paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0101128 | In the two examples we showed that the even shift (one sided even shift) can not be represented as a RES (IES) in our standard model: the Bakers map (the doubling map). To prove REF we will show that if the even shift is representable as a RES for some Axiom A diffeomorphism then it is in fact an RES for the Bakers map. The proof of part one is similar. Let MATH be the Bakers map with the standard NAME partition. We suppose by way of contradiction that MATH is a RES for some Axiom A diffeomorphism MATH with respect to the NAME partition MATH. Consider the SFT MATH defined by MATH and denote the projection MATH. By assumption there is a rectangle like hole MATH, that is the set MATH of points which never fall into MATH satisfies MATH. By the proof of REF we can find an rectangle like hole MATH which yields MATH as a RES for MATH. Let MATH be the invariant set of points which never fall into MATH and let MATH be the projection. Consider MATH and MATH. Since the maps MATH and MATH preserve local product structure the set MATH is a rectangle like hole. If MATH is open then MATH is a RES in the Bakers map, a contradiction. If MATH is not open then the interior of MATH is also a rectangular like hole, which using the convention defined in REF shows that MATH is an RES for the Bakers map, a contradiction. |
math/0101128 | We begin by describing a one dimensional construction from CITE. Consider the dynamical system MATH given by MATH . Suppose MATH and consider the ``partition" MATH and MATH. Given MATH we say MATH if there exists a positive integer MATH such that MATH for MATH and MATH. It is well known that the code MATH of the orbit of MATH satisfies MATH for all MATH . The set of all sequences which are the code of some orbit is called the one sided MATH - shift, it is characterized by CITE MATH . Let MATH be defined by MATH mod MATH . Inspired by the above a number MATH is called a MATH - number if MATH for all MATH . If MATH is a MATH - number then the set MATH is conjugate to the one sided MATH - shift. If MATH is the interval MATH then the set of points whose forward MATH orbit never falls in MATH is exactly MATH . Now let MATH be the Bakers map. Fix MATH a MATH number and let MATH be the rectangular hole MATH . Let MATH . Since the hole stretches from the bottom to the top of MATH it is easy to see that MATH if and only if MATH for all MATH and all MATH . Thus the exclusion shift in this example is exactly the natural extension of the one sided MATH - shift. We remark that the two sided MATH - shift is of finite type, sofic, or not sofic if and only if the one sided one has the same property. It is well known that the MATH - shift is of finite type if the binary expansion of MATH is finite, it is sofic if it eventually periodic and otherwise it is not sofic CITE. |
math/0101128 | Let MATH be an exclusion subshift, that is, MATH is an Axiom A map, MATH a fixed proper, generating NAME partition, and MATH an open hole with compact boundary. Then MATH is also a proper, generating NAME partition. Let MATH . Clearly MATH and MATH. Furthermore, since MATH is proper we have MATH as MATH thus MATH . The ``exclusion" system MATH is a SFT. Clearly MATH and MATH . Clearly any topologically transitive component which is a subset of MATH is non boundary supported since periodic points are dense in SFTs. On the other hand if MATH then MATH is not an accumulation point of periodic points MATH which avoid MATH since any such periodic point necessarily belongs to MATH for some MATH. Thus we have shown that every non boundary supported topologically transitive component is contained in MATH. If each MATH was topologically transitive then MATH would be coded. Let MATH be the topologically transitive components of MATH. The MATH form a filtration in the sense that for each MATH there exists MATH such that MATH. In other words the transitive components MATH form an at most countable union of directed trees with each nodes out degree is exactly one. The equality in REF implies that each topologically transitive component of MATH contains a set MATH where the union is taken over a path in one of the trees (we call this a path limit). Noticing that such a path is uniquely defined by the root of the tree since the out degree is always one, implies that there are at most countably many such paths and thus MATH has at most countably many topologically transitive components. This finished the proof of the countability of the claim of the theorem. Suppose MATH is a topologically transitive component of MATH. To see that MATH is coded we will define a new filtration. Since MATH is a subshift of finite type it has finitely many topologically transitive components MATH which are pairwise disjoint. Consider those components MATH which are strictly contained in MATH. We can assume that the MATH are so ordered that MATH for all MATH . We only need to show that MATH . The rest of the proof is devoted to establishing this equality. If REF is not true there is a MATH such that for all MATH we can find another MATH which we denote without loss of generality MATH such that MATH for all MATH but MATH. In the terminology introduced above this means we can find two disjoint paths in the trees whose path limits are both contained in MATH. Fix MATH . For MATH consider a finite word MATH where each symbol and each transition which characterize MATH appear in MATH . Since MATH is topologically transitive there is a point MATH where the words MATH and MATH both appear in MATH . Thus we can find MATH (which we assume positive without loss of generality) so that MATH . Consider the point MATH with symbolic coding MATH . We can assume without loss of generality that MATH for MATH . From our assumptions for any sufficiently large MATH and for MATH the rectangles MATH defined by MATH are disjoint from MATH. We can form a new SFT, MATH, with alphabet the MATH and all the transitions made by the orbit segment MATH allowed. This is clearly a topologically transitive SFT. It also contains MATH for MATH since any legal transition in these sets is a legal transition in MATH. But, by the construction of MATH we have MATH for sufficiently large MATH and thus MATH and MATH a contradiction. |
math/0101128 | CASE: Let MATH denote the topologically transitive component. If MATH is isolated and the orbit of MATH is dense then MATH for some MATH. The point MATH is also isolated, for if not then by continuity MATH is not be isolated. If the orbit of MATH is not periodic then since MATH is isolated we have MATH and MATH is not forward invariant. This contradiction implies that MATH is periodic. Since MATH's orbit is dense it must coincide with the topologically transitive component. CASE: A component which is not finite can not contain any isolated points. It is a simple exercise to show that a closed set without isolated points can not be countable. CASE: If MATH is finite then this is clear. Suppose MATH is uncountable and the forward orbit of MATH is dense in MATH. Since MATH is not isolated the forward orbit of MATH must come arbitrarily close to MATH to be dense. |
math/0101128 | For all MATH let MATH be the maximal interval containing MATH which is a subset of MATH, we refer to MATH as a gap. Let MATH be the smallest positive integer such that MATH . Let MATH be the index such that MATH . Let MATH be the set of MATH such that the orbit of MATH falls into the same component MATH of the maximal hole at the same time, that is, MATH and MATH. We claim that the continuity of MATH implies that MATH . Indeed, if this is not the case, then MATH . Since MATH this contradicts the definition of the gap MATH. Until further notice we assume that MATH is a SFT. Each topologically transitive component MATH is closed, thus we can define MATH and MATH. Since MATH is a SFT it has only a finite number of topologically transitive components MATH and these components are disjoint. Until further notice we also suppose that all the points MATH and MATH have unique coding (that is, the orbits of MATH and MATH do not intersect MATH). With this additional assumption the disjointness of the MATH implies that MATH and MATH do not belong to any transitive component other than MATH. This implies that there are gaps MATH REF on the left (right) side of MATH (MATH). Fix MATH. As we saw above all MATH between MATH and MATH fall into the same hole at the same time MATH. By the continuity of MATH this implies that MATH. A similar statement holds for MATH. Consider the ordering on MATH which is compatible with the ordering on MATH. This ordering can always be defined in an inductive manner by simply considering the relative order of the elements of MATH. (if MATH is locally order preserving then this is simply the lexicographical order on MATH ). Fix a MATH . Consider the symbolic coding of MATH and MATH. Call these codings MATH and MATH (here the dependence on MATH is suppressed since MATH is fixed). We claim that if MATH is uncountable then MATH is not a preimage of MATH and vice versa MATH is not a preimage of MATH. To see this fix a higher block coding which defines NAME transition graph of MATH . The fact that MATH is defined via a minimum implies that if there are several followers of the symbol MATH in the NAME graph, then MATH is minimal follower in the sense that in the ordering it is smaller than all other followers. In a similar fashion the sequence MATH is maximal. If MATH is a preimage of MATH or vice versa, then MATH and MATH are eventually maximal and minimal at the same time. This means that from the point on that they agree there the maximal follower of MATH is also the minimal follower of MATH, so there is only one follower of MATH. Thus MATH and MATH are eventually periodic and the MATH is simply this periodic orbit. In particular MATH is finite, finishing the proof of the claim. If MATH is uncountable, then we have just shown the disjointness of the codes of MATH and MATH. Since we are still assuming that the points MATH and MATH have unique coding this implies the disjointness of their MATH - orbits. Thus at least two point of MATH lie on MATH. On the other hand if MATH is at most countable, then by REF it is finite and consists of a single periodic orbit. The points MATH and MATH lie on this orbit. Under the above assumptions we have shown that for every uncountable MATH there are at least two points of MATH on the border of MATH and for every at most countable MATH there is at least one such point. Thus, since the endpoints can only belong to a single MATH we have shown MATH . Since MATH is at most REF to REF, if we drop the assumption that the MATH and MATH have unique coding then (at most) two symbolically disjoint topologically transitive components can share such a point when projected to the interval MATH. This can happen at most MATH times yielding the formula in the statement of the theorem in the case that MATH is a SFT. Finally if MATH is not a SFT we approximate MATH by NAME holes MATH in the same way as in REF . Arguing similarly to the proof of REF we have MATH and the difference consists of BSCs. Note that the number of transitive component of MATH is less than or equal to the limsup of the number of components of the approximating sequence. It remains to bound the number of BSCs. Let MATH be the number of points in MATH which are boundary points of non BSCs. By the definition of a BSC such a point can not belong to a BSC. We claim that the number of BSCs is at most MATH. Let MATH be the set of points in MATH which belong to a BSC. By definition of BSCs MATH must be disjoint from MATH or else they would be approximable by periodic