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math/0011152 | If MATH is amenable and MATH is an affine onto MATH-map between compact convex MATH-spaces, then for every MATH-fixed point MATH there exists a MATH-fixed point MATH such that MATH. Indeed, the set MATH is a compact convex MATH-space and hence contains a MATH-fixed point. Let MATH be an onto homomorphism. Assume that MATH is uniquely amenable. Applying the remark of the previous paragraph to the sets of probability measures on MATH and MATH, we see that every left-invariant measure on MATH is the image under MATH of a left-invariant measure on MATH. Since the latter is unique, it follows that the left-invariant measure on MATH is also unique. |
math/0011152 | The preceding discussion shows that REF - REF are equivalent. Applying REF to MATH, we see that MATH. The implication MATH can be established directly. We omit the proof, since we show below REF that REF actually implies that MATH is precompact, and for precompact groups REF is straightforward: just note that the MATH-space structure on a compact space MATH extends to a MATH-space structure. Of course one has to keep in mind that in our proofs the implication MATH is never used. |
math/0011152 | Let MATH be a w.a.p. topological group, and let MATH be an onto homomorphism. If MATH, then MATH. The map MATH is an isometric embedding for the norm topologies and hence a homeomorphic embedding for the weak topologies. Since MATH sends the orbit of MATH to the orbit of MATH and the orbit of MATH is weakly relatively compact, so is the orbit of MATH. Thus MATH. Now let MATH be a subgroup of MATH and MATH. A uniformly continuous bounded function defined on a subspace of a uniform space can be extended to a uniformly continuous bounded function defined on the entire space CITE, CITE. Since the right uniformity of MATH is induced by the right uniformity of MATH, there exists a function MATH such that MATH. By assumption, MATH. The restriction map MATH sends weakly relatively compact subsets of MATH to weakly relatively compact subsets of MATH, while the image of the MATH-orbit of MATH contains the MATH-orbit of MATH. We conclude that MATH. |
math/0011152 | Let MATH be non-precompact. There exists an infinite set MATH and a neighbourhood MATH of unity such that the family MATH is disjoint. The subgroup of MATH generated by MATH is countable and non-precompact. Thus we may assume that MATH is countable. Every countable group, being MATH-bounded, is isomorphic to a subgroup of the product MATH of countable metrizable groups. We may assume that the projections of MATH to the factors MATH are onto and hence all the MATH's are in MATH. Since the class of precompact groups is closed under products and subgroups, at least one of the factors MATH's has to be non-precompact. |
math/0011152 | Only MATH requires a proof. Let MATH. As an immediate consequence of this assumption and NAME REF , the topological group MATH is uniquely amenable. Assume first that MATH is separable metrizable. According to NAME 's REF , MATH is precompact. The general case is now being reduced to the case of a separable metrizable group with the aid of REF . |
math/0011153 | The proof of REF is standard, but we sketch it here for the sake of completeness. Let MATH be the quotient map. Since MATH, there is a closed curve MATH curve, not necessarily simple, on MATH that represents an infinite order element of the orbifold fundamental group of MATH. Let MATH in MATH be the subset of MATH that consists of all the fibers in MATH corresponding to points of MATH. Then, MATH is an incompressible REF-torus in MATH, though not necessarily embedded. However, this is sufficient to guarantee that there exists a MATH subgroup of MATH, namely the fundamental group of MATH. |
math/0011153 | The proof of REF is essentially contained in Soma CITE; we include it here solely for the sake of completeness. We begin by considering the prime decomposition of MATH. That is, write MATH as the connected sum MATH, where each MATH is a prime REF-manifold. (Note that we are including in this discussion the case that MATH is itself prime, and so has trivial prime decomposition.) Since simplicial volume behaves additively with respect to connected sums (compare CITE), the hypothesis that MATH has zero simplicial volume implies that each MATH has zero simplicial volume as well. Since the connected sum of graph manifolds is again a graph manifold (compare Soma CITE), it suffices to show that each MATH is a graph manifold. Since each MATH is prime, it is either irreducible or diffeomorphic to MATH, which is a NAME fibered space. So, we may assume without loss of generality that MATH is irreducible. Let MATH be the torus decomposition of MATH. Recall that MATH is assumed to be geometrizable. If MATH is empty, then MATH admits a geometric structure other than the one modelled on MATH (which is excluded by the assumption on the simplicial volume of MATH), and so MATH is a graph manifold, by REF . If MATH is non-empty, then MATH is sufficiently large, and so NAME 's geometrization conjecture holds for MATH. Since MATH, each component of MATH is a NAME fibered space, as no piece can be hyperbolic, by REF . It follows that MATH must be a graph manifold. |
math/0011153 | We start by showing that if MATH admits a geometric structure modelled on one of these REF geometries, then the minimal entropy problem for MATH can be solved. Observe first that if MATH admits a geometric structure modelled on MATH, then the minimal entropy problem can be solved by the results of CITE. It follows immediately from REF that if MATH admits a geometric stucture modelled on one of the seven geometries MATH, MATH, MATH, MATH, MATH, MATH, or MATH, then MATH is a graph manifold. Hence by REF and the chain of REF , we have that for such a MATH, the minimal entropy satisfies MATH. We now show that if MATH admits a geometric structure modelled on one of MATH, MATH, MATH, or MATH, then the minimal entropy problem for MATH can be solved. To do this, we need to show that MATH admits a smooth metric MATH with MATH. CASE: MATH, MATH, MATH: All the NAME fields in these geometries grow at most linearly (in the case of MATH they are actually bounded), and hence all the NAME exponents of every geodesic in MATH are zero. It follows from NAME 's inequality CITE that all the measure entropies are zero. Hence, by the variational principle, the topological entropy of the geodesic flow of MATH must be zero. CASE: MATH: This geometry can be described as MATH with the metric MATH . Here, not all the NAME fields grow linearly, but they certainly grow polynomially. Again this implies that all the NAME exponents of every geodesic in MATH are zero and hence the topological entropy of the geodesic flow of MATH must be zero. Since we have assumed that MATH admits a geometric structure, we complete the proof by showing that if MATH admits a geometric structure modelled on one of remaining geometries, namely MATH, MATH, and MATH, then MATH cannot admit a metric of zero topological entropy. To do this, we use the next lemma, together with the fact described in Subsection REF, that if MATH grows exponentially, then MATH for any smooth metric MATH on MATH. Let MATH be a closed orientable REF-manifold, and suppose that MATH admits a geometric structure modelled on one of MATH, MATH, or MATH. Then MATH grows exponentially. In the case that MATH admits a geometric structure modelled on MATH or MATH, we start by recalling from REF that MATH is then a NAME fibered space. The base orbifold of the NAME fiber space admits a hyperbolic structure, and so the orbifold fundamental group of the base orbifold contains a free subgroup of rank REF, and hence so does MATH. Hence, MATH grows exponentially. In the case that MATH admits a geometric structure modelled on MATH, we have that MATH is finitely covered by the mapping torus MATH of a hyperbolic automorphism of a REF-torus. Note that a hyperbolic automorphism of a REF-torus is an NAME diffeomorphism, and so the suspension flow on MATH is an NAME flow. It is known that the fundamental group of a REF-manifold with an NAME flow has exponential growth (see for example CITE). This completes the proof of REF . |
math/0011153 | Let us show that REF implies REF . Suppose then that MATH has zero simplicial volume and that the minimal entropy problem for MATH can be solved. We show that MATH must then admit a geometric structure modelled on either MATH or MATH. Since the fundamental group of MATH contains a MATH subgroup, REF ensures that either MATH contains an incompressible embedded REF-torus or MATH is a NAME fibered space. We now split the proof into two cases: CASE: Suppose first that MATH contains an incompressible embedded REF-torus, and so is sufficiently large. Since we have assumed that MATH, REF yields that MATH is a graph manifold. Hence, by REF , we have that MATH. However, using work of CITE, specifically REF, we see that either MATH contains a free subgroup of rank REF or MATH is finitely covered by a REF-torus bundle over MATH. In the former case, MATH grows exponentially and therefore the minimal entropy problem cannot be solved for MATH. In the latter case, MATH admits a geometric structure modelled on one of MATH, MATH, or MATH (compare REF). However, in the case that MATH admits a geometric structure modelled on MATH, we know from REF that the minimal entropy problem cannot be solved for MATH. Hence, if the minimal entropy problem can be solved for MATH and if MATH contains an incompressible embedded REF-torus, then MATH admits a geometric structure modelled on either MATH or MATH. CASE: The other case is that MATH is a NAME fibered space. Here, REF ensures that MATH possesses a geometric structure modelled on one of MATH, MATH, MATH, MATH, MATH or MATH. Since the fundamental group of MATH admits a MATH subgroup, the geometric structure on MATH cannot be modelled on MATH or MATH. Since we have assumed that the minimal entropy problem can be solved for MATH, REF yields that MATH must admit a geometric structure modelled on either MATH or MATH, as desired. To see that REF implies REF , recall from REF that if MATH admits a geometric structure modelled on MATH or MATH, then MATH admits a smooth metric MATH with MATH. Finally to prove that REF implies REF , observe that if MATH admits a smooth metric MATH with MATH it then follows from REF that MATH has zero simplicial volume. This completes the proof of Theorem A. |
