimated by REF . This proves the second inequality in REF. |
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 compute the derivative of MATH, we first note that MATH . Differentiating the second line of REF, we get MATH where we have used the facts that MATH and that MATH is decreasing for MATH. Now, REF follows from REF. |
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 estimate MATH . The second term can be estimated by integration by parts, see REF . Let MATH be any upper bound on MATH satisfying the hypotheses on MATH in that lemma. Since MATH, we may assume that MATH. Application of REF with MATH shows that the second term in REF is bounded by MATH . We have thus obtained the inequality MATH . CASE: Consider now the case MATH. Since MATH is an odd function, MATH follows. Hence we may assume that MATH. We consider the linear SDE REF with initial condition MATH, and denote by MATH the time of the first return of MATH to zero. Then we have MATH and REF yields MATH . The second term in REF can be estimated using the density of the random variable MATH, for which REF gives the bound MATH . We obtain MATH . CASE: Before inserting the estimate REF for MATH, we shall introduce some notations and provide bounds for certain integrals needed in the sequel. Let MATH and MATH. Then MATH where we used the changes of variables MATH in REF and MATH in REF. CASE: Now we are ready to return to our estimate on MATH, compare REF. Inserting the bound REF on MATH yields two summands, the first one being MATH where we used REF to bound the integral. The second summand is MATH where we used REF again. We can now collect terms. Introducing the abbreviations MATH the previous inequalities imply that MATH . CASE: We will now iterate the bounds on MATH. This will show the existence of two series MATH and MATH such that MATH . To do so, we need to assume that MATH . By our choice REF of MATH, this condition reduces to MATH which is satisfied for small enough MATH by our REF on MATH, provided MATH. Using the trivial bound MATH in REF, we find that REF holds with MATH and MATH. Inserting REF into REF again, we get MATH . By induction, we find MATH as a possible choice, where we have used the fact that MATH by REF . Taking the limit MATH, and using MATH, we obtain MATH . In order to obtain also a bound on MATH, we insert the above bound on MATH into REF, which yields MATH by REF. This proves the proposition, and therefore REF , by taking the sum of the above estimates on MATH and MATH. |
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 immediately obtains that MATH satisfies the ODE MATH where MATH with MATH, MATH and MATH. We first consider the particular solution MATH of REF starting at time MATH in MATH. By REF, we know that MATH for all MATH. We will use the fact that MATH where MATH. We have used the transformation MATH, introduced MATH and bounded the last integral by MATH. We now introduce the first exit time MATH. For MATH, we have MATH . Since MATH, the term in brackets can be assumed to be larger than MATH. Hence REF shows that MATH which implies MATH . Since MATH, we obtain MATH, and thus MATH. This shows MATH CASE: Let MATH and MATH be given. Let MATH and MATH be solutions of REF with initial conditions MATH and MATH, respectively. Then there exists a MATH such that the difference MATH satisfies MATH where we have used REF. It follows that MATH which proves REF in particular. If MATH, we can use the relation MATH to show that MATH which proves REF for MATH. Finally, if MATH, we can use the fact that MATH to prove that REF holds for some constant MATH. |
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 last inequality is obtained by splitting the integral at MATH. Using the fact that MATH for all MATH, we reach the conclusion that this integral is bounded by a constant times MATH, which completes the proof of REF. CASE: We now use the fact that MATH solves the ODE MATH . Then, REF is an immediate consequence of this relation, and REF is obtained from the fact that MATH whenever MATH, and MATH whenever MATH. Here we used REF and the monotonicity of MATH for small MATH. |
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 straightforward consequence of the same estimates. |
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 under that action are in a one-one correspondence with the nonequivalent MATH-diagrams. A REF-factor of MATH can be represented by a MATH matrix whose entries belong to MATH and are all distinct: each row corresponds to an edge, two row entries being the end points of the edge. This correspondence is not unique. It is defined to whithin an equivalence induced on the set of such matrices by independently permuting entries in each row and permuting the rows bodily. This amounts to the action of the wreath product MATH. The action of MATH on the set of REF is equivalent to the action of MATH on the set of MATH-matrices. We thus arrive at the following setting: given two sets MATH and MATH, consider the set of bijective mappings MATH. The wreath product MATH acts on the set MATH by the rule MATH where MATH, MATH, and the group MATH acts on the set MATH. Two mappings MATH are equivalent if there exist a MATH and a MATH such that MATH for all MATH. The equivalence classes of mappings are in a one-one correspondence with the orbits of MATH. In this setting, an argument of CITE applies which he used to prove a theorem on the number of classes of bijective mappings. The proofe is complete. |
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 the number of permutations of the same cycle type MATH in MATH we rewrite REF as follows MATH . To determine MATH we first establish a relationship between the cycles of MATH and the cycles of MATH. If MATH is a cycle of MATH then its projection onto MATH is a cycle of MATH. If MATH is a cycle of MATH then, for any MATH, the length of a cycle of MATH induced by MATH depends only on MATH, MATH. Let MATH where MATH is the identity permutation. If MATH is odd then MATH induces one cycle of length MATH. If MATH is even then MATH induces two cycles both of length MATH. Let MATH be odd. For MATH to have cycle type MATH the permutation MATH must have cycle type MATH. The number of such MATH equals MATH . Now we fix MATH and count the number of MATH such that MATH has cycle type MATH. Each cycle MATH of MATH induces two cycles of MATH of the same length MATH. Hence MATH must be even for each MATH. There are MATH choices for MATH. Clearly, MATH uniquely determines MATH for all MATH. For different cycles of MATH the choices are independent, so we have MATH different MATH. Multiplying the expressions for MATH and MATH we get MATH for MATH odd. Let now MATH be even. For MATH to have cycle type MATH the permutation MATH must have cycle type MATH with MATH, MATH. The number of such MATH equals MATH . We fix MATH and count the number of corresponding MATH. Each MATH-cycle of MATH induces one cycle of MATH of the length MATH. Hence MATH must be odd for each MATH-cycle of MATH. There are MATH choices for MATH. Each MATH-cycle of MATH induces two cycles of MATH, both having length MATH. Hence MATH must be even for each MATH-cycle of MATH, and we again have MATH choices for MATH. It follows, there are MATH different MATH such that MATH has cycle type MATH. Multiplying the expressions for MATH and MATH and summing up over all admissible MATH we obtain MATH for MATH even. It remains to substitute REF into REF . Setting MATH we have MATH for MATH odd and MATH for MATH even which completes the proof. |
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 coefficients we obtain MATH for MATH whence MATH for MATH even. Comparing upper bounds REF we conclude that MATH for MATH, MATH. Going back to REF we see that MATH as MATH and the result follows. |
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 permutations of cycle type MATH. So we can write MATH where MATH is the number of MATH of cycle type MATH and MATH is the number of MATH of cycle type MATH. Applying the analysis in the proof of REF we see that for each MATH of cycle type MATH, MATH, MATH, MATH there are MATH permutations MATH such that MATH has cycle type MATH. Multiplying and summing up over all admissible MATH and simplifying we get MATH . Substituting the expressions for MATH and MATH into REF we obtain MATH . Denoting MATH and substituting REF into REF we get the statement of the theorem. |
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 calculate that MATH have degree REF; MATH have degree REF; and MATH have degree MATH. We can easily compute the differential on MATH: MATH . The NAME DGA MATH associated to MATH has the same generators and grading; its differential is the same as the above differential, except with each monomial reversed. Note that MATH; that is, there exist MATH and MATH such that MATH. (Choose MATH and either MATH or MATH.) Since MATH is simply MATH with monomials reversed, we conclude that MATH. We claim that MATH, which implies that MATH and MATH are not isomorphic as graded algebras. Suppose to the contrary that there exist MATH of degree MATH respectively, so that MATH. Replace MATH by MATH and MATH by MATH, as generators of MATH; the differential on MATH becomes MATH . Then MATH is in the two-sided ideal of MATH generated by MATH. Let MATH be the free associative unital algebra over MATH generated by two variables MATH, with grading which assigns degree MATH to MATH and REF to MATH. We have a grading-preserving algebra map MATH which sends MATH to MATH, MATH to MATH, and all other generators to REF. Write MATH and MATH. Then MATH is in the two-sided ideal MATH of MATH generated by MATH; by inspection, we see that MATH is generated by the single expression MATH. We conclude that MATH in the algebra MATH. But MATH is generated as a vector space by monomials of the form MATH, where MATH. Since MATH has degree MATH, we can write MATH (for MATH), and thus any monomial MATH in the expansion of MATH will satisfy MATH. In particular, MATH cannot equal REF in MATH. This contradiction proves that MATH, as desired. |
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 (namely, if we think about MATH as embedded in MATH, MATH, MATH, MATH where MATH, MATH, MATH are the unit imaginary quaternions) and let MATH, MATH, MATH be their duals with respect to the canonical metric of MATH. We regard the MATH as vector fields on MATH. Let MATH be their dual forms with respect to the metric MATH; their restrictions to any fibre coincide with the MATH. The usual splitting of MATH into vertical and horizontal parts is now refined to MATH where MATH represents the horizontal vector fields orthogonal to MATH. We now define an almost complex structure MATH on MATH by: CASE: MATH, CASE: MATH, CASE: MATH for any MATH orthogonal to MATH. As for MATH, MATH is a section of MATH and the restriction of MATH to MATH is an endomorphism of MATH, the last item in the definition is consistent. One easily shows that MATH and is compatible with MATH. |
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 they are zero. Let us compute the NAME derivative of the metric MATH on the total space in the direction MATH. We obtain successively: MATH because MATH and the brackets in the last two terms are vertical. MATH as MATH does not depend on vertical directions and again because the brackets in the last two terms are vertical. MATH . Here MATH, MATH and MATH. Hence MATH . MATH . We obtained that MATH are Killing fields iff MATH and MATH are horizontal. |
