paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0010223 | Once we show that REF MATH, then REF the fiber MATH is an AR by REF , so REF the principal bundle has a global section and it is trivial and REF follows from REF . It remains to prove (MATH). CASE: MATH and REF (under REF ): We can apply REF to MATH. We can verify REF as follows: CASE: By the assumption MATH, MATH, MATH, MATH, MATH for each MATH. CASE: If MATH is a component of MATH, then MATH contains a component of MATH. (Also, MATH meets both MATH and MATH (if MATH then MATH is a compact component of MATH, a contradiction), so MATH is not connected. Hence if MATH is a disk or a NAME band, then MATH is a disjoint union of arcs.) By REF (Compact case) it follows that MATH and this implies (MATH). CASE: MATH and REF : Since MATH and MATH meets both MATH and MATH in REF MATH, it follows that MATH is a disjoint union of disks, thus MATH by REF . This implies (MATH). CASE: The remaining REF MATH: It follows that REF MATH is a disjoint union of disks MATH and an annulus MATH and REF MATH is an arc, MATH is a disjoint union of arcs REF and MATH, MATH are the boundary circles of MATH. Since MATH is an annulus and MATH, from REF it follows that MATH. Each MATH is isotopic rel MATH to MATH. Since MATH is isotopic to id rel MATH, it follows that MATH and so MATH. This completes the proof. |
math/0010223 | By REF each MATH is an AR. Thus by REF MATH is also an AR and it strongly deformation retracts onto the single point set MATH. Hence the principal bundle MATH is trivial and MATH strongly deformation retracts onto the fiber MATH. In particular, MATH is connected and MATH. |
math/0010223 | By REF , for each MATH there exists a strong deformation retraction MATH REF of MATH onto MATH. A strong deformation retraction MATH REF of MATH onto MATH is defined as follows: MATH . Since MATH, the map MATH is continuous. (In REF , the same conclusion follows from REF by taking the end compactification of MATH.) |
math/0010223 | By REF for each MATH we have the trivial principal bundle MATH . It follows that MATH admits a section MATH, and the map MATH is fiber preserving homotopic to MATH over MATH. Since each fiber of MATH has MATH, this homotopy is a MATH-homotopy. Since MATH is an ANR REF , by REF MATH is also an ANR. |
math/0010223 | Since MATH is locally compact and locally connected, MATH is a topological group and MATH is a closed subgroup of MATH. Since MATH is locally compact and second countable, MATH is also second countable. A complete metric MATH on MATH is defined by MATH for MATH, where MATH is a complete metric on MATH with MATH. Since MATH CITE, MATH is not locally compact. Finally, by REF . REF MATH is an ANR. This completes the proof. |
math/0010224 | The assertion will be verified in the next four steps: CASE: For each MATH there exists a map MATH such that MATH and MATH. CASE: There exists a homotopy MATH such that REF MATH and MATH and REF each MATH admits a neighborhood MATH such that MATH for some MATH. CASE: There exists a function MATH such that MATH, MATH and MATH is continuous for each MATH. CASE: There exists a homotopy MATH such that MATH and MATH. |
math/0010224 | CASE: Since the restriction map MATH is a trivial bundle with an AR fiber and MATH, inductively we can construct a section MATH of MATH such that MATH. The section MATH is defined by MATH. CASE: We replace the interval MATH by MATH. For each MATH let MATH. We have MATH and MATH. Since MATH is a trivial bundle with an AR fiber, using the "fiber preserving" absolute extension property, we can inductively construct a sequence of homotopies MATH such that MATH, MATH, MATH and MATH. Hence we can define a homotopy MATH by MATH. Since MATH for MATH, we can continuously extend G by MATH. Since MATH, the homotopy satisfies the required conditions. CASE: Since MATH, from REF (Compact case) it follows that MATH is an ANR and MATH has the h.n. complement in MATH. Inductively we can construct a sequence of homotopies MATH such that MATH, MATH and MATH. In fact, since MATH is an ANR, given MATH, we have a homotpy MATH such that MATH and MATH. Let MATH be an absorbing homotopy of MATH into MATH and take a map MATH with MATH. The desired homotopy MATH is defined by MATH. Since MATH and MATH, the required function MATH is defined by MATH. CASE: We define a function MATH by MATH. By REF it follows that MATH is continuous and MATH. Choose a small map MATH such that MATH REF and define the homotopy MATH by MATH . Then MATH is continuous and it follows that MATH and MATH. This completes the proof. CASE: In this case MATH is an ANR, and the conclusion follows from the next lemma. |
math/0010224 | We verify the following assertion: CASE: Every MATH admits an open neighborhood MATH and a homotopy MATH such that MATH is the inclusion MATH and MATH. This easily implies that MATH itself admits an absorption homotopy into MATH (compare CITE). Consider the bundle MATH (over its image) (compare Proof of CITE, CITE). From the assumption it follows that MATH is also an ANR, in which MATH is open, and MATH for some MATH since MATH. Let MATH and consider the restriction map MATH. By REF MATH for some MATH. Since the map MATH is locally trivial, it has a section MATH over an open neighborhood MATH of MATH in MATH. Then MATH is an open neighborhood of MATH in MATH and for any MATH we have MATH. By REF has the h.n. complement in MATH. Since MATH, by REF (Compact case), MATH also has the h.n. complement in MATH. Let MATH and MATH are the corresponding absorbing homotopies. Then the required absorbing homotopy MATH is defined by MATH. |
math/0010224 | CASE: Choose a point MATH and let MATH (MATH). Since MATH is proper, MATH is compact. For MATH and MATH consider the subspace MATH . Since MATH, it suffices to show that each MATH is compact. For each MATH we have MATH, MATH, where MATH. By the NAME theorem, the space of MATH-Lipschitz maps from MATH to MATH, MATH, is compact, and the map MATH is a closed embedding. Thus MATH is compact as required. |
math/0010224 | We define a compatible metric MATH of MATH by MATH where MATH is the length of a path MATH in MATH. First we note that MATH is a locally NAME homeomorphism. This follows from the following observations: CASE: We may assume that MATH is compact. CASE: In general, if MATH and MATH are compact connected Euclidean polyhedra and MATH is a NAME, then MATH is a NAME map. CASE: Take any regular neighborhood MATH of MATH in MATH and a NAME MATH. Then MATH contains a MATH-neighborhood MATH for some small MATH. It follows that MATH is NAME and NAME for MATH with MATH, so that MATH is also NAME. It remains to show that MATH is a locally NAME homeomorphism. Let MATH denote the MATH-triangulation of MATH. Subdividing MATH if necessary, we may assume that MATH is simplicial on each simplex of MATH. Since MATH is a MATH-triangulation, for each MATH-simplex MATH the restriction MATH is MATH-diffeomorphism and there is a constant MATH such that for each piecewise MATH-curve MATH in MATH . We show that MATH is locally NAME. A similar argument shows that MATH is also locally NAME. Let MATH be any point of MATH and MATH be a strongly convex open neighborhood of MATH in MATH (compare CITE) containd in a compact subcomplex MATH of MATH. Let MATH. Any points MATH and MATH in MATH can be joined by a unique shortest geodesic MATH of MATH contained in MATH. For any MATH we can find a piecewise MATH-curve MATH in MATH joining MATH and MATH such that MATH and MATH decomposes into subcurves MATH so that each MATH is contained in a MATH-simplex MATH (compare CITE). Then MATH. Since MATH and MATH, it follows that MATH. |
math/0010224 | CASE: The value MATH satisfies the following equations: MATH . Thus MATH and MATH, and so MATH. |
math/0010224 | CASE: The assumption implies that MATH and MATH. CASE: Set MATH and MATH. |
math/0010224 | REF follows from REF and the definition of MATH. REF follows from the next facts: CASE: If MATH (MATH), then MATH is a MATH-Lipschitz homeomorphism CITE. CASE: If MATH, then MATH is a NAME. CASE: MATH. |
math/0010224 | We choose any homeomorphism MATH with MATH and a similar affine embedding MATH, where MATH is a principal simplex of in MATH of MATH. For MATH, MATH and MATH we obtain the stability homeomorphism MATH. The following pairs MATH REF satisfy REF - REF : MATH . In fact, REF follow from REF respectively and REF is obvious. Thus MATH induces the required stability homeomorphism of tuples. Since MATH, it follows that MATH for each pair MATH. |
math/0010224 | Choose a NAME MATH onto a Euclidean polyhedron and let MATH. By REF MATH induces a homeomorphism of tuples, defined by MATH: MATH where MATH. It suffices to show that the latter tuple is MATH-stable. Let MATH, MATH and MATH be the maps given in the proof of REF for MATH. Then MATH and MATH. Since MATH, the data MATH, MATH and MATH satisfy REF - REF , and this implies that MATH. This completes the proof. |
math/0010224 | We choose REF a triangulation MATH of MATH with MATH for each simplex MATH of MATH, REF a sequence of principal simplices MATH of MATH such that MATH, MATH and MATH (MATH), and REF similar affine embeddings MATH such that MATH and MATH for some MATH. For MATH, MATH and MATH we obtain the stability homeomorphism MATH. By the choice of MATH and REF the following pairs MATH REF satisfy REF - REF : MATH . Note that for MATH, REF if MATH (MATH), then MATH and also MATH are MATH-Lipschitz homeomorphisms compare CITE, and REF if MATH for each MATH, then MATH is a MATH-Lipschitz homeomorphism for MATH by REF and the choice of MATH and MATH. Thus MATH induces the required stability homeomorphism of tuples. Since MATH, the homeomorphism MATH preserves the identity components of the homeomorphism groups. |