math/0011153 | Let us prove that REF implies REF . Suppose that MATH has zero simplicial volume and that the minimal entropy problem for MATH can be solved. Since MATH is geometrizable and its simplicial volume vanishes, REF tell us that MATH is a graph manifold. Hence, by REF , MATH has zero minimal entropy. Since we are assuming that the minimal entropy problem can be solved for MATH, the fact that MATH has zero minimal entropy in turn implies there exists a smooth metric on MATH with zero topological entropy. This in turn implies, by the discussion in REF, that MATH does not have exponential growth. However, it is a fact from combinatorial group theory (which follows immediately from the existence of normal forms for free products, for instance) that if MATH and MATH are two finitely generated groups, then the free product MATH contains a free subgroup of rank two unless MATH is trivial or MATH is trivial, or MATH and MATH are both of order two. Since the fundamental group of a connected sum is the free product of the fundamental groups of the summands, we conclude that either the prime decomposition is trivial or there are only two summands both of which have fundamental group MATH. In the former case, it follows that MATH must be either irreducible or MATH, while in the latter case MATH must be MATH, where MATH is REF-dimensional real projective space. Since MATH and MATH both admit a geometric structure modelled on MATH, we may assume from now on that MATH is irreducible. There are now several cases, depending on MATH. Suppose first that MATH is finite. Since MATH is geometrizable, we have that MATH admits a geometric structure modelled on MATH. In the case that MATH is infinite and contains a MATH subgroup, the assumption that the simplicial volume of MATH is zero, together with the fact that the minimal entropy problem can be solved for MATH, allows us to apply Theorem A to see that MATH admits a geometric structure modelled on MATH or MATH. The remaining case is that MATH is infinite and does not contain a MATH subgroup. Since MATH is geometrizable, either MATH admits a hyperbolic structure or MATH is NAME fibered. (Since MATH does not contain a MATH subgroup, MATH cannot admit a geometric structure modelled on MATH, as MATH manifolds are finitely covered by REF bundles over the circle.) However, since MATH, MATH cannot admit a hyperbolic structure. Note though that MATH cannot admit a geometric structure modelled on MATH, MATH, MATH, or MATH, as such manifolds always have a MATH in their fundamental groups, by REF . Hence, the only possibilities remaining are that MATH admits a geometric structure modelled on either MATH or MATH, as desired. To see that REF implies REF , recall from REF that if MATH admits a geometric structure modelled on MATH, MATH, MATH, or MATH, then MATH admits a smooth metric MATH with MATH. Finally to prove that REF implies REF , observe that if MATH admits a smooth metric MATH with MATH, it then follows from REF that MATH has zero simplicial volume. This completes the proof of Theorem B. |
math/0011158 | See CITE. |
math/0011158 | Let MATH be a point satisfying REF . For large MATH we have MATH by definition of non-uniform expansion on random orbits. Fixing MATH we see that REF implies MATH for every MATH in a neighborhood MATH of MATH. Now we take MATH so that MATH and let MATH be small enough to get MATH which is possible after REF of slow approximation to MATH. Moreover, fixing MATH sufficiently large in order that it be also an upper bound for for the set MATH, then the set MATH is such that MATH for all MATH and MATH that is, MATH, in particular MATH for all MATH. Hence, defining MATH it holds MATH for MATH, and REF imply MATH . Since MATH we deduce MATH . By the previous arguments we may apply REF to the sequence MATH with MATH and MATH (we may suppose MATH too by increasing MATH if needed). Thus there are MATH and MATH times MATH such that MATH for every MATH. We observe that REF is just the first part of the requirements on MATH-hyperbolic times for MATH if MATH. Now we apply again REF , this time to the sequence MATH, where MATH is small enough so that for MATH with MATH we have by REF MATH . Defining MATH, MATH, MATH and MATH . REF ensures that there are MATH times MATH satisfying MATH for every MATH, MATH. Let us note that the condition on MATH assures MATH. So if MATH, then there must be MATH and MATH for which REF both hold. This means that for MATH and MATH we have MATH and hence these MATH are MATH-hyperbolic times for MATH, with MATH and MATH. It follows that for MATH almost every MATH there are (positive frequency of) times MATH for which MATH for every MATH. Now the conclusion of the lemma is a direct consequence of REF . |
math/0011158 | We are assuming MATH since MATH is a MATH-hyperbolic time for MATH. This means that MATH . Either way it holds MATH because MATH and MATH for all MATH in the ball of radius MATH around MATH. Therefore REF implies MATH . But MATH and MATH so MATH and thus the right hand side of the last expression is bounded from above by MATH. The assumptions on MATH assure this last bound to be smaller than MATH, which implies the statement. |
math/0011158 | The proof will be by induction on MATH. First we show that there is a well defined branch of MATH on a ball of small enough radius around MATH. Now we observe that REF gives for MATH because MATH is a MATH-hyperbolic time for MATH. This means that MATH is a MATH-dilation in the ball of radius MATH around MATH. Consequently there is some neighborhood MATH of MATH inside the ball of radius MATH that is diffeomorphic to the ball of radius MATH around MATH through MATH, when MATH is a map with critical set satisfying REF . For MATH let us suppose that we have obtained a neighborhood MATH of MATH such that MATH is a diffeomorphism onto the ball of radius MATH around MATH with MATH for all MATH and MATH. Then, by REF and under the assumption that MATH is a MATH-hyperbolic time for MATH, MATH for every MATH on the ball of radius MATH around MATH whose image MATH is in MATH (above we convention MATH for MATH). This shows that the derivative of MATH is a MATH-dilation on the intersection of MATH with the ball of radius MATH around MATH, and hence there is an inverse branch of MATH defined on the ball of radius MATH around MATH. Thus we may define MATH as the image of the ball of radius MATH around MATH under this inverse branch, and recover the induction hypothesis for MATH. In this manner we get neighborhoods MATH of MATH as above for all MATH. |
math/0011158 | For MATH the distance between MATH and either MATH or MATH is smaller than MATH which is smaller than MATH. So, by REF we have MATH and it is enough to take MATH, recalling that MATH and also REF . |
math/0011158 | Is is enough to obtain that each connected component MATH is periodic under the action of MATH for MATH, in the sense that MATH for some MATH and all MATH. There are components MATH with nonempty interior, since the interior of MATH is nonempty. So we may take a component MATH that contains some ball MATH. Then we have MATH and so MATH. NAME Recurrence Theorem now guarantees there is MATH such that the MATH-orbit of MATH has the same MATH as an accumulation point. We see that there must exist some MATH such that MATH. In view of the independence of the permutation on the choice of MATH, we conclude that MATH is sent inside itself by MATH for all MATH. |
math/0011158 | See CITE. |
math/0011158 | The proof relies in showing that NAME 's Lemma can be applied to this partially ordered set and that minimal domains are pairwise separated. See CITE. |
math/0011158 | Let MATH be a weak-MATH accumulation point of MATH. We may write MATH for each continuous MATH. Moreover dominated convergence ensures that we may exchange the limit and the outer integral sign and, by definition of MATH, we get MATH according to the definition of MATH. Thus REF must hold and MATH is stationary. Noting that MATH is dense in MATH with the MATH norm, we see that REF holds for all MATH-integrable functions MATH. In particular, if MATH is such that MATH, then MATH . This process may be iterated to yield MATH and, since MATH by nondegeneracy REF , we must have MATH. |
math/0011158 | Let us assume MATH for simplicity (see CITE for the general case) and let us consider a stationary absolutely continuous probability measure MATH with MATH. We first show the ergodicity of MATH, in the sense that MATH is MATH-ergodic. It turns out that to be MATH-ergodic it suffices that either MATH or MATH for every NAME set MATH satisfying MATH for MATH almost every MATH (compare CITE and CITE). So let us take MATH such that MATH and MATH satisfies the left hand side of REF . Then it must be MATH because MATH and there is a closed set MATH such that MATH and also MATH. Hence MATH also satisfies the left hand side of REF because of nondegeneracy REF (with MATH), since MATH . This means that when MATH we have MATH for MATH almost all MATH. Since a set of MATH measure REF is dense in MATH (we are supposing MATH to be positive on open sets) and MATH varies continuously with MATH, we see that MATH for all MATH because MATH is closed. We then have that the interior of MATH is nonempty by REF on random perturbations and we may apply the methods of decomposition into connected components as before REF . In this manner we construct an invariant domain inside MATH which, in turn, is inside a minimal invariant domain. This contradicts minimality and so we conclude that MATH must contain MATH. Thus we have MATH proving MATH to be MATH-ergodic. Now, given MATH continuous we consider the map MATH from MATH to MATH, where MATH is the natural projection. The Ergodic Theorem then ensures MATH for MATH almost all MATH, which is just the same as MATH for MATH almost all MATH. Finally considering the ergodic basin MATH, defined as the set of points MATH for which MATH for all MATH and MATH almost every MATH, it is easy to see that MATH satisfies REF in the place of MATH and we must have as before MATH. This shows that if another stationary absolutely continuous probability measure MATH is such that MATH, then the basins of MATH and MATH must have nonempty intersection. Thus these measures must be equal. Moreover MATH and so, by absolute continuity, MATH and thus MATH is a physical probability. |
math/0011158 | Take MATH and let MATH be a weak-MATH accumulation point of the sequence MATH defined in REF . We will prove that this is the only accumulation point of REF by showing that the values of the MATH in decomposition REF depend only on MATH and not on the subsequence that converges to MATH. The definition of the average in REF implies that there is a subset of parameter vectors MATH with positive MATH measure for which there is MATH such that MATH. We define for MATH . We clearly have MATH and MATH for all MATH, since the supports of stationary measures are themselves invariant. In addition, since MATH is a regular REF probability measure, we may find for each MATH an open set MATH and a closed set MATH such that MATH with MATH and MATH. In fact, there is an at most countable number of MATH-neighborhoods of MATH whose boundaries have positive MATH measure, and likewise for the compacts coinciding with the complement of the MATH-neighborhood of MATH. Then, taking MATH we have MATH for some sequence of integers MATH, and likewise for MATH where MATH is arbitrary. This shows MATH . We also have MATH which shows that the MATH depend only on the random orbits of MATH and not on the particular sequence MATH. Thus we see that the sequence of measures in REF converges in the weak-MATH topology. Moreover the sets MATH are pairwise disjoint by definition and their total MATH measure equals MATH, thus forming a MATH modulo zero partition of MATH. We observe that if MATH, then MATH for some MATH and MATH. This means this MATH modulo zero partition of MATH satisfies the first item of the proposition. Now fixing MATH, for all MATH (the ergodic basin of MATH) it holds that MATH for MATH almost every MATH. Recall that MATH by the definition of physical measure. Using dominated convergence and integrating both sides of the above equality twice, first with respect to the NAME measure MATH, and then with respect to MATH, we arrive at the statement of REF . Recall that up until now the noise level MATH was kept fixed. For small enough MATH the measures MATH depend on the noise level, but we will see that the number of physical measures is constant. Fixing MATH we let MATH in the interior of MATH be such that the orbit MATH has infinitely many hyperbolic times. Recall that MATH is non-uniformly expanding (possibly with criticalities). Then there is a big enough hyperbolic time MATH so that MATH, by REF , where we take MATH. Since MATH and MATH is invariant under MATH for all MATH, we must have MATH where MATH is the constant given by REF and MATH is the ball of radius MATH around MATH. On the one hand, we deduce that the number MATH is bounded from above by some uniform constant MATH since MATH is compact. On the other hand, since each invariant set must contain some physical measure (by REF ), we see that for MATH there must be some physical measure MATH with MATH. In fact MATH is invariant under MATH for every MATH. This means the number MATH of physical measures is a nonincreasing function of MATH. Thus we conclude that there must be MATH such that MATH is constant for MATH, ending the proof of the proposition. |
math/0011158 | We start by observing that if MATH then every weak-MATH accumulation point of a family MATH of physical measures when MATH must equal the unique SRB measure MATH for MATH. Hence the weak-MATH limit of MATH when MATH exists and equals MATH for all MATH. Then there must be a single physical measure MATH for all small enough MATH. In fact, let us assume there are distinct families MATH and MATH of physical measures as given by REF . Take MATH and a sequence of continuous maps MATH such that MATH for all big MATH, where MATH denotes the ball of radius MATH around MATH for each MATH. If we fix MATH, then MATH. Thus for all small enough MATH we must have MATH and MATH. Hence the distance between MATH and MATH is smaller than MATH. Since MATH may be arbitrarily large we see that MATH and MATH get arbitrarily close when MATH. Thus MATH and MATH must coincide for small MATH because REF and its proof show that these families, when distinct, are at a uniform distance apart (since MATH and MATH are nested families of pairwise disjoint compact sets). In general, if MATH then the weak-MATH accumulation points of a family MATH for MATH when MATH are convex linear combinations MATH of the MATH SRB measures of MATH. The preceding argument implies that two distinct families MATH and MATH of physical measures cannot have weak-MATH accumulation points expressed as linear convex combinations MATH with both MATH and MATH nonzero for some MATH (take MATH and repeat the arguments in the above paragraph). Hence MATH. |
math/0011158 | We will use the following auxiliary result: Let MATH be a compact metric space, MATH a closed (compact) subset and MATH a curve in MATH (not necessarily continuous) such that all its accumulation points (as MATH) lie in MATH. Then for every open neighborhood MATH of MATH there is MATH such that MATH for every MATH. Indeed, supposing not, there is a sequence MATH with MATH when MATH such that MATH. Since MATH is compact this means that MATH has some accumulation point in MATH, thus outside MATH, contrary to the assumption. Now, the space MATH of all probability measures in MATH is a compact metric space with the weak-MATH topology, and the convex hull MATH of the (finitely many) SRB measures of MATH is closed. Hence, considering the curve MATH in MATH, we are in the context of the above result, since we are supposing MATH to be stochastically stable. A metric on MATH topologically equivalent to the weak-MATH topology may be given by MATH where MATH and MATH is a dense sequence of functions in MATH, see CITE. Let MATH continuous be given and let us fix some MATH. There must be MATH such that MATH and, by the auxiliary result in the beginning of the proof, there exists, for some MATH and every MATH, a probability measure MATH for which MATH. This in particular means that MATH by the definition of the distance MATH, which implies MATH . Hence we get MATH which completes the proof of the lemma. |
math/0011158 | Take MATH given by REF . It is sufficient to prove that there is some uniform constant MATH such that if MATH is a NAME set in MATH with diameter smaller than MATH then MATH . Let MATH be a NAME set in MATH with diameter smaller than MATH and MATH an open ball of radius MATH containing MATH. We may write MATH where MATH is a (possibly finite) family of two-by-two disjoint open sets in MATH. Discarding those MATH that do not intersect MATH, we choose for each MATH a point MATH. For MATH let MATH be the neighborhood of MATH in MATH given by REF . Since MATH is contained in MATH, the ball of radius MATH around MATH, and MATH is a diffeomorphism from MATH onto MATH, we must have MATH (recall that by our choice of MATH we have MATH). As a consequence of this and REF , we have for every MATH that the map MATH is a diffeomorphism with bounded distortion: MATH for all MATH. This finally gives MATH where MATH is a constant only depending on MATH, on the volume of the ball MATH of radius MATH, and on the volume of MATH. |
math/0011158 | Let MATH be some NAME set in MATH. We have for each MATH . (in this last inequality we used that MATH is MATH-invariant and estimate REF above). Let now MATH be some fixed small number. Since we are assuming MATH with uniform MATH-tail this means that there is some integer MATH for which MATH . We take MATH and MATH . For this last measure we have MATH . On the other hand, it follows from the definition of MATH-hyperbolic times that there is some constant MATH such that MATH for MATH. Defining MATH as the set of those points in MATH whose distance to MATH is greater than MATH, we have MATH and this last measure has density bounded by some uniform constant, as long as we take the maps MATH in a sufficiently small neighborhood of MATH in the MATH topology. |
math/0011158 | Let us fix MATH and MATH. For a sequence MATH we write MATH and define MATH. We start by observing that for MATH the number of sequences MATH such that MATH is bounded by MATH . Using NAME 's formula (compare CITE) the expression on the right hand side is bounded by MATH, where MATH depends only on MATH and MATH when MATH. REF ensure MATH (recall that MATH). Hence the measure of the union MATH of all the sets MATH with MATH is bounded by MATH . Since MATH we may choose MATH so small that MATH. Then MATH with MATH for big enough MATH. Note that MATH and MATH do not depend on MATH. Setting MATH we also have MATH for all big MATH and for every NAME probability MATH on MATH, by NAME 's Theorem. Since MATH then NAME 's Lemma implies MATH and this means that MATH almost every MATH satisfies REF . |
math/0011158 | Let MATH be the constant provided by REF . We fix MATH sufficiently small so that MATH holds for some MATH and take MATH satisfying REF . REF now imply MATH for large enough MATH. This means REF holds for MATH almost every MATH. We observe that if MATH, then MATH cannot be hyperbolic times for MATH. Hence MATH for all MATH. In particular MATH and MATH. |
math/0011158 | This will be obtained as a consequence of the fact that the set of vertical lines constitutes a normally expanding invariant foliation for MATH. Let MATH be the space of continuous maps MATH endowed with the sup norm, and define the map MATH by MATH . Note that MATH is well-defined, since MATH for small MATH and MATH. Moreover, MATH is a contraction on MATH: given MATH and MATH then MATH . This last quantity can be made smaller than MATH, as long as MATH and MATH are chosen sufficiently small. This shows that MATH is a contraction on the NAME space MATH, and so it has a unique fixed point MATH. It is no restriction for our purposes if we think of MATH as being equal to MATH for some small MATH. Note that the map MATH depends continuously on MATH and for MATH small enough the fixed point of MATH is close to the zero constant map. This holds because we are choosing MATH close to MATH, MATH and MATH close to MATH. Then, for MATH small enough, we have MATH uniformly close to MATH and it is not hard to check that MATH is precisely the map whose integral leaves of the vector field MATH give the invariant foliation MATH associated to MATH. Since this foliation depends continuously on the dynamics and for MATH we have MATH (see CITE), we finally deduce that MATH is uniformly close to zero for small MATH. We have defined MATH in such a way that if we take MATH, then for every MATH and MATH . Now, for fixed MATH, we take MATH to be the set of integral curves of the vector field MATH defined on MATH. Since the vector field is taken of class MATH, it does not follow immediately that through each point in MATH passes only one integral curve. We will prove uniqueness of solutions by using the fact that the map MATH has a big expansion in the horizontal direction. Assume, by contradiction, that there are two distinct integral curves MATH with a common point. So we may take three distinct nearby points MATH such that MATH, MATH, MATH and MATH have the same MATH-coordinate. Let MATH be the horizontal curve joining MATH to MATH. If we consider MATH for MATH, where MATH is the projection from MATH onto MATH, we have that the curves MATH are nearly horizontal and grow in the horizontal direction (when MATH increases) by a factor close to MATH for small MATH and MATH, see CITE. Hence, for large MATH, MATH wraps many times around the cylinder MATH. On the other hand, since MATH and MATH are always tangent to the vector field MATH on MATH, it follows that all the iterates of MATH and MATH have small amplitude in the MATH-direction. This gives a contradiction, since the closed curve made by MATH, MATH and MATH is homotopic to zero in MATH and the closed curve made by MATH, MATH and MATH cannot be homotopic to zero for large MATH. Thus, for fixed MATH we have uniqueness of solutions of the vector field MATH, and from REF it follows that MATH is a MATH-invariant foliation of MATH by nearly vertical leaves. |