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 complex NAME manifolds MATH and MATH, respectively. On all these fibers MATH a simultaneous parallelization is induced by a choice of a REF-Sasakian structure on MATH and a NAME structure on MATH. From this point of view, the above mentioned complex structure on the NAME manifold is by construction given by the choice of the standard complex structure on the fibers MATH and by the lift of the complex structure of the NAME bases. The integrability of the whole complex structure was insured by the computations described above. Observe now that these same computations, leading to MATH, can be carried out even if the complex structure on the fibers is not defined in the standard way (that is, MATH), but according to formulas like: MATH where the matrix MATH, whose entries are real and constant, has trace MATH and determinant MATH. The complex structures defined in this way on MATH correspond to all the possible choices of the generator MATH, and it can be shown that all these complex structures on the NAME surface are inequivalent (compare CITE, p. REF). |
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 finite partition of unity MATH subordinate to MATH . Then we have a locally finite open cover of MATH that refines MATH that is, for each MATH there is MATH such that MATH . Fix MATH . Suppose MATH and MATH . Then MATH . Hence MATH . As MATH and MATH are convex, so is the set MATH. Therefore a convex combination of MATH is a well defined subset of MATH. Then the map MATH given by MATH is well defined and belongs to MATH . |
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 u.s.c. by CITE. Thus MATH . Now we apply REF with MATH for all MATH by REF . Therefore there exists MATH such that for all MATH . We know that MATH are admissible. Therefore by REF , so is MATH . Now MATH is compact and MATH is an ANR as a closed neighborhood in a locally convex space. Hence by REF , MATH has a fixed point MATH . Therefore by REF , there exists MATH such that MATH is a fixed point of MATH . |
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 compact in MATH by CITE. But MATH hence MATH is precompact. Thus MATH where MATH is precompact, so MATH is compact by REF . |
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 MATH by REF . Second, since every MATH is u.s.c., MATH implies MATH . Hence MATH is MATH-invariant. Assume now that MATH is dense in MATH for each MATH . Since MATH is u.s.c., MATH is closed. Therefore we can choose a closed convex neighborhood MATH of some MATH such that MATH and MATH . In addition we have MATH . Now we apply REF with MATH as follows. We let MATH . Then MATH . We can also rewrite REF as MATH . By REF , there is a compact MATH such that MATH . Let MATH. Since MATH is compact, MATH is precompact. Therefore by REF , for any neighborhood MATH of MATH has a fixed point in MATH . Therefore there is MATH such that MATH has a fixed point MATH . Thus MATH . Now MATH REF , MATH is a linear subspace of MATH REF , and MATH is MATH-invariant REF . Suppose now that MATH is a bounded neighborhood of MATH. Since MATH is compact by REF , we have MATH where MATH is compact and MATH is finite dimensional. Therefore MATH is finite dimensional. |
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 linear subspace MATH which is either MATH-invariant or MATH-weakly-invariant. Since MATH is MATH-weakly-invariant. |
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 has an eigenvector corresponding to some MATH. Therefore the eigenspace MATH is nonzero, closed and not equal to the whole MATH . Then for each MATH we have MATH . Hence MATH . |
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 metric fibrations over bushy trees MATH, such that the fibers of MATH and MATH are contractible MATH-manifolds for some MATH. Let MATH be a quasi-isometry. Then there exists a constant MATH, depending only on the metric fibration data of MATH, the quasi-isometry data for MATH, and the constant MATH, such that: CASE: For each hyperplane MATH there exists a unique hyperplane MATH such that MATH. CASE: For each horizontal leaf MATH there is a horizontal leaf MATH such that MATH. REF is an immediate consequence. To obtain REF , consider a horizontal leaf MATH of MATH. By applying REF we obtain a horizontal leaf MATH of MATH uniformly NAME close to MATH. But MATH is uniformly NAME close to MATH and so MATH is uniformly NAME close to some horizontal leaf MATH. Finally, the closest point projection MATH moves points a uniformly bounded distance, and so MATH is a uniformly NAME close to its closest point projection in MATH. |
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 in which each of the singular solv-manifolds MATH sits, as follows. First, note that each point of MATH corresponds to (the isotopy class of) a normalized MATH-structure MATH on MATH. We may assemble these structures into a MATH-equivariant MATH-bundle over MATH, each fiber equipped with a MATH-structure in the appropriate isotopy class, so that the MATH-strucures vary continuously as the base point in MATH varies. Restricting to MATH we obtain a MATH-equivariant MATH-bundle MATH. The universal cover of the MATH-bundle over MATH is a MATH-bundle over MATH, with smoothly varying MATH-structures on the fibers, on which the MATH extension of MATH acts, namely the once-punctured mapping class group MATH. Restricting to MATH we obtain a MATH-equivariant MATH-bundle MATH. Restricting to any geodesic MATH we obtain the fibration MATH. Note that the foliation of MATH by geodesics lifts to a foliation of MATH by REF: REF-manifold over the geodesic MATH is MATH. The fiberwise MATH-structures vary continuously in MATH, and the arc length parameter on geodesics of MATH varies continuously; as noted earlier, these data determine singular solv-metrics on each MATH, and these metrics vary continuously in MATH. Now lift the MATH-equivariant map MATH to a MATH-equivariant continuous map of MATH-bundles, MATH, taking each fiber of MATH homeomorphically to the corresponding fiber of MATH, and taking each singular solv manifold MATH to the corresponding singular Riemannian manifold MATH. By cocompactness of the MATH actions on MATH and MATH, and by continuity of the MATH-structures on fibers of MATH and the MATH structures on fibers of MATH, it follows that MATH induces quasi-isometries from MATH fibers to MATH fibers with uniform quasi-isometry constants; in particular, we get uniform quasi-isometries from the horizontal sets of MATH to the horizontal sets of MATH, over all geodesics MATH in MATH. Moreover, since the map MATH is uniformly quasi-isometric from a geodesic MATH in MATH to the corresponding geodesic MATH in MATH, it follows that MATH induces a uniform quasi-isometry from the MATH term in the metric on MATH to the MATH term in the metric on MATH. This implies that the family of maps MATH is uniformly quasi-isometric. |
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 quasi-isometries between any two leaves, and moreover the composition of two such quasi-isometries is another one. Each leaf is quasi-isometric to the hyperbolic plane MATH and its NAME boundary is a circle, so there is a canonical identification of all the circles at infinity to a single circle which we denote MATH. These facts were noted by NAME in CITE. It is well-known that a quasi-isometry MATH induces a homeomorphism MATH between the circles at infinity, and that coarsely equivalent quasi-isometries induce the same boundary map. Any horizontal respecting quasi-isometry MATH therefore induces a homeomorphism MATH between their respective circles at infinity. The underlying idea of our proof is to find additional quasi-isometrically invariant structures on MATH and MATH which encode the stable and unstable foliations, and use this information to prove quasi-isometric invariance of these foliations. To proceed with the proof we need some notation. Let MATH denote the plane MATH. The plane MATH comes equipped with a MATH-structure, as explained in REF. For each MATH let MATH be the map MATH, MATH. In other words, MATH flows along vertical flow segments of MATH from MATH to MATH; each such flow segment has length MATH, and so MATH is a quasi-isometry in the canonical coarse equivalence class from MATH to MATH as discussed above. In fact, the map MATH is a MATH-affine homeomorphism with stretch factor MATH, implying that MATH is MATH-bilipschitz. Note that MATH, for all MATH. The MATH-metric on MATH is a MATH metric, and it is also NAME hyperbolic because MATH is quasi-isometric to MATH. The boundary MATH is a circle, and each quasi-isometry MATH induces a homeomorphism MATH. For any two points MATH there exists a MATH-geodesic with endpoints MATH, and any two such geodesics are the boundary of an isometric embedding of MATH for some MATH (see CITE). Since MATH is not the Euclidean plane but has a cocompact isometry group, there is an upper bound on MATH independent of MATH. This bound is moreover independent of MATH, because coboundedness of the NAME MATH and compactness of the fibers of MATH together imply that the collection of locally MATH metric spaces MATH lies in a compact space of locally MATH metrics (this is the one place in the proof where we use coboundedness of the NAME geodesics in the weak convex hull of MATH). Let MATH be a MATH-independent bound for MATH. For each bi-infinite geodesic MATH on MATH, the set MATH is called a vertical plane in MATH. Note that if MATH then MATH, and so to MATH there is associated a unique pair of points in the circle MATH which we call the endpoints of MATH. Moreover, the discussion in the previous paragraph shows that the NAME distance between any two vertical planes of MATH with the same endpoints MATH is at most MATH. The first structure which MATH must coarsely respect is the collection of vertical planes. Let MATH be a horizontal-respecting quasi-isometry. If MATH is any vertical plane in MATH, then MATH is a bounded NAME distance from some vertical plane MATH in MATH. To prove REF , let MATH be a bijective quasi-isometry of MATH such that MATH is NAME close to MATH, with a uniform NAME constant; we assume that the parameterizations are chosen so that MATH under this quasi-isometry. Composing the map MATH with a uniformly finite distance map MATH, we obtain a quasi-isometry MATH whose quasi-isometry constants are independent of MATH. Fix a vertical plane MATH and for each MATH consider the geodesic MATH. Its image quasigeodesic MATH is uniformly NAME close in MATH to some geodesic MATH, in particular MATH is close to MATH. We need only show that for each MATH, MATH is uniformly NAME close to MATH, for then we can set MATH and it follows that MATH is uniformly NAME close to MATH. Since any two geodesics in MATH which are NAME close are MATH-Hausdorff close, we need only show that the NAME distance between MATH and MATH is finite, in other words these two geodesics have the same endpoints in MATH. But this is an immediate consequence of the coarse commutativity of the following diagram: MATH . This finishes the proof of REF . We now find finer structures which must be coarsely respected by MATH. Each vertical plane MATH comes equipped with a horizontal foliation, obtained by intersecting the plane with the horizontal foliation MATH of MATH. Denote the leaves of this foliation by MATH so that MATH for any MATH. The horizontal foliation MATH has a transverse orientation, pointing in the direction of