math/0010224 | We may assume that MATH. For each MATH we choose a MATH-open neighborhood MATH of the origin MATH in MATH such that the exponential map MATH maps MATH diffeomorphically onto a small open neighborhood MATH of MATH in MATH with MATH. Since MATH is isometric at the origin of MATH, if MATH is so small, then MATH is a REF-Lipschitz homeomorphism. We choose any MATH-similar linear isomorphism MATH (MATH small) and consider the embeddings MATH. If MATH is small, then MATH and MATH for some MATH. Note that MATH, MATH (MATH, MATH), is a MATH-similar homeomorphism onto a shortest geodesic MATH. For MATH, MATH and MATH we obtain the stability homeomorphism MATH. By the choice of MATH and REF the following pairs MATH REF satisfy REF - REF : MATH . Note that for MATH, REF if MATH (MATH), then MATH is a MATH-Lipschitz homeomorphism, and REF if MATH for each MATH, then MATH is a MATH-Lipschitz homeomorphism for MATH. For REF , we observe that MATH, where MATH is a REF-Lipschitz homeomorphism, MATH is MATH-similar, and MATH is the extension of MATH by the identity, which is a MATH-Lipschitz homeomorphism since MATH is so. REF follows from REF and the choice of MATH, MATH and MATH. Therefore MATH induces the required stability homeomorphism of tuples. Since MATH, the homeomorphism MATH preserves the identity components of the homeomorphism groups. |
math/0010224 | CASE: MATH is a MATH-set of MATH itself since MATH REF are MATH-sets of MATH and MATH. Note that MATH. We identify MATH with MATH by taking any linear ordering of MATH. Take an appropriate sequence MATH of disjoint infinite subsets of MATH so that MATH satisfies REF . It follows that REF MATH and REF MATH. CASE: The assertion follows from the next REF . The required closed embedding MATH is defined by MATH, where MATH, MATH, MATH is a homeomorphism of quadruples. |
math/0010224 | The proof is a modification of the argument of CITE. CASE: We can write MATH, where MATH is a MATH subset of MATH, and MATH, where MATH is a closed subset of MATH and MATH. By CITE there exist REF a map MATH such that MATH and REF maps MATH such that MATH, MATH and MATH. The required map MATH is defined by MATH. CASE: Since MATH is MATH-compact, there exists a closed embedding MATH such that MATH CITE. Since MATH, we can write MATH so that each MATH is compact, MATH and MATH. Since REF MATH is a MATH-set of MATH, REF MATH is a MATH-set of MATH and REF MATH has the h.n. complement in MATH, there exists a map MATH such that MATH, MATH is an embedding, MATH, MATH, MATH and MATH, where MATH and MATH is a fixed complete metric on MATH. By CITE the map MATH is a closed embedding such that MATH, MATH and MATH. We regard as MATH. For each MATH, there exists a map MATH such that REF MATH, REF MATH maps MATH homeomorphically onto MATH, REF MATH and REF if MATH then MATH so that MATH and MATH. This follows from an easy shrinking argument based on the next claim. For any neighborhoods MATH of MATH and MATH of MATH in MATH, there exists a homeomorphism MATH such that MATH on MATH, MATH, MATH and if MATH, then MATH for each MATH. The homeomorphism MATH is obtained as MATH, where MATH is sufficiently large and MATH shrinks MATH radially towards MATH in MATH. By the repeated application of Claim, we can construct a sequence of homeomorphisms of MATH which converges to the desired map MATH. The map MATH is a closed map and MATH, MATH. Hence the map MATH is a closed map and MATH. It remains to show that REF MATH, REF MATH and REF MATH is injective. CASE: If MATH, then MATH for some MATH and it follows that MATH and hence MATH for each MATH. CASE: MATH. CASE: Suppose MATH and MATH. If MATH, MATH then we may assume that MATH and MATH. The choice of MATH implies that MATH and MATH. By the choice MATH we have MATH and so MATH. If MATH, then since MATH is injective, we have MATH for some MATH, and since MATH we have MATH. Again, by the choice MATH we have MATH and MATH. This completes the proof. |
math/0010224 | Let MATH. By REF there exists a closed embedding MATH such that MATH, MATH and MATH. Note that MATH. There exists a map MATH such that MATH and MATH CITE. The required closed embedding MATH is defined by MATH, where the homeomorphism MATH is defined by MATH. |
math/0010224 | Since MATH, the product MATH satisfies both MATH- and MATH-stability. Hence by REF MATH is both a MATH- and a MATH-manifold. Thus by REF we have MATH . |
math/0010225 | First of all, if MATH has an atom then MATH is supported on a periodic orbit, hence the result is trivial because in this case MATH for MATH and MATH otherwise. We may assume then that MATH has no atoms, consequently MATH as MATH for all MATH. At first we suppose that MATH and assume that MATH is small enough that MATH. Note that this implies MATH. For MATH let MATH be the MATH - first return time of MATH to MATH. By NAME 's Theorem and ergodicity of MATH we have MATH for MATH-a.e. MATH . Let MATH . Clearly MATH . For all MATH and for all MATH there exists MATH such that MATH for all MATH . Let MATH where we have suppressed the obvious MATH dependence on the set MATH. We choose MATH such that MATH . By REF for all MATH and all MATH. Thus MATH for some MATH with MATH . It immediately follows that MATH with MATH whenever MATH and MATH . Next we define MATH . By the NAME Density Theorem, MATH and thus MATH . For each MATH there exists MATH such that MATH for all MATH . Thus if MATH is sufficiently small then MATH, where MATH . Let MATH be the set of points MATH for which MATH as MATH. By REF. Hence MATH. Denote MATH and set MATH. Assume additionally that MATH. The limiting distribution MATH is continuous at MATH and MATH, hence if MATH is sufficiently small MATH . Thus for all MATH, MATH . A similar computation with MATH yields MATH . For MATH sufficiently small, MATH. Thus REF implies: MATH . Define MATH . For each point MATH of continuity of MATH, the function MATH as MATH. Combining REF yields: MATH . We just showed that for any MATH, there exists a set MATH with MATH and a real number MATH such that for all MATH, REF holds whenever MATH. The conclusion of the theorem for MATH-a.e. MATH follows from the remark that MATH-almost every MATH is contained in the set of full measure MATH. Finally we want to prove the uniform convergence in MATH in the case MATH is continuous. Since MATH as MATH, it is uniformly continuous; hence MATH as MATH. Moreover, MATH is bounded by MATH by NAME 's inequality. Hence REF gives MATH . If MATH then MATH while if MATH then MATH . |
math/0010225 | It is enough to check that the variation of MATH is finite. Since MATH is MATH on the interior of MATH, the variation is estimated by MATH . |
math/0010225 | In the proof we use the estimate given by REF to apply REF. We recall the quantities considered there (MATH is any integer). MATH . We denote by MATH the density of the measure MATH with respect to MATH. Let MATH be an interval in MATH and set MATH. We now compute an upper bound for MATH and MATH. Obviously MATH, and for each MATH . We used for the last inequality the estimate on decay of correlations REF given by REF , with MATH and MATH. Since MATH is an interval we have MATH, hence the summation on MATH gives MATH . We consider now MATH. The decay of correlations given by REF yields (with MATH and MATH) MATH . REF together with REF yield (with MATH) MATH for some constants MATH and MATH independent of the interval MATH. Since MATH has no atoms the countable union MATH has zero measure. Hence for MATH-a.e. point MATH, the iterates MATH are well defined and MATH. One easily sees then that MATH as MATH provided MATH is not periodic. Consider the set MATH . The NAME Density Theorem tells us that MATH is MATH-almost everywhere finite because the density MATH, MATH-a.e. Since MATH is aperiodic (it is ergodic with no atoms), we conclude that MATH. Moreover, for all MATH, using REF we get MATH as MATH. We conclude then by REF that for MATH-almost all MATH, MATH uniformly in MATH. |
math/0010225 | Let MATH. Obviously MATH. Let MATH be any recurrent point in MATH; by the NAME Recurrence Theorem, this concerns MATH-a.e. MATH. Let MATH be the component of MATH containing MATH and MATH be any integer such that MATH. As MATH, there exists a neighborhood MATH of MATH such that MATH maps MATH monotonically onto MATH. Let MATH be the first return map to MATH. By taking MATH sufficiently large, we can ensure that MATH is compactly contained in MATH and that MATH for all MATH. Let MATH be any maximal interval on which MATH is monotone. Since MATH, it follows that MATH is onto, and therefore contains a fixed point, say MATH. The derivative MATH is uniformly bounded away from MATH, see CITE. Write MATH and MATH. Let MATH be the partition of MATH into the sets MATH. We will show that MATH is a MATH-map. Given intervals MATH, MATH is said to contain a MATH-scaled neighborhood of MATH if both components of MATH have length MATH. Let MATH be the distance between MATH and MATH. For each branch MATH (where MATH is the return time), there is an interval MATH such that MATH maps MATH monotonically onto MATH. It follows that MATH contains a MATH-scaled neighborhood of MATH. More precisely, if MATH, then MATH contains a MATH-scaled neighborhood of MATH. This allows us to use the NAME Principle: Let MATH be intervals and assume that MATH is a MATH diffeomorphism. If MATH contains a MATH-scaled neighborhood of MATH, then there exists MATH such that the distorsion MATH . CASE: If MATH satisfies REF, then we can choose MATH. In the general MATH case, we have MATH, where MATH is a nonnegative number such that MATH as MATH. Note that for our purpose, MATH is a first return map. Hence MATH for all MATH and MATH the return time of MATH. For the proof of the proposition and remark, see CITE. CASE: By taking MATH small, we can choose MATH as close to MATH as we want. It follows that MATH, that is, MATH is expanding. Since MATH is MATH-conformal, MATH . By REF, MATH . This shows that MATH. Let MATH be intervals such that MATH is monotone, and let MATH and MATH be the components of MATH. Then MATH expands the cross ratio if MATH . REF is equivalent to REF (see CITE) which is easily seen to carry over to the iterates MATH. It follows that MATH is convex on each branch and therefore MATH is nondecreasing. Hence, each branch domain MATH contains at most one point MATH at which MATH changes sign. We get MATH . Therefore MATH . Thus MATH is a MATH-map and it satisfies the assumptions of REF . By REF , the conclusion follows because the set of MATH satisfying the above assumptions have full measure. |