points. Also by definition every MATH must belong to MATH. If two holes MATH and MATH amalgamate to a single MATH then two points in MATH belong to MATH and (that is, their orbit falls into MATH) and thus can not belong to MATH. Thus the cardinality of MATH is at most MATH. Clearly any BSC MATH contains at least one of the points MATH. We will associate with each BSC MATH a subset MATH of MATH of ``insertable boundary points". We claim that two distinct BSCs must have nonintersecting sets MATH. Once this claim has been established it immediately follows that the number of BSCs is at most MATH which will complete the proof of the theorem. We turn to the proof of the claim. Fix MATH a BSC. Let MATH. Note that MATH is uniquely defined by the convention we made in REF. Fix MATH such that the orbit of MATH is dense in MATH. Consider any MATH such that MATH. For each MATH we define MATH in the following way: MATH is the longest initial block of one of the MATH which agrees with the initial block of the same length of MATH. We call MATH the flag of MATH at time MATH. We remark that MATH if MATH. By REF either MATH is an isolated periodic point which hits MATH or it visits MATH only a finite number of times. In the second case since by the proposition MATH is recurrent, we can assume by replacing MATH by MATH for sufficiently large MATH that the orbit of MATH does not visit MATH at all. We remind that MATH is fixed and thus we drop the MATH dependence of the notations. We say that a flag MATH is covered by another flag if there is a MATH such that MATH (see REF ). If a flag is not covered by any other flag we say it is uncovered. The motivation for this terminology is the following fact: if the flag of MATH is uncovered at time MATH then we can concatenate MATH's initial segment of length MATH with any orbit in MATH whose initial segment agrees with the next MATH entries of MATH and the concatenated orbit belongs to MATH (it may be wandering and thus not belong to MATH ). We claim that there are infinitely many indices MATH such that the flags at these times are uncovered. To see this this consider MATH. Clearly there must be a MATH such that MATH. Let MATH be the smallest such MATH, this flag must be uncovered. Arguing inductively the flags at times MATH are uncovered. For each MATH we define the color REF of MATH at time MATH to be the set of MATH such that the initial blocks of length MATH of MATH and MATH coincide. We assume that we start with a fine enough NAME partition that each MATH belongs to a different element of the time zero partition. This implies that the color of MATH is unique for each MATH. Let MATH be the set of colors which occur infinitely often at uncovered times. Clearly this set is nonempty. We are now ready to prove the above claim. Consider MATH and MATH which have nonempty intersection, and let MATH belong to this intersection. We will recursively construct a point MATH such that the orbit of MATH is dense in MATH. To construct MATH fix a positive sequence MATH. Consider an initial segment of MATH of length MATH where MATH is the smallest MATH such that the orbit segment MATH is MATH dense in MATH. Consider the smallest MATH such that MATH and MATH agree at at least MATH places. Furthermore consider am initial segment of MATH of length MATH where MATH is the smallest MATH such that the orbit segment MATH is MATH dense in MATH. Consider the smallest MATH such that MATH and MATH agree at at least MATH places. These definitions guarantee that the sequence MATH belongs to MATH. Here the symbol MATH mean concatenation and MATH. Recursively repeating this construction produces the sequence MATH . By construction MATH and the orbit of MATH is clearly dense in MATH. In particular MATH in nonwandering and thus MATH. This finishes the proof of the claim. |
math/0101128 | Fix a generating NAME partition MATH and let MATH. Let MATH. If MATH and MATH then since MATH is generating by continuity there is a MATH such that MATH. Since MATH is compact we can cover MATH by a finite collection of the sets MATH to obtain a neighborhood MATH of MATH such that MATH . Since we used a finite collection of MATH, the hole MATH consists of a finite union of element of MATH for some sufficiently large integer MATH and thus MATH is a SFT. |
math/0101128 | Consider the set MATH . Since MATH is totally disconnected there are holes MATH such that MATH. Since MATH such a MATH satisfies the requirements of REF and so defines a SFT. Next we will show that an open set MATH of holes satisfy the requirements of REF . By compactness of MATH we can find an open set MATH with MATH and a positive integer MATH such that for all MATH there exists an integer MATH, satisfying MATH such that MATH . Then, just as in the proof of REF there is a neighborhood MATH of MATH such that for each MATH for some MATH satisfying MATH we have MATH . This immediately implies that we can choose a small neighborhood MATH of MATH such that for any hole MATH the boundary of MATH satisfies the requirements of REF and thus the corresponding shift MATH is a SFT. |
math/0101128 | Consider an arbitrary hole MATH and the associated invariant set MATH. If MATH is a generic point in the sense that it visits (in both forward and backwards time) any cylinder set (defined by the NAME partition) with the correct frequency, then MATH and MATH completely fall into MATH and thus are both disjoint from MATH. Both MATH and MATH are curves which are dense in MATH, thus since MATH is two dimensional the complement of their union is totally disconnected (that is, no two points are in the same connected component). Since MATH is a subset of this set it is also totally disconnect. Apply REF finishes the proof. |
math/0101128 | Consider the set MATH of generic points in the same sense as in the proof of REF . The set MATH is of full NAME measure. Suppose MATH is a rectangle like hole with the property that every other corner point (those which are noted in the description of MATH) is in MATH . Such a hole satisfies the requirements of REF and thus MATH is a SFT. Clearly the set of such holes is dense and of full measure. The proof of openness is the same as in the proof of the previous theorem. |
math/0101128 | We call a polyhedral hole MATH large if the NAME dimension of the associated invariant set MATH is less than or equal to one. We note that if MATH then MATH . Thus if MATH is a large hole then so is MATH . Our proof is local. Fix an open set MATH. Consider the set MATH of points which never fall into MATH. Suppose MATH is large enough that MATH . This implies that MATH for almost every MATH REF . Here MATH denotes the orthogonal projection from MATH onto MATH, the line through the origin in the direction MATH . The term almost every MATH refers to the NAME measure on the MATH. Consider the set of MATH - dimensional polyhedra contained in a co-dimension one hyperplane with normal direction MATH. Such a polyhedron plays the role of a face of a polyhedral hole. Let MATH be the parameter on MATH . The projection of any such face onto MATH is simply a point MATH. The pair of parameters MATH determine a family of faces which lie in a common co - dimension one hyperplane. Suppose MATH is a generic direction, that is, MATH. Then for a.e. MATH, any polyhedral face normal to MATH will be disjoint from MATH . Now consider any polyhedron MATH which contains MATH in it's interior and for which all the faces are generic in the above sense, that is, they do not intersect the set MATH. This immediately implies that the boundary MATH satisfies the requirements of REF . The set of such polyhedra is clearly open, and locally dense, and locally of full measure. |
math/0101133 | Observe that MATH and MATH for all MATH. So one gets for every MATH . Next we have MATH . So it is clear that we can define a normal, unital and injective MATH-homomorphism MATH by MATH for all MATH, and then MATH satisfies MATH for MATH and MATH. It is obvious that MATH for all MATH. Further we have MATH . Both statements together give MATH for all MATH. Hence MATH is indeed an action of MATH on the NAME algebra MATH. The uniqueness statement is obvious. |
math/0101133 | By REF, so MATH . Next we have MATH . Both computations together give the formula stated in the proposition. |
math/0101133 | Since MATH, MATH commutes with the elements MATH if and only if MATH. Since MATH is faithful this equality is valid if and only if MATH. |
math/0101133 | Define MATH as above. Then one has for all MATH . Hence MATH is an action of MATH on MATH. Considering MATH as an isomorphism of MATH, we clearly have MATH for all MATH and MATH . So MATH; similarly one proves the converse inclusion. Hence MATH is an isomorphism of MATH onto MATH. Recalling the definition of the dual actions MATH and MATH and the unitary MATH implementing them, we only have to observe that MATH and MATH commute to get the formula MATH. |
math/0101133 | It is easy to check that MATH. Next we have to prove that MATH . So we have to prove that MATH . Starting the computation on the left hand side we get MATH . This computation proves our result. |
math/0101133 | First let MATH be stabilizable with a unitary MATH. Using the action MATH and the isomorphism MATH of REF we get the first two statements, because they are valid for ordinary actions REF . Recalling that MATH and using that MATH we get MATH . Now we define the following linear subspace of MATH . Choose MATH, MATH and an orthonormal basis MATH for MATH. Then we have with MATH-strong-MATH convergence MATH . So MATH and hence MATH. This proves the result for stabilizable actions. From REF the general result now follows. |
math/0101133 | Let MATH act on MATH. Choose MATH, MATH, a vector MATH and an orthonormal basis MATH for MATH. Define MATH . Then we get MATH . Because MATH is normal we get that MATH . This final computation proves the lemma. |
math/0101133 | Let MATH act on MATH by MATH, hence we may assume that MATH. Suppose first that MATH is stabilizable with the unitary MATH. Then we can construct the action MATH and the isomorphism MATH as in REF . Because MATH we get immediately that MATH, where MATH is the dual weight of MATH on MATH. Let MATH be the canonical NAME for MATH as defined in REF . Define for MATH . Then MATH is a NAME for MATH. The isomorphism MATH and REF give MATH is a MATH-strong-MATH - norm core for MATH and MATH for all MATH and MATH. By REF we can define a unique isometry MATH such that MATH for all MATH and MATH. Later we will show that MATH to get the second statement of the proposition. Let MATH and MATH the domain of the MATH-strong-MATH - norm closure of the restriction of MATH to MATH. Choose MATH, MATH, MATH and an orthonormal basis MATH in MATH. Define MATH and for every finite subset MATH . Then every MATH belongs to MATH and, using REF , we get MATH . So we observe that MATH converges in norm to MATH . Further MATH converges in MATH-strong-MATH topology to MATH. So we may conclude that MATH and MATH . In the beginning of the proof we saw that the elements MATH span a MATH-strong-MATH - norm core for MATH, so MATH is a MATH-strong-MATH - norm core for MATH and by REF MATH so MATH. Hence we have also proved the second statement of the proposition for stabilizable cocycle actions. Consider now the general case. Choose a n.s.f. trace MATH on MATH. Then MATH is a n.s.f. weight on MATH, and MATH is a stabilizable cocycle action of MATH on MATH. Writing MATH, it is clear that MATH and MATH. Then it is clear that the dual weight of MATH equals the tensor product weight MATH. Choose now a NAME MATH for MATH and let MATH be the canonical NAME for the tensor product weight MATH. By the previous part of the proof we get a NAME MATH for the dual weight MATH, such that CASE: MATH for all MATH, MATH and MATH. CASE: MATH is a MATH-strong-MATH - norm core for MATH. To deduce the second statement above from the first part of the proof, observe that the elements MATH with MATH and MATH span a MATH-strong-MATH - norm core for MATH. Choose now an arbitrary NAME MATH for the weight MATH. The lemma following this proposition shows that MATH is a MATH-strong-MATH - norm core for MATH. By REF there is a unique isometry MATH such that MATH for all MATH and MATH. Because the elements MATH span a core for MATH, MATH is unitary and we can define a new NAME MATH for MATH by MATH and MATH. Of course, the elements MATH still span a core of MATH and MATH for all MATH and MATH. Let MATH be the tensor product NAME for the weight MATH. Then MATH and MATH agree on a core for both MATH and MATH. Hence MATH, from where MATH. Then finally MATH, which concludes the proof. |