increasing MATH. Note that the NAME distance in MATH between MATH and MATH is exactly MATH. There is a projection MATH with MATH. Define a quasivertical line in MATH to be a subset MATH such that the projection MATH is a quasi-isometry. We divide the collection of vertical planes into three types: stable, unstable, and doubly unstable. A stable vertical plane is one which is contained in a leaf of the stable foliation on MATH; thus it is either a regular leaf of the stable foliation, or it is a union of two half-planes of a singular leaf. Similarly an unstable vertical plane is one contained in an unstable leaf. All other vertical planes are called doubly unstable vertical planes. The three types of vertical planes - stable, unstable, and doubly unstable - can be distinguished from each other by horizontal respecting quasi-isometries which respect the transverse orientation, by observing the asymptotic behavior of quasivertical lines. To make this more precise, say that two quasivertical lines MATH are upward NAME close if the NAME distance between MATH and MATH is finite; downward NAME close is similarly defined using MATH. Let MATH be a vertical plane. Then: CASE: If MATH is a stable plane, then any two quasivertical lines in MATH are upward NAME close but not downward NAME close. CASE: If MATH is an unstable plane, then any two quasivertical lines in MATH are downward NAME close but not upward NAME close. CASE: If MATH is a doubly unstable plane then there exist two quasivertical lines in MATH which are neither upward NAME close nor downward NAME close. To prove REF when MATH is stable or unstable, observe first that MATH with its horizontal foliation MATH is isometric to the hyperbolic plane MATH in the upper half plane model, with the ``horizontal" horocyclic foliation centered on the point MATH. If MATH is stable (respectively, unstable) then the transverse orientation points towards MATH (respectively, away). Next observe that any quasivertical line in MATH is a quasigeodesic with one endpoint at MATH, and so any two quasivertical lines are NAME close in the direction of MATH. To prove REF when MATH is doubly unstable, observe that MATH is a MATH-geodesic in MATH which is not contained in a leaf of either the MATH-foliation or the MATH-foliation on MATH. There exists, therefore, two points MATH which do not lie in the same leaf of either the MATH-foliation or the MATH-foliation on MATH. The vertical flow lines MATH, MATH in MATH are evidently neither upward nor downward NAME close in the singular solv-metric on MATH, and so the same is true in MATH. This proves REF . The same discussion holds, of course, in MATH. Now consider a horizontal-respecting quasi-isometry MATH. By reversing the upward orientation in MATH if necessary, we may assume that MATH respects the upward orientation. Let MATH be any vertical plane in MATH. We have shown in REF that MATH is NAME close to some vertical plane MATH in MATH. Composing the map MATH with any finite distance map MATH, we obtain therefore a quasi-isometry MATH. Each horizontal leaf of MATH (respectively, MATH) is a coarse intersection of MATH with a horizontal leaf of MATH (respectively, MATH), and it follows that MATH coarsely respects the horizontal foliations and their transverse orientations. REF shows manifestly that stable vertical planes are coarsely respected by MATH, and similarly for unstable vertical planes. To finish the proof of REF , given two stable (respectively, unstable) vertical planes MATH in MATH or in MATH, we must give a quasi-isometrically invariant property which characterizes MATH lying in the same weak stable (respectively, unstable) leaf. Namely, MATH lie in the same leaf if and only if, for any MATH, the triple coarse intersection MATH is unbounded. |
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 such that MATH is a bounded subset of MATH. Since MATH acts properly on MATH, there exist MATH such that MATH. It follows that MATH, and so MATH is periodic in MATH. To prove REF implies REF , choose a horizontal leaf MATH of MATH and so we have the split exact sequence MATH where MATH is an infinite cyclic subgroup of index MATH in MATH. We also have a finite index inclusion MATH. Since MATH acts by automorphisms of MATH it follows that MATH has a finite index subgroup stabilizing MATH; this subgroup is clearly identified with a subgroup of MATH and so the latter is infinite. |
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 prove that MATH is surjective: every horizontal respecting quasi-isometry of MATH is a bounded distance from an isometry. NAME 's geometrization theorem for pseudo-Anosov mapping tori gives a hyperbolic structure on REF-dimensional orbifold MATH. This yields a properly discontinuous, cocompact, isometric, faithful action MATH and a quasi-isometric, MATH-equivariant homeomorphism MATH . For each singular vertical geodesic MATH, the image MATH is a quasigeodesic in MATH, which by the NAME Lemma is a bounded NAME distance (not depending on MATH) from a unique geodesic in MATH; let MATH denote the set of all such geodesics in MATH. Let MATH be the group of isometries of MATH which permute the collection MATH. Also, let MATH be the foliation of MATH obtained by pushing forward via MATH the horizontal foliation of MATH, and let MATH be the subgroup of MATH which coarsely respects MATH. Note that MATH. Now we show that MATH factors through MATH. For each horizontal respecting quasi-isometry MATH of MATH, we obtain a quasi-isometry MATH of MATH, inducing a homomorphism MATH. Since MATH coarsely respects singular vertical geodesics and horizontal leaves in MATH, it follows that MATH coarsely respects MATH and MATH. Since MATH coarsely respects MATH, since MATH is invariant under the cocompact isometry group MATH, and since there are only finitely many orbits of the action of MATH on MATH, we may directly apply the main theorem of CITE which says in this setting that MATH is a bounded distance (not uniformly so) from a unique isometry of MATH which strictly respects MATH. It follows that the image of MATH is contained in MATH, and in fact in MATH. It's evident that the composition MATH is identical with the homomorphism MATH. The homomorphism MATH is obviously injective, and so to prove surjectivity of MATH it suffices to show that MATH. Consider the short exact sequence MATH . Since MATH is normal in MATH, and since MATH is identified via MATH with a finite index subgroup of MATH (both being discrete and cocompact on MATH), it follows that MATH has a normalizer of finite index in MATH; choose coset representatives MATH of the normalizer. Each leaf of MATH is coarsely equivalent to MATH, and MATH coarsely respects MATH, and so each conjugate subgroup MATH is coarsely equivalent to MATH. Now we apply an elementary lemma of CITE which says that for a finite collection of subgroups in a finitely generated group, the coarse intersection of those subgroups is coarsely equivalent to their intersection. The intersection of the above conjugates of MATH is therefore coarsely equivalent to MATH. Another elementary lemma of CITE says that in a finitely generated group, given subgroups MATH, if MATH are coarsely equivalent, then MATH has finite index in MATH. It follows that the intersection of the conjugates of MATH in MATH has finite index in MATH. Thus we obtain a normal subgroup MATH of finite index in MATH. The quotient group MATH is a finite index supergroup of MATH, and so the quotient is virtually cyclic. Since MATH is a REF-orbifold group it follows that the quotient MATH is either MATH or MATH. The orbifold MATH therefore fibers over the circle or the interval orbifold, with generic fiber MATH, and with MATH identified with MATH. Lifting this fibration to MATH we obtain a MATH equivariant fibration coarsely equivalent to MATH. We may therefore replace MATH with this fibration, and so MATH is strictly invariant under MATH. The monodromy map on MATH is pseudo-Anosov. There is a finite index covering map MATH, taking fibration to fibration. By uniqueness of pseudo-Anosov homeomorphisms in their isotopy classes CITE, it follows that the stable and unstable measured foliations for the monodromy map of MATH lift to the stable and unstable measured foliations for the monodromy map on the generic fiber MATH of MATH. But this shows that MATH acts isometrically on MATH, proving that MATH. This proves REF . |
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 the identity map on MATH. We claim that the induced map MATH is the identity. It follows that the induced quasi-isometry MATH is a bounded distance from the identity, and so MATH takes each horizontal leaf of MATH a bounded distance from itself; and since the induced boundary map of that leaf is the identity, it follows that MATH takes each point a bounded distance from itself, proving the proposition. For proving the claim (and for later purposes) we review the well-known embedding of MATH into the space of MATH-invariant measures on the ``NAME band beyond infinity" of the hyperbolic plane. That is, consider the double set of the circle, MATH; with respect to the NAME model MATH, the usual duality gives a bijection between MATH and the NAME beyond infinity MATH. Let MATH be the space of NAME measures on MATH with the weak-MATH topology, and let MATH be the space of projective classes of elements of MATH. The space MATH embeds into the space of MATH-invariant elements of MATH, by lifting a measured foliation on MATH to a MATH-invariant measured foliation on MATH, and then identifying each leaf of the lifted foliation with the correspond pair of endpoints in MATH. The space MATH therefore embeds as MATH-invariant elements of MATH. Given MATH let MATH be the corresponding projective class of measures, let MATH be the support, and let MATH be the union of all the pairs in MATH. Note that MATH may also be described as the endpoint set of MATH, the set of endpoints in MATH of the leaves of the measured foliation MATH on MATH obtained by lifting any measured foliation MATH on MATH that represents MATH. By REF , for all MATH and all quasi-isometries MATH of MATH we have MATH . From our assumption that MATH is the identity on MATH it follows that MATH, for all MATH. Note however that if MATH then MATH, because the projective classes of MATH in MATH are connected by a NAME in MATH, and so the measured foliations MATH can be chosen to be transverse in MATH; the lifted foliations MATH are therefore transverse in MATH and so their endpoint sets MATH in MATH are disjoint. It follows that MATH for all MATH. |