math/0010225 | Let MATH and for MATH . Fix MATH and let MATH . Let MATH be the first return map to MATH. We then define a partition of MATH by MATH, where MATH. One easily check that MATH is a partition into intervals, and each branch of MATH is monotone and onto. Let MATH be the normalized NAME measure restricted to MATH and let MATH when MATH for some MATH and MATH otherwise. In the proof of REF it is proved that for some constant MATH for all MATH. A straightforward computation shows that MATH for all MATH and MATH. Since MATH is increasing and convex on each MATH, we get MATH . Taking into account that MATH for any MATH, we get MATH . Therefore, since MATH is MATH in the interior of MATH and MATH, MATH . Finally, it is obvious that MATH, hence MATH is a MATH-map. By REF MATH-almost all points inside MATH have exponential return time statistics, and the conclusion follows since MATH has full NAME measure. |
math/0010225 | First let us make the following remark. As MATH is expanding, a straightforward calculation shows that REF implies that MATH and MATH uniformly in MATH. Let MATH be the NAME operator MATH . Let us write MATH for the seminorm MATH. If MATH, we have MATH . Here we summed over the pairs MATH in the same atom MATH with MATH and MATH. As MATH is onto for every MATH, these pairs are well-defined. The first term in the above expression is bounded by MATH . Next we use the Mean Value Theorem and REF to estimate MATH for some MATH. The constant MATH is an upper bound in REF. This gives for the second term MATH . Let MATH. Then MATH . Obviously MATH. Therefore also MATH . By construction MATH uniformly in MATH. This allows us to use the NAME theorem, which shows that MATH is a quasicompact operator. Since each branch is onto, MATH is a simple eigenvalue with NAME eigenvector MATH, and is the unique eigenvalue on the unit circle. Consequently, for any NAME function MATH we have MATH for some MATH and MATH. It follows that the correlations between NAME functions MATH and MATH functions decays exponentially fast: if MATH we find MATH for MATH. |
math/0010225 | We can rewrite REF as MATH where MATH is NAME and bounded away from below (see CITE), and MATH is the NAME operator as defined in the previous proof. Recall also that MATH. Since MATH is one-to-one and onto when MATH, the above facts imply that MATH . The lemma follows now from REF by taking MATH and MATH. |
math/0010225 | Let us assume for simplicity that MATH. We denote by MATH the interior of a subset MATH in the induced topology. Since MATH is expanding, there exists MATH such that MATH for any MATH and integer MATH. By the NAME property, the bounded distortion and the conformality of the map MATH we can find some constant MATH such that the following property holds: For any integer MATH and cylinder MATH there exists MATH and MATH such that MATH for any MATH different from MATH. Given MATH, we have MATH; in particular, MATH. Let MATH, MATH and consider the following partition MATH . Let MATH. Any MATH is a subset of the annulus MATH. Thus there exists MATH disjoint balls or radius at least MATH inside MATH. Since the area of MATH is equal to MATH when MATH we get MATH . Obviously when MATH we also have MATH. Since the measure MATH is MATH-conformal and the map itself is conformal and NAME we have for some constant MATH, MATH for any MATH for some MATH. The previous inequalities imply (observe that MATH if MATH and recall that MATH) MATH . Taking MATH sufficiently small gives the result. |
math/0010225 | The proof closely follows the one of REF , we use the same notation MATH and MATH. Let MATH and MATH to be chosen later on. By REF MATH . Similarly, we get by REF and (twice) REF MATH for all MATH. Taking MATH and MATH gives MATH . For all non periodic points MATH we have MATH, which implies MATH as MATH. Since this concerns MATH-almost all points in MATH, the theorem is proved. |
math/0010225 | Let MATH be arbitrary, and let MATH be open disks such that MATH and MATH is compactly contained in MATH. Let MATH be the modulus of MATH. Assume also that MATH is convex-like in the sense that MATH, where MATH is as in REF. The strategy is to find a subset MATH of MATH such that MATH for all MATH, and then we can invoke REF . If MATH is a NAME set, we can assume that MATH is contained in the NAME set MATH, which is the basin of MATH in this case. Moreover, there are no neutral or stable periodic orbits. Thus each point in MATH, in particular MATH, converges to MATH. It follows that MATH intersects MATH for at most finitely many MATH. Let MATH be the component of MATH containing MATH. Then MATH for all MATH as required. Since MATH consists of the intersection and difference of at most finitely many, convex-like disks, we find MATH. Assume now that MATH is connected and by NAME 's results also locally connected. Let MATH, MATH, be the periodic components of the NAME set, with MATH the basin of MATH. Since MATH is a polynomial (with exceptional point MATH), MATH. There are only finitely many such components, and by NAME 's Theorem (see for example, CITE), every MATH is eventually mapped into MATH. We construct a special forward invariant subset MATH of MATH. CASE: Consider the renormalization MATH of the highest possible period MATH. (If MATH is not renormalizable, then we just take MATH.) It is known that MATH contains a MATH-periodic point MATH with at least two external rays, say MATH and MATH. The existence of such external rays (and the arcs MATH defined later on) is guaranteed by results initiated by NAME, see CITE and references therein. Let MATH. CASE: If MATH, MATH, contains a stable periodic point, let MATH be a disk compactly contained in MATH such that MATH and MATH. There is at least one MATH-periodic point MATH in the boundary of MATH. Let MATH be a smooth compact arc connecting MATH and MATH such that MATH. Since MATH, there is also an external ray MATH landing at MATH. CASE: If MATH contains a parabolic point MATH in its boundary such that each MATH converges to MATH, let MATH be a disk such that MATH, MATH, and MATH. Let MATH be an external ray landing at MATH. Let MATH . Then MATH is connected, and for some MATH, MATH is forward invariant. We start a NAME puzzle construction by putting, for MATH, MATH . For each MATH, MATH maps any element of of MATH into an element of MATH. Using the arguments in CITE, one can show that the diameters of the elements MATH tend to MATH as MATH, unless MATH eventually intersects a NAME disk. We can assume that the point MATH. Moreover, since MATH densely fills the boundary of any NAME disk, and MATH, we can assume that MATH does not lie on the boundary of a NAME disk. Find MATH so large that the element MATH of MATH containing MATH is contained in MATH. Let MATH. Then MATH for all MATH. Note also that MATH is bounded by finitely many smooth curves of MATH, MATH and MATH. At worst these curves end in a logarithmic spiral, namely as they approach the hyperbolic periodic points MATH. Therefore also MATH is convex-like, and obviously simply connected. It follows that MATH. The rest of the argument works for both MATH locally connected and MATH a NAME set. Let MATH be the first return map to MATH. Then MATH is defined on a countable collection MATH of disks MATH. The modulus MATH, and for each branch MATH there exists a disk MATH such that MATH maps MATH univalently onto MATH. It follows from the NAME MATH-Theorem (for example, CITE) that the distortion of MATH is uniformly bounded: MATH . More precisely, take MATH and let MATH be a straight arc containing MATH such that MATH varies little on MATH. Then MATH . To estimate this, let MATH be the maximal round disk centered at MATH contained in MATH. Let MATH be the radius of MATH; we have MATH. Define MATH. Then MATH is a univalent map on the unit disk with MATH. By REF, we obtain MATH provided MATH is close to MATH. Therefore MATH, proving that MATH . The distortion bound MATH applies by the same argument also to iterates MATH of the induced map. Therefore each domain MATH is not much less convex than MATH: MATH for some MATH. We can also assume that MATH and hence MATH is so small that MATH. Thus MATH is hyperbolic. Recall that MATH is the MATH-conformal measure of MATH. As MATH is a first return map, it is straightforward to show that MATH is a MATH-conformal map for MATH. In fact, MATH can also be constructed using NAME 's techniques CITE. Now we can invoke REF with the invariant measure MATH. |
math/0010227 | As the algebra is finite dimensional, it follows from the weight space decomposition that there exists at least one weight MATH which is "extreme" in the following sense : MATH . Then for any vector MATH we have MATH, where MATH denotes the adjoint operator in MATH. Thus the vector is central, and the corresponding vertex of MATH is isolated. |
math/0010227 | As MATH for a weight system MATH of MATH, we know that MATH contains at least one isolated point, call it MATH. Thus, in MATH is adjacent to any other point. If MATH for MATH, then the path MATH is a geodesic joining them. Thus MATH. If one of the points equals MATH, then the distance is one. |
math/0010227 | As we have imposed MATH for any weight MATH, we can reorder the weights of MATH in such manner that a relations MATH corresponds to a sum MATH, where MATH. Thus the maximal number of sums of weights equals the number of possibilities MATH with MATH and MATH. It is easily seen that this number is precisely MATH. |
math/0010227 | The assertion is obviously true for MATH. Suppose it is also for MATH. Then we have MATH CASE: if MATH then we have MATH, and as MATH we have MATH . As MATH and MATH , it follows MATH . CASE: if MATH then MATH. In this case MATH and as before MATH. |
math/0010227 | Let MATH with MATH. From the previous result we know that MATH . Now MATH for MATH, and MATH, with equality when MATH is odd. Then MATH which shows that MATH . |