math/0011159 | Let MATH denote the complement of a MATH-neighborhood of the diagonal in MATH . Let MATH denote a smooth extension of MATH to the whole of MATH . Since MATH and MATH are formally adjoint we have MATH . The last equality is due to REF and to the fact that MATH is a fundamental solution of MATH . Let MATH be the integral in the second summand. Since MATH can be chosen to be arbitrarily small we have proved the lemma. |
math/0011159 | Let MATH . For MATH let MATH be disjoint one-dimensional submanifolds. We denote the MATH-ball in MATH by MATH. Choose disjoint tubular neighborhoods MATH of MATH such that the diffeomorphism is given by the geodesic exponential map restricted to the normal bundle of MATH in MATH . Coordinates on MATH respectively MATH compatible with this product decomposition will be denoted by MATH respectively MATH . As described in CITE, we choose representatives of the NAME duals MATH of MATH such that for every MATH . Since MATH we can choose a form MATH satisfying MATH . Then by REF there is a function MATH with the property MATH . The summand containing MATH is zero by the NAME 's theorem. Since MATH and MATH are disjoint, we can treat MATH like a smooth form in this integral. The last expression is therefore equal to MATH . Because of the special choice REF of the NAME duals MATH follows. This completes the proof because MATH can be arbitrarily small. |
math/0011159 | The set of paths MATH fulfills REF and the continuity condition in REF by construction. Let MATH denote the function MATH on MATH . To prove that MATH has REF , we show first that MATH holds in the MATH-sense. This corresponds to REF. Let MATH . Let MATH denote the distance of two points MATH. Because of the remarks preceding REF there is a smooth double form MATH on MATH such that for MATH and MATH we have MATH . By REF of MATH an analogous statement about MATH is true, that is, there is a continuous function MATH on MATH such that MATH . Hence MATH is bounded above by a constant MATH which does not depend on MATH. Let MATH denote the volume of MATH with respect to the volume form MATH, the diameter of MATH is denoted by MATH . The geodesics are parameterized by MATH hence their velocity vector MATH is no longer than MATH . By NAME 's theorem we find MATH . The last expression becomes arbitrarily small if MATH . This proves REF. With an analogous argument one can show REF. To prove REF observe that the corresponding integral in REF remains bounded when two straight lines approach each other. The same is true for geodesics on the Riemannian manifold MATH . The set MATH meets REF again by construction. |
math/0011159 | By REF of the short path system, the linking numbers are almost always well defined. Using the linking form and REF we find MATH . The last limit exists by REF. This theorem also implies that the limit function is MATH, just like the double form MATH itself. Hence MATH is well defined. |
math/0011159 | Let MATH be another REF-form with the property MATH . Hence MATH is closed. Since MATH , we can choose a function MATH on MATH such that MATH . By the NAME 's theorem we find MATH . Hence MATH is well defined. Choose a REF-form MATH such that MATH . Since MATH symmetry follows from the NAME 's theorem and the fact that REF-forms commute with differential forms of any degree. The last statement follows from MATH . |
math/0011159 | Recall that MATH and MATH . Using REF we find MATH . Because MATH is closed, the summand containing MATH does not contribute to the integral by the NAME 's theorem, hence MATH . By the second part of REF, the right hand side of REF is equal to MATH and this is equal to MATH by REF. |
math/0011165 | By REF . In the above, we've used the skew-symmetry property, for example, MATH . The coefficients in the summation of index MATH is obtained as follows: for each MATH we can either choose MATH or MATH but not both, the same for MATH. For any appropriately fixed MATH, there are MATH ways to choose MATH from the former and MATH ways from the later. Once MATH are chosen, MATH are determined. We can now show that MATH by comparing the coefficient of MATH of the following polynomials in MATH: MATH . The proposition follows at once. |
math/0011165 | We can use symmetry to bring REF into the form in which at least MATH holomorphic MATH appear together with at most MATH anti-holomorphic MATH. |
math/0011165 | Choose a volume form MATH. Denote by MATH the dual volume form in MATH. Then for any vectors MATH we have MATH. It is easy to check that MATH where MATH is the NAME vector field in MATH. It follows from this that MATH . Now let us calculate REF. Using REF we get MATH . So it remains to show that MATH which follows by applying the differential MATH to the identity MATH . We now can finish the proof by observing that MATH . |
math/0011165 | Let MATH and MATH . From MATH one has MATH . Using REF and observing that MATH we can rewrite the integrand as MATH . |
math/0011165 | By REF MATH . To get REF we use the following observations. Let MATH and MATH be holomorphic functions on a complex curve MATH and MATH is a real valued function. Then since MATH we have, taking the real and imaginary parts respectively, MATH . One can easily show that (by using REF ) MATH . This is a special case of REF in the Appendix. Writing REF as MATH and subtracting from this MATH which is zero (to check this use the skewsymmetry with respect the alternations MATH and MATH), we see, using REF, that REF is equal to MATH by REF. To calculate REF we compute, in two different ways, the expression MATH . CASE: The integral over MATH, and hence the whole expression, is zero because MATH where both parts are understood as currents. Notice that MATH and MATH have integrable singularities and thus provide currents on MATH. CASE: Using formulas MATH and MATH we see that MATH is equal to MATH . Since this expression is skewsymmetric with respect to the transposition MATH exchanging the indices MATH and MATH we can write REF as MATH . We got last line by using transposition MATH. Now REF is exactly REF since the other three possible terms vanish due to alternation MATH (or MATH). Therefore REF is equal to MATH . We got this line by noting that MATH. Notice that MATH since the expression is unchanged under the transposition MATH, we see that REF equals to MATH by transposition MATH followed by MATH. This finishes the proof of the proposition. |
math/0011165 | Here is the algebraic reason behind this lemma. There is the following exact sequence of MATH-vector spaces MATH . It is a special case of the NAME complex. The map MATH admits a natural splitting MATH . If MATH then there is a map MATH . Now the proposition is an immediate corollary of the following lemma. |
math/0011165 | Let us show that that REF equals to MATH . Indeed, using REF we get MATH . Using this we write REF as a sum of the following MATH terms: MATH . Notice that a priori REF is a sum of REF terms corresponding to REF terms in REF. However REF of them disappear after the alternation. For instance, MATH since the involution MATH does not change the expression. Computing the sign of the appropriate permutation, we see that this sum is equal to REF. Indeed, the terms MATH, MATH, MATH and MATH provide the first summand, the terms MATH, MATH, MATH, MATH the second summand, and the rest the third summand. The third term in REF vanishes by using the involution MATH. This involution also bring the first summand in REF into the following form MATH . The second term in REF contributes MATH . The lemma, and hence REF , are proved. |
math/0011165 | CASE: Because MATH, for any MATH one has MATH . Denote by MATH the left side of REF multiplied by MATH. Taking the imaginary part of REF and alternating MATH we get MATH where the sum is over MATH such that MATH. Indeed, denote the expression inside of MATH in REF by MATH. An alternation of REF contributes to MATH if and only if MATH, so that we can specify MATH terms from MATH and make them contribute the MATH part of MATH, which, together with MATH, contribute the log part in MATH. Let MATH be the left hand side of REF considered for arbitrary MATH. Let us multiply it by MATH and take a sum over MATH. Since MATH for any MATH we have MATH for some polynomial MATH in MATH whose coefficients are MATH-linear combinations of MATH's. Using the identity MATH we have MATH . Replacing MATH by MATH we get MATH. The left hand side is an odd polynomial of degree MATH; it has a zero of order MATH at MATH; therefore it must have a zero of order MATH at MATH, so its degree is at least MATH. Thus it is a zero polynomial. CASE: Denote the left side of REF by MATH. Taking the real part of REF and alternating we get MATH . By definition it is clear that MATH. Multiplying REF by MATH and taking sum over MATH we have MATH for some polynomial MATH. Using combinatorial identity REF we write it as MATH . Changing MATH to MATH and noticing that the left hand side is an even polynomial we get MATH . Therefore MATH is a constant. Thus we finally have MATH (MATH). The lemma is proved. |
math/0011165 | The first expression follows directly from REF . Now we prove the second. By REF MATH where MATH. Now let us look at the terms in the expansion of REF which correspond to the term MATH . By REF , each term inside the sum of REF with MATH as the first factor contributes to REF as many as MATH times. So the total contribution to REF from REF is MATH . Therefore MATH where MATH-id and for MATH . Now we observe that for any MATH we have MATH . Therefore MATH which together with REF yields the second equality. To prove the last equality in our proposition it suffices to observe that MATH . |
math/0011166 | That MATH is reflexive and antisymmetric is obvious. To prove transitivity suppose that MATH and MATH. REF for MATH is easily seen to be satisfied. To see that REF for MATH is satisfied note that MATH and that MATH which shows that MATH, as required. To show that REF hold, let MATH. Given any pair MATH, it may, without loss of generality be assumed that MATH and MATH are successive elements of MATH. Hence, either MATH or MATH. In the second case, it follows immediately from the fact that MATH that MATH. Since MATH it follows that MATH and so MATH. On the other hand, in the first case MATH since MATH and MATH . Moreover, using REF and MATH it is possible to conclude that MATH. Hence, in either case MATH which establishes that REF holds for MATH. To see that REF holds for MATH let MATH. It follows from MATH and MATH that MATH and, since MATH, that MATH. Hence, MATH establishing REF . |