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 induces a permutation on the set of periodic hyperplanes of MATH: given a periodic geodesic MATH in MATH, if we let MATH be the geodesic in MATH coarsely equivalent to MATH, then MATH takes MATH to MATH and MATH to MATH. It follows that the conjugation action of MATH on subgroups of MATH takes MATH to MATH. Any isometry of MATH which preserves each leaf of the horizontal foliation is conjugated by MATH to a similar isometry of MATH, and so conjugation by MATH takes MATH to MATH. In other words, conjugation by MATH preserves the collection of subgroups MATH . Intersecting this collection it follows that conjugation by MATH in the group MATH preserves the subgroup MATH, acting as an automorphism of that subgroup. In other words, in MATH we have MATH. Now we show that the image MATH lies in MATH. We need the following fact: Let MATH be a closed orbifold and MATH a NAME subgroup of MATH. Then the relative commensurator MATH of MATH in MATH is equal to the subgroup of MATH stabilizing the limit set MATH, and MATH has finite index in MATH. Let MATH be the subgroup of MATH stabilizing MATH, and so MATH. Let MATH be the weak convex hull of MATH (see REF ). Clearly MATH is also the subgroup of MATH stabilizing MATH. By REF the group MATH acts cocompactly on MATH, and so MATH acts cocompactly on MATH. Clearly MATH acts properly on MATH. It follows that MATH has finite index in MATH, which immediately implies MATH. For the opposite inclusion, suppose MATH, and so MATH. The limit set of the NAME subgroup MATH is MATH. Since MATH is closed, MATH is open in MATH. Since fixed points of pseudo-Anosov elements of MATH are dense in MATH, there exists a pseudo-Anosov element MATH having a fixed point not in MATH. Infinitely many powers of MATH are therefore not in MATH, and so MATH has infinite index in MATH. We showed above that MATH is a NAME subgroup of MATH and so by REF it remains to show that MATH acting on MATH leaves the set MATH invariant. For this purpose it suffices to show that the action of MATH leaves the set MATH invariant. However, we know that this set equals MATH, and we also know that MATH leaves this set invariant: MATH permutes the elements of this set which are endpoints of periodic geodesics in MATH, by REF , but the endpoints of periodic geodesics are dense. This completes the proof of the inclusion MATH. For the opposite inclusion, consider the following commutative diagram of short exact sequences: MATH . The left vertical arrow is a finite index injection because MATH is a finite covering. The right vertical arrow is a finite index injection by REF using the isomorphism MATH. It follows that the middle vertical arrow is a finite index injection. But this shows that MATH is a finite index supergroup of MATH in MATH, and so MATH. This completes the computation of MATH. To complete the proof of REF , it remains to prove that an element of MATH is in the normal subgroup MATH if and only if it coarsely preserves each horizontal leaf of MATH. Evidently MATH coarsely preserves each horizontal leaf. Suppose conversely that the quasi-isometry MATH coarsely preserves each horizontal leaf; let MATH be the coarse equivalence class of MATH. It follows that MATH coarsely respects each hyperplane of MATH, in particular each periodic hyperplane MATH, and so we have MATH for each periodic line MATH in MATH. But we can say more, as follows. The inclusion map MATH, being injective and with finite index image, is a quasi-isometry, and we therefore obtain a quasi-isometry MATH. Using this quasi-isometry we may conjugate the left action of MATH on itself to obtain a quasi-action of MATH on MATH. In particular, we obtain a sequence of uniform quasi-isometries MATH, with MATH. Each of the MATH coarsely preserves each horizontal leaf of MATH, and note that the coarseness constant is uniform independent of MATH and of the leaf, by application of REF using uniformity of the quasi-isometry constants of MATH. It follows that each MATH coarsely preserves each periodic hyperplane MATH, and MATH coarsely preserves each horizontal leaf of MATH, again with uniform coarseness constants independent of MATH. This implies that MATH. We therefore have MATH . |
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 side takes on at most MATH values. This implies that the centralizer of MATH in MATH has index at most MATH. Intersecting all subgroups of MATH of index MATH gives a finite index subgroup MATH which commutes with each MATH. From the above equation it follows that MATH is a homomorphism on MATH. Since MATH is finite it follows that MATH has finite index kernel MATH in MATH, and so MATH has finite index in MATH and in MATH. In other words, MATH on the finite index subgroup MATH of MATH, and so MATH represents the identity element of MATH. |
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 of a finite collection of finite index subgroups of MATH, whose intersection is a finite index subgroup corresponding to a surface group MATH which is a cover of MATH. The group MATH lifts to a subgroup MATH, which projects to a NAME subgroup of MATH. The group MATH is therefore a (surface)-by- REF group with finite index in MATH, and so also in the original group MATH. |
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 from symmetry and convexity that MATH. Thus it is sufficient to prove REF for MATH. Hence we may assume MATH is on or below MATH. Now let MATH be the reflection of MATH across MATH, and let MATH be the point where MATH intersects MATH. By the above assumptions on MATH and simple geometry, we have MATH and MATH. Using symmetry and convexity we therefore obtain MATH . Since MATH, it follows that MATH as desired. |
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, MATH . Next, provided MATH is small, an application of REF gives MATH . Together, REF complete the proof. |
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 dual site MATH, let MATH denote the event that there exist open dual paths MATH. By REF , MATH . We wish to show that the second term on the right side of REF is at most half of the first term on the right side. Let MATH. We have MATH . By REF , provided MATH is large, MATH . Fix MATH. We decompose the event MATH according to whether there is a MATH-near dual connection from MATH to MATH in MATH, where MATH is from REF and MATH is to be specified. Let MATH. Using REF , provided MATH and MATH are large we obtain MATH where MATH is from REF . Since MATH, provided we choose MATH large enough (depending on MATH) this gives MATH . Next, let MATH and for MATH let MATH. We have MATH . Now for MATH, MATH and similarly MATH . Hence provided MATH is large enough (depending on MATH), we obtain using REF that MATH . Combining this with REF , provided MATH and MATH are large we get MATH and then MATH . Combining REF we obtain MATH . Then from REF , provided MATH is large, MATH . Let MATH and MATH be dual paths of (minimal) length MATH from MATH to MATH and from MATH to MATH, respectively, in MATH. Let MATH denote the event that all dual bonds in MATH and MATH are open. From the FKG inequality, MATH and the lemma follows. |
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 MATH occurs on MATH and MATH occurs on MATH in MATH. If MATH for some MATH, then MATH since MATH. Therefore MATH and hence MATH. For MATH let MATH be the event that MATH occurs on some set MATH with MATH. We call MATH-sufficient if MATH for all MATH implies MATH. Let MATH denote the set of all finite MATH-sufficient subsets of MATH. Since the event MATH occurs on MATH for all MATH, we have by ratio weak mixing MATH . Combining REF proves REF , provided MATH is sufficiently large. Under REF of weak mixing and the near-Markov property for open circuits, the proof through the first inequality of REF is still valid, but we need to modify the rest of REF as follows. Fix MATH and let MATH, MATH, MATH. Let MATH be the set of all circuits (of regular bonds) in MATH which surround MATH and let MATH be the event that some circuit MATH is open; for MATH there is a unique outermost open circuit in MATH, which we denote MATH. Note that MATH and MATH . Let MATH . By weak mixing we have for MATH: MATH . Define the set of ``good" circuits MATH and let MATH . From REF and the near-Markov property for open circuits, if MATH is sufficiently large we obtain MATH where MATH for MATH as in REF . We need to bound MATH. To do this, we decompose MATH into REF pieces by intersecting it with MATH and MATH (the latter meaning there is no MATH.) We then show that the first piece is approximately bounded by MATH, and the other two pieces are negligible relative to the size of the full event. Specifically, from REF , MATH . Next, one application of REF yields MATH this and a second application of REF , with NAME 's inequality, yield MATH . Finally, similarly to REF we have using REF MATH where MATH. Combining REF we obtain MATH . Summing over MATH yields MATH which substitutes for REF . |
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 and MATH for all MATH, where MATH consists either of all odd MATH or of all even MATH. For MATH, we call the segment of MATH with endpoints MATH and MATH an interior gap. Let MATH be a dual path from MATH to MATH in MATH with MATH. We can extend MATH to a doubly infinite path MATH by adding on (possibly non-lattice) paths MATH from MATH to MATH and MATH from MATH to MATH, both in MATH. The path MATH divides the plane into two regions, MATH to the left of MATH and MATH to the right. Replacing MATH with MATH in the definition of MATH, we obtain another doubly infinite path MATH. The path MATH is not necessarily self-avoiding, but MATH has exactly two unbounded components MATH and MATH, to the left and right of MATH, respectively. Since MATH, there exist sites MATH and MATH for which MATH and hence MATH. Let MATH be a (possibly non-lattice) path from MATH to MATH in MATH. Then MATH must intersect MATH, and hence must intersect MATH, necessarily in some interior gap. Thus every MATH from MATH to MATH in MATH must cross at least one interior gap, so there exists an interior gap MATH which separates MATH and MATH, that is, exactly one of MATH is in MATH. It follows that MATH is a clean MATH-bottleneck. Since MATH, the proof is complete. |
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 classes: right offspring, which intersect MATH, and left offspring, which intersect MATH. Also, every point of MATH is either in a left offspring, in a right offspring, or in MATH. The diameter of MATH is at most MATH, while the diameters of MATH and MATH are at least MATH, so the right and left classes each include at least one (MATH)-large offspring,. Further, if MATH then REF follows from REF . Let MATH and MATH be sites of MATH with MATH. At least one of these points is not in MATH, so we may assume MATH is in some MATH. There are now four possibilities. First, if also MATH, then REF holds. Second, if instead MATH, then there exists a (MATH)-large offspring MATH, and we have MATH and again REF holds. Third, suppose MATH for some MATH and there exists a third (MATH)-large offspring MATH with MATH. Let MATH. Then MATH so once more, REF holds. The fourth possibility is that MATH for some MATH and MATH are the only (MATH)-large offspring; each is necessarily actually (MATH)-large. From REF we have MATH . Using this and the fact that MATH (since the unit square encloses unit area) we obtain MATH . Taking square roots yields REF . |
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 MATH. We next show that MATH . If not, there exist MATH, MATH and a dual path MATH from MATH to MATH in MATH with MATH. Let MATH be the last site of MATH in MATH, and MATH the first site of MATH after MATH which is in some segment MATH with MATH. Since all sites MATH are extreme points, we must have MATH. We claim that MATH is a MATH-bottleneck. By REF this is a contradiction, so our claim will establish REF . Suppose MATH; the proof if MATH is similar. We have MATH and MATH so MATH. Therefore MATH contains a segment in MATH which includes MATH and has diameter at least MATH. Similarly since MATH contains a segment in MATH which includes MATH and has diameter at least MATH. This proves the claim and thus REF . From REF we have MATH . REF shows that it is optimal to take MATH of order REF in our choice of MATH, so we now set MATH. For MATH and MATH for each MATH, let MATH be the event that for each MATH there is an open dual path MATH from MATH to MATH in MATH, with MATH for all MATH. Then we have shown MATH . Provided MATH is sufficiently large, REF and induction give MATH which with REF yields MATH . The number of possible MATH in REF is at most MATH, which with REF yields MATH provided MATH, and hence MATH, is large enough. It is easily verified that, provided MATH is small enough, MATH by considering two cases according to which of MATH and MATH is larger. This and REF establish REF for MATH; as we have noted, this and REF establish REF as given. |