math/0010234 | Simply observe that every pliable foursome from MATH is also a pliable foursome in MATH and hence that the restrictions of the elements of MATH to MATH form a pliable family for MATH. |
math/0010234 | By normality we can extend, for every MATH, any partition in MATH between MATH and MATH to a partition in MATH between MATH and MATH. Now apply the assumption that the full family is essential. |
math/0010234 | REF implies immediately that MATH is hereditarily indecomposable: for every MATH the projection MATH is as required. To see that MATH is infinite-dimensional we observe that MATH is a partition between the even-numbered faces MATH and MATH of MATH - indeed: MATH . By REF this implies that MATH is an essential family in MATH; so MATH is even strongly infinite-dimensional. |
math/0010234 | As observed in the previous proof the traces of the odd-numbered faces of MATH on MATH form an essential family in MATH. One can therefore find a component MATH of MATH such that the traces from that family on MATH also form an essential family. Now let MATH be (the restriction of) the projection onto the first MATH coordinates. Consider the monotone-light factorization of MATH, that is, write MATH, where MATH is a monotone surjection and MATH is a light map, compare CITE. Since MATH is light we have MATH, compare CITE. For odd MATH let MATH and MATH and observe that MATH and MATH. From these last equalities it follows that the MATH and MATH form an essential family in MATH and so MATH. Because MATH we may conclude that MATH contains a MATH-dimensional continuum MATH. Since MATH is monotone and MATH is hereditarily indecomposable, so is MATH. |
math/0010234 | We use the partition MATH and the open sets MATH and MATH from REF again. Let MATH be a pliable family for MATH. Choose open neighbourhoods MATH and MATH of MATH and MATH respectively with disjoint closures. Whenever MATH and the open sets MATH and MATH with disjoint closures are found apply NAME 's lemma to get a continuous function MATH such that MATH and MATH and set MATH where MATH. Because the closures of MATH and MATH are disjoint the closures of MATH and MATH are disjoint as well. Furthermore, because MATH we have MATH for MATH, MATH. In the end the sets MATH and MATH are disjoint open neighborhoods of MATH and MATH, respectively. A direct application of REF shows that MATH is hereditarily indecomposable: for every MATH the function MATH is a suitable partner for MATH. |
math/0010234 | We use the proof of REF . Let MATH be a pliable family of continuous functions on the NAME cube MATH. As before let MATH, where MATH and MATH is the projection onto the MATH-th coordinate. The odd-numbered faces of MATH induce an essential family on MATH; it is also essential on some component of MATH. This component is the required continuum. |
math/0010234 | Let MATH and MATH be disjoint open sets around MATH and MATH respectively such that MATH. Let MATH and MATH and MATH. It is clear that MATH and that MATH, so that we are done in case MATH. If MATH then we can find an increasing sequence MATH and a decreasing sequence MATH such that MATH and MATH for all MATH. Because the sets MATH and MATH are open and because MATH has uncountable cofinality we can find a MATH such that MATH and MATH for all MATH. It follows that MATH. |
math/0010234 | Let MATH be a continuum of weight MATH and assume that MATH is embedded into the NAME cube MATH. Let MATH be the hereditarily indecomposable continuum from REF and let MATH be an essential map. For every finite subset MATH of MATH we consider the map MATH and the continuum MATH, where MATH is the projection of MATH onto MATH. Because MATH is essential we may apply REF from CITE to find a subcontinuum MATH of MATH such that MATH. Because MATH is compact the net MATH has a convergent subnet in MATH; its limit MATH is a subcontinuum of MATH that maps onto MATH. |
math/0010234 | We now assume that our continuum MATH is embedded in MATH, which we consider to be a subset of MATH. We take the continuum MATH from REF and let MATH be an essential map. For every finite subset MATH of MATH we let MATH denote the projection of MATH onto MATH. An application of REF yields for every finite set MATH a subcontinuum MATH of MATH such that MATH and the restriction MATH is weakly confluent. As before we take a convergent subnet MATH of MATH with limit MATH; then MATH and it remains to show that MATH is weakly confluent. To this end let MATH be a subcontinuum of MATH and choose for every MATH a subcontinuum MATH of MATH such that MATH. The subnet MATH of MATH has a convergent subnet MATH with limit MATH; it should be clear that MATH and MATH. |
math/0010234 | This is a straightforward generalization of the proof of REF. Let MATH denote the standard NAME set in MATH. For MATH put MATH-if MATH then MATH and set MATH. Just as in CITE one verifies that MATH is a closed and connected subset of MATH; to see that MATH is one-dimensional one only has to realize that every basic open cover lives on a finite subset of MATH and hence that it can be given an open refinement of order MATH. Finally, the map MATH, where MATH is the NAME step function, is a monotone map. |
math/0010234 | By the previous lemma we can find a one-dimensional continuum MATH of the same weight as MATH and a monotone surjection MATH. Next find a hereditarily indecomposable continuum MATH and a weakly confluent surjection MATH. As in the proof of REF we take the monotone-light factorization of MATH, that is, a space MATH, a monotone map MATH and a light map MATH so that MATH. Because MATH is monotone the space MATH is hereditarily indecomposable, because MATH is light it is one-dimensional and because MATH is weakly confluent so is MATH and because MATH is monotone the map MATH is weakly confluent. |
math/0010234 | Consider the continuum MATH constructed in REF and its essential family MATH. Partition MATH into MATH many sets MATH of size MATH and let MATH be a base for MATH. For each MATH let MATH be the union of all components of MATH on which MATH is not essential. Observe that MATH is open in MATH and that the family MATH is not essential on any compact subset of MATH. Next let MATH for each MATH and put MATH. The set MATH is open in MATH and the family MATH is not essential on any compact subset of MATH: if MATH is such a set cover it by finitely many MATH and use the disjointness of the sets MATH to make a set of partitions whose intersection misses MATH. It follows that every partition between MATH and MATH must meet MATH and hence that MATH contains a non-trivial continuum MATH. Let MATH be any subcontinuum of MATH, let MATH and fix MATH such that MATH and MATH. Consider the component MATH of MATH in MATH; because MATH is hereditarily indecomposable we have MATH. But then MATH and so MATH is essential on MATH and hence on MATH. |
math/0010234 | We only sketch the argument. Let MATH be an embedding and define MATH by ``MATH is the unique point in MATH". It is straightforward to check that MATH is onto and that MATH for all MATH. |
math/0010234 | Let MATH be a lattice-base for the closed sets of MATH (of minimal size) and identify it with its copy MATH in MATH. By the NAME theorem CITE there is a lattice MATH, of the same cardinality as MATH, such that MATH and MATH is an elementary substructure of MATH. The space MATH is as required. |
math/0010234 | That MATH is a chain follows from hereditary indecomposability of MATH. It is clear that MATH is complete: if MATH then MATH is the supremum of MATH in MATH. To see that MATH has no jumps take MATH and MATH in MATH with MATH and fix an open set MATH such that MATH and MATH. Now the component MATH of MATH that contains MATH meets the boundary of MATH, so MATH, and is contained in MATH, so MATH. The set MATH is closed in MATH: its complement, MATH, is a basic open set. Likewise the sets MATH and MATH are closed in MATH; this shows that the order topology on MATH is contained in the subspace topology. Because both topologies are compact NAME they coincide; because this topology is metric we find that MATH is isomorphic and homeomorphic to MATH. |
math/0010234 | Apply hereditary indecomposability. |
math/0010234 | The function MATH is continuous and, for every MATH, the set MATH is an arc that connects MATH and MATH; it follows that MATH. |
math/0010234 | Striving for a contradiction, assume that for MATH such as in the formulation of the lemma, every continuum in MATH has diameter less than MATH. Since MATH is finite, there clearly is a finite disjoint collection MATH of closed subsets of MATH of mesh less than MATH such that MATH is contained in the interior MATH of MATH. Since by assumption each component of MATH has diameter less than MATH, the set MATH can also be covered by a finite disjoint collection MATH of closed sets with diameter less than MATH. For MATH and MATH let MATH be a closed set in MATH separating the disjoint closed sets MATH . Then, clearly, MATH, which in turn contradicts our assumption that the pairs MATH and MATH form an essential family. |
math/0010234 | Put MATH and assume that MATH is a finite closed cover of MATH with mesh less than MATH such that each element of MATH meets at most MATH other elements of MATH. We shall associate to each element MATH a compact subset MATH of MATH such that CASE: MATH meets every element of MATH, CASE: MATH, CASE: MATH whenever MATH and MATH are distinct elements of MATH. Assume that MATH is already defined on a subfamily MATH of MATH, and take MATH in MATH; we show how to extend MATH to MATH (this then means that we can define MATH on all of MATH in finitely many steps). The set MATH of all elements of MATH that meet MATH has, by assumption, cardinality less than MATH; because MATH iff MATH it suffices sure that MATH does not meet MATH for any MATH in MATH. For each MATH in MATH let MATH be the closed MATH-ball about MATH and fix a MATH. We shall show that MATH does not cover MATH. Indeed, otherwise, because MATH is connected, we could arrange this family into a sequence MATH, MATH, with MATH for each MATH. But then we could find an upper bound for MATH, thus: MATH . This would contradict our choice of MATH as he minimum diameter of the elements of MATH. Take MATH and let MATH be the closed MATH-ball about MATH. We set MATH. If MATH then MATH and so MATH meets MATH; clearly MATH, so MATH has the first two required properties. Finally, if MATH then MATH. The collection MATH is finite, disjoint and has mesh less than MATH. To reach our final contradiction consider any continuum MATH in MATH, take MATH that intersects MATH and fix MATH with MATH. Now, MATH meets MATH and MATH does not so MATH; but then MATH, because MATH is hereditarily indecomposable. This, however, means that MATH . This contradicts REF . |