math/0011166 | To prove that the set REF is dense let MATH and MATH be given. Let MATH be such that MATH. Using REF it follows that MATH and, hence, MATH is a permutation of MATH. Letting MATH it follows that MATH. Observe for later reference, that is has actually been shown that MATH . To prove that the set REF is dense let MATH and MATH be given. Since it may as well be assumed that MATH, it is possible to find MATH so large that MATH and for all MATH and MATH . Using the set REF choose MATH such that MATH and MATH. From REF it can be assumed that MATH and that MATH. Now, let MATH. That REF for the relation MATH is satisfied follows from the choice of MATH while REF is obvious. REF has no content in the case of MATH since MATH. Now use transitivity and the fact that MATH. Observe for later reference, that is has actually been shown that MATH . There is no problem in proving that the set REF is dense. |
math/0011166 | This is standard using REF for REF for REF. |
math/0011166 | This is an immediate consequence of REF . |
math/0011166 | Proceed by induction on MATH to show that MATH. Note that the cases MATH or MATH a limit pose no problem. Given that MATH let MATH be enumerated, in order, by MATH. If follows directly from REF and the definition of MATH that CASE: MATH CASE: MATH if MATH. In particular, noting that there is some MATH such that MATH, MATH and MATH . Hence, MATH. |
math/0011166 | Proceed by induction on MATH noting that the cases MATH and MATH a limit are trivial. Therefore, suppose that REF has been established for MATH and that MATH. Without loss of generality it may be assumed that MATH. Choose MATH and MATH such that MATH and MATH and, moreover, MATH and, for each MATH the condition MATH decides the value of the least ordinal MATH such that MATH. Furthermore, it may, without loss of generality be assumed that MATH and that MATH for MATH. Then use the induction hypothesis to find MATH extending MATH and such that MATH witnesses that MATH. Without loss of generality, MATH. Now, the fact that MATH implies that MATH is a permutation of MATH. Hence it is possible to define MATH and MATH. Then define MATH so that MATH and note that MATH and MATH. |
math/0011166 | This is immediate from REF . |
math/0011166 | Notice that since MATH the restrictions on extension do not apply and it is easy to check that the condition MATH belongs to MATH. Since MATH it follows that MATH and, hence, that MATH. That MATH is immediate from the definition. |
math/0011166 | Let MATH the maximum member of MATH. Define MATH by MATH and define MATH by MATH (The only difference is to be found in the last lines of the two definitions.) It is easy to check that MATH and MATH and that both MATH and MATH belong to MATH. Moreover MATH. Hence MATH and MATH are compatible. |
math/0011166 | This is standard. If the cofinality of MATH is less than MATH then choose a countable, cofinal subset MATH of MATH. Using REF it follows that the set of all MATH such that MATH is dense. It follows that MATH has a MATH-centred dense subset. On the other hand, if the cofinality of MATH is greater than MATH and MATH it is possible to choose MATH such that MATH for each MATH. Using REF it may be assumed, by extending each condition, that MATH for each MATH. Now there are MATH and MATH and an uncountable set of MATH such that MATH and MATH. Any two of these are easily seen to be compatible. |
math/0011166 | Proceed by induction on MATH. If MATH the result is immediate and if MATH is a limit the result follows from the induction hypothesis and the finite support of the iteration. Therefore consider the case MATH and assume that the countable chain condition has already been established for MATH. Next, observe that if MATH is not cofinal in MATH then the induction hypothesis is easily applied since, in this case, MATH has a maximal element below MATH and so MATH has the countable chain condition by REF . If MATH has cofinality different from MATH then, once again REF implies that MATH has the countable chain condition; therefore, in either case, so does MATH. Hence it remains to consider the case that MATH is cofinal in MATH and has cofinality MATH. From the hypothesis on MATH, and the fact that MATH must be the limit of MATH it follows that MATH. By appealing to REF and extending the conditions in question, it is possible to guarantee that each MATH is in MATH and that this is witnessed by MATH . As well, by thinning out, it may be assumed that there is MATH such that MATH for each MATH, and there is a pair MATH such that MATH for each MATH and that MATH form a MATH-system with root MATH. Using the fact that MATH is a limit, choose some MATH such that there is some uncountable MATH such that if MATH and MATH then MATH. It may as well be assumed that MATH for each MATH. Then use REF to extend each MATH to some MATH. Now, using the induction hypothesis, find MATH such that MATH and MATH is compatible with MATH. Choose MATH extending both MATH and MATH. Let MATH be generic over MATH and containing MATH. Observe that MATH as interpreted in MATH is a partial order like MATH in MATH except that the first factor of MATH is MATH. Since MATH it follows that MATH, in MATH. It follows that MATH and MATH belong to MATH. Now use REF in MATH to conclude that MATH and MATH are compatible. Hence, so are MATH and MATH. |
math/0011166 | Let MATH and suppose that MATH for some integer MATH. From REF it can be assumed that MATH. Now choose MATH so large that there exist distinct integers MATH and MATH in MATH such that MATH if and only if MATH for all MATH. It is possible to choose a bijection MATH such that MATH. However, letting MATH it follows that MATH for each MATH. Hence MATH and MATH. |
math/0011166 | This is a standard use of genericity. |
math/0011166 | Let MATH be a closed unbounded set such that MATH is unbounded. Construct a finite support iteration MATH so that NAME 's Axiom is forced to hold by the iteration on MATH and such that a tower of permutations MATH is generically constructed along MATH such that MATH where MATH is MATH generic over MATH. Then let MATH be the automorphism of MATH defined by MATH if and only if there is some MATH such that MATH and MATH. To see that MATH is nowhere trivial suppose that MATH is induced by MATH on MATH. Let MATH be an ordinal large enough that MATH and MATH both belong to MATH. Using REF it is possible to find in MATH an infinite set MATH such that MATH. However, there is no guarantee that MATH belongs to MATH. Let MATH be the first member of MATH greater than MATH. Then MATH does belong to MATH. From REF it follows that MATH for infinitely many MATH belonging to MATH. Therefore MATH is infinite contradicting that MATH induces MATH on MATH and MATH. |
math/0011166 | Let the iteration be obtained from the sequence MATH where each successor constructed by using one of the mentioned partial orders and let the MATH-name MATH be given. Let MATH be a countable elementary submodel containing MATH and MATH. Let MATH, let MATH enumerate MATH and let MATH. Standard fusion arguments will allow the construction of a family MATH such that: CASE: If MATH for each MATH then MATH. CASE: If MATH then MATH decides the value of MATH. CASE: If MATH is defined to be MATH then MATH. It follows that, for each MATH there is some MATH such that MATH and MATH if and only if, letting MATH, the set of REF has a proper lower bound. For NAME and NAME forcings checking for a lower bound is easily seen to be MATH in the parameter defining the fusion sequence; in fact, it is NAME. |
math/0011166 | Assuming the lemma is false, it is possible to find a condition MATH and a MATH-name for a set of integers MATH such that MATH where MATH is a name for the generic subset of MATH added by MATH. Let MATH be the characteristic function of MATH. Using the fact that MATH is sufficiently NAME, find MATH such that MATH and MATH is analytic. Let MATH be an infinite set of integers disjoint from the domain of MATH such that MATH has infinite complement in the integers. It follows that if MATH is defined to be MATH then MATH is analytic. There are two possibilities. First, suppose that the domain of MATH is all of MATH. In this case it is possible to find a continuous function MATH defined on a comeagre subset of MATH such that MATH holds for all MATH in the domain of MATH. Now let MATH be defined by MATH and observe that MATH is continuous. Furthermore, MATH. To see this, let MATH be any condition witnessing that MATH holds. Observe that not only does MATH force MATH but also MATH. Hence, MATH. Therefore, since MATH and MATH are sets in the ground model, it follows from the absoluteness of MATH that MATH. But now it follows that MATH is trivial on MATH by REF. In the other case there is some MATH such that there is no MATH such that MATH holds. This implies that for every MATH there are infinitely many integers in MATH whose membership in MATH is not decided by MATH. NAME implies that MATH which is a contradiction to the assumption that MATH. |
math/0011166 | Any automorphism can be reflected on a closed unbounded subset of MATH consisting of ordinal of uncountable cofinality. If MATH is any such ordinal such that MATH then REF can be applied to show that the automorphism can not be extended any further. |
math/0011170 | Let MATH be a MATH-regular subgraph that is connected and does not separate (MATH). Now REF if the complement vertex set MATH satisfies REF , then it is a product set, MATH. REF at work in a product graph Now REF if we can choose MATH for MATH, then this yields a vertex MATH that in MATH has graph-theoretic distance at least MATH from the complement set MATH (symbolized by the vertex MATH in our figure). Thus this vertex of MATH has degree MATH, which is impossible. Hence we get that the MATH-regular, induced, connected and non-separating subgraphs all have the form MATH or MATH, where MATH is MATH-regular, induced, connected and non-separating. By NAME 's conjecture, which is valid for MATH, it follows that MATH is the graph of a facet of MATH, and hence that MATH is the graph of a facet of MATH. |
math/0011170 | The vertex set of MATH may be decomposed as MATH, into the vertices that lie in the face MATH respectively, not in the bottom facet respectively, not in the top facet of the wedge. In REF one then verifies that if MATH meets both MATH and MATH, then it contains always ``none or both" from a pair of corresponding top and bottom vertices. |
math/0011170 | To be derived from NAME 's evenness criterion; see CITE or CITE. |
math/0011170 | We work with the duals, MATH. By REF plus REF , we need only treat the case where MATH is even, and MATH. CASE: The graph MATH has a simple combinatorial description, via NAME 's evenness criterion: Its vertices MATH correspond to those subsets MATH which split into a union of MATH adjacent pairs modulo MATH (that is, we identify the elements of MATH with MATH). The splitting into adjacent pairs is always unique. We shall call a ``block" any non-empty union of adjacent pairs that is contiguous, that is, without a gap. Two vertices are adjacent if they differ in a single element, that is, if one arises from the other by moving one block by ``one unit." This also provides a canonical orientation on each edge: we put a directed edge MATH if we get from MATH to MATH by moving a block ``up" (mod MATH). The resulting digraph is, of course, not acyclic. One single pair is moved, thus we get an edge of MATH, and of MATH, corresponding to MATH. We shall also consider the subgraph MATH, which has the same vertex set as MATH, but only retains those directed edges that correspond to moves of single pairs. (In REF below, this is the graph formed by the straight edges only.) The structure of the digraph MATH is closely linked to the poset MATH equipped with componentwise partial order. This poset is a distributive lattice. Moreover, reduction modulo MATH defines a surjective, and locally injective, digraph map MATH, from the NAME diagram of MATH onto the digraph MATH. (ii.) Now assume REF that MATH satisfies REF . Every move of a block can be decomposed into a sequence of moves of pairs. Thus REF implies that also MATH is connected (as an undirected graph): every directed arc in MATH canonically corresponds to a sequence of directed arcs in MATH. Thus MATH is acyclic if and only if MATH is acyclic. (iii.) Next we treat the case that MATH contains a directed cycle. Every such cycle lifts into a two-way infinite, maximal chain MATH in the lattice MATH. Every element MATH is contained in a finite interval MATH between elements MATH. Now MATH restricts to a maximal chain MATH in the interval MATH, while MATH will lie on some other maximal chain MATH of this interval. But in a distributive (and hence semimodular) lattice one can move from any maximal chain to any other one by one-element exchanges, see CITE. This implies via REF not only the elements of MATH, but also all elements of chains that we can move to, belong to MATH. Thus we get MATH, and hence MATH. This depicts the digraph MATH, and the lattice MATH, for MATH, MATH. The digraph MATH is finite, the ends are to be identified according to the capital letters. It is not planar, but for MATH it embeds into a NAME band. The lattice MATH is infinite. (iv.) Thus we may assume that MATH is acyclic. We can then identify MATH with an induced subgraph of the NAME diagram of MATH: every connected component of MATH is isomorphic to MATH. With this REF , and lower and upper semi-modularity of the lattice MATH, imply that MATH corresponds to an interval in MATH (see again CITE); that is, there are elements MATH such that MATH is isomorphic to the NAME diagram of the interval MATH. This corresponds to unique vertices MATH such that MATH consists of all vertices of MATH that lie on a directed path of minimal length from MATH to MATH. CASE: Assume that MATH contains no ``no-gap vertex," whose set MATH consists of one single block of size MATH. (In the case MATH, compare REF , this means that MATH contains no vertex on the border of the NAME strip.) Then every no-gap vertex needs to have exactly one neighbor in MATH, which is a one-gap vertex consisting of MATH elements from a block of MATH adjacent vertices. Every such block corresponds to a MATH-clique; these cliques contain two no-gap vertices each, and each no-gap vertex is contained in two of these MATH-cliques. (In REF , this corresponds to the chain of triangles in the boundary of the NAME strip.) Now we lift the situation to the lattice MATH, where the chain of cliques gives rise to MATH distinct chains, which are disjoint by our assumption MATH. The interval MATH can contain at most one element from each of the chains. Thus it can contain at most MATH elements from the chains. Thus at most MATH no-gap vertices are adjacent to a vertex in MATH. Projecting this back to MATH, we find that at most MATH no-gap vertices are adjacent to a vertex in MATH. But they all have to be, so we get MATH, a contradiction. (vi.) If MATH contains a one-block vertex, then we see, using REF , that both MATH and MATH must be one-block vertices. By symmetry, we may then assume that MATH and MATH for some MATH. This ends REF : and REF - the identification of those parameters MATH for which each vertex in MATH has exactly one neighbor in MATH - is now easy. |
math/0011170 | We specify a pure MATH-dimensional subcomplex MATH of MATH which will comply with our conditions (the definition is due to NAME): Every tetrahedron MATH has a unique vertex MATH opposite to its free face. A triangle MATH belongs to MATH if both its neighboring tetrahedra MATH belong to MATH, and if MATH. If the grey MATH-complex is MATH, then the black MATH-dimensional subcomplex is its core. CASE: The complex MATH has no free edge. Suppose that MATH is a free edge of the triangle MATH. The tetrahedra of MATH which contain MATH are cyclically ordered: MATH, as in REF . Figure We can assume that MATH and MATH. For, if MATH and MATH were both not in MATH, then this would imply MATH, and hence MATH. Because MATH, we have MATH. This vertex lies in MATH, and it cannot be the vertex of MATH that is opposite to MATH; thus we get that MATH. In particular, this means that MATH is not the free face of MATH, and thus MATH. Iterating these arguments, we see that MATH and MATH. So, MATH is another triangle in MATH. CASE: The complex MATH is empty if and only if MATH is a vertex star. If MATH is empty, as MATH is dually connected, all the vertices MATH are equal, say, MATH. Hence MATH is part of MATH. But then MATH must be the whole star, because there is only one free face per tetrahedron. CASE: If MATH separates then so does MATH. NAME duality implies that a subcomplex of a MATH-sphere separates if and only if it has non-trivial MATH-dimensional homology. The MATH-dimensional complex MATH collapses down to a MATH-dimensional subcomplex MATH, which contains MATH. Since MATH has dimension MATH, it is clear that the (relative) homology group MATH vanishes. Hence, the map MATH in the exact sequence MATH is injective, and the claim is proved. |
math/0011170 | The dual graph of the star of MATH in MATH corresponds to a (simple) MATH-dimensional polytope, and is therefore MATH-connected. Any dual path form MATH to a tetrahedron in a different component has to pass through MATH. |
math/0011170 | MATH is covered from both sides with tetrahedra MATH and MATH. The vertices MATH are both vertices of MATH, so that every edge of MATH contains at least one of them. |
math/0011170 | If MATH is a regular vertex (all incident edges have degree MATH in MATH), then the corresponding local subgraph of MATH is a cycle: It is the dual graph of the star of MATH in MATH. If MATH is a singular vertex (two incident edges have degree MATH in MATH), then the subgraph of MATH is a union of three paths between two nodes. |
math/0011170 | Remove two nodes/triangles MATH from MATH, and denote MATH the two nodes which we want to join by a MATH-avoiding path. In MATH, the intersection MATH contains at most two vertices. Now the MATH-skeleton of MATH is MATH-connected (e. g. by NAME 's lemma). Thus there is a MATH-avoiding vertex-edge-path in MATH from a vertex of MATH to some vertex of MATH. That is, there are vertices MATH of MATH such that MATH, MATH, and the MATH's are edges of MATH. Figure Choose triangles MATH adjacent to the edges MATH. Figure By REF , consecutive triangles MATH and MATH can be joined within the star of MATH, avoiding MATH. |
math/0011170 | Because MATH is induced, MATH is a face of MATH. If MATH is a triangle, remove the corresponding node from MATH. If MATH is an edge, remove the edge from MATH, which is cut by MATH. By REF , we can assume that we want to join two tetrahedra MATH which intersect MATH in triangles MATH and MATH respectively. By REF we can find a path from MATH to MATH in the remaining graph. This path can be lifted to a dual path which joins MATH and MATH, and which avoids MATH. |
math/0011173 | We may rescale MATH. By the Fundamental theorem of Calculus we have MATH . The claim follows from the rapid decay of MATH. |
math/0011173 | Observe that when MATH, then MATH can be factorized as the product of MATH and a multiplier whose symbol is a bump function adapted to MATH, and is therefore disposable. Thus it suffices to prove the claims in the first paragraph. In fact, it suffices to verify the claim for MATH, as the other projections are linear combinations of this type of multiplier. We may rescale MATH. The symbol MATH of MATH supported on the region MATH, is smooth except when MATH or MATH, and obeys the derivative bounds MATH for arbitrarily many indices MATH, MATH away from the singular regions MATH, MATH. Standard calculations then show that the kernel MATH obeys the bounds MATH which is integrable as desired. |
math/0011173 | It suffices to check MATH. By scaling we may take MATH. We then decompose MATH . The first term is disposable by REF , so by conjugation symmetry it suffices to demonstrate the boundedness of MATH. It suffices to show that the multiplier with symbol MATH is bounded on MATH. However, the transformation MATH given by MATH is easily seen to be an isometry on MATH (indeed, for each time MATH, MATH is just the unitary operator MATH). The claim then follows by conjugating by MATH and observing that the multiplier with symbol MATH is bounded on MATH. |
math/0011173 | We may rescale MATH. We may assume that MATH since the claim follows from NAME 's REF otherwise. From the NAME summation formula we can construct a NAME function MATH whose NAME transform is supported on interval MATH and which satisfies MATH for all MATH. From NAME we have MATH . On the other hand, from the MATH . NAME estimate (see for example, CITE) and the NAME support of MATH we have MATH and the claim follows by square-summing in MATH. |