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 nature of MATH, MATH is empty if any of the inequalities in REF fails. Note that for some MATH, MATH . Our induction hypothesis is that for some constants MATH, for all MATH, all MATH satisfying the basic assumptions, and all MATH satisfying REF , for any enclosure event MATH, we have MATH if the size condition is also satisfied, then in addition MATH with MATH from REF . We wish to verify these hypotheses for MATH. For MATH it suffices to consider MATH and REF is REF (together with REF ), while REF follows easily from the first inequality in REF , if MATH is large. Hence we may assume MATH and fix MATH. Let MATH denote the event that MATH occurs with MATH a primary MATH-bottleneck in MATH of type MATH, and let MATH. Let MATH be an enclosure event; it is easy to see that bottleneck surgery cannot destroy MATH, that is, MATH . (This is the reason for considering only enclosure events, not general increasing events.) Since MATH, we then have from the bounded energy property: MATH . Fix MATH and for MATH and MATH in MATH, let MATH denote the event that there exist disjoint exterior open dual circuits MATH such that: CASE: MATH surrounds MATH, MATH and MATH for all MATH; CASE: letting MATH denote the open dual circuit enclosing maximal area among all offspring of all MATH, we have MATH an offspring of MATH satisfying MATH and MATH ; CASE: there is no open dual path connecting MATH to MATH for MATH. We supress the parameters in MATH when confusion is unlikely. Then considering only MATH-large offspring we see that MATH where the union is over all parameters satisfying MATH . Here REF and the first inequality in REF follow from REF above and REF ,and MATH is from REF . Temporarily fix such a set of parameters and let MATH be circuits with MATH . Define events MATH . Then MATH . Note that as in REF , MATH . Define events MATH and let MATH denote the event that MATH occurs on MATH. Then MATH and MATH . The relation between area MATH and diameter MATH tells us roughly whether the circuit MATH (or its collection of descendants) is regular or irregular; we thereby subdivide the circuits into ``large regular," ``small regular" and ``irregular" categories as follows: MATH where MATH is chosen so that MATH . Let MATH (compare REF .) Now MATH is an enclosure event so by the induction REF , summing over MATH gives MATH . (Note that the size condition can be used here for MATH.) Since MATH for all MATH, from REF , provided MATH is sufficiently large we get MATH and MATH . Combining REF - REF we obtain MATH . Summing over MATH (which appears via MATH), dividing by MATH and iterating this (taking MATH and MATH to be the full space and MATH, at the last iteration step) we obtain using REF MATH . We now want to sum REF over all parameters of MATH allowed in REF . We first view MATH as fixed and allow the other parameters to vary. Note that the number of parameter choices is at most MATH, and the number of possible MATH is at most MATH, for some MATH. Suppose we can show that, under the basic assumptions, MATH and, if the size condition is also satisfied, MATH . Then from REF , MATH and if the size condition is satisfied, MATH . Thus, summing over MATH, then over MATH, and using MATH and REF , we obtain REF for MATH. Now REF is a direct consequence of REF , so we turn to REF and assume that the size condition holds. Let MATH and set MATH. We claim that MATH . Observe that MATH and hence MATH . This yields MATH which proves REF for MATH. For MATH we can use REF and a convex combination of the lower bounds REF and MATH for MATH to obtain MATH which proves REF for MATH. From REF , MATH and hence by REF MATH which proves REF for MATH. We need slightly different estimates for MATH. If MATH then using REF we obtain MATH which proves REF for MATH. If MATH then by REF , MATH which again proves REF for MATH. Finally if MATH then by REF remains valid with MATH in place of MATH, and REF/REF subtracted from the right side, once again proving REF for MATH. Thus REF holds in all cases. The next step is to sum REF over MATH. There are REF cases. CASE: MATH. Then using REF , MATH which proves REF . CASE: MATH. By REF , MATH . This and the size condition imply MATH if MATH is small enough, with MATH as in REF , and then MATH . Let MATH . Provided MATH is small enough, we have by REF MATH . It is easily checked that MATH . Choose MATH satisfying MATH . (If MATH or MATH is empty then the corresponding MATH or MATH is undefined.) We now consider two subcases. CASE: MATH or MATH. From the definition of MATH if MATH, and from REF if MATH, we have MATH where MATH. Therefore by REF with MATH, MATH . Hence by REF with MATH, and REF , MATH which gives REF . CASE: MATH. Let us relabel MATH as MATH. We have MATH while from REF , provided MATH is small enough, MATH so MATH which implies MATH . Using this, and using REF twice (with MATH and with MATH), we get MATH . If MATH then REF remains valid. This, with REF , shows that, whether MATH or not, REF (with MATH if MATH) still holds. The proof of REF , and thus of REF , is now complete. Taking MATH and MATH the full configuration space in REF , and summing over MATH satisfying REF shows that MATH . This proves REF , with MATH. It remains to prove REF . This is similar to the proof of REF , so we will only describe the changes. Again fix MATH. We make the same induction hypothesis, except that REF is replaced by MATH and the requirement that the size condition be satisfied is removed. This hypothesis is true for MATH, where only MATH is relevant; hence we fix MATH and MATH. Let MATH . In place of MATH we use MATH . In place of REF and their multi-case proofs, we have simply, using the first half of REF , MATH and hence using REF MATH . This leads directly to REF , as in the proof of REF . In place of REF we have MATH . For MATH we have MATH, so provided MATH is large enough, this proves REF . |
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 that MATH . Iterating this, and using REF and MATH, we obtain MATH . |
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 MATH . Note that if MATH or MATH is nonempty we must have MATH. If MATH we have MATH and hence MATH. Therefore MATH . If MATH we have MATH so that MATH which implies MATH . Therefore MATH . We can now use REF to sum over MATH and MATH in REF , obtaining REF . |
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 sufficiently large, MATH where the sums are over MATH. Now MATH can be bounded as in REF , so if we take MATH sufficiently large, REF prove that REF holds with conditional probability approaching REF as MATH. |
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. Let MATH be a constant to be specified, and suppose that MATH satisfies REF , but for some MATH and some MATH we have MATH and MATH. By REF we have MATH. Let MATH denote the region bounded by MATH and by the segment of MATH from MATH to MATH. Let MATH denote the line through MATH and MATH. There exists a site MATH with the following property: a line tangent to MATH at MATH passes through MATH. This MATH is a projection of MATH onto MATH and satisfies MATH . Let MATH . Write MATH, etc. for the intersection of MATH, etc. with MATH. We have several cases. CASE: MATH contains a MATH-near dual connection from MATH to MATH outside MATH. Here MATH is from REF . Then this event and the event MATH occur at separation MATH, so by REF , MATH . CASE: MATH contains no MATH-near connection from MATH to MATH outside MATH, and MATH. In this case, if MATH did not intersect MATH, then at least half of MATH would be contained in MATH. (This requires REF , which shows that we cannot have MATH.) Then from REF , MATH contradicting the definition of MATH. Thus MATH must intersect MATH. Let MATH be the first site of MATH in MATH, and let MATH be the last site of MATH in MATH. Note that by the definition of REF , we have MATH . We have two subcases within REF . CASE: MATH and MATH contain MATH-near connections from MATH to MATH and from MATH to MATH, respectively, both outside MATH. Then these near-connections occur at separation MATH from each other and from the event MATH, so by REF , since MATH, MATH . (Here we have actually used a trivial extension of REF , since the near-connection events occur not necessarily on the dual cluster of a single MATH but rather on the union of two such clusters.) Case REFb. MATH does not contain a MATH-near connection from MATH to MATH outside MATH. (The other alternative to REF within REF , symmetric to this one, is that MATH does not contain a MATH-near connection from MATH to MATH outside MATH; the proof is similar.) In this case the events MATH outside MATH outside MATH and MATH outside MATH occur at separation MATH from each other. Further, we have for MATH, using REF , MATH . Hence by REF , MATH . CASE: MATH contains no MATH-near connection from MATH to MATH outside MATH, and MATH. As in REF there are two symmetric alternatives within this, and we need only consider one, so we assume MATH. Then the events MATH and MATH occur at separation MATH and we have MATH . Hence MATH . Now let MATH denote the event that there is an open dual path from MATH to MATH containing a site MATH with MATH. Combining the three cases we obtain MATH and then analogously to REF , MATH . Analogously to REF - REF , this leads to MATH . We will need the following straightforward extension of REF , under the conditions of REF : MATH . It is easy to see (compare the proof of REF ) that MATH . Let MATH, with MATH from REF . From REF , MATH . Here the restriction to MATH is permissible as in the proof of REF . Provided we take MATH large enough, with REF this completes the proof. |