math/0010234 | Let MATH be an essential family in the normal space MATH. Apply NAME 's lemma to get continuous functions MATH such that MATH and MATH for all MATH and take the diagonal map MATH. If MATH is a partition between the faces MATH and MATH of MATH for each MATH, then MATH is a partition between MATH and MATH and so MATH; but then MATH as well. |
math/0010236 | Denote MATH. Then we have the triple MATH of cell complexes. The lemma immediately follows from the long exact homological sequence for this triple: MATH . Since MATH, MATH is an injection. If MATH is connected then MATH is REF-dimensional, and, from comparing dimensions, we see that MATH is a surjection. Hence MATH and MATH is an injection. We need to notice only that MATH is generated by (co)edges in MATH. |
math/0010236 | Notice first that the map MATH has at least one basis of cardinality MATH. To construct it, take a spanning tree MATH in the graph MATH (MATH can be empty), then MATH is rather obviously a basis of MATH and has cardinality MATH CITE. If now MATH are the (co)edges in MATH then the linear functionals MATH on MATH are linearly independent, hence the MATH functionals MATH on MATH defined by the rule MATH are also linearly independent. Therefore MATH, and being an isotropic subspace in MATH, it is a Lagrangian subspace. Notice that if we represent MATH by a MATH-matrix MATH in the basis MATH, the columns of MATH will represent the functionals MATH. By REF the columns of MATH are linearly independent if and only if the correspondent set of (co)edges is independent in MATH. Hence the bases of the Lagrangian matroid associated with MATH are exactly the bases of the map MATH. |
math/0010236 | Indeed, let MATH be a small cycle around a covertex MATH (for vertices the proof is analogous). Let MATH be the oriented edges exiting from or entering MATH; notice that we count loops with both ends at MATH twice, but with opposite signs. The corresponding edges MATH form the boundary of the country MATH around MATH; again, edges corresponding to loops MATH appear twice, with opposite orientations. Now if MATH is an arbitrary cycle in MATH then MATH . Without loss of generality we can change the orientation of edges and co-edges in such a way that MATH for all coedges MATH, MATH which are not loops and MATH is the boundary cycle MATH of the country MATH around MATH. Then the right hand side of the previous equation is exactly the index of intersection of MATH and MATH, and hence equals MATH. |
math/0010236 | REF depict the process of contraction. When contracting the edge MATH we wish to keep all changes restricted to the closure of the union of two countries of the dual map MATH containing the endpoints MATH and MATH of the contracted edge MATH. REF is pre-contraction and REF is post-contraction. The horizontal edge MATH has been contracted to a point MATH combining the vertices at its ends, and the corresponding coedge MATH has disappeared. The cycles MATH and MATH are canonical images of MATH and MATH in MATH. As a result, MATH, in comparision with MATH, has lost its intersections with the coedge MATH and edge MATH but has attained intersections with every edge MATH which had exited from MATH prior to contraction; notice, however, that the sign of the index of intersection has changed: MATH . Hence MATH . If MATH is a small circle around MATH, as shown on REF , then MATH and MATH . Analogously MATH . Notice that MATH by REF and MATH by definition of the scalar product MATH . Now we can compute: MATH . But MATH hence we can simplify: MATH . |
math/0010237 | Two vertices of a polytope are adjacent exactly if there exists a linear functional which takes equal, maximal values on the vertices, and smaller values on all other vertices of the polytope. Such a functional in the first case is MATH by inspection. In the second case, we have the condition that not both of the two sets listed are bases. Suppose without loss of generality that MATH is not a basis. Then MATH is clearly a required functional. |
math/0010237 | Choose a fundamental basis. Orient it with positive sign, and where MATH is the height function relative to this basis, assign each other basis MATH the sign MATH. Note that this is MATH, as MATH is always even. REF is satisfied, since horizontal edges connect only bases of the same height; REF is satisfied as vertical edges connect bases of height differential MATH. There are no short edges, so REF is satisfied, and REF was specifically stated above, so this defines an orientation relative to the fundamental basis chosen. It is immediate from consideration of REF that no other orientation is possible. |
math/0010237 | This can be checked, case by case, by considering the restrictions that the axioms place on each of the possible face types listed in the previous section. We exhibit the first the case of the rectangle nsRect. Notice that short edges connect two bases whose heights differ by REF; thus, the four vertices of the face take at least two different heights. Let a vertex of the face of minimal height correspond to the basis MATH. Then the other vertices are MATH for some MATH satisfying MATH. The short edges are MATH and MATH, the long edges MATH and MATH. Suppose there are two vertices of the same height. Then both long exchanges are horizontal, and MATH, MATH where MATH is the fundamental basis, by REF . If MATH also, the case becomes trivial, so assume otherwise. As paths must be non-decreasing in height, a path through this face travels exactly one short edge if it moves from one of MATH to one of MATH and no short edges otherwise. As sign changes take place exactly when traversing short edges, the case follows. If there are not two vertices at the same height, then both long edges are vertical, and thus induce sign changes. Now MATH. The only journey for which there is a choice of two increasing paths is MATH, which may take place as MATH or MATH; if there is no choice of increasing paths, the case is trivial. REF gives MATH and MATH. If MATH then we obtain MATH and both paths contain one sign change. Alternatively, if MATH then we obtain MATH and both paths contain two sign changes. This establishes the result in this case. We now exhibit the case of the triangle iTri. Let MATH be the bases at the ends of the long edge and MATH the third basis. If the long edge is vertical, with say MATH lowest, the only possible paths are MATH and MATH. Since the edge is vertical, MATH is of the opposite sign to MATH, and so whichever sign MATH takes, either path has one sign change. Suppose now MATH is horizontal. Thus MATH and MATH take the same sign, and since any pair of paths must either both enter or both leave at MATH (depending on whether it is higher or lower than MATH), the result is clear. Similar reasoning proves the lemma for the other types of two-dimensional faces. |
math/0010237 | First cut MATH by the hyperplane MATH which is constant for MATH and passes through MATH, letting MATH be the resulting polytope containing MATH and MATH. Now let MATH be the vertex figure of MATH at MATH. Now MATH has a facet MATH contained in MATH, and vertices MATH and MATH corresponding to the original edges MATH and MATH. Now all we want is a path of edges from MATH to MATH which misses MATH; those edges will then correspond to the desired two-dimensional faces in MATH. This now follows easily from REF. |
math/0010237 | Set MATH to be the assertion that the theorem holds for some particular MATH as above, and now attempt to prove MATH for every vertex in the polytope (which is exactly the theorem) by an induction we shall call MATH. We must prove: CASE: MATH holds. CASE: Given that MATH holds for all vertices MATH with MATH we may deduce MATH. The basis is clearly trivial. The previous lemma tells us that we can find a path of two-dimensional faces MATH of the polytope where each face MATH has two edges MATH incident to MATH, MATH and MATH, and all the MATH's increase going away from MATH. Let MATH be the assertion that MATH holds when the number of faces above is exactly MATH. If we can prove MATH for all k under the inductive hypothesis of MATH, the theorem is complete. We thus assume MATH for all vertices MATH with MATH, and proceed by induction MATH on MATH. CASE: MATH holds. CASE: MATH implies MATH. Let MATH be the point at the end of MATH. If MATH, the two paths have the same first edge, and since MATH we have MATH; but the two paths differ only after MATH, and so MATH. Thus the basis of MATH holds. Take now MATH. If MATH is the vertex of maximal MATH in the face MATH, then there is an increasing path segment from MATH through MATH around the edge of the face to MATH, which then gives us an alternative path replacing MATH through a transformation facewise with MATH replaced by MATH, completing the induction MATH. If not, we can replace the path from MATH to MATH with one that goes through the point of maximal MATH in the face, using again MATH, and now continue as above. This again gives a situation where MATH is decreased by one after a transformation facewise, and so the induction is complete. |