math/0101133 | Choose a NAME MATH for MATH. Because all NAME for MATH are isomorphic, we may assume that MATH . Choose now MATH. Fix an element MATH, MATH and fix elements MATH such that MATH. Choose then a net MATH in MATH such that MATH . Then MATH will be a net in MATH, converging MATH-strong-MATH to MATH. Further it is easy to verify that MATH and this converges in norm to MATH . So, indeed MATH is a MATH-strong-MATH - norm core for MATH. |
math/0101133 | Let us first suppose that the second statement is true and take a corresponding unitary MATH. Define MATH . Then we get for all MATH . So MATH. Next we define the unitary MATH in MATH by MATH from where MATH . Hence MATH. To show that MATH is a cocycle action of MATH on MATH, we compute, for all MATH: MATH . Further we have MATH . On the other hand, MATH . Hence we obtain the equality MATH . So, MATH is a cocycle action of MATH on MATH. We claim that the map MATH is such that MATH. To show this we use REF . First let us check that MATH. We have for all MATH: MATH . Hence we get MATH. To prove our claim it now suffices to observe that MATH for all MATH, where we used the fact that MATH in the last line of the computation. Next, for MATH we clearly have MATH and further MATH so we may conclude that MATH. Thus, we have proved that MATH. To finish the first part of the proof we only have to verify the formula MATH, but for MATH we get MATH . Here we used the notation MATH and observe that MATH for all MATH and MATH. So we have proved that the second statement of the proposition implies the first one. To prove the converse statement we have to show that for every cocycle action MATH of MATH on MATH there exists a unitary MATH such that MATH . Taking MATH an easy computation proves the result. |
math/0101133 | Denoting by MATH the maximal operator space tensor product, it follows from REF that the completely bounded maps from MATH to MATH can be identified with MATH. Combining this with REF it follows that the completely bounded maps MATH from MATH to MATH are in (completely isometric) one-to-one correspondence with elements MATH by the formula MATH . Let first one of the statements of REF be satisfied. Then take such a unitary MATH satisfying MATH. Define MATH for all MATH. Let MATH be an orthonormal basis for MATH and MATH. Then MATH so MATH . Further we have MATH . Thus, MATH which concludes the first part of the proof, because the remaining equality can be obtained in a similar way. Vice versa, let us have a map MATH. Then there exists an element MATH such that MATH for all MATH. The same computations as above prove that MATH, and that MATH is a unitary. |
math/0101133 | Choose MATH. We will prove that MATH. Observe that MATH . Using the previous lemma we immediately get that MATH. |
math/0101133 | Using the lemma we get MATH . The definition of MATH and the fact that MATH and MATH are matched imply MATH . From this we may conclude that MATH . Using the first formula of the proof we get MATH . Because for all MATH we have MATH we get MATH . Combining this with REF we get MATH . Finally MATH and hence we get MATH . |
math/0101133 | The first inclusion follows from REF and from the definition of MATH. The second equality is equivalent to MATH. REF imply MATH . But we know that MATH . So, MATH . Finally we use the equality MATH to conclude that MATH . Combining this with REF we get MATH . The third and fourth statement of the corollary immediately follow from the second one. |
math/0101133 | It suffices to prove the second equality, then the first one follows as usual. Choose MATH, MATH and MATH. Let MATH be an orthonormal basis for MATH. It follows from REF that, with MATH-strong-MATH convergence MATH . Next choose MATH and an orthonormal basis MATH for MATH. Then with MATH-strong-MATH convergence we have MATH . Denote by MATH the above expression summed over MATH where MATH and MATH are finite subsets. Then it follows from REF that all MATH belong to MATH and MATH . Taking the limit over MATH and using the fact that MATH is MATH-strong-MATH - norm closed we get that for all finite subsets MATH the element MATH defined by MATH belongs to MATH and MATH . Taking the limit over MATH we get that the element MATH defined by MATH belongs to MATH and MATH . Because MATH we get MATH . By REF the elements MATH span a core for MATH. So we can conclude that for all MATH and all MATH the element MATH belongs to MATH and MATH . From this it follows that for all MATH and MATH we have MATH and MATH . |
math/0101133 | Let us introduce the notations MATH and MATH for the modular operators of the weights MATH and MATH. We first claim that for all MATH . To prove this claim, choose MATH and MATH. Then, using REF , one gets MATH . Because MATH belongs to MATH we have that MATH belongs to the domain of MATH and MATH . Because MATH is a core for MATH we get the first claim. Using REF rather then REF one has, for all MATH: MATH . Rewriting REF we get, for all MATH, MATH and MATH: MATH . Rewriting REF we get, for all MATH, MATH and MATH: MATH . Next we claim that MATH. To prove this, choose MATH and MATH. Using the two previous equations one has MATH . From this chain of equalities the second claim follows. Then MATH. Using this and REF we get MATH for every MATH, MATH, MATH and MATH. This means that MATH . |
math/0101133 | Define the following subspaces of MATH. MATH . Using REF we get MATH . It is clear that MATH is self-adjoint. Because MATH is a multiplicative unitary we get that MATH is an algebra. Using the previous computation we get that MATH is a MATH-algebra, closed in MATH-strong-MATH topology. We have MATH and because MATH is faithful we may conclude that MATH. In particular MATH and hence MATH is a NAME subalgebra of MATH. Then the bicommutant theorem gives, for all MATH: MATH . In particular, we get that MATH belongs to MATH. REF implies MATH . Because MATH it follows that MATH . Because MATH is faithful we may conclude that MATH belongs to MATH for all MATH. Together with MATH we get that MATH. But then also MATH which concludes the proof. |
math/0101133 | MATH is a MATH-anti-homomorphism from MATH into MATH. REF gives MATH for all MATH. Because MATH is continuous in MATH-strong-MATH topology, it follows from REF that MATH. But clearly MATH and so MATH. Hence MATH is a MATH-anti-automorphism of MATH. Let MATH be an orthonormal basis for MATH. Because MATH is a multiplicative unitary we get for all MATH, with MATH-strong-MATH convergence MATH . So we also get for all MATH . Because both MATH and MATH are continuous in the MATH-strong-MATH topology, it follows from REF that MATH . |
math/0101133 | Let MATH be as above, then MATH is a n.s.f. weight on MATH which is right invariant. So MATH is a l.c. quantum group, and REF gives that MATH is its multiplicative unitary. It follows from REF that the underlying NAME algebra of the dual locally compact quantum group is MATH . Applying REF to the cocycle matching MATH of MATH and MATH, we get that the dual l.c. quantum group is precisely MATH. |
math/0101133 | Let MATH be an orthonormal basis for MATH. Then we have, with MATH-strong-MATH convergence MATH . Denote by MATH the same expression summed over a finite subset MATH. We claim that for all MATH the normal functional MATH belongs to MATH and MATH . To prove this claim, we choose MATH. Then MATH . But then for all finite subsets MATH we have MATH and MATH . Because MATH is MATH-strong-MATH - norm closed the claim of the lemma follows. |
math/0101133 | Denote, for all MATH and MATH. Then MATH and MATH are bounded and MATH . So we can define for every MATH an operator MATH as MATH . Considering MATH as an operator in MATH and MATH as an operator in MATH, we get MATH for all MATH, so MATH is continuous in the MATH-strong-MATH topology. We claim that MATH for all MATH. Indeed, using the formula MATH, proved in REF , we have MATH from where the claim follows. For all MATH we have MATH, hence MATH and MATH . If now MATH, there is a net MATH in MATH such that MATH in the MATH-strong-MATH topology and MATH in norm. Because MATH is continuous in MATH-strong-MATH topology and MATH is MATH-strong-MATH - norm closed, we get MATH and MATH . This concludes the first part of the proof. Now suppose that we also have MATH such that MATH. Then we get, for all MATH: MATH . If MATH we take the same net MATH in MATH as above and obtain that MATH . |
math/0101133 | Consider the l.c. quantum group MATH with left invariant weight MATH, NAME MATH for MATH and multiplicative unitary MATH. Applying REF to the l.c. quantum group MATH we obtain a canonical left invariant weight MATH on the dual MATH of MATH, with canonical NAME MATH. In view of REF it suffices to prove that MATH and MATH. By the uniqueness of left invariant weights, there exists a positive number MATH such that MATH. So MATH. Then we can define a unitary operator MATH on MATH such that MATH for all MATH. Because both NAME involved are the identical representations, we get MATH. From the dual version of REF we get, for all MATH and MATH: MATH . But from REF it follows, for all MATH and MATH: MATH . Both formulas together imply MATH. Because we already have MATH, it follows that MATH. Hence we can take MATH such that MATH. We have to prove that MATH. We can already conclude that the modular group MATH of MATH equals the modular group MATH of MATH and that MATH. Choose now MATH, MATH and MATH such that there exist elements MATH and MATH in MATH with MATH and MATH bounded and MATH . Remark that such elements MATH form a dense subspace of MATH in the MATH-strong-MATH topology. Because MATH and MATH we can take elements MATH such that MATH is analytic with respect to MATH, MATH is analytic with respect toMATH and MATH, MATH for all MATH. Now REF gives that MATH . Using the relation MATH we get that MATH is analytic with respect to MATH and MATH . Finally observe that MATH, so defining MATH and MATH, where MATH is an orthonormal basis for MATH, we have REF. Because MATH, it follows from REF that the element MATH defined by MATH belongs to MATH and MATH . Choose now MATH and MATH. Then MATH . Now apply REF to the l.c. quantum group MATH. Because MATH for all MATH, and because MATH we get that MATH satisfies the conditions of this lemma. Hence there exists a MATH-strong-MATH continuous map MATH from MATH to MATH, such that for all MATH we get MATH and MATH . Because MATH and MATH is implemented by MATH we get, for all MATH: MATH . Since MATH, it follows from REF that MATH . If we now take MATH, then MATH . So MATH. Now combine REF to get MATH . Because this is valid for enough elements MATH and MATH we get that MATH which concludes the proof. |
math/0101133 | From REF it follows that MATH from where the first statement follows. The second statement follows from the first one and from the observation that MATH is a cocycle bicrossed product of MATH and MATH. |
math/0101133 | Observe that MATH . Then one can conclude that MATH and MATH . Now one can check that MATH, that is, MATH is an action of MATH on MATH, and MATH. We get similar formulas for MATH from MATH and MATH (this follows from the NAME algebraic version of REF ). So it remains to prove the final statement. Using CITE, choose MATH and MATH. Then MATH . We claim that MATH. For this we choose MATH and compute MATH . From this the claim follows, so MATH and MATH . Combining this with REF and the fact that the elements MATH form a MATH-strong-MATH - norm core for MATH, we get the final statement. |
math/0101133 | REF shows that MATH, defined in REF , is the dual l.c.quantum group of MATH, so MATH maps MATH faithfully into MATH. REF implies that MATH maps MATH faithfully into MATH and REF shows that MATH. Because the construction of MATH corresponds to the cocycle matching MATH, REF also shows that MATH. Let MATH and MATH be the modular conjugations of MATH and MATH in the NAME MATH and MATH respectively (see REF ). From REF we know that MATH is the canonical left invariant weight on the dual MATH with canonical dual NAME MATH. So, the action MATH of MATH on MATH constructed in REF satisfies MATH for all MATH where MATH . Up to a flip map, MATH is the cocycle crossed product MATH and MATH is the dual weight of MATH with accompanying dual weight NAME MATH. Hence it follows from REF that MATH is implemented by MATH. Then we get MATH . So it follows from REF that MATH, therefore REF gives MATH . Since MATH, REF show that MATH is a cleft extension. |
math/0101133 | CASE: The definition of MATH and MATH. CASE: Represent MATH, MATH and MATH on the NAME of a left NAME weight and denote by MATH the NAME space on which MATH acts, by MATH the multiplicative unitary of MATH and by MATH and MATH the modular conjugations of the invariant weights on MATH and MATH. Let MATH be the action of MATH on MATH constructed with MATH, as in REF , then MATH . One also has MATH . Applying now REF to MATH we get an action MATH of MATH on MATH by MATH . The action MATH satisfies MATH . Because MATH and MATH turn MATH into an extension, one has MATH. We claim that also MATH. Indeed, REF gives MATH . Similarly we have MATH . So MATH . But for all MATH we have MATH and so MATH. Therefore, MATH . Since MATH we get that MATH is a two-sided coideal in MATH in the sense of CITE, NAME REF Then it follows from CITE, NAME REF that MATH . Because MATH one has MATH which proves the claim. Since MATH and MATH turn MATH into a cleft extension, there exists a unitary MATH satisfying MATH . From REF it follows that MATH is a cocycle action of MATH on MATH, where MATH . Because MATH, we can define MATH and then MATH is a faithful and normal MATH-homomorphism, MATH is a unitary and MATH . Still using REF we can define a MATH-isomorphism MATH satisfying MATH . From now on, we denote MATH. CASE: To show that also MATH has a cocycle crossed product structure we will find a cocycle action of MATH on MATH. Let us define a unitary MATH in MATH by MATH . Then we have MATH . So MATH and we define MATH. Observe now that MATH . From this we may conclude that MATH . Hence REF shows that MATH has a cocycle crossed product structure and we can define MATH . Then MATH is a faithful and normal MATH-homomorphism, MATH is a unitary in MATH and MATH . If MATH is the NAME algebra acting on MATH generated by MATH and the elements MATH, it also follows from REF that there is a unique MATH-isomorphism MATH satisfying MATH . CASE: The definition of MATH. CASE: The comultiplication MATH on MATH makes of MATH a l.c. quantum group and then, MATH is an isomorphism. We claim that, writing MATH one has MATH . Indeed, observe that, by definition of MATH and REF , we have MATH . From this it follows that MATH . On the other hand, MATH . Both computations together yield MATH . But MATH . Then one can conclude, applying MATH to REF , that MATH . Apply MATH to this expression, then MATH and that is precisely the claim. The formula MATH implies MATH which together with the previous formula determines completely the comultiplication MATH on MATH. CASE: Repeating the proof of REF and using REF, we prove that the dual weight MATH on MATH of the weight MATH on MATH, with canonical NAME MATH in the sense of Terminology REF, is left invariant on the l.c. quantum group MATH. Defining MATH where MATH flips MATH and MATH, we get that MATH . So MATH is the multiplicative unitary associated with the l.c. quantum group MATH and the NAME MATH for its left invariant weight MATH. CASE: The comultiplication MATH makes MATH a l.c. quantum group and MATH is an isomorphism. Exactly as above for MATH, now starting from MATH one can prove that MATH where MATH. Analogously as above, we get that the dual weight MATH on MATH of the weight MATH on MATH, with canonical NAME MATH, is left invariant on MATH, and MATH . Then we may conclude that MATH is precisely the dual of MATH. CASE: Apply REF to the l.c. quantum group MATH to get a canonical left invariant weight MATH on MATH with canonical NAME MATH. We already have MATH and we have the left invariant weight MATH on MATH with NAME MATH. Just as in the beginning of the proof of REF , we get MATH such that MATH. Denote by MATH the modular conjugation of MATH in the NAME MATH. Putting MATH we get that MATH is the modular conjugation of MATH in the NAME MATH. Combining this with REF , we have MATH for all MATH. Further we get MATH. If MATH is the modular conjugation of MATH in the NAME MATH, it follows from REF that MATH for all MATH. Further we get MATH. Define a unitary MATH on MATH by MATH . For all MATH one has, using that MATH, MATH . From REF we get the existence of a complex number MATH such that MATH . Using this and the fact that MATH we get, for all MATH, MATH . Taking both computations together we can define a faithful MATH-homomorphism MATH satisfying MATH for all MATH and MATH for all MATH. CASE: The proof of the compatibility relations. First we prove the relation MATH for all MATH. Recall that MATH . Hence we obtain, for all MATH, MATH . Using that MATH for all MATH, and using the previous formula, we get for all MATH . Because MATH is a cocycle action of MATH on MATH we have MATH . Using this, we can continue the previous computation as MATH . Because MATH is faithful we may conclude that MATH for all MATH, from where the needed relation follows. Using now the formula MATH for all MATH and the fact that MATH is a cocycle action of MATH on MATH we obtain similarly, for all MATH, MATH . CASE: Let us prove the relation MATH . From REF one has MATH and so MATH . Using now that for all MATH we have MATH we obtain MATH . Using now that MATH for all MATH we get MATH . Before we continue this computation, we use the cocycle property of MATH to obtain MATH . Then we can continue the computation above and obtain MATH . Because MATH is faithful it now follows that MATH . Because MATH this gives us MATH so that finally MATH . CASE: It follows from REF that MATH is indeed the cocycle bicrossed product of MATH and MATH. We also have the isomorphism MATH satisfying MATH. Let MATH be the canonically associated isomorphism MATH characterized by MATH . Since MATH, one has MATH . So we obtain MATH and MATH, which concludes the proof. |
math/0101133 | First let MATH be an isomorphism of MATH onto MATH satisfying MATH and MATH. Recall that MATH satisfies MATH where MATH and MATH denote the multiplicative unitaries of MATH and MATH respectively. Concerning the dual actions MATH and MATH of MATH on MATH and MATH respectively, it follows from REF that MATH . Hence we get MATH . So MATH. As before we define the unitaries MATH and MATH as MATH . Then we define a unitary MATH. REF and the formula above imply MATH . Therefore, MATH. REF shows that there is a unitary MATH such that MATH, hence MATH . Let MATH and MATH be the canonical left invariant weights on MATH and MATH with NAME MATH and MATH. We claim that for all MATH we have MATH and MATH . To prove this, choose MATH and an orthonormal basis MATH in MATH. Define the element MATH . So, with MATH-strong-MATH convergence MATH is given by MATH . If we denote by MATH the sum over a finite subset MATH, we get that MATH and MATH . Since MATH is MATH-strong-MATH - norm closed, we get that MATH and MATH . Because such elements MATH form a MATH-strong-MATH - norm core for MATH, the claim is proved. Similarly MATH if and only if MATH. From the formula MATH for all MATH one can conclude that MATH . Then also MATH for all MATH and MATH for all MATH. Therefore, MATH for all MATH. So it follows from REF that MATH and hence MATH . It remains to prove the formula for MATH. We have that MATH . Because MATH and because of REF , we get that MATH . From the faithfulness of MATH it follows that MATH and so MATH . This concludes the proof of the first statement. Vice versa, let MATH be a cocycle matching of MATH and MATH, let MATH be a unitary in MATH and define MATH by the formulas mentioned in the proposition. Then it is straightforward to check that MATH is a cocycle matching of MATH and MATH. Let MATH and MATH be the corresponding cocycle bicrossed products and let MATH and MATH be the corresponding multiplicative unitaries. Then one can check that MATH . So, defining MATH, we get an isomorphism of the l.c. quantum groups MATH and MATH such that MATH. Hence we get that MATH and MATH, which concludes the proof. |
math/0101133 | We first claim that there exist MATH and MATH such that MATH. If this is not the case, then for every MATH and MATH we have MATH. Hence, using REF MATH . Summing over MATH, it then follows that MATH which contradicts the faithfulness of MATH. So one can take MATH and MATH such that MATH. Since MATH, we have MATH. Fix MATH and define MATH . Suppose MATH. One has MATH for all MATH. We claim that MATH. Indeed, MATH . For all MATH and so MATH . From this the claim follows. But then MATH . Applying MATH we obtain MATH that is, a contradiction with the faithfulness of MATH. Therefore, MATH, which concludes the proof. |
math/0101133 | Let MATH. Then MATH . This proves the inclusion MATH. Vice versa, let MATH and suppose MATH. Then MATH, which gives the converse inclusion. |
math/0101133 | Define MATH . Then MATH and further, using the previous lemma, MATH . Hence MATH and one can take MATH such that MATH. Then MATH . From this we have MATH and similarly MATH, which concludes the proof. |
math/0101133 | If there are MATH, both satisfying the condition above, one can take MATH such that MATH . We may suppose that MATH. By REF MATH (respectively, MATH) is a partial isometry with initial projection MATH and final projection MATH (respectively, MATH). Hence MATH is a partial isometry with initial projection MATH and final projection MATH. On the other hand, REF gives MATH so MATH. Choose MATH with MATH. Then MATH so MATH. |
math/0101133 | By the previous lemma there is at most one MATH such that MATH. On the other hand, if such a MATH does not exist, then MATH for all MATH and summing over MATH we get MATH, which contradicts REF . So, let MATH be the unique element in MATH satisfying MATH . Then MATH for all MATH. Suppose now that MATH and MATH. By REF , both MATH and MATH are partial isometries with initial projection MATH and final projection MATH. REF shows that all elements MATH satisfy MATH, and hence MATH. But MATH is a partial isometry with initial and final projection equal to MATH. Hence MATH and so MATH. Hence MATH is one-dimensional and generated by a partial isometry MATH with initial projection MATH and final projection MATH. Also we get MATH. Fix now MATH and consider the map MATH. First of all this map is injective. Take MATH and suppose that MATH. Writing MATH, we can take MATH such that MATH . Taking the adjoint we get MATH and this contradicts REF . Therefore, the map MATH is injective. To prove its surjectivity, choose MATH. By the above reasoning we can take MATH and MATH such that MATH . Taking the adjoint we get MATH and this implies that MATH. |
math/0101133 | Recall that the action MATH introduced in REF satisfies MATH for all MATH. In order to prove that MATH is a cleft extension, it suffices, by REF , to show the existence of a unitary MATH satisfying MATH. Applying MATH we have to prove the existence of a unitary MATH in MATH satisfying MATH. Choose partial isometries MATH as in the previous proposition. Fix MATH. Then MATH is a family of partial isometries in MATH with initial projections MATH and final projections MATH. Because the map MATH is a bijection of MATH we can define a unitary MATH in MATH by the sum MATH converging in the MATH-strong-MATH topology. Since MATH for all MATH, we get MATH. Identifying MATH with MATH we can define MATH by MATH. Then MATH is unitary and MATH . This concludes the proof. |
math/0101133 | Let MATH and MATH. Then we know that MATH . So we get that MATH . From this we may conclude that MATH and MATH, MATH. The other statements of the lemma are proved analogously. |
math/0101133 | For any MATH we define MATH by MATH and MATH . We claim that MATH is a left invariant integral on MATH. To prove this, choose MATH. Choose MATH and define MATH by MATH. We have to prove that MATH. Choose MATH. Let MATH. Then MATH . Because MATH we get that MATH for all MATH. Hence MATH . Hence it follows that MATH is a left invariant integral on MATH. Then we can normalize the NAME measure on MATH in such a way that the first equality of the lemma holds for all continuous functions on MATH with compact support in MATH. The same equality then follows for all positive NAME functions as usual. The second equality follows from the first and the formula MATH . |
math/0101133 | REF shows that MATH preserves NAME sets of measure zero. Now MATH has complement of measure zero in MATH and its image under MATH is MATH. Hence MATH is an open subset of MATH with complement of measure zero. Using twice REF , we get for every continuous function MATH on MATH with compact support in MATH where we have used twice the NAME theorem and the invariance of the NAME measure on MATH to obtain the last equality. The NAME theorem can be applied because the integrand has compact support in MATH. Then the same equality is valid for all positive NAME functions MATH on MATH. Taking MATH of the form MATH we get the result. |
math/0101133 | If MATH for all MATH and MATH and MATH has its support in MATH for certain compact sets MATH and MATH, then MATH and MATH, where MATH is the NAME measure on MATH. Hence MATH defines a bounded operator. If now MATH with MATH and MATH, then MATH and the result follows in this special case. When MATH is continuous on MATH and equals MATH elsewhere we can find a sequence MATH of linear combinations of functions MATH where MATH is continuous on MATH and MATH is continuous on MATH, converging uniformly to MATH. It is clear from the inequality above that MATH converges in norm to MATH and that MATH converges in MATH to MATH. Hence MATH and MATH, because MATH is closed. Finally, if MATH is NAME, if its support is contained in MATH and if MATH for all MATH and MATH, we can use the NAME theorem to obtain a sequence MATH of functions continuous on MATH and equal to MATH elsewhere, satisfying MATH for all MATH and MATH, and such that MATH where MATH is the NAME measure on MATH. Using REF one can check that MATH in the MATH-strong-MATH topology and MATH in MATH. Because MATH is MATH-strong-MATH - norm closed, the result follows. |
math/0101133 | Define a unitary MATH by MATH . We claim that MATH. In view of REF it suffices to prove that MATH which amounts to prove, for almost all MATH, the equality MATH . But applying REF to MATH and to MATH we obtain this last formula. Now we can take MATH. The second statement can be proved similarly. Finally, to prove that MATH for almost all MATH it suffices to check that MATH for almost all MATH, which is clear. Similarly we get MATH for almost all MATH. |
math/0101133 | Let MATH be the set of bounded NAME functions on MATH whose support is contained in a compact subset of MATH and let MATH. Using REF one can check that for all MATH where MATH. Because MATH is a homeomorphism of MATH we get MATH. Then it follows that MATH and MATH. When both MATH and MATH belong to MATH, one can check that MATH where MATH . We see that MATH. From REF it follows that MATH is a MATH-strong-MATH - norm core for MATH. The computations above show that it is also a MATH-algebra, so MATH is a core for MATH. Let now MATH and MATH be in MATH. Then we get, using REF and the notation MATH introduced there, MATH . In this computation we used twice the NAME theorem on a bounded NAME function whose support has compact closure. Since the functions MATH form a core for MATH, the computation above shows that MATH and MATH . Then for all MATH we get MATH and, with MATH as in the statement of the proposition, MATH . Denote by MATH the strictly positive multiplication operator with the function MATH. Because MATH we see that MATH for all MATH. Because the subspace MATH of MATH is invariant under MATH for all MATH, it is a core for MATH. Hence it follows that MATH, and so MATH because both are self-adjoint. From the formulas for MATH and MATH we get the formula for MATH. Next REF give that MATH is also a cocycle bicrossed product. If we define MATH then MATH and MATH become a matched pair of groups and the maps MATH and MATH are given by MATH . The maps MATH and MATH satisfy REF. Hence we can make the cocycle bicrossed product, and REF shows that this equals, up to a flip map, MATH. Then the formulas for MATH and MATH follow from the formulas for MATH, MATH and MATH. Finally, we get immediately that MATH and MATH commute strongly. Hence it follows from REF that the scaling constant equals MATH. |
math/0101133 | Using REF we get that MATH and MATH for all MATH, MATH and MATH. From REF it follows that MATH. If MATH is the multiplication operator with the function MATH defined in the proposition, one can compute that the closure of MATH is the multiplication operator with the function MATH which commutes with MATH. Hence MATH for all MATH and MATH. Similarly we prove that MATH implements MATH for all MATH. Because MATH and MATH generate MATH, the operators MATH and MATH are proportional. Because MATH, it follows that MATH. From REF it follows that MATH is the closure of MATH, from where the formula for MATH follows. The formula for MATH can be obtained in the same way. |
math/0101133 | MATH is trivial if and only if MATH or if and only if MATH almost everywhere. The continuity of MATH on MATH gives the first statement. REF shows that MATH where the function MATH on MATH is defined by MATH. Then MATH is affiliated to the center of MATH if and only if MATH REF , and the second statement follows. To prove the final statement, observe that MATH is a NAME algebra if and only if MATH is trivial and MATH is affiliated to the center of MATH. If MATH is a NAME algebra, then by the previous statements, MATH for all MATH and MATH . Dividing this equation by the formula for MATH REF we get MATH . Rewriting this we get the equality MATH . Clearly we can also go the other way around to complete the proof. |
math/0101133 | Let us suppose that MATH is trivial. Then for all MATH we have MATH. Hence we get that MATH for all MATH. Because MATH is trivial, we also have MATH . Applying this equation to MATH and MATH we get MATH for all MATH. Combining REF with the equality MATH we get that MATH for all MATH. Because MATH is trivial this gives the equation MATH . Using the previous proposition we obtain that MATH is a NAME algebra. |
math/0101133 | Because MATH preserves the NAME measure of MATH we get MATH for all MATH. Because MATH and MATH preserve the modular functions, we get MATH for all MATH. Then we can apply REF . |
math/0101133 | The necessary and sufficient conditions for MATH to be a NAME algebra given in REF are independent of MATH and MATH. |
math/0101133 | REF show that the dual MATH is again a bicrossed product, obtained by interchanging MATH and MATH. Now defining the isomorphism MATH, one verifies that MATH for all MATH. Hence, interchanging MATH and MATH, we get an isomorphic matched pair and so an isomorphic l.c. quantum group. |
math/0101133 | REF shows that MATH is non-compact and non-discrete. The modular functions of MATH and MATH are trivial and MATH. Using the notation introduced in REF we get MATH . Because MATH is non-trivial, the scaling group of MATH is non-trivial, so MATH is not a NAME algebra. Because MATH is non-trivial, the modular automorphism group of the left NAME weight MATH is non-trivial. Hence MATH is not a trace. Because the right NAME weight MATH on MATH is given by MATH, also MATH is not a trace. Finally, REF gives MATH and hence MATH is non-unimodular. |
math/0101133 | The proof is completely similar to the one of REF and uses again REF . Let us mention only that the groups MATH and MATH are unimodular, MATH and we get that MATH . |
math/0101133 | First, one can compute directly that MATH where MATH is an arbitrary function on MATH, is a solution of REF . We will call such a solution trivial. Because the solutions of REF form a linear space, we will only determine a general solution of REF modulo a trivial solution. In the course of the proof we use the subscript notations MATH, MATH, MATH to denote the partial derivative of MATH to the first variable and to the first and third variable respectively. Take the derivative to MATH of a sufficiently smooth solution MATH of REF and evaluate it in MATH and MATH. Then we get MATH where MATH. If MATH is a smooth function such that MATH, this gives MATH where MATH is some smooth function. Hence, modulo a trivial solution, MATH that is, MATH does not depend on its second variable: MATH . Now REF gives MATH . Taking the derivative to MATH and evaluating it in MATH we get MATH . Taking the derivative to MATH it follows that MATH . With MATH we get that MATH and hence MATH for some smooth function MATH. Plugging this into REF we get MATH and so there exists a number MATH such that MATH. Then it follows that MATH for some smooth function MATH. Using REF we have MATH . If MATH is a primitive for the function MATH we get MATH for some smooth function MATH. Since MATH, we have, modulo a trivial solution, MATH or MATH for some smooth function MATH. Plugging this into REF we get MATH . With MATH it follows that MATH does not depend on its second variable. Then there is a number MATH such that MATH. But also MATH is a trivial solution with MATH. Hence we may conclude that MATH modulo a trivial solution, where MATH. One can check directly that this is indeed a solution. It will be more convenient, taking MATH, to get a general sufficiently smooth solution of REF of the form MATH where MATH and MATH is an arbitrary smooth function. |
math/0101135 | The implication MATH is trivial. MATH. Consider MATH the set of finite sums of self-adjoint commutators of MATH. Note that MATH is a real vector subspace of MATH. Put MATH. We show that MATH. Suppose the contrary, that is, MATH. This is equivalent to MATH . It follows that MATH is a real bounded functional on MATH of norm MATH. By the NAME - NAME theorem it can be extended to a norm-MATH functional on MATH and furthermore to a bounded complex functional on MATH, denoted also by MATH. Observe that MATH is necessarily a tracial state on MATH, which contradicts our hypothesis. Because MATH, there exist some elements MATH such that MATH. In particular we have MATH . Let MATH and MATH. Then we have MATH but on the other hand, by REF we have also MATH . The required properties are now fullfilled with MATH and MATH. MATH. Suppose that MATH are as in MATH. Define MATH. Then MATH is a bounded positive map on MATH with norm MATH. It follows that MATH is invertible in the NAME algebra MATH of bounded maps on MATH. Let MATH . Note that MATH, thus MATH is positive too. By definition of MATH, for any MATH we have MATH so MATH is a finite sum of almost MATH commutators If moreover MATH is a positive element in MATH then MATH so MATH is a finite sum of at most MATH self-adjoint commutators. |
math/0101137 | We note that, because MATH is a MATH-derivation, MATH . From this it follows that MATH is in the domain of the MATH operator of NAME theory, and moreover that MATH. Hence MATH. One also has MATH since the joint distributions of MATH and MATH is the same as MATH and MATH, for any MATH. It follows that MATH. |
math/0101137 | Recall that MATH (and MATH) satisfy the KMS condition: for all MATH (or MATH), there exists a (unique) function MATH, analytic on the strip MATH, and so that (writing MATH for either MATH or MATH ) MATH . Fix MATH and let MATH be as above, so that MATH . Then MATH where in the last step we used the fact that MATH intertwines MATH and MATH. Using the NAME for MATH, we then get MATH . Since MATH we get, setting MATH, that MATH as claimed. |
math/0101137 | We may assume, by linearity, that MATH is a monomial, that is, MATH, for MATH. In this case, we have MATH the last equality by REF . This implies the statement of the Lemma. |
math/0101137 | Assume first that MATH exists. Set MATH. We must verify that REF holds. By linearity, and the fact that MATH, it is sufficient to consider the case when MATH, with MATH and MATH. Then MATH, where MATH , and MATH, with MATH, MATH. Note that for MATH, MATH, and MATH. Using this , we get MATH . Conversely, assume that MATH satisfying REF exists. Since the argument above is reversible, it is sufficient to prove that MATH is in the image of MATH. Let MATH be the dual action of MATH on MATH, given by MATH, MATH, MATH. It is sufficient to prove that MATH, since MATH consists precisely of those vectors, which are left fixed by MATH. It is sufficient to prove that MATH if MATH. Since MATH is assumed to be in the closure of MATH, it is sufficient to check this for MATH. But then by REF , MATH which gives the desired result, since MATH acts trivially on MATH. |
math/0101137 | Let MATH be a positive element, so that MATH, where MATH is a normal faithful trace on MATH, and MATH is an unbounded operator on MATH, affiliated to MATH. The modular group of MATH is then given by MATH, MATH. Denoting by MATH the element MATH, we then get MATH . Consider MATH . Let MATH and MATH be two elements in the domain of MATH, so that MATH. Then we get, writing MATH: MATH . Now, for all MATH, we have MATH so that MATH . It follows that MATH . Now fix MATH and choose MATH in the domain of MATH, MATH, so that MATH . One can choose MATH, for example, to be elements of the algebra MATH. Then MATH . Since MATH, for MATH sufficiently close to zero (so that MATH), we get that MATH . Thus MATH . Differentiating this in MATH, and noting that MATH gives MATH . Since MATH is anti-self-adjoint, this implies that MATH, so that MATH, because MATH is faithful. This means that MATH for all MATH. |
math/0101137 | Clearly, the second statement implies the first. To get the second statement, write MATH and apply the theorem to conclude that MATH. Since MATH generate MATH, MATH must be in the center of MATH, which must consist of multiples of identity, since MATH is a factor. But then MATH is a scalar multiple of identity, so that MATH and MATH are proportional. |
math/0101137 | Let MATH be a copy of MATH, free from MATH. Since MATH, and since the centralizer of MATH is a factor CITE, we can find a projection MATH, which is fixed by the modular group, and so that MATH, and MATH is rational and close to MATH. We may moreover find a family of matrix units MATH, MATH, fixed by the modular group, and so that MATH . Denote by MATH the algebra generated in MATH by MATH. Note that MATH, the algebra of MATH matrices. The restriction of MATH to MATH is the usual matrix trace. Then MATH . Write MATH. Then the inequality MATH implies that MATH since MATH and MATH are orthogonal. Hence MATH . It follows that MATH . Denote by MATH the state MATH on MATH. Then MATH . Arguing exactly as in CITE, we get that MATH . Since MATH was a rational number, arbitrarily close to MATH, we get the desired estimate for MATH. |
math/0101137 | In view of REF , we may assume that MATH is a trace, MATH. Assume that MATH is not a factor. Then there exists a central projection MATH of some trace MATH, MATH. Moreover, MATH for all MATH. Since MATH increase to all of MATH, given MATH, there is a large enough MATH and a projection MATH, so that MATH. Then for any MATH, MATH . Now applying REF , we deduce that for any MATH, MATH . Hence MATH, which is a contradiction, since MATH was arbitrary. |
math/0101137 | Assume that MATH is not a factor. Then there exists a central projection MATH, MATH, MATH. Moreover, MATH for all MATH. Since automatically MATH and MATH increase to all of MATH, given MATH, there is a large enough MATH and a projection MATH, so that MATH. Then for any MATH, MATH . |
math/0101138 | First we will compute MATH. Fix a reference point MATH at which to compute the curvature. Consider the plane MATH. MATH is isometric to the euclidean plane, so the NAME curvature of MATH is REF. The reflection through the plane MATH preserves MATH and MATH, so the lines of principal curvature of MATH at MATH are invariant under this reflection. Thus, they must be parallel to the vectors MATH and MATH. To compute the principal curvatures, we can compute the geodesic curvatures of MATH - geodesics in the MATH and MATH directions. Consider the curve MATH. The length of MATH on the interval MATH is MATH. If we do a variation of this curve normal to MATH, the length is MATH. Since the reflection in the plane MATH preserves MATH, the osculating plane of MATH must be the plane MATH, so that the normal variation lies in this osculating plane. The geodesic curvature is then the logarithmic derivative of the length of the variation (compare to the euclidean case), MATH, which gives the principal curvature of MATH in the direction of MATH. The other principal curvature is MATH. Thus, the NAME curvature of MATH is MATH, so MATH. To compute MATH, notice that the plane MATH is geodesic. So the sectional curvature is the NAME curvature of the plane MATH. Consider the annulus MATH. By the NAME - NAME theorem, MATH . We have MATH . Thus MATH. Similarly, MATH. |
math/0101138 | We will construct the metric MATH by smoothing MATH on the region MATH, by smoothing the functions MATH and MATH using the construction given in the following lemma. Suppose MATH where MATH and MATH are MATH on MATH, and MATH, MATH. Then we may find MATH functions MATH on MATH for MATH such that CASE: there is a MATH such that MATH and MATH for MATH, MATH for MATH, and CASE: MATH . The idea is to smoothly interpolate between the functions MATH and MATH. If we were to just use a partition of unity to interpolate between them, we would not control the first and second derivatives of the interpolation. So instead, we use a partition of unity to interpolate between MATH and MATH to get a function MATH. Integrating twice, we must add bump functions in order to get the correct properties. Define MATH . Then MATH is a MATH bump function. Let MATH. Then MATH, and MATH is supported on MATH. Let MATH, if MATH, and MATH. Let MATH REF . Then MATH for MATH, and MATH for MATH, and thus MATH interpolates between MATH and MATH. Let MATH. MATH for MATH, but it will differ from MATH for MATH by a constant REF , so we must adjust MATH to obtain a function which interpolates between MATH and MATH. Let MATH, then MATH. Let MATH REF . Now MATH for MATH, and MATH uniformly as MATH. We have MATH which goes to REF uniformly as MATH, since MATH is uniformly bounded. Let MATH, and define MATH. Let MATH REF . We choose MATH so that the bump function has small enough second derivative, that is MATH as MATH. Let MATH. It is clear from the construction that MATH, for MATH, and MATH, for MATH. We also have MATH. Now, since MATH as MATH REF , we see that MATH satisfies the second condition of the lemma. Continuing with the proof of REF , one may see by integrating that under the conditions of REF , we have MATH and MATH uniformly as MATH REF . For the functions MATH and MATH defining MATH, we can use REF to construct a sequence of functions MATH and MATH such that MATH, MATH as MATH, and MATH. Similarly MATH, MATH and MATH, where MATH comes from the construction in REF . Moreover, for MATH small enough, MATH and MATH for all MATH, so that MATH is negatively curved for small MATH REF . One may use these properties to show that MATH, MATH, and MATH as MATH. From this, we deduce that MATH. |
math/0101145 | For a crossing coherent about a MATH, the rotation number of MATH is MATH for some integer MATH (see REF ); then by definition, MATH . The proof for the other case is identical. |
math/0101145 | Suppose that MATH has MATH non-convex corners. Let MATH be the closed curve in the diagram of MATH which is the union of the oriented boundary MATH and the paths MATH. Then the rotation number of MATH is, by definition, MATH. On the other hand, the rotation number of MATH is also the sum of the rotation numbers of its pieces. We may assume that the crossings of the diagram of MATH are orthogonal. The sum of the rotation numbers of the smooth pieces of MATH is simply MATH, because MATH is traversed counterclockwise, and each corner contributes a rotation of MATH if convex, and MATH if non-convex. Also, the rotation number of MATH is MATH, and similarly for MATH. Thus the rotation number of MATH is MATH . This implies that MATH, as desired. |
math/0101145 | It suffices to prove that homology does not change under stabilizations. Consider the natural inclusion and projection MATH . On one hand, MATH. We need to prove that MATH is chain homotopic to MATH, that is, that there exists some linear map MATH that satisfies MATH . It is not hard to check that the following MATH satisfies these requirements: MATH . |
math/0101145 | The only place where we use the non-commutativity of MATH in proving invariance is in the definition of the map MATH in REF . In the abelianized case over MATH, we can redefine MATH so that it is still a chain homotopy. If MATH is even, then we define MATH as follows: MATH if MATH is odd, MATH . It is easy to check that this MATH works in the abelianized version of the proof of REF . |
math/0101145 | We prove this on generators of MATH. There is nothing to prove for the MATH since, by REF , MATH contains only terms involving MATH with MATH. On the other hand, direct computation shows that MATH and MATH. |
math/0101145 | By REF , it suffices to prove equality on the generators MATH. Let MATH denote the sum of the terms which appear in both MATH and MATH; let MATH denote the sum of the terms in MATH involving MATH; and write MATH and MATH. The terms in MATH do not contain MATH or MATH, and so MATH. The terms in MATH must involve MATH; since MATH, we have MATH. Now consider the terms in MATH; these arise from disks of the type shown on the right in REF . There is a one-to-one correspondence between these disks and pairs of disks in MATH, one with positive corner at MATH and a negative corner at MATH, and one with positive corner at MATH. Thus MATH is the result of taking MATH and replacing every occurrence of MATH by MATH; in other words, MATH. We conclude that MATH as desired. |
math/0101145 | Let MATH. Construct a NAME surface MATH for the inverse of MATH by analytic continuation. Think of MATH as a branched cover over the image of MATH. Let MATH be the lifting of MATH to a map from MATH to MATH. Since the inverse of MATH is single-valued on MATH, MATH must be an homeomorphism. Using this construction, we must prove first that MATH is the unique map in MATH with a given configuration of branch points in the image, and second that any small variation of the images of the branch points can be accomplished inside MATH. For the first part, suppose that MATH is an element of MATH whose branch points have the same images as those of MATH. Then MATH lifts as a biholomorphism to MATH. The map MATH is an automorphism of the disk, and hence MATH and MATH differ only by a reparametrization. Projecting down to MATH, we see that MATH and MATH differ only by a reparametrization, or in other words, MATH in MATH. For the second part, let MATH be the NAME surface obtained by perturbing the image of the branch locus of MATH in MATH. Note that MATH projects to a different region in MATH when a boundary branch point is perturbed along the diagram of MATH. Since small perturbations of the image of the branch locus do not change the topology of MATH, the Uniformization Theorem provides a biholomorphism MATH that projects to an element of MATH. Finally, each interior branch point has two degrees of freedom, whereas a boundary branch point contributes but one; REF follows. |
math/0101145 | We prove this theorem in two steps. We show first that MATH, and then that the projection from MATH to MATH is a homeomorphism. To prove MATH, we clearly need only to show that MATH . To this end, let MATH be an immersion satisfying the conditions for MATH. Use MATH to pull back the complex structure from MATH. Then the NAME Mapping Theorem provides a conformal equivalence MATH between the new complex structure and the standard structure on the interior of MATH. Hence MATH is holomorphic on the interior of MATH. That MATH lies in MATH now follows from the proof of REF and the fact that holomorphic maps preserve orientation. We are left to show that MATH is a homeomorphism. It is clear from the discussion above that the image of MATH lies in MATH. The interesting part of the proof lies in the construction of an inverse MATH to MATH that lifts maps in MATH to MATH. To do this, note that a few simple manipulations of the MATH REF give, for MATH: MATH is harmonic. Thus, given MATH, we define MATH by solving the NAME problem with a (discontinuous) boundary condition which may be formulated as follows: suppose that MATH. By assumption, MATH lies in the diagram of MATH. Away from the crossings, let MATH be the MATH coordinate of the knot MATH that lies above MATH. This defines the boundary condition MATH uniquely on MATH. Once we have determined MATH, a similar manipulation of the MATH REF yields: MATH . Combined with the NAME Lemma, REF tells us that we can find a MATH (unique up to an additive constant) so that MATH lifts MATH and solves the MATH equations on the interior of MATH. In fact, the solutions extend continuously to the boundary away from the punctures MATH. The lift MATH clearly satisfies condition MATH of the definition of MATH. The first and last two conditions, namely that of finite energy and of asymptotic approach to NAME strips at the punctures, need proof. We tackle the last condition first. We begin by describing a local model for the lifting MATH near a positive puncture in the special case where the crossing that is the image of MATH is bounded by the MATH and MATH axes. Further, suppose that the lift MATH of the MATH axis has constant MATH coordinate MATH and that the lift MATH of the MATH axis has constant MATH coordinate MATH. In the MATH plane, the exponential map takes the strip MATH to the lower right-hand quadrant. Consider the following lifting MATH of MATH: MATH . It is straightforward to check that this map is MATH-holomorphic and tends asymptotically to the NAME chord over the origin as MATH goes to MATH. It is not hard to generalize this model to the case where MATH and MATH are arbitrary straight lines whose projections MATH and MATH pass through the origin. Note that, in this case, the original map MATH is MATH, where MATH, MATH, MATH, and MATH for some integer MATH. REF and NAME shows that every positive puncture is MATH-close to a straight-line model in the MATH-plane: Let MATH and MATH be curves in MATH that pass through the origin. Let MATH be a holomorphic map that satisfies: CASE: MATH for MATH. CASE: MATH uniformly in MATH. Then there exist constants MATH, MATH, and MATH such that MATH with MATH, MATH, MATH, and MATH for some integer MATH. Given a crossing with curves MATH and MATH, let MATH be a holomorphic curve that satisfies the hypotheses of REF . Let MATH be a straight-line solution with respect to the lines MATH and MATH. The theorem asserts that these two solutions differ by MATH. Further, a simple calculation using the fact that the MATH coordinates of MATH are determined from their MATH coordinates shows that the boundary conditions for the NAME problem differ by MATH. By the maximum principle, the MATH liftings of MATH and MATH differ by at most MATH. Furthermore, by explicitly writing out the formula for the liftings to the MATH coordinate that come from the proof of the NAME Lemma (see CITE, for example), we see that the difference there is MATH. Thus, on the strip MATH, the lifting of the straight-line model is exponentially close to the lifting of the general case. Since the straight-line case tends asymptotically to a NAME chord, so must the general case. Finally, we have to show that MATH has finite energy. Let MATH be a small half-disk around the puncture MATH in the disk MATH. Similarly, for each MATH, let MATH be a half-disk around MATH. Let MATH be the boundary of MATH. NAME 's Theorem gives: MATH . As the MATH get smaller, MATH approaches MATH and so, using the straight-line model and REF , we have: MATH . This completes the proof that the inverse map MATH is well-defined. Using the same methods as for the characterization of the asymptotic behavior of MATH, it is clear that MATH is a continuous inverse to MATH. The theorem follows. |
math/0101146 | Let MATH be the algebra generated by MATH and MATH. The condition on cumulants implies that MATH is valued in MATH. It follows from the moment-cumulant formula that MATH is valued in MATH, and hence that MATH . Since the moment-cumulant formula determines MATH, it follows that MATH . |
math/0101146 | If MATH, it follows that the MATH-valued cumulants of MATH, restricted to MATH are valued in MATH. Hence by REF , the MATH-valued cumulant series of MATH are the same as the restriction of the MATH-valued cumulant series; hence the only scalar-valued cumulant of MATH which is nonzero is the second cumulant MATH, so that the distribution of MATH is the semicircle law. Conversely, assume that the distribution of MATH is the semicircle law. Let MATH, MATH. Then we have MATH so that MATH . By the NAME inequality, we have that if MATH, MATH which is a contradiction. Hence MATH. |
math/0101146 | We prove the theorem in the first formulation. Assume that REF is satisfied by the cumulant series of MATH. Let MATH be as in REF. Since the freeness of MATH from MATH with amalgamation over MATH is a condition on the MATH-moment series of MATH, and MATH have the same MATH-moment series as MATH, it is sufficient to prove that MATH are free with amalgamation over MATH from MATH. Since REF is satisfied, MATH and hence MATH belong to the algebra MATH generated in MATH by MATH and MATH, MATH, MATH and MATH. Therefore, it is sufficient to prove that MATH is free from MATH with amalgamation over MATH. Let MATH, so that MATH, and let MATH, so that MATH (allowing also MATH and/or MATH to be equal to MATH). We must prove that MATH . Note that the factorization REF as well as the definition of the generators of MATH (see REF) imply that MATH has values in MATH. It follows that we may assume that MATH (since MATH). By the definition of MATH, its kernel is spanned by irreducible non-trivial words in the generators MATH and MATH. Then MATH is again a linear combination of words in the generators MATH and MATH. By linearity, we may reduce to the case that MATH is a single word. If MATH is irreducible, it must be non-trivial (since each MATH is non-trivial), hence MATH, so that MATH. So assume that MATH is not irreducible. Since each MATH is irreducible, this means that MATH contains a sub-word of the form MATH . Using the relation REF and the factorization REF , we get that MATH since MATH. Thus in any case, MATH. We have therefore seen that the factorization condition implies freeness with amalgamation. To prove the other implication, assume that MATH are free with amalgamation over MATH from MATH. Let MATH be MATH-valued random variables, so that MATH where MATH denote MATH-valued cumulants. (Note that the first occurrence of MATH is actually redundant, as MATH). Then by the first part of the proof, MATH are free from MATH with amalgamation over MATH. Moreover, by REF , the MATH-valued distributions of MATH and MATH are the same. By assumption, MATH are free with amalgamation over MATH from MATH. This freeness, together with the MATH-valued distribution of MATH, determines their MATH-valued distribution. Indeed, the freeness assumptions determine MATH which in view of the assumptions on MATH determines MATH . It follows that the MATH-valued distributions of MATH and MATH coincide. Hence the MATH-valued cumulants of MATH satisfy REF . |
math/0101146 | Since MATH is free from MATH with amalgamation over MATH, we have that for all MATH and MATH, MATH . Since MATH is free from MATH with amalgamation over MATH, we get similarly that for all MATH, MATH . Applying this with MATH and combining with the previous equation gives MATH since MATH. Hence MATH is free from MATH with amalgamation over MATH. |
math/0101147 | Let MATH denote a vector norm of MATH and consider the function MATH such that MATH for MATH, MATH for MATH and MATH linearly interpolates between MATH and MATH on MATH. Clearly, for any MATH, MATH . For any MATH, we can find MATH such that MATH . By hypothesis, for any MATH and MATH we have MATH . Therefore, for any MATH, MATH for all sufficiently large MATH. For any MATH, we have MATH and, therefore, for all sufficiently large MATH we have MATH where MATH denotes the compact set MATH . Since MATH is continuous and MATH is compact, MATH is uniformly continuous on MATH and hence there exists MATH such that MATH whenever MATH and MATH. We can choose MATH large enough so that MATH . Collecting all estimates, we obtain MATH for all sufficiently large MATH. Since MATH is arbitrary, the Lemma follows. |
math/0101147 | The enumeration of MATH is obtained by setting MATH, MATH, in NAME 's REF . Given a vertex marked tree MATH with MATH vertices, one can mark its edges in MATH ways. The vertex marking can then be removed by dividing by MATH which gives MATH. The remaining formulas are obvious. |
math/0101147 | There exits a simple bijection between rooted forests that we want to enumerate and trees with vertex set MATH such that the vertex MATH is MATH-valent. We just add new edges which join MATH to the roots of the forests. Now we apply REF . |