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 . The idea now is to split MATH into two halves and approximate the probability on the right side of REF by the product of the probabilities of the two halves. With this independence the two halves can in effect be pulled apart from one another to increase the area enclosed by MATH at only a small cost in increased boundary length. To accomplish this we first need some definitions. Let MATH be a path from MATH to MATH in the slab MATH. Let MATH and MATH denote the regions to the left and right, respectively, of MATH in MATH. The right-side area determined by MATH is MATH evaluated for MATH large enough that MATH contains MATH. (Note that for such MATH, the right-side area does not vary with MATH. Also, in our definition the path MATH must be oriented so that MATH.) The left-side area MATH is defined similarly using the left side of MATH. Let MATH and MATH be the points of MATH of maximum and minimum second coordinate, respectively, using the leftmost if there is more than one. Then MATH where MATH is MATH traversed in the direction of negative orientation. Let MATH and let MATH be the first and last lattice sites, respectively, of the segment of MATH containing MATH. Let MATH be the first site in MATH for which MATH and let MATH be the closest site to MATH in MATH. MATH and MATH are defined similarly with subscripts REF interchanged. Suppose now that MATH is MATH-bottleneck-free and satisfies MATH for MATH of REF . Then MATH and MATH are disjoint, and since MATH the absense of bottlenecks implies MATH . It follows that MATH REF and MATH . When MATH and MATH are paths such that the endpoint of MATH is the initial point of MATH, we let MATH denote the path obtained by concatenating MATH and MATH. Then MATH since the paths differ only inside MATH. Again we may interchange REF and MATH. Let MATH and let MATH for MATH. Presuming MATH is large enough, REF implies that MATH . Hence using REF , MATH . Let us assume for convenience that MATH is an integer (if not, the necessary modifications are simple), and let MATH and MATH be the lattice sites which are MATH units to the right of MATH and MATH, respectively. We now ``pull apart" the two halves of MATH by replacing each of these four sites by its right-shifted counterpart in the first probability on the right side of REF . Specifically, by the FKG property, MATH . From REF and the bounded energy property we obtain MATH completing the proof. |
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 MATH and define the event MATH . Note that MATH. Let MATH be the event that there exist MATH for which MATH and MATH via an open dual path outside MATH. Then for some MATH, MATH and MATH . Here the last inclusion follows from the fact that if MATH and MATH then MATH must surround or intersect MATH. The events MATH and MATH necessarily occur at separation MATH, so by REF we have MATH . We want to replace MATH with MATH on the right side of REF . We have MATH . Let MATH be a bond configuration on MATH. Conditionally on MATH, MATH is an increasing event (since MATH requires MATH, meaning MATH is part of MATH) and MATH is a decreasing one, so using REF again, MATH . Therefore MATH. With REF this shows that MATH . By REF and translation invariance, there exists a site MATH such that MATH . But the last event implies that MATH surrounds MATH, which contains MATH. Hence the last probability in REF is bounded by the first probability in REF , so that MATH . Since MATH, this completes the proof of REF . |
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 and MATH. Let MATH denote the event that MATH is the unique exterior dual circuit in MATH satisfying both MATH for some MATH and MATH. From REF , we have for MATH . Also, from the near-Markov property for open circuits, the FKG property, and REF , provided MATH is large, MATH so by REF , for MATH, using the MATH-compatibility of MATH, MATH . Combining these facts we obtain MATH . Here the second inequality uses the fact that MATH can occur in each bond configuration for at most MATH sites MATH. With REF , provided we take MATH large enough, REF gives MATH which completes the proof of REF . The proofs of REF are essentially the same, except that for REF , in place of REF we have the following. From REF and the proof of REF , for some MATH and for MATH sufficiently large: MATH . With REF this proves REF . The modification is similar for REF . For REF we must do more, because MATH may be much smaller than the order of the error term MATH in REF . Let MATH, with MATH still to be specified. From REF we have MATH . From REF , similarly to REF we have MATH . Let MATH denote the event that MATH and MATH is the unique exterior dual circuit in MATH satisfying MATH. Similarly to REF we have MATH . From REF we have MATH . But the last event, which says roughly that MATH approximates the appropriate NAME shape, implies that MATH surrounds some MATH-compatible site MATH. (Take MATH to be the closest site to REF in MATH.) Here MATH is from REF . Thus provided MATH is small, REF give MATH . We can repeat the argument of REF - REF (excluding the last inequality of REF ) once more to get MATH . Provided MATH is small, together, REF prove the lemma. |
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 satisfying MATH the last inequality following from REF and subadditivity of the square root. For MATH, we need only consider MATH and REF is immediate from the definitions, while REF follows easily from the first inequality in REF and the fact that MATH is increasing, and REF follows from REF . Hence we may assume MATH and fix MATH. Let MATH. There may be open dual paths MATH in MATH each connecting MATH to another open dual circuit MATH. Since MATH is exterior, for each such MATH there is a unique bond MATH with MATH which is part of such a MATH. We denote the set of all such bonds by MATH. Thus MATH. Let MATH . When this event occurs, the only MATH-large open dual circuit in MATH is MATH. From the above and the bounded energy property we have MATH where MATH is from REF . Given a circuit MATH define the event MATH . Note that MATH. It follows easily from the near-Markov property and the first inequality in REF that provided MATH is large, given MATH with high probability there are no MATH-large open dual circuits outside MATH; more precisely, MATH . Hence as in the proof of REF , conditioning on MATH and using the near-Markov property and REF for MATH gives MATH . Here the sums include MATH and MATH. Now REF establishes REF for MATH; we next establish REF . We have by REF MATH . Hence provided MATH is large enough, the right side of the next-to-last inequality in REF is bounded by MATH which proves REF for MATH. Turning to REF , from REF for MATH, and REF , the right side of the fourth inequality in REF is bounded by MATH . We now have two cases. CASE: MATH. From REF we have MATH . We can use this, REF with MATH, REF with MATH, and the fact that the maximum exceeds any convex combination to conclude that, provided MATH is sufficiently large, REF is bounded above by MATH . CASE: MATH. In this case it is easily checked that we must have MATH and MATH for MATH to be nonempty. But then MATH so REF follows from REF . The proof of REF for MATH is now complete. Summing REF as in REF , MATH . If MATH and MATH are large enough, then REF show that MATH and MATH . Therefore MATH . To prove REF , then, we need to bound the last probability in REF . We will sum as in REF , but this time using REF instead of REF . Let MATH. By REF , we need only consider MATH . Therefore using REF , MATH . By REF , MATH . With REF (taking MATH) this shows that, if MATH is sufficiently large, MATH . With REF this proves REF . Then REF prove REF - REF . It remains to establish REF . Let MATH to be specified, let MATH, and let MATH. Then provided MATH is small enough, we have MATH with MATH from REF . We have MATH . Here the second inequality uses REF , the third inequality uses REF , and the fourth inequality follows from REF . With REF this proves REF . |
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-near connection from MATH to MATH in MATH, or there is not; here MATH is to be specified. In the first case, there is also an open dual path from MATH to MATH, so we can apply REF (assuming MATH is large) to obtain MATH provided MATH is chosen small enough and MATH large enough. In the second case, there exists dual sites MATH just outside MATH and open dual paths MATH and MATH occuring at separation MATH. It follows easily from REF and the fact that MATH is close to MATH that MATH . Also, since MATH, MATH . Therefore by REF , MATH . Now REF , summed over MATH with MATH, prove the lemma. |
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 (compare REF .) Given a configuration MATH, there necessarily exists an open dual path from MATH to MATH in MATH which contains only one bond in MATH and one bond in MATH; we denote these two bonds by MATH and MATH, respectively, making an arbitrary choice if more than one choice is possible. If the configuration MATH is in MATH for some MATH, then we can modify at most MATH bonds (those in MATH) to obtain a configuration in MATH; in the resulting configuration we have MATH. Therefore from the bounded energy property we have MATH . This and REF prove the lemma. |
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 disjoint. (Our labeling as usual is cyclical: MATH.) For some MATH to be specified, let us call a pair MATH from the skeleton very short if MATH, short if MATH and long if MATH. In what follows, very short pairs can be handled quite trivially but tediously, so for convenience we will assume there are no very short pairs. For long pairs we define MATH and MATH to be the points on the line segment MATH at distance MATH from MATH and from MATH, respectively, and let MATH be dual sites in MATH within distance MATH of MATH and MATH, respectively. For short pairs we let MATH be a dual site in MATH within distance MATH of the midpoint of MATH. With minor modification of the definition of the MATH-hull skeleton, we may assume the set MATH has lattice symmetry, that is, for each MATH the reflection of MATH across the horizontal or vertical axis is another MATH, and analogously for the sites MATH and MATH. For each MATH we let MATH denote a dual lattice path of minimal length from MATH to MATH outside MATH. We call such a MATH a short link. Let MATH denote the bond boundary of MATH in MATH. Let MATH be the vertical line through MATH. Let MATH and MATH denote the open half planes to the left and right, respectively, of the vertical line through MATH. Let MATH and MATH denote the open half planes above and below the horizontal line through MATH, respectively. (In general we use the convention that subscripts MATH refer to left, right, upper and lower halfspaces, respectively, with combinations, such as MATH, referring to quadrants.) Let MATH denote the open slab between the vertical lines through MATH and MATH. Let MATH be the integer part of MATH, MATH the integer part of MATH and MATH the integer part of MATH. Let MATH . (Note each of these intersections is a single point.) We call these REF points determining points. Lattice symmetry yields corresponding determining points with appropriate subscripts in the other three quadrants. We may assume that MATH is one of the sites MATH of the MATH-hull skeleton of MATH (if not, we add MATH to the skeleton), and analogously for the other sites just defined. Let MATH be the second closest dual site above MATH in MATH, and analogously for MATH. If MATH, for some MATH, we define MATH to be MATH, and again analogously for the other determining points. Loosely, the idea is to remove from MATH its intersection with each of the width-MATH vertical slabs MATH, MATH, then raise or lower the segments of MATH between these slabs to adjust the area as desired, then reconnect these segments to make a new circuit enclosing area MATH. To do this we must first ensure that MATH intersects each vertical line bounding any of these four slabs only twice. We refer to REF width-MATH vertical slabs above as removal slabs. We call REF regions MATH (whose closures together form the complement of REF removal slabs) retention regions. By a retained segment we mean a connected component of the intersection of MATH with a retention region. Each retained segment has the form MATH for some MATH; we call MATH an initial determining point and MATH a final determining point, and call MATH a retention pair. We let MATH denote the set of all REF retention pairs. For each initial determining point MATH, in the boundary of some retained region MATH, we let MATH be a dual lattice path from