math/0010237 | Notice first that upon obtaining the form MATH as above, bases of maximal height correspond to non-zero minors of MATH of maximal size. This means exactly that rank of MATH is this height. We utilise two results from the standard theory of quadratic forms (see, for example, CITE for a particularly clear treatment. Note, however, that NAME defines index as the number of positive terms, rather than negative). If MATH is the determinental minor consisting of the first MATH columns and rows of a square, symmetric matrix MATH then we can rearrange by swapping of columns and of the corresponding rows so that not both of MATH and MATH are zero for MATH, and MATH is non-singular, where MATH. A matrix where this holds is called regularly arranged. Notice that putting MATH into a regular arrangement, and permuting column labels when permuting columns (of A), does not alter the oriented Lagrangian matroid represented by MATH. When a matrix is in regular arrangement, NAME 's method tells us that the index is exactly the number of changes of sign in the sequence MATH, discounting any zeroes. We may assume that MATH is in a regular arrangement. We choose an increasing path by taking the list of bases corresponding to the non-zero MATH; this is clearly increasing, and in fact has no horizontal edges. The theorem is immediate. |
math/0010237 | We first consider contradiction-freeness. The result is trivial when the Lagrangian matroid polytope is one-dimensional or smaller. If not, it is enough to show the procedure contradiction - free on REF - dimensional faces, which is a simple check. For part two, as observed when REF - REF were stated, it is enough to check axioms on these faces also, and this too is immediate. It remains only to show that choosing a different fundamental basis gives results respecting REF . The only possible points of concern are around short edges, since long edges give no freedom of sign choice if the result is a oriented Lagrangian matroid, as we have shown it must be. Notice that where MATH are the ends of a short edge, that is MATH for some MATH, REF yields MATH regardless of what signs we are given relative to MATH. Fix some MATH, and take all signs as being calculated from the signs relative to MATH decided as above through REF . Let us suppose the edge is directed from MATH to MATH. We first show MATH regardless of signs relative to MATH. Now, MATH . But if MATH is lower than MATH, then MATH and MATH, as the edge is non-inducing, so MATH is positive as required. If MATH is higher then MATH, MATH and MATH, so MATH remains positive. Now consider some basis MATH as fundamental. To complete the proof of this point, we must show that MATH if and only if MATH is lower than MATH relative to MATH. But MATH from the above and REF . But notice, as before, that MATH is REF when MATH lies below MATH relative to MATH and REF otherwise, which completes our proof. Finally, for part three, observe once more that only short edges need give us concern. From REF we obtain MATH whenever MATH are the ends of a short edge, and so we direct the edge according to which of the two is positive. This then obviously gives the correct oriented polytope. |
math/0010239 | By the inductive construction in REF above we showed that any MATH is of the form MATH for MATH as in the proposition (and that these MATH's all lie in MATH). We also showed that MATH for all MATH, so in particular MATH, since MATH. Therefore MATH is precisely the set of these MATH. |
math/0010241 | We prove REF together by showing that for MATH as in REF and MATH, if MATH then MATH and MATH for some MATH. With these hypotheses, choose any vertex MATH such that MATH is maximum. Then, since MATH we see that MATH . Since there are MATH terms on the right side (considering MATH as the multiplicity of the term MATH) and each of these has absolute value at most MATH, it follows that MATH and MATH for all MATH such that MATH. Now, we may repeat this argument with any such vertex MATH in place of MATH, et cetera. Since MATH is strongly connected, it follows that MATH and MATH, completing the proof. |
math/0010241 | If MATH has a single vertex then the result is trivial, so assume MATH. Now, since MATH is strongly connected, the matrix MATH is invertible. Since MATH is one-dimensional, MATH is also one-dimensional; hence there is a unique (nonzero) vector MATH such that MATH and MATH. Then MATH, so MATH is the stationary distribution of the NAME chain represented by the stochastic matrix MATH. Since MATH is strongly connected, every state of this NAME chain is recurrent, so every entry of MATH is positive. Since MATH solves the system MATH, which has integer coefficients, every entry of MATH is rational. Finally, there is a unique positive integer multiple of MATH which gives the vector MATH with the desired properties. |
math/0010241 | Let MATH be a list of the strong components of MATH such that if there is a directed edge from MATH to MATH, then MATH. We prove the claim by induction on MATH. For the basis of induction (MATH), and for the induction step (MATH), we may assume that MATH for all MATH such that MATH is a non-terminal strong component of MATH. If MATH is a terminal strong component of MATH then there is nothing to prove. Otherwise, there is at least one directed edge with initial vertex in MATH and terminal vertex not in MATH. Therefore, MATH for some nonzero diagonal matrix MATH of nonnegative integers. Now, since MATH for all MATH such that MATH non-terminal, MATH and by REF we conclude that MATH. |
math/0010241 | Let MATH. By REF , MATH if MATH is not a terminal strong component of MATH. Hence, if MATH is a terminal strong component of MATH, then MATH so that MATH . Conversely, if MATH for each terminal strong component of MATH then the MATH defined by MATH is in the kernel of MATH. |
math/0010241 | Let MATH be a reduction sequence in MATH, and let MATH be the Laplacian matrix of MATH. The first claim is that we may apply elementary row and column operations to MATH, involving only rows MATH and columns MATH, so that these rows and columns of the resulting matrix induce a MATH-MATH identity matrix. We prove this by induction on MATH, the basis MATH being clear. For the induction step, since MATH is a reduction sequence of length MATH, the induction hypothesis gives elementary operations involving only rows MATH and columns MATH, which, when applied to MATH, result in a matrix MATH in which these rows and columns induce a MATH-square identity matrix. By the last condition defining a reduction sequence, we have either MATH for all MATH, or MATH for all MATH. Examining the way in which MATH was obtained from MATH, we see that either MATH for all MATH, or MATH for all MATH. Now, elementary row or column operations can be used to cancel any nonzero entries MATH or MATH with MATH, and the value of MATH is left unchanged. Finally, multiplying row MATH by MATH if necessary produces the MATH-MATH identity submatrix, as claimed. Having produced this MATH-MATH identity submatrix, we use elementary column operations to zero out all entries in rows MATH except for the MATH-s in the MATH positions MATH. Then we use elementary row operations to zero out all entries in columns MATH except for the MATH-s in the MATH positions MATH. The result is a matrix, equivalent to MATH, which is also equivalent to a matrix with the block structure MATH for some MATH-square matrix MATH. Therefore, MATH occurs at least MATH times in the NAME normal form of MATH. Considering a reduction sequence of maximum length MATH completes the proof. |
math/0010241 | To show that MATH is injective, we must show that if MATH is such that MATH, then MATH. Accordingly, assume that MATH and MATH satisfy MATH. Let MATH. Then MATH . The entries of MATH are rational numbers, but we need to find MATH such that MATH. To do this, notice that MATH so that MATH. Since MATH is strongly connected, the kernel of MATH is MATH, and this implies that MATH for some MATH. Now MATH, and since every entry of MATH is an integer, every entry of MATH is an integer. Finally, since MATH it follows that MATH, which shows that MATH and completes the proof. |
math/0010241 | Let MATH be the vector of vertex activities of MATH. Use elementary column operations to add column MATH to column MATH for all MATH. Then use elementary row operations to add MATH times row MATH to row MATH for all MATH. The resulting matrix MATH is equivalent to MATH and has the required form, since MATH. |
math/0010241 | Let MATH be the submatrix of MATH obtained by deleting the row and column indexed by MATH; so we have MATH for some row vector MATH and column vector MATH. Since MATH, we have MATH . Similarly, let MATH be the submatrix of MATH obtained by deleting the row and column indexed by MATH; so we have MATH for some row vector MATH and column vector MATH. Since MATH is a simple vertex of MATH, REF implies that MATH. Now, since MATH we see that MATH. Also, since MATH is a simple vertex of MATH we see that MATH (here, MATH). Since MATH we have MATH, from which the result follows. |
math/0010241 | For MATH, let MATH be the row vector with entries MATH for each MATH, and let MATH be the column vector with entries MATH for each MATH. For MATH, let MATH be the row vector with entries MATH for each MATH, and let MATH be the column vector with entries MATH for each MATH. Then the matrices MATH and MATH have the block forms as shown: MATH . Here, MATH is the submatrix of MATH induced by rows and columns in MATH, MATH is the submatrix of MATH induced by rows and columns in MATH, and MATH for MATH. Since both MATH and MATH are simple vertices of MATH, an argument analogous to the proof of REF shows that MATH and MATH are equivalent to MATH respectively, in which MATH. Let MATH denote the vector of activities of MATH, so that MATH. Then the sum over all MATH of MATH times the MATH-th row of MATH is equal to MATH, since both MATH and MATH are simple in MATH. Since the columns of MATH sum to MATH, the columns of MATH sum to MATH. Now, for MATH, add MATH times row MATH of MATH to the row indexed by MATH. Then, for MATH, add column MATH of the resulting matrix to the column indexed by MATH. The result is the matrix MATH in which MATH . Finally, by multiplying the last MATH rows and columns of MATH by MATH, we obtain the matrix MATH. This shows that MATH and MATH are equivalent, so that MATH and MATH. |