math/0011173 | MATH-REF . Construction of MATH. Define MATH to be the space of functions MATH whose norm MATH is finite. Observe that MATH is the dual of MATH. We also define the plane wave space MATH to be the atomic NAME space whose atoms are functions MATH with MATH for some MATH. We now define MATH by the norm MATH . The estimates REF are now straightforward and are left to the reader. It remains to prove REF, and REF. MATH-REF . Proof of the product estimates REF. The estimate REF will follow from REF, and duality, so we need only prove REF. By REF it suffices to show MATH . But this follows by reducing MATH to a MATH atom, then using NAME and NAME. MATH-REF . Proof of the square-summability estimate REF. Fix MATH. By REF we may rescale MATH. By REF we may replace the MATH norms on the right-hand side by MATH. We expand the left-hand side of REF using REF and treat the three components separately. MATH-REF . The contribution of MATH. By triangle inequality and NAME we can bound this contribution by MATH . Since the caps in MATH are finitely overlapping, we have MATH, and the claim then follows from REF. MATH-REF . The contribution of MATH. This follows from NAME and the observation that the MATH have finitely overlapping MATH-support as MATH varies. MATH-REF . The contribution of MATH. Suppose that MATH is a direction such that MATH. Then from elementary geometry we see that the functions MATH have finitely overlapping MATH-Fourier support as MATH varies. We thus have from NAME that MATH for all MATH. Taking suprema in MATH, then taking suprema in MATH and using REF we obtain the claim. MATH-REF . Proof of the embedding REF. By REF we may rescale MATH. By the atomic nature of MATH and conjugation symmetry we may assume that MATH has NAME support in the region MATH for some integer MATH, in which case we need to show MATH . Fix MATH. We now prove the claim for the three components of REF separately. MATH-REF . The estimation of the MATH norm. By NAME and NAME we have MATH . For fixed MATH, the function MATH has MATH-support in an interval of length MATH. The claim then follows by using NAME and NAME. MATH-REF . The estimation of the MATH norm. Let MATH be such that MATH. By NAME and NAME we have MATH . From elementary geometry we see that for fixed MATH, the function MATH has MATH-support in an interval of length MATH. The claim then follows by using NAME and NAME, and then taking suprema in MATH. MATH-REF . The estimation of the MATH norm. From the NAME inversion formula in polar co-ordinates and the NAME support of MATH we have MATH . From NAME 's inequality and the definition of MATH we thus have MATH . The MATH factor is bounded and can be discarded. Since MATH, we can estimate the previous by MATH . By NAME this is bounded by MATH . By NAME this is bounded by MATH . Undoing the polar co-ordinates we can estimate this by MATH . By NAME we thus see that the contribution of the MATH norm is acceptable. This finishes the proof of REF. |
math/0011173 | We shall just prove these claims for MATH, as the MATH claims then follow from the factorization MATH. If MATH then we factorize MATH, which is disposable by REF . Similarly for MATH and MATH. Now suppose that MATH. It suffices to show that MATH is disposable, since the remaining claims then follow by taking suitable linear combinations of these multipliers. By conjugation symmetry we may take MATH. It is then an easy matter to verify that the symbol of MATH is a bump function adapted to the parallelopiped MATH . The kernel is then rapidly decreasing away from the dual of this parallelopiped, and the claim follows. |
math/0011173 | By REF it suffices to show MATH . By the atomic nature of MATH we may assume that MATH has NAME support in the region MATH for some integer MATH. But then the claim then follows from NAME 's theorem and the finite overlap of the MATH . NAME supports. |
math/0011173 | We may rescale MATH. By the atomic nature of MATH and conjugation symmetry we may assume that MATH has NAME support in the region MATH for some integer MATH, in which case we need to show MATH . To control the first term of REF, we use REF to obtain MATH . Since MATH is supported in an interval of length MATH, the claim then follows from NAME and NAME. The second term of REF is trivial, so it remains only to show MATH for all MATH and signs MATH. Fix MATH, MATH. We may assume that MATH since the left-hand vanishes otherwise. In particular we have MATH. By REF we may estimate the left-hand side by MATH and the claim follows from NAME. |
math/0011173 | This estimate is easily verified for the first two factors of REF. Thus it only remains to show MATH for all MATH and signs MATH. Fix MATH, MATH. We split MATH. To treat the second term we use REF to estimate this contribution by MATH which by NAME is bounded by MATH . But this is acceptable by the second term in REF. It remains to estimate MATH . From the construction of MATH and the NAME support of MATH we observe the identity MATH . Discarding the MATH we then see that this contribution is acceptable by the third term in REF. |
math/0011173 | If MATH does not vanish, then by the NAME transform there must exist MATH such that MATH and MATH for MATH, and that MATH for MATH. Similarly for MATH. From our hypothesis MATH we see that MATH for MATH. By conjugation symmetry we may assume MATH. By symmetry it suffices to consider three cases. MATH-REF . (MATH case) MATH and MATH. Observe the identity MATH . The left hand side is MATH, but the left hand side can at most be MATH. Since we are assuming MATH, this case is therefore impossible. MATH-REF . (MATH case) MATH and MATH, MATH. This forces MATH and MATH. Now observe the identity MATH . The right-hand side has magnitude MATH, and in the imbalanced case has magnitude MATH. The left-hand side can be rewritten as MATH and therefore MATH . By the sine rule we thus have MATH . Combining this with the previous we see that we are in either REF or REF. MATH-REF . (NAME case) MATH and MATH. If MATH, then MATH. Since MATH, we must then have MATH, at which point we could swap MATH and MATH. Thus we may assume that MATH. If MATH, we have the identity MATH . The right-hand side has magnitude MATH, and in the imbalanced case has magnitude MATH. The left-hand side can be rewritten as MATH so that MATH . Combining this with the previous we see that we are in either REF or REF. If MATH, we have the identity MATH . The right-hand side has magnitude MATH, and in the imbalanced case has magnitude MATH. The left-hand side can be rewritten as MATH so that MATH . By the sine rule we thus have MATH . Combining this with the previous we see that we are in either REF or REF. |
math/0011173 | By symmetry we may assume MATH. By scaling we may take MATH. By conjugation symmetry we may assume that MATH, MATH are real-valued. We may assume that either MATH (low-high interaction) or that MATH and MATH (high-high interaction). By scaling we may take MATH. We may of course replace MATH by MATH. The left-hand side is MATH which by duality is MATH for some MATH in the unit ball of MATH. By REF this is MATH . Suppose we are in the low-high interaction case MATH. By REF we can estimate REF by MATH which is acceptable by the normalization of MATH. Now suppose we are in the high-high interaction case MATH, MATH. By REF we may estimate REF by MATH which is acceptable by the normalization of MATH. |
math/0011173 | By symmetry we may assume MATH. We may then assume that MATH since the expression vanishes otherwise. The estimate REF follows from REF , so we turn to REF. This shall be a variant of REF. We may assume that MATH since the claim follows from REF , and REF otherwise. Consider the contribution of MATH. By REF and NAME we can bound this contribution by MATH which is acceptable by REF. Now consider the contribution of MATH. We may then freely insert MATH in front of MATH. By REF and NAME we can then bound the previous by MATH which is acceptable by REF. |
math/0011175 | We start by taking out factors of the determinant: MATH . Now we are in the position to apply REF with MATH, MATH, MATH, MATH. After a little simplification we get the claimed result. |
math/0011175 | The proof follows from REF with MATH and MATH. The sign of MATH cancels exactly with the sign obtained from the reordering of the columns of MATH in the determinant. |
math/0011179 | Comparing REF shows that MATH is the translation by MATH of the SL REF-fold MATH of REF. Hence MATH is a nonsingular special Lagrangian MATH if MATH, and an SL MATH-cone singular at MATH if MATH. It is also easy to see that MATH is well-defined, continuous, piecewise smooth, and not smooth on MATH. One can show from REF that if MATH then MATH and MATH . Thus, if MATH dividing by MATH and rearranging yields MATH . Using the equations MATH and MATH to rewrite these expressions gives the second case of REF when MATH and the third when MATH. If MATH, REF implies that MATH, so MATH, giving the first case of REF , the second when MATH and the third when MATH. So, if MATH then we can recover MATH from MATH as in REF - REF . Conversely, for any MATH in MATH, defining MATH by REF - REF and reversing the proof above, we find that MATH. Hence MATH, and MATH is a special Lagrangian fibration of MATH. |
math/0011179 | As MATH has the symmetries MATH, uniqueness in REF gives MATH, and REF follow by taking partial derivatives. REF holds as REF depends only on MATH rather than MATH, and MATH is independent of MATH. An important part of the proof of REF was to derive a priori estimates for MATH and MATH in terms of MATH. This was done by using functions of the form MATH as super- and subsolutions for MATH at each point of MATH, and so derive a bound for MATH on MATH. As the maxima of MATH occur on MATH, this implies bounds for MATH on MATH. Now in our case this can be done uniformly in MATH. That is, if MATH on MATH then MATH on MATH for all MATH, and therefore MATH on MATH for all MATH, and a similar statement for supersolutions. Following the proof of CITE, we easily deduce REF . Suppose MATH, MATH, MATH and MATH. Define MATH . Then MATH satisfy REF in MATH, and MATH at MATH. Furthermore, MATH, MATH come from REF with boundary data MATH . Thus MATH is a nontrivial linear combination of MATH, and has exactly REF local maximum and REF local minimum on MATH. Applying REF with MATH shows that MATH has no zeroes in MATH. But this contradicts MATH at MATH. So given MATH and MATH, whenever MATH we have MATH. Since MATH depends continuously on MATH it follows that MATH is either a strictly increasing or a strictly decreasing function of MATH, and which of the two is independent of MATH. Then REF shows it is increasing, proving REF . For MATH, we can consider MATH to be a solution of REF on MATH, coming from REF with MATH and boundary data MATH. Now MATH has at most REF local maxima and REF local minima on MATH by CITE. Hence, applying REF with MATH shows that MATH has at most two zeroes in MATH, for any MATH. We shall use this to show that CASE: Suppose MATH and MATH. Then MATH. CASE: Suppose MATH. Then MATH. REF gives MATH. So if MATH and MATH then MATH at MATH and MATH, by REF . As there are at most two such points we must have MATH, proving REF . Similarly, REF gives MATH. Thus if MATH and MATH then by REF we see that MATH at the four points MATH and MATH, a contradiction. This proves REF . Now from REF and the continuity of MATH it is not difficult to see that either REF hold, or REF hold, but swapping `increasing' and `decreasing' throughout. But MATH so the average of MATH on the MATH-axis is MATH, and MATH . Therefore MATH is greater on the MATH-axis than on the MATH-axis, and so REF hold, rather than their opposites. |