MATH to MATH in MATH, of minimal length, and let MATH be the bond boundary of MATH in MATH. For each final determining point MATH we let MATH be a dual lattice path from MATH to MATH in MATH, of minimal length, and define MATH analogously to MATH. We refer to MATH and MATH as the endpaths of the retention pair MATH. For each retention pair MATH let MATH and let MATH denote the event that REF for each MATH, we have MATH-cylindrically in MATH, REF for each MATH we have MATH open and all bonds in MATH closed, and REF both endpaths of MATH are open and all bonds in MATH are closed. These REF component events are denoted MATH, MATH and MATH. For a configuration in MATH, the paths MATH together with the short links MATH and the two endpaths form an open dual path from MATH to MATH outside MATH, and there is no open dual connection from this path to any point of the retention region boundary except MATH and MATH. By REF , provided MATH is large we have MATH . From the bounded energy property, MATH which with REF yields MATH . For a configuration MATH, and for MATH the retention region with MATH, we can associate an area MATH as follows. There is a unique outermost open dual path MATH from MATH to MATH in MATH. If MATH is a halfspace MATH or MATH, then MATH is the area of the region between MATH and MATH. If MATH is a slab and MATH, then MATH is the area of the region in MATH below MATH. If MATH is a slab and MATH, then MATH is the area of the region in MATH above MATH. We define corresponding nonrandom areas MATH similarly but using MATH in place of MATH. Then MATH. It is not hard to see that for some MATH we have MATH . In fact, if this were false we could obtain extra area MATH almost ``for free" in REF ; more precisely, one could replace MATH with MATH on the left side in the conclusion of that theorem. But, assuming MATH is large, this would contradict REF . It follows from REF that for each retention pair MATH there exists MATH such that MATH . Let MATH where the sum is over REF retention pairs. We call MATH a removal pair if MATH is a connected component of the intersection of MATH with some removal slab. Let MATH denote the set of all REF removal pairs. For each removal pair MATH and corresponding removal slab MATH, let MATH be a dual path from MATH to MATH in MATH, and let MATH denote the event that all bonds in MATH are open, while all bonds in the bond boundary of MATH in MATH are closed. We call MATH a long link. There are REF long links in each removal slab, one each in the upper and lower half planes. Let MATH be the total area in REF removal slabs, between the upper and lower long link in each slab. Assuming the long links are chosen to have length of order MATH (say, at most MATH), we have from the bounded energy property that MATH . Define the event MATH . For a configuration MATH, there is an open dual circuit surrounding MATH satsifying the constraint that it include all of the short links MATH, long links MATH and endpaths MATH. There is a unique outermost such circuit subject to this constraint, obtained by taking the outermost MATH-cylindrical connection from MATH to MATH for each MATH; we denote this circuit MATH. Because of the cylindrical nature of these connections and the closed state of the bond boundaries of the short and long links, we have MATH, unless MATH and MATH are disjoint with MATH surrounding MATH and no open dual path connecting MATH to MATH. It therefore follows from the near-Markov property that MATH . By REF , MATH . Using these facts with REF (which is still valid here), we obtain MATH . Let MATH be the upward shift of a configuration MATH by REF unit, and for an event MATH let MATH. Let MATH be the set consisting of REF retention pairs corresponding to segments of MATH in the lower half plane. Given a constant MATH, for MATH we can replace MATH with MATH for all MATH throughout the argument leading to REF at the expense of only a possible increase in MATH, provided we alter REF long links MATH in the outer REF removal slabs to connect to the appropriate shifted sites MATH instead of to MATH. (The possible increase in MATH reflects a possible reduction in the probabilities of the events MATH, resulting in an increase of MATH in REF .) We can readily keep the area MATH fixed when we so alter the long links. We thereby obtain MATH . Provided MATH is large, we can choose MATH so that MATH . We can then repeat this entire argument, but shift upward (by some amount MATH) only the event MATH for the central of REF retention pairs in MATH. This gives MATH . We can choose MATH so that MATH . But it is easy to see that one can alter the long links to change MATH by any amount up to MATH, so that MATH at the expense of only a possible increase to MATH in REF . With REF this completes the proof. |
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 MATH large, the probability that a random normalized height function on MATH does not lie in the open set around MATH is exponentially small in MATH. |
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 triangle of side length MATH with MATH, then on that triangle we have the approximation property we want, because there the two functions MATH and MATH (the unique linear function that agrees with MATH on the corners of the triangle) both lie within MATH of MATH. Given MATH, let MATH be the set of all MATH such that MATH (depending on MATH as above) can be taken to be at least MATH. Take MATH small enough that the measure of MATH is at least MATH times MATH. (We can do that since MATH is differentiable almost everywhere.) Now take MATH. Look at any MATH-mesh, and the piecewise linear approximation MATH we get from it. If MATH is sufficiently small, then all but a MATH fraction of the mesh triangles lie entirely within the region. At least a MATH fraction of them must intersect MATH, which proves that the desired approximation property holds on at least a MATH fraction of the triangles. For the second property, we will apply a result on metric density (see REF). Let MATH be open subsets covering the set of possible tilts such that any two tilts contained within the same subset differ by at most MATH, and for MATH let MATH. It follows from the theorem on metric density that (for each MATH) if MATH is sufficiently small, then for all but a MATH fraction of the points MATH, at least a MATH fraction of the ball of radius MATH about MATH lies in MATH (and thus the tilt at those points differs by at most MATH from MATH). Now we can take MATH small enough that this result holds for all MATH from MATH to MATH, and then as above it follows that for MATH, a MATH fraction of the mesh triangles in any MATH-mesh lie entirely within MATH and satisfy the second property. Thus, if MATH is sufficiently small, at least a MATH fraction of the triangles satisfy both properties (since a MATH fraction satisfy each). |
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 piecewise linear approximation MATH coming from a MATH-mesh satisfies MATH on all but a MATH fraction of the triangles of the mesh, and MATH as MATH. Let MATH. Then if MATH, MATH on all but a MATH fraction of the triangles, and by REF the contribution to MATH from the non-exceptional triangles is at most MATH greater than the corresponding contribution to MATH. Of course, the remaining MATH fraction of the triangles contribute a total of MATH. Hence, MATH . By choosing MATH sufficiently small, one can make the MATH term less than MATH. Thus, MATH is upper semicontinuous. |
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, the limit of the subsequence must have maximal entropy. Now uniqueness is easy. Suppose that MATH and MATH are two asymptotic height functions that maximize entropy, with MATH. Their derivatives cannot be equal almost everywhere, so for some MATH we must have MATH on a set of positive measure, by the strict concavity of MATH. It follows that MATH which contradicts the assumption that MATH and MATH were maximal. (Notice that since MATH and MATH are asymptotic height functions, so is MATH.) Therefore, only one asymptotic height function can maximize entropy. |
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 functions are equal modulo MATH and thus that the lattice operations are well defined.) Then all values of MATH are within MATH of what one would get simply by un-normalizing MATH, because if one takes any lattice point, assigns it a value that is correct modulo MATH and at least MATH below the un-normalized value of MATH, and looks at the minimal height function taking that value there, then the NAME constraint on MATH implies that one stays below MATH. The height function MATH will typically not have the correct boundary values on MATH. Instead, it will have boundary heights corresponding to some different boundary height function MATH, although they will be within MATH of the actual boundary heights for MATH. To fix this problem, let MATH be the minimal height function on MATH, and let MATH be the maximal one. Then consider MATH, which is a height function with the correct boundary values for MATH. It follows from REF that MATH and MATH differ by only MATH from the minimal and maximal extensions MATH and MATH of MATH, respectively. Thus, since MATH, it follows that MATH differs by only MATH from MATH. Because the normalization of MATH differs from MATH by only MATH, we see that the normalization of MATH differs from MATH by at most MATH, as desired. |
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 two sides they alternate between MATH and MATH. |
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 point on the boundary of MATH, then the average height function for these tilings is given (exactly) by a linear function of the two position coordinates, so that its graph is a plane. If we select a random tiling from this distribution, then with probability differing from MATH by an exponentially small amount, the heights along the border of the patch stay within MATH of the plane, by REF. (It is not hard to check that all the large deviation results from REF, such as REF , apply to random tilings generated this way, not just from the uniform measures on tilings of finite regions. All that matters is conditional uniformity, in the sense that two tilings agreeing everywhere except on a subregion are equally likely to occur; conditional uniformity follows from MATH.) Consider all possible boundary height functions on MATH that stay within MATH of a plane. The average of the corresponding average height functions, weighted by how likely they are to occur in the weighted probability distribution, is within MATH of the plane (for MATH large). By REF, all boundary height functions that stay within MATH of the plane have average height functions within MATH of each other. Since the average is within MATH of the plane, each must be within MATH of it. This completes the proof. |