math/0010241 | In particular, the naturality condition must hold when MATH and MATH is in the automorphism group MATH of MATH. In this situation, the assignment MATH gives a representation of MATH acting on MATH and the assignment MATH gives a representation of MATH acting on MATH. (All the representation theory we need is in REF.) Commutativity of the diagram means that MATH is a MATH-equivariant isomorphism, so these two representations are linearly equivalent. A representation of a finite group is determined up to linear equivalence by its group character, so MATH exists if and only if the characters MATH and MATH of these two representations are equal. Since these are permutation representations, their characters are given by counting fixed points; that is, for each MATH, MATH and MATH respectively. Thus, to show that a natural construction of MATH is impossible, it suffices to find a connected undirected graph MATH and automorphism MATH such that MATH. This is easy: let MATH be a circulant graph with at least three vertices, and let MATH be a cyclic permutation of all of MATH. Every spanning tree of MATH has both leaves and non-leaf vertices, and so it can not be left fixed by MATH. Thus, MATH. On the other hand, the MATH element of MATH is such that MATH, since MATH is a group automorphism. Thus, MATH, and since MATH it follows that MATH does not exist. |
math/0010241 | If MATH is nonnegative and MATH is legal for MATH, then MATH is nonnegative. From this it follows that if MATH is nonnegative then every configuration in MATH is nonnegative. More generally, for any configuration MATH, define the configuration MATH by MATH and MATH. If MATH is a legal sequence for MATH which does not contain MATH, then it is a legal sequence for MATH. Since MATH is nonnegative, it follows that MATH is nonnegative. That is, MATH is nonnegative for all MATH. For each MATH, let MATH denote the length of a shortest directed path from MATH to MATH in MATH, and let MATH. For each configuration MATH, define the `label' of MATH to be MATH, in which MATH. Define a total order MATH on MATH as follows: MATH if and only if either MATH, or MATH and MATH, MATH,MATH, MATH for some MATH. Notice that MATH has the same order type as MATH. Also notice that if MATH and MATH is legal for MATH, then MATH . It follows that, for any configuration MATH, the set MATH is finite. But for any MATH, the set of configurations MATH is also finite. These two observations suffice to show that MATH is finite. |
math/0010241 | This follows immediately from the facts that the off-diagonal elements of MATH are nonpositive, and that a vertex MATH is legal for a configuration MATH if and only if MATH. |
math/0010241 | We proceed by induction on MATH, the length of MATH. For the basis of induction, MATH, assume that MATH is legal for MATH. If MATH then the empty sequence is legal for MATH, as required. Otherwise, write MATH for some sequence of vertices MATH. Since MATH, MATH does not occur in the sequence MATH. The previous lemma and induction on MATH now show that MATH is legal for MATH, as required. For the induction step, assume the result for sequences of length MATH, and consider MATH. First, assume that MATH, and consider MATH and MATH. Then MATH and MATH is legal for MATH. Applying the induction hypothesis to MATH, MATH, and MATH, we see that MATH is legal for MATH. For the remaining case, assume that MATH, so that MATH. As in the basis of induction, since MATH is legal for MATH, MATH is legal for MATH. Now apply the induction hypothesis to MATH, MATH, and MATH to conclude that MATH is legal for MATH. Hence, MATH is legal for MATH, completing the induction step and the proof. |
math/0010241 | Let MATH be the graph with vertex-set MATH and directed edges MATH when MATH is legal for MATH. Then MATH is a nonempty graph, and, by the proof of REF , since MATH for all MATH contains no directed cycles. Therefore, MATH has at least one sink vertex, which is a stable configuration on MATH. Now suppose that MATH and MATH are two stable configurations in MATH. Let MATH and MATH be sequences of vertices which are legal for MATH, do not contain MATH, and are such that MATH and MATH. Let MATH and let MATH. The hypothesis of REF is satisfied, so produce the subsequence MATH of MATH as in that lemma. Now, since MATH is a legal sequence for MATH which does not contain MATH, and since MATH is stable, it follows that MATH is the empty sequence. From the construction of MATH, it follows that MATH for each MATH. By symmetry, we may repeat this argument with MATH and MATH interchanged, and deduce that MATH for each MATH. Finally, since MATH we conclude that MATH, finishing the proof. |
math/0010241 | For REF , let MATH be a critical configuration, and let MATH be a nonempty legal sequence of vertices for MATH, such that MATH. Then MATH is a nonzero vector of nonnegative integers such that MATH, so that MATH. By REF , it follows that MATH is a positive integer multiple of the vector MATH of vertex activities. Every MATH is fired in the sequence MATH exactly MATH times, hence at least once, and so MATH for all MATH. For REF , we use the strategy of the proof of REF , but with a different definition of the label of a configuration. For a configuration MATH, define MATH as in the proof of REF . If MATH, then for each MATH, let MATH denote the length of a shortest directed path which begins at MATH and ends at a vertex MATH such that MATH, and let MATH be the maximum value of MATH for all MATH such that MATH. For MATH, let MATH, and define the `label' of MATH to be MATH. Define a partial order on the set of all labels as follows: MATH if either MATH and MATH is nonnegative (except at MATH), or MATH (in which case MATH) and MATH, MATH,, MATH, MATH for some MATH. Notice that for any configuration MATH with MATH nonnegative, there are only finitely many stable configurations MATH such that MATH (in fact, the number of them is MATH). It follows that the set of all stable configurations, partially ordered by the order MATH on their labels, has no infinite ascending chains. Now, if MATH is legal for MATH and MATH, then MATH, even when MATH. Hence, if MATH is stable but not nonnegative, then MATH. Since there are no infinite ascending chains, there is a nonnegative integer MATH such that MATH is maximal, and this must be a stable, nonnegative configuration. For REF , let MATH be any stable configuration. By REF , we may assume that MATH is nonnegative. There are exactly MATH nonnegative stable configurations on MATH, and the successor function acts as an endofunction on this set. Since this set is finite, for any MATH in it there exists a nonnegative integer MATH and positive integer MATH such that MATH. That is, MATH is critical. |
math/0010241 | That coevalence is an equivalence relation is easy to see. Let MATH be a critical configuration on MATH, let MATH be a nonempty sequence of vertices which is legal for MATH, such that MATH, and as short as possible subject to these conditions, and let MATH. As in the proof of REF , we see that MATH for some positive integer MATH. Since the bank vertex occurs MATH times in the sequence MATH it follows that MATH is coeval with exactly MATH critical configurations. |
math/0010241 | Let MATH be the vector of vertex activities of MATH. Let MATH be any critical configuration of MATH, and let MATH be a nonempty sequence of vertices which is legal for MATH and such that MATH. Let MATH be the multiplicity vector of MATH; as in REF we have MATH for some positive integer MATH. Since MATH is stable, MATH. If this is the only occurrence of MATH in the sequence MATH, then MATH and it follows that the coevalence class of MATH is the singleton MATH. Otherwise, let MATH be all the indices from MATH to MATH such that MATH, and let MATH. Since MATH and MATH, we see that MATH. Now, MATH is a legal sequence for MATH which does not contain any occurrence of MATH. Applying REF with MATH, let MATH be the sequence so produced. This sequence is legal for MATH, and since MATH is stable, it follows that MATH is the empty sequence. Therefore, for each MATH, the multiplicity of MATH in MATH is at most MATH. Since MATH is critical, we may repeat the argument of the above paragraph, using MATH and MATH. As above, we conclude that for each MATH, the multiplicity of MATH in MATH is at most MATH. For each MATH, let MATH. Applying the above argument for each sequence MATH, we see that MATH for each MATH and MATH. But MATH, from which it follows that MATH. Therefore, MATH, so that MATH, and it follows that the coevalence class of MATH is the singleton MATH, completing the proof. |
math/0010241 | For MATH the element MATH of MATH is in MATH if and only if MATH for some positive integer MATH. Since every column MATH of MATH satisfies MATH, if MATH is in MATH then MATH. Conversely, if MATH then MATH for some rational MATH-indexed vector MATH, since the rank of MATH is MATH. Hence, MATH for some positive integer MATH, so that MATH is in MATH. This proves that MATH. For the second claim, consider two configurations MATH and MATH on MATH, so that MATH. If MATH then there are nonnegative integers MATH such that MATH. Therefore, there are MATH-indexed vectors MATH of nonnegative integers such that MATH, so that MATH is in MATH. This proves that the map MATH is injective. Since this map is clearly surjective, the theorem is proved. |
math/0010242 | Note that MATH is invertible on MATH and MATH. Let MATH with MATH, then MATH. In the same way we get MATH, hence the matrices MATH are invertible and uniformly bounded by MATH and MATH . REF yields that MATH strongly. |
math/0010243 | Since MATH is positive definite we have MATH and by REF operators CITE MATH . We denote by MATH the set of all formal series MATH for which MATH . It follows from REF that MATH is a NAME algebra with respect to the multiplication (discrete convolution) MATH where MATH and MATH. By REF on page REF an element of MATH has an inverse in MATH if it is not contained in a maximal ideal of MATH. Any maximal ideal of MATH consists of elements of the form (compare REF) MATH where MATH with MATH and MATH . Due to REF we get MATH, hence MATH. Thus a necessary and sufficient condition for an element in MATH to be not contained in a maximal ideal of MATH is MATH for all MATH. By REF is positive definite, hence MATH for all MATH and REF follows. REF are now clear, since in both cases we can easily find a weight function such that REF are satisfied. |
math/0010243 | We have MATH . To prove REF we note that by REF there exists a MATH such that MATH with a constant MATH depending only on MATH and on the condition number of MATH. Since MATH we get MATH for all MATH. That means we can choose MATH independently of MATH. Write MATH and note that MATH for MATH. Since the non-zero entries of MATH decay exponentially, it is easy to show that there exists a MATH with MATH such that MATH for some constant MATH. It is trivial that MATH also decays exponentially for MATH. We absorb MATH and the other constants in the constant MATH and get the desired result. For the proof of REF we proceed analogously to above by applying REF to conclude that MATH for some constant MATH depending only on MATH and on MATH. The norm MATH can be estimated via MATH similarly for MATH where MATH. Since all arising constants - absorbed in one constant MATH - depend only on MATH and on the exponent MATH, the proof is complete. |
math/0010243 | CASE: It is well-known CITE that the eigenvalues of MATH are given by MATH . For REF that means they are regularly spaced samples MATH of the function MATH, which immediately yields the left hand side of REF . In order to prove the right hand side of REF it is sufficient to estimate MATH where MATH and MATH. Define the sequence MATH by MATH if MATH and MATH if MATH. There holds MATH . Relations REF follow now from this estimate. REF can be proved similarly by using MATH and applying the corresponding decay properties to REF. |
math/0010243 | CASE: Set MATH and MATH. By REF operators CITE MATH is given by MATH . By REF we can easily find a MATH such that MATH is invertible for all MATH, which implies that MATH. In this case by the properties of circulant matrices CITE the entries MATH of MATH can be computed as MATH . Now consider MATH . We estimate the expression above in two steps: CASE: We first consider REF. Note that MATH and MATH . By setting MATH we get MATH and MATH . Hence MATH . By REF MATH implies MATH. Hence we get for MATH . CASE: Now we estimate REF: MATH . By REF we can easily find for any MATH a MATH such that for all MATH there holds MATH. Thus MATH for some MATH and some constant MATH. By combining REF, and REF and hiding expressions as MATH in the constant MATH, we obtain estimate REF. CASE: The proof of REF is similar to the proof of REF. The only steps that require a modification are REF. By REF MATH implies MATH. Hence we can estimate MATH as follows. MATH . By adapting REF to the case of polynomial decay we can estimate REF by MATH . Combining REF with REF yields the desired result. |
math/0010243 | Similar to REF we write MATH . Using the NAME complement CITE we can write MATH as MATH . We will first show that the entries of the matrix MATH are exponentially decaying off the corners of the matrix. Note that REF implies MATH where MATH. We analyze the decay behavior of MATH in two steps by considering first MATH and then MATH. Recall that MATH and note that MATH. By REF we know that MATH for some MATH, where MATH depends on MATH and on MATH and MATH, but is independent of MATH. We set MATH. There holds MATH . For simplicity we will absorb any constants arising throughout this proof that depend solely on MATH (or MATH) in the constant MATH. We analyze the sum REF further by splitting it up into three parts and in addition consider first the entries MATH with MATH. CASE: MATH: MATH . CASE: MATH: MATH . CASE: MATH: MATH . Similar expressions can be obtained for the case MATH by interchanging the roles of MATH and MATH in the derivations above. Thus MATH . We now estimate the decay of the entries of MATH. Since MATH is hermitian, it is sufficient to consider only the entries MATH with MATH. There holds MATH . As before we proceed by splitting up this sum into three parts. CASE: MATH: CASE: MATH: CASE: MATH: MATH . Hence, by combining REF , and REF we get MATH . The entries of MATH satisfy MATH . By REF there exists a MATH and a constant MATH depending on MATH and on MATH and MATH such that MATH . Hence MATH satisfies MATH. Set MATH. After some lengthy but straightforward computations we get MATH for some MATH. Hence MATH, which together with REF completes the proof. |
math/0010243 | First note that by REF we can always find a MATH such that MATH exists for all MATH. There holds MATH where MATH is a MATH finite section of MATH. The result follows now by applying REF . |
math/0010243 | The proof is similar to that of REF . To avoid unnecessary repetitions we only indicate the modifications, that are required. By REF MATH is invertible. Note that MATH is a matrix with three bands, one band is centered at the main diagonal, and the two other bands of width MATH are located at the lower left and upper right corner of the matrix. It follows from REF that the entries of MATH decay exponentially off the diagonal and off the lower right and upper left corner. More precisely, MATH where MATH is as in REF with MATH. With this result at hand it is easy to show that the entries of MATH decay exponentially. MATH has a simple form, it is a NAME matrix with first row MATH . When we compute MATH the non-zero entries of MATH are multiplied by exponentially decaying entries due to the exponential decay of MATH. This leads to the estimate MATH. |
math/0010243 | By definition of MATH we have MATH where MATH. We know from REF in the proof of REF that the entries of MATH can be bounded by MATH . For MATH and MATH we define the orthogonal projection MATH by MATH and identify the image of MATH with MATH. We set MATH. In words, MATH is obtained from MATH by taking only the central MATH submatrix of MATH and setting the other entries surrounding this block equal to zero. Then MATH has MATH ``full" rows and MATH ``sparse" rows, where each of the latter rows has non-zero entries only at the first MATH and last MATH coordinates. Thus the dimension of the space spanned by the sparse rows is at most MATH. Hence MATH. Due to the decay properties of MATH it is easy to see that MATH . Using REF we get after some straightforward calculations MATH . It is obvious that for each given MATH we can find a MATH such that for all MATH. Since MATH is hermitian, we have MATH. Thus MATH . Hence for large MATH the spectrum of MATH lies in MATH. By the NAME interlace theorem we conclude that at most MATH eigenvalues of MATH have absolute value exceeding MATH. |
math/0010244 | First we show that REF implies that the entries of MATH satisfy MATH for some MATH and some constant MATH. It is clear that MATH and MATH . Now MATH for some MATH. Set MATH, then MATH for some constant MATH depending on MATH. Similarly MATH . By combining REF and taking the square root we get MATH . This together with REF yields REF. Now consider MATH . Note that MATH is a matrix that has a finite number of columns and a biinfinite number of rows, of which the rows with index MATH are zero. By REF the entries of MATH and the nonzero rows of MATH decay exponentially off the diagonal. Relation REF implies MATH . Using REF it follows that there exists a MATH and a constant MATH depending on MATH and on the condition number of MATH, such that MATH and the same is true for MATH. Now it is easy to see that there exist MATH and MATH such that MATH . |
math/0010244 | Set MATH and MATH. Since MATH is positive definite we have MATH and by REF operators CITE MATH . The property MATH is equivalent to MATH. By REF on page REF an element of MATH has an inverse in MATH if it is not contained in a maximal ideal. Any maximal ideal of MATH consists of elements of the form (compare REF) MATH where MATH with MATH and MATH . Due to REF we get MATH, hence MATH. Thus a necessary and sufficient condition for an element in MATH to be not contained in a maximal ideal is MATH for all MATH. By REF is positive definite, hence MATH for all MATH, consequently MATH. |
math/0010244 | The assumption MATH implies MATH and MATH for MATH. Denote MATH . MATH is periodic in MATH with period MATH. Since MATH we also get MATH, compare REF . In words, MATH is a block NAME operator. The frame property implies that MATH is hermitian positive definite and MATH . The submultiplicativity of the function MATH implies that the spaces MATH, MATH, and MATH are NAME algebras under convolution (compare CITE). Hence MATH, and MATH. By REF MATH . The dual MATH can be expressed as (see for example, CITE) MATH . This equation together with REF and the assumption MATH yields that MATH. |
math/0010244 | CASE: The proof of this result is similar to the proof of REF . We only indicate the necessary modifications. Let MATH be the matrix-valued defining function of a hermitian positive definite NAME operator MATH and denote MATH. It follows from the basic properties of block NAME operators that MATH for all MATH and MATH . The fact that MATH is self-adjoint positive-definite implies that MATH is in the same algebra as MATH, see CITE. Thus if MATH (and consequently MATH) for weights satisfying REF, then MATH. The rest follows now by repeating the steps in REF and using the remark following REF . CASE: In this case the frame operator MATH is a block NAME operator (see CITE). The assumption MATH implies that MATH. Along the same lines as above and in REF it follows that MATH and MATH and consequently MATH, similar for MATH. |
math/0010245 | Without loss of generality assume MATH, otherwise we can always set MATH. Note that MATH where MATH and MATH. Define MATH and MATH. Observe that MATH and MATH . For fixed MATH there holds MATH as MATH. Furthermore, for each MATH we have that MATH for MATH. This together with REF implies that MATH strongly for MATH. |
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