math/0101147 | The same distribution of trunk heights is obtained if, instead of edge trees, we consider random elements of MATH. The notion of trunk and its height have an obvious analog for such trees. Given a tree MATH with MATH and MATH vertices, we can associate to it a forest with MATH components by deleting the trunk path MATH from MATH. This forest comes with an additional structure, namely, an ordering on the components of the forest. Since there are MATH possible orderings, we conclude using REF that the probability to have MATH equals MATH . If MATH then, as MATH, we have MATH hence the probability REF is asymptotic to MATH, which completes the proof. |
math/0101147 | As in proof of REF , we can replace random edge trees by random elements of MATH. We can construct elements of MATH with given root component of size MATH as follows: partition the MATH vertices into sets of order MATH and MATH, take an element of MATH and an element of MATH, join their roots by an edge, and choose the root of the first tree to be the root of the union. It follows that the probability that the root component has size MATH equals MATH where the asymptotics follow immediately from the NAME REF . |
math/0101147 | This can be seen, for example, from the equation MATH satisfied by the function MATH which is the generating function for MATH and is, essentially, the same as the NAME. The equation implies that MATH . |
math/0101147 | Since, by REF , MATH and the right-hand side converges to MATH in distribution by REF , it suffices to prove that MATH converges to the MATH-uniform random variable. Consider the subset of MATH formed by trees with the root component of fixed cardinality MATH. Clearly, on this subset, MATH is uniformly distributed on the interval MATH and, hence, on this subset, MATH converges, in distribution, to the MATH-uniform random variable. Now REF concludes the proof. |
math/0101147 | It suffices to prove the Proposition for the sets MATH of the form MATH . It is clear, that in REF we can replace the summation over MATH by the summation over MATH for any MATH, because the contribution of any particular value of MATH is suppressed by the factor MATH. By REF , choosing MATH sufficiently large, we can make the distribution of MATH be arbitrarily close to the uniform distribution on MATH, whence MATH . The measure MATH, by REF , just counts the number of trees of size MATH, or, more concretely, MATH where the second equality uses the NAME formula (see for example, CITE) MATH and the last equality is by the definition of the integral. This determines the measure MATH uniquely and concludes the proof. |
math/0101148 | Let MATH be the union of mutually disjoint MATH - compressing disks for MATH as in REF . Let MATH (MATH respectively) be the union of the components of MATH which are disjoint from MATH (which intersect MATH respectively). Let MATH be a component of the manifold obtained from MATH by cutting along MATH, and MATH. Let MATH be the union of the components of MATH that are contained in MATH. Let MATH be the manifold obtained from MATH by cutting along MATH, and MATH. Then we have the following cases. CASE: In this case, MATH, and we have MATH. By the definition of MATH - compression body REF , we see that MATH is a trivial compression body such that MATH. CASE: By the definition of MATH - compression body, we see that each component of MATH is a MATH - ball intersecting MATH with MATH. Hence each component of MATH is a solid torus, say MATH, such that MATH is an annulus which is a neighborhood of a longitude of MATH. This implies that each component of MATH is a trivial compression body such that the union of the MATH boundaries is MATH. Since MATH is recovered from MATH by identifying pairs of annuli corresponding to MATH, we see that the triviality can be pulled back to show that MATH is a trivial compression body with MATH, where MATH. In fact, we see that either MATH is a MATH - ball or MATH is a solid torus with MATH a core circle. CASE: By the definition of MATH - compression body, for each component MATH of MATH, we have either MATH is a MATH - ball intersecting MATH with MATH, or MATH is a trivial MATH - compression body such that the MATH boundary is a component of MATH. In either case, MATH is a trivial compression body such that MATH. Hence MATH is a union of trivial compression bodies such that the union of the MATH boundaries is MATH. Since MATH is recovered from MATH by identifying pairs of annuli corresponding to MATH, we see that the triviality can be pulled back to show that MATH is a trivial compression body with MATH, where MATH. By the conclusions of REF , we see that MATH is recovered from a union of trivial compression bodies by identifying the pairs of disks in MATH boundaries, which are corresponding to MATH, and this implies that MATH is a compression body (see REF ). Moreover, since the MATH boundary of each trivial compression body MATH is MATH, we see that MATH. |
math/0101148 | If there is a compressing disk MATH for MATH, then by compressing MATH along MATH, we obtain two REF - spheres, each of which intersects MATH in one point. This contradicts the existence of a REF - fold branched cover of MATH with branch set MATH. |
math/0101148 | Suppose that MATH is parallel to an annulus MATH in MATH. Let MATH. Note that MATH is a (possibly empty) union of annulus. Let MATH. Then MATH. Since MATH is MATH - incompressible, we see that MATH. Since no component of MATH is a MATH - disk, each component of MATH is an annulus. Let MATH be a component of MATH such that MATH contains a component of MATH. Let MATH be the component of MATH such that MATH. Note that MATH is a component of MATH and is also a component of MATH. Let MATH be the component of MATH such that MATH is the component of MATH other than MATH. Then MATH is an annulus which is either a component of MATH, or a component of MATH. If MATH is a component of MATH, then the component of MATH corresponding to MATH is a MATH - sphere, hence, a component of MATH or MATH is a MATH - sphere, a contradiction. If MATH is a component of MATH, then the component of MATH corresponding to MATH is a MATH - disk, contradicting the assumption of REF . |
math/0101148 | Suppose that there is a MATH - compressing disk MATH for MATH in MATH. If MATH, then we have a contradiction as in the proof of REF . Suppose that MATH. Let MATH be the disk obtained from MATH by MATH - compressing MATH along MATH such that MATH. Since MATH is MATH - essential in MATH, this shows that MATH is a MATH - meridian disk of MATH. Hence MATH is MATH - reducible, a contradiction. |
math/0101148 | By using the argument as in REF of REF, we can show that MATH is incompressible in MATH. Suppose that MATH is parallel to an annulus MATH in MATH. Let MATH, MATH be as in REF of REF. Let MATH be the component of MATH such that MATH, and MATH the component of MATH such that MATH. By using the argument of the proof of REF of REF, we see that MATH is disjoint from MATH, hence MATH. Hence MATH is an annulus, and this shows that MATH bounds a MATH - disk in MATH, a contradiction. |
math/0101148 | Suppose that there is a REF - sphere component MATH of MATH. We note that MATH is a union of non-empty meridian disks of MATH. Let MATH. Note that MATH is a planar surface in MATH. Let MATH be a union of mutually disjoint arcs properly embedded in MATH such that MATH, and that MATH is an annulus, say MATH. Let MATH be a REF - sphere obtained from MATH by pushing MATH into MATH, and MATH into MATH such that MATH. It is clear that MATH consists of a disk, say MATH, obtained from MATH by banding along MATH. Claim-MATH is a meridian disk of the compression body MATH. Suppose, for a contradiction, that either MATH or MATH, say MATH, is not a meridian disk, that is, there is a disk MATH in MATH such that MATH. Note that we have either MATH, or MATH. If MATH, then MATH is recovered from MATH by banding along arcs properly embedded in MATH. This shows that MATH, and this implies that each component of MATH is not a meridian disk, a contradiction. On the other hand, if MATH, then MATH. This shows that MATH, and this implies that each component of MATH is not a meridian disk, a contradiction. Since MATH and MATH are parallel in MATH, we see by Claim that MATH is reducible. |
math/0101148 | Recall that MATH is the union of the components of MATH that are contained in MATH. We see, from the definition of MATH, that each MATH is obtained as in the following manner. CASE: Take a component MATH of MATH (MATH or MATH, say MATH) such that there exists a component MATH of MATH such that MATH. Let MATH-the components of MATH intersecting MATH. Then MATH-the union of components MATH of MATH such that MATH. It is clear that this construction process gives conclusion REF. If MATH is a component of MATH, then we have conclusion REF. |
math/0101148 | Since the argument is symmetric, we may suppose that MATH satisfies conclusion REF . Since MATH is a compression body, it is clear that MATH is a compression body. Let MATH. There is a union of mutually disjoint meridian disks, say MATH, of MATH such that MATH, and each component of the manifold obtained from MATH by cutting along MATH is homeomorphic to either a REF - ball or MATH, where MATH is a component of MATH with MATH corresponding to MATH. Hence MATH is homeomorphic to a manifold obtained from MATH by attaching REF - handles along the simple closed curves corresponding to MATH in MATH, and capping off some of the resulting REF - sphere boundary components. By the definition of compression body REF , this implies that MATH is a compression body, unless there exists a REF - sphere component MATH of MATH, which is disjoint from MATH. However such MATH must be a component of MATH, a contradiction. |
math/0101148 | Suppose that some compression body is trivial. By changing subscripts if necessary, we may suppose that MATH is trivial. Then we claim that MATH, that is, we have conclusion REF . In fact, if we have conclusion REF , then MATH. However this expression obviously implies MATH is not trivial, a contradiction. Hence MATH, and this implies that MATH is a trivial compression body such that MATH is a component of MATH. Let MATH be any component of MATH. Then by extending MATH vertically to MATH, we obtain a disk MATH properly embedded in MATH. Since each component of MATH is not a REF - sphere, MATH is not contractible in MATH. Hence MATH is a compressing disk of MATH, a contradiction. |
math/0101148 | We prove this by using an argument of CITE. If MATH is reducible, then by CITE, we see that any NAME splitting of MATH is reducible. Hence we may suppose that MATH is irreducible. If a component of MATH is a REF - sphere, then MATH is reducible REF . Hence we may suppose that each component of MATH is not a REF - sphere (hence, MATH is a NAME splitting of MATH). Since the argument is symmetric, we may suppose that the pair MATH, MATH satisfies conclusion REF . By CITE, there exists an incompressible REF - sphere MATH in MATH such that MATH intersects MATH in a circle. Let MATH. Note that MATH is a meridian disk of MATH. Since MATH is irreducible, MATH bounds a REF - ball MATH in MATH. Let MATH be the closure of the component of MATH such that MATH. Since MATH, we see that MATH is a handlebody, that is, MATH. Let MATH be a spine of MATH, and MATH. It is clear that MATH is an incompressible REF - sphere in MATH, and MATH is a NAME surface of MATH. Hence, by CITE, there exists an incompressible REF - sphere MATH in MATH such that MATH intersects MATH in a circle. It is obvious that REF - sphere MATH gives a reducibility of MATH. |
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