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 the number of south-going dominos bisected by it is MATH, as one can see by considering the total height change along the segment. Similarly, on a horizontal segment of length MATH, the number of west-going dominos bisected minus the number of east-going ones bisected is MATH. If we add up all of these quantities, the error term becomes MATH which is MATH since MATH. The total number of dominos in any tiling of MATH is MATH. By the results of the previous paragraph, this quantity is also twice the number of east-going or south-going dominos plus MATH. We have MATH. Hence, the total number of east-going or south-going dominos is MATH. Every tiling is determined by the locations of its east-going and south-going dominos. (To see why, recall that superimposing the matchings corresponding to two tilings gives a collection of cycles. Any disagreement between the two tilings yields a cycle of length at least MATH, which must contain an east-going or south-going edge that is in one tiling but not the other.) Hence, there are at most MATH tilings, since there are MATH possibilities for the number of east-going or south-going dominos, at most MATH places to put them, and at most MATH ways to choose which are east-going. It is not hard to check that NAME 's formula implies that MATH . Therefore, the entropy of MATH is MATH. |
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 entropy. Without loss of generality we can assume that MATH, since otherwise we can go from MATH to MATH and from MATH to MATH. Given any extension MATH of MATH, let MATH be the infimum of MATH and the maximal extension of MATH, so that MATH is an extension of MATH. Similarly, given any extension MATH of MATH, let MATH be the supremum of MATH and the minimal extension of MATH, so that MATH is an extension of MATH. The maps MATH and MATH are not inverses of each other, but they come fairly close to being inverses. Given an extension MATH of MATH, MATH agrees with MATH at every lattice point except those with heights less than or equal to their heights in the minimal extension of MATH. By REF , the minimal extension of MATH is within MATH of the minimal extension of MATH, so these points have heights within MATH of their minimal heights. Similarly, given an extension MATH of MATH, MATH agrees with MATH at all lattice points that are not within MATH of their maximal heights. By assumption, the tilt MATH of the plane satisfies MATH. Thus, the height difference between any two points in the plane is bounded by MATH times the sup norm distance between them. Therefore, REF imply that the extreme heights at a point differ from the heights on the plane by at least MATH times the sup norm distance to the boundary. Look at all lattice points not within sup norm distance MATH of the boundary. By REF , at any such point the average height for extensions of MATH or MATH is within MATH of the plane (notice that because MATH, being MATH is the same as being MATH), and the extreme heights differ from the plane by at least MATH, so the extreme heights differ from the average heights by an amount on the order of MATH, for MATH small. By REF, the probability that any such point will have height within MATH of its extreme heights is exponentially small in MATH. Thus, given any point not within sup norm distance MATH of the boundary, the probability that MATH or MATH will not be the identity at that point is exponentially small. It follows that with probability nearly MATH, MATH and MATH are the identity except on at most MATH lattice points. Thus, the numbers of extensions of MATH and MATH differ by at most a factor of MATH so the entropy of extensions of MATH to MATH differs from that of extensions of MATH by MATH. |
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 .) The torus is obtained by identifying opposite sides of MATH, so that tilings of MATH give tilings of the torus, but not vice versa. Keep in mind that because of the weighted edges in the torus, the probability distribution on its tilings will not be uniform. However, the equation MATH implies conditional uniformity (as mentioned earlier), so if we fix the behavior on the boundary of MATH, then the conditional distribution on extensions to the interior will be uniform. In REF, we will define a set MATH depending on MATH. By REF , for sufficiently large even MATH, either MATH or MATH is in MATH. In REF , we show that the entropy of MATH tori with tilt MATH converges to MATH as MATH in MATH. First we will suppose that MATH, so that the entropy of the MATH torus is MATH. It follows from REF that with probability exponentially close to MATH, in a random tiling of the torus, the heights on the boundary of the square will be fit to within MATH by the average height function, which is a linear function with tilt MATH. The number of toroidal boundary conditions is exponential in MATH, and by REF each has about the same entropy (except ones that are not nearly planar, but they are very unlikely to appear). REF tells us that the entropy of the torus equals the average of the entropies for the different boundary conditions, plus a negligible quantity for large MATH (since we have normalized by dividing by half of the area MATH). Because all the nearly planar boundary conditions have the same entropy, the torus must as well, so since we know it has entropy MATH as MATH, each of the nearly planar boundary conditions must have entropy MATH. (Since MATH is fixed as MATH, we can absorb the MATH into the big-MATH term.) Now it is easy to deal with the case of MATH. If MATH, then MATH and MATH. The entropies for tilings of MATH, MATH, and MATH squares with nearly planar boundary conditions are nearly the same. (To prove that, embed a MATH square into a MATH one, and a MATH one into a MATH one, extending the boundary conditions arbitrarily. Then the number of tilings increases with each embedding, so the entropy of the MATH square is caught between the other two, to within a MATH factor coming from the differing areas. By REF , this result holds for all nearly planar boundary conditions.) Note that the same argument as above goes back from squares to the torus, thus proving entropy convergence for all MATH (not just MATH). The claim about free boundary conditions follows easily (since the number of boundary conditions that stay within MATH of the plane is only exponential in MATH, and all of them have about the same entropy). |
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 squares, such that their boundary heights are within MATH of being planar. (Of course, those near the edges will stick out over the boundary, but if MATH is large enough, we can make the squares small enough compared to the triangle that only a MATH fraction of the squares will cross the boundary.) Except for an error of MATH from the squares that cross the boundary, the entropy of the triangle is at least the entropy of the squares, which is what we wanted. An analogous argument (involving tiling a square with triangles) shows that the entropy of the triangle is at most what we expect. |
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). Then, as is pointed out after the statement of REF , MATH. We will take MATH, and show that the entropy we want to compute is MATH. We know that MATH on all but at most a MATH fraction of the triangles in the mesh. Those triangles can change the entropy by only MATH, so we can ignore them. We can also ignore the MATH fraction of the triangles that do not lie within the normalization of MATH (which change the entropy by MATH). We will call the triangles within the normalization of MATH on which MATH the included triangles, and the others the excluded triangles. Let MATH be any element of MATH. The entropy of MATH is bounded below by the sum over all included triangles of the entropy of MATH restricted to that triangle, plus the MATH contribution from the excluded triangles. It is bounded above by the same sum (including the MATH), but with free boundary conditions on the included triangles (subject to the condition of staying within MATH of MATH). We can now apply REF . It tells us that each included triangle's contribution to the entropy of MATH is approximately equal to its contribution towards the entropy of MATH. It follows that our upper and lower bounds for the entropy of MATH both equal MATH. This gives us the desired conclusion. |
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 mapping to MATH). When MATH, MATH degenerates to a segment; still, MATH maps the punctured open unit disk injectively onto the exterior of the segment. The case of MATH is similar. |
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 . We then have MATH and similar expressions for MATH and MATH (see REF - REF ). The quantity MATH is a NAME sum for the integral of MATH. The function MATH is continuous, hence NAME integrable, on the complement of a small neighborhood of its two (possible) singularities MATH and MATH, defined in REF. We break the proof into four cases: the case MATH, the case MATH but not both MATH and MATH, the case MATH and MATH, and finally the case MATH. If MATH, then MATH is continuous everywhere on MATH and so the NAME sums converge to MATH. In the case MATH but not both MATH and MATH, there are two singularities. It suffices to show that the NAME sums MATH for each MATH are small on a small neighborhood of the singularities. This is not quite true since for some MATH the product MATH may have a factor in which MATH lands close to a singularity; as a consequence this MATH may be very small. However, we will show that this can happen for at most one of the four products MATH. For each term in the four products REF - REF , MATH is of the form MATH, for integers MATH. Furthermore for each pair of integers MATH exactly one of the four products has a term with MATH. Thus at most one of the four products has a term closer than MATH to the singularity MATH. The same MATH will have the term closest to the other singularity MATH. Fix a small constant MATH and let MATH be the MATH-neighborhood of a singularity. In MATH we use the NAME expansion MATH and by REF below, the ratio of MATH and MATH is not real. The sum of those terms in MATH for which MATH is MATH for constants MATH with MATH. (We used here the fact that MATH for small MATH; we then could take the big-O term out of the summation because of the MATH factor.) To bound this sum, note that MATH for some positive constants MATH since MATH. We use polar coordinates around the singularity. In the annulus around MATH of inner radius MATH and outer radius MATH, there are at most MATH points MATH which contribute to the sum, and each such point contributes MATH to the sum. Therefore the sum on MATH (for those MATH without terms within MATH of the singularity) is bounded by MATH . For the MATH which does have a term closer than MATH to the singularity, the above calculations give an upper bound on MATH. (Including in the factor close to the singularity only decreases the product.) Thus we have shown that MATH converges to MATH. This proves the convergence of MATH to MATH. In the case MATH and MATH we have only one singularity MATH, and MATH. For the other MATH's, when MATH is close to MATH we have MATH . An argument similar to the previous case holds: on MATH we have MATH . Now MATH. Summing over annuli as before gives the bound. Finally, suppose MATH. For any MATH we have MATH (recall that coefficients of the polynomial MATH are non-negative). For fixed MATH, the limits MATH both exist, and (as we shall see in REF below) converge to the same value as MATH. Thus MATH exists and converges to this same value. This completes the proof. |
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 these for MATH and plugging in to the above gives MATH which is zero only if the real and imaginary parts of the numerator and denominator are in proportion: either MATH or (clearing denominators) MATH . Neither of these is possible (recall that MATH only when MATH and MATH), so the proof is complete. |
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