math/0011179 | This follows quickly from REF using the Intermediate Value Theorem, except for the multiplicities of the singular points. To find these we apply REF to MATH. We have MATH, with at most REF local maxima and REF local minima on MATH by CITE. So REF shows that there are at most two singularities of MATH in MATH, counted with multiplicity. Multiplicity REF follows at once, and multiplicity REF from REF , as MATH is of maximum type. |
math/0011179 | Since MATH are continuously differentiable and satisfy REF , we have MATH using REF . Hence MATH is independent of MATH, and REF holds for some unique MATH. |
math/0011179 | Regard MATH and MATH as functions on the compact annulus MATH, as they are invariant under the MATH-action REF . Now using the maximum principle for elliptic equations of a certain form, CITE shows that the maximum and minimum of MATH on a domain MATH occur on MATH, and CITE that the maximum of MATH occurs on MATH. These proofs are also valid on the compact annulus MATH. But MATH and MATH on MATH. Maximizing and minimizing then gives REF . Next we estimate MATH on MATH, following CITE. Observe that if MATH with MATH then MATH since MATH by REF and MATH. It follows that we can treat REF as a quasilinear elliptic equation on MATH, which is uniformly elliptic on MATH, with constants of ellipticity depending only on MATH and MATH. Now CITE show that if MATH satisfies a quasilinear equation MATH of the form REF on a domain MATH and MATH, then MATH for some MATH depending only on MATH, upper bounds for MATH and MATH, and certain constants to do with MATH, which ensure that MATH is uniformly elliptic and MATH not too large. As this is a local result, it is enough for the conditions to hold within distance MATH of MATH. So there exists MATH depending only on MATH and MATH such that MATH. But MATH determines MATH by REF , and we easily deduce a bound for MATH. Thus there exists MATH depending only on MATH and MATH such that MATH and MATH. Finally we estimate MATH. Since MATH is periodic in MATH with period MATH and MATH, we see that the variation of MATH is at most MATH. Similarly, the variation of MATH is at most MATH. Therefore MATH varies only a bounded amount from its average value on MATH, and the same on MATH. We need to bound the difference between these average values. To do this we use the method of CITE to bound MATH on MATH. Define MATH. Then CITE shows that MATH . The proof is that MATH by NAME 's Theorem, and using REF to rewrite the l.h.s. gives REF . Therefore MATH where the second line follows from NAME 's inequality, and the third from REF and the inequality MATH, which follows from REF . Using MATH, MATH and the methods of CITE we may derive a priori estimates depending only on MATH and MATH for the third line of REF and the second integral on the first line. This gives an upper bound, MATH say, for MATH. But MATH . Therefore the difference in the average values of MATH on MATH is at most MATH. But the variation of MATH on MATH and on MATH is at most MATH, from above. It follows that the variation of MATH on both lines MATH together is at most MATH, which depends only on MATH and MATH. Now MATH satisfies a maximum principle by CITE, and so the maximum and minimum of MATH on the compact annulus MATH occur on MATH. Hence the difference between the maximum and minimum of MATH is the variation of MATH on both lines MATH together, and is at most MATH. But MATH, so the maximum is nonnegative, and the minimum nonpositive. Therefore the maximum is at most MATH and the minimum at least MATH, and MATH, completing the proof. |
math/0011179 | When MATH, existence and uniqueness of MATH comes from REF , and of MATH from REF . To extend this to the case MATH by taking the limit MATH we follow CITE, using the a priori estimates of REF . There are few significant changes, and the problems caused by singular points on the boundary in CITE are absent in this case, as there are no points MATH on MATH. The final part is proved as in CITE. |
math/0011179 | Clearly we have MATH for each MATH, where MATH and MATH are as in REF . Let MATH. We shall show that there exists a unique MATH such that MATH. Let MATH, MATH and MATH. Consider the function MATH. By REF and the last part of REF this is a continuous function. If MATH then REF gives MATH, and so REF gives MATH on MATH. Thus MATH is strictly increasing. And as the maximum and minimum of MATH on MATH is achieved on MATH and so is a value of MATH, REF implies that MATH as MATH. Hence by the Intermediate Value Theorem there exists a unique MATH such that MATH, and then MATH is given by MATH. It is easy to see from REF that MATH, and that this is the only MATH for which this holds. Therefore distinct MATH are disjoint, and MATH. The remainder of the proof follows REF . |
math/0011179 | By symmetries of the MATH the MATH satisfy MATH and MATH. Hence MATH for MATH. Also MATH. Thus MATH for MATH and MATH. Now by CITE, MATH satisfies a kind of maximum principle. Applying this on the rectangle MATH and using the fact that MATH is zero on two sides of the rectangle and negative on the other two, we find that MATH on MATH when MATH and MATH. Similarly, MATH on MATH when MATH and MATH. This implies REF when MATH. Also, by REF we see that MATH on MATH and MATH on MATH. By symmetries of the MATH the MATH satisfy MATH, and so MATH. REF for MATH then follows by integration on the line segment MATH. Taking the limit MATH in REF shows that MATH is decreasing on MATH and increasing on MATH. If it were not strictly decreasing or increasing then it would be constant on some subinterval MATH of MATH with MATH. Then MATH is constant on MATH, and CITE implies that it is constant on MATH, a contradiction. This completes REF . Similarly, taking the limit MATH in REF shows that MATH on MATH and MATH and MATH on MATH and MATH. But if MATH at any interior point of these regions, then we can derive a contradiction using the strong maximum principle CITE. This completes REF . |
math/0011179 | Let MATH. As in the proof of REF , the function MATH is continuous, strictly increasing, and tends to MATH as MATH. Thus by the Intermediate Value Theorem there exists a unique value MATH such that MATH. Since MATH by definition, this means that MATH by REF , and so MATH by definition of MATH. Hence MATH exists, and is continuous as MATH depends continuously on MATH. Similarly, considering the function MATH we find a unique value MATH such that MATH, and MATH exists and is continuous. Also MATH is given in REF , so MATH follows from REF . Let MATH, and consider the function MATH. By REF this has period MATH with a maximum at REF and a minimum at MATH. But MATH is strictly increasing and zero when MATH. Hence the maximum of MATH is negative when MATH, zero when MATH and positive when MATH. In particular, if MATH then MATH for all MATH, so that MATH has no singularities. This proves half of REF . By similar arguments using REF and the Intermediate Value Theorem, we easily find that the singularities of MATH, which are the zeroes of MATH, are as given in REF - REF , and the type of each singularity also follows from REF . Finally, to identify the multiplicity of each singularity MATH we can use REF , as knowing the sign of MATH constrains the winding number of MATH about REF along MATH in REF . |
math/0011183 | See CITE. |
math/0011183 | See CITE. |
math/0011183 | See CITE. |
math/0011183 | See CITE. |
math/0011183 | We start by proving that given any MATH, the sequence MATH has accumulation points (in the MATH-norm) in MATH. Let MATH and take a sequence MATH in MATH converging to MATH in the MATH-norm. It is no restriction to assume that MATH for every MATH and we do it. For each MATH we have MATH for large MATH. So, taking MATH large enough we have MATH . Using the well known fact that the transfer operator does not expand MATH-norms, we also have MATH for every MATH. It follows from REF that there exists some MATH and a sequence MATH for which MATH and, moreover, MATH. Now we apply the same argument to the sequence MATH in order to obtain a subsequence MATH such that MATH converges in the MATH-norm to some MATH with MATH. Since MATH there is some sequence MATH for which MATH . On the other hand, MATH and this last term goes to REF as MATH. This finally implies that MATH thus proving that the sequence MATH has accumulation points in MATH. Now let MATH be a MATH-invariant set with positive NAME measure. Considering MATH in the previous argument, we obtain some sequence MATH and MATH for which MATH . Moreover, MATH (here we use that MATH is normalized). Considering MATH already multiplied by a constant factor in order to have MATH-norm equal to REF, we take MATH. Since MATH we have that MATH gives full weight to MATH, thus concluding the proof of the result. |
math/0011183 | Let MATH be a MATH-invariant set with positive NAME measure and MATH a measure as in REF . Since MATH (recall REF ) and MATH gives full weight to MATH, it follows from NAME 's inequality that MATH . We take MATH. |
math/0011183 | We will prove that MATH is ergodic, and so an SRB measure for MATH in MATH. Let MATH be any MATH-invariant set with MATH. We have MATH and so the set MATH is also MATH-invariant. Since we are taking MATH with MATH, it follows from REF that MATH. As a consequence, MATH can be covered by a finite number of MATH-invariant sets MATH intersecting MATH in a positive NAME measure set, and which are minimal: for MATH there is no MATH-invariant set MATH with MATH. It follows from REF that for each MATH there is an absolutely continuous MATH-invariant measure MATH giving full weight to MATH. Take MATH and denote MATH. Since MATH is MATH-invariant, we have that MATH is also MATH-invariant, and so MATH . Hence, assuming MATH normalized we have that each MATH is an absolutely continuous MATH-invariant probability measure giving full weight to MATH. The minimality of each MATH implies that MATH is ergodic for MATH, and so it coincides with the SRB measure MATH. This in particular implies that MATH. The fact that MATH is a minimal MATH-invariant set that contains MATH implies that MATH is an ergodic absolutely continuous MATH-invariant probability measure, thus coinciding with MATH. |
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