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math/0106158 | This follows immediately from REF applied to each bisimplicial set MATH, where MATH is any rigid hypercover of MATH. |
math/0106158 | For simplicity of notation, let MATH be the pro-space MATH. As described in REF , MATH is a colimit of a diagram of strict maps such that the diagram has no loops and each object is the source of only finitely many arrows. Moreover, each of the categories MATH has finite limits because of the existence of rigid limits (see REF). This allows us to apply the method of CITE to compute MATH. The index set MATH for MATH is the product category MATH . For each MATH in MATH, the space MATH is the coequalizer of the diagram MATH . In this diagram, the upper map is induced by the maps MATH and the projections MATH, while the lower map is induced by the maps MATH . The forgetful functor MATH is cofinal by REF . Therefore, we might as well assume that MATH is the indexing category for MATH. If MATH is a rigid hypercover of MATH, then MATH is the coequalizer of the diagram MATH . For every MATH, the rigid hypercover map MATH gives us a map MATH. Since MATH is actually a directed set, this means that MATH is isomorphic to MATH. It follows that MATH is isomorphic to the coequalizer of the diagram MATH . In other words, MATH is MATH. This is precisely the definition of MATH. |
math/0106158 | As in CITE, we can write MATH as a disjoint union of simplicial schemes MATH, where each MATH is connected in the sense that the simplicial set MATH is connected. Since MATH, MATH, and realization all commute with disjoint unions, it suffices to assume that MATH is connected. We choose any basepoint in MATH. Now both MATH and MATH are pointed connected pro-spaces. By CITE, it suffices to show that the natural map MATH induces an isomorphism of pro-homotopy groups in all dimensions. By REF , we may as well consider the map MATH to study the homotopy groups in dimension less than MATH. This map induces an isomorphism on pro-homotopy groups by REF . Since MATH was arbitrary, the map MATH is a pro-isomorphism for all MATH. |
math/0106158 | By REF , the map MATH is a weak equivalence. By CITE, the map MATH is a weak equivalence. Thus, the composition of these two maps is also a weak equivalence. |
math/0106158 | Every finite limit can be expressed in terms of finite products and fiber products, so it suffices to consider a diagram of schemes MATH such that the three vertical maps are étale (respectively, separated). We want to show that the induced map MATH is also étale (respectively, separated). We prove the lemma for étale maps. The proof for separated maps is identical. See CITE for the necessary properties of separated maps. Recall that base changes preserve étale maps CITE. Let MATH be the map in question. Factor MATH as MATH . The second and third maps are étale because they are base changes of MATH and MATH respectively. It remains to show that the first map is also étale. The diagram MATH is a pullback square, where MATH is the diagonal map. It suffices to observe that MATH is étale CITE. |
math/0106158 | The category of connected pointed schemes has finite limits. To construct such limits, just take the basepoint component of the usual limit of schemes. The proposition now follows from this observation and the universal property of pullbacks of schemes. |
math/0106158 | We need to show that every geometric point MATH of MATH lifts to MATH. Let MATH be the composition of MATH with the projection map MATH. Since each MATH is a rigid cover of MATH, there exist canonical lifts MATH of each MATH to MATH. They assemble to give a geometric point MATH of MATH because MATH is a diagram of rigid cover maps. |
math/0106158 | The map MATH factors as a local isomorphism MATH followed by the map MATH. The latter is étale and separated by REF , so the composition is also étale and separated. The other parts of the definition of a rigid cover are satisfied by construction. |
math/0106158 | As in the proof of REF , it is important that the category of connected pointed schemes has finite limits. The lemma now follows from this observation and the universal property of limits. |
math/0106158 | Consider the functor MATH that takes a rigid hypercover MATH of MATH to the hypercover MATH of MATH given by the formula MATH. This functor is full and faithful, so it suffices to show that every rigid hypercover of MATH belongs to the image of this functor. Let MATH be an arbitrary rigid hypercover of MATH. Then MATH is a simplicial diagram in the category MATH. There is at most one rigid hypercover map between any two rigid hypercovers of MATH, so the map MATH is the identity map for all MATH. It follows that all of the maps MATH are isomorphisms; in fact, they are all the same isomorphism for all maps from MATH to MATH. |
math/0106158 | Let MATH be a hypercover of MATH. By induction and REF , each MATH and each MATH are étale schemes over MATH. Thus, MATH is a simplicial object in the category of étale schemes over MATH. The remark after CITE finishes the argument. The proof of the second claim is similar. Instead of considering étale schemes over MATH, we must consider disjoint unions of étale schemes over MATH. |
math/0106158 | This follows from REF and induction. Because MATH, MATH, and MATH are all split by REF , the degeneracy maps take care of themselves. |
math/0106158 | First note that MATH . Thus REF gives us the surjectivity. REF finishes the proof. |
math/0106158 | This follows from REF and induction. The degeneracy maps take care of themselves because MATH, each MATH, and MATH are all split by REF (for MATH, one also needs REF ). |
math/0106158 | For convenience, let MATH be the category MATH . Since each MATH is actually a directed set, so is MATH. The category MATH is also a directed set, so it suffices to show that for every object MATH of MATH, there is an object MATH of MATH and a rigid hypercover map MATH over MATH for every MATH. For each MATH, define MATH to be MATH . The idea is that MATH is a ``rigid right NAME extension". The rigid limit is finite because MATH is at most MATH. The functoriality of rigid limits as expressed in REF assures us that MATH is in fact a rigid hypercover of MATH. The projections MATH are the desired maps. |
math/0106158 | We show that both functors MATH and MATH have the same right adjoint. The right adjoint of MATH is the functor taking a space MATH to the simplicial space MATH. On the other hand, the right adjoint of MATH is the functor taking a space MATH to the MATH-th coskeleton of the simplicial space MATH. For formal reasons, this last simplicial space is isomorphic to the simplicial space MATH. To show that MATH and MATH are isomorphic, use adjunction and the fact that MATH is isomorphic to MATH for every MATH and MATH. |
math/0106158 | When MATH, the MATH-th homotopy group of MATH only depends on MATH. Hence REF gives the result. |
math/0106158 | The proof is basically the same as the proof of REF . One just needs to check that the ingredients used there also apply to pro-spaces. |
math/0106158 | When MATH, the MATH-th homotopy pro-group of MATH only depends on MATH. Hence REF gives the result. |
math/0106161 | We have MATH and the second claim follows by taking adjoints. |
math/0106161 | If MATH we have MATH unless MATH, so MATH. Because MATH, this gives MATH. Repeated calculations of this form show that MATH unless either MATH extends MATH or MATH extends MATH. Suppose for the sake of argument that MATH extends MATH. Then calculations as above show that MATH. |
math/0106161 | We only give an outline here as the argument closely follows that of CITE and CITE. Let MATH and let MATH be the space of functions of finite support on MATH. The set of point masses MATH forms a basis for MATH. By thinking of MATH as MATH we may use REF , and the relation MATH to define an associative multiplication and involution on MATH such that MATH is a MATH-algebra. As a MATH-algebra MATH is generated by MATH and MATH. From the way we have defined multiplication, the elements MATH have the property that MATH and MATH for all MATH with MATH. Let us mod out by the ideal MATH generated by the elements MATH for all MATH with MATH and the elements MATH for all MATH. Then the images MATH of MATH and MATH of MATH in MATH form a NAME MATH-family that generates MATH. The triple MATH has the required universal property, though MATH is not a MATH-algebra. A standard argument shows that MATH is a well-defined, bounded seminorm on MATH. The completion MATH of MATH is a MATH-algebra with the same representation theory as MATH. Thus if MATH and MATH are the images of MATH and MATH in MATH, then MATH has all the required properties. Now for each MATH let MATH be an infinite-dimensional NAME space. Also for each MATH let MATH if MATH is not a sink, and let MATH be an infinite-dimensional NAME space if MATH is a sink. Let MATH and for each MATH let MATH be the partial isometry with initial space MATH and final space MATH. Finally, for MATH define MATH to be the projection onto MATH, where this is interpreted as the zero projection when MATH. Then MATH is a NAME MATH-family. By the universal property there exists a homomorphism MATH. Since the MATH's and MATH's are nonzero, it follows that the MATH's and MATH's are also nonzero. |
math/0106161 | Recall that MATH contains MATH for all MATH and MATH for all MATH and is closed under finite intersections and finite unions. Hence the right hand side of the above equation is contained in MATH. To see the converse note that the right hand side contains MATH for all MATH and MATH for all MATH and is closed under finite intersections and finite unions. Furthermore, since MATH is a finite subset we may always choose it to be disjoint from MATH simply by discarding any unwanted vertices in MATH. |
math/0106161 | Since MATH is a singleton set for all MATH, we see that MATH equals the collection of all finite subsets of MATH. If MATH is a NAME MATH-family CITE in MATH, then we may define MATH for all MATH. (Note that this will be a finite sum.) Then MATH is a NAME MATH-family. Conversely, if MATH is a NAME MATH-family, then it restricts to a NAME MATH-family MATH. The result follows by applying the universal properties. |
math/0106161 | Let MATH. If MATH, then consider the projection MATH. For any MATH and MATH we have MATH. Similarly, MATH. Since MATH is dense in MATH, it follows that MATH is a unit for MATH. Conversely, suppose that MATH is unital. List the elements of MATH and MATH. Note that these sets are either finite or countably infinite. For MATH define MATH. Then MATH for all MATH and MATH. Also MATH is an approximate unit since for any MATH we may choose MATH large enough so that MATH acts as the identity on MATH. Now if MATH for some MATH, then there exists MATH and MATH so MATH and MATH. Since MATH is unital, we must have MATH in norm. But the only way that this could happen is if MATH is eventually constant. Thus MATH for some MATH. Furthermore, MATH must be all of MATH for if there were MATH, then MATH contradicting the fact that MATH. Hence MATH and since MATH we are done. |
math/0106161 | Simply use the fact that MATH is the disjoint union of its singleton sets. |
math/0106161 | We shall induct on the number of elements in MATH. If MATH, then the claim holds trivially. Assume the claim is true for sets containing MATH elements. Let MATH and choose MATH. If we let MATH, then MATH where MATH. |
math/0106161 | By definition the MATH's have mutually orthogonal range projections. Furthermore, MATH so the initial projections commute. Furthermore, if MATH, then by REF MATH . Now we shall show that REF of NAME algebras holds. Suppose that MATH and MATH are finite subsets of MATH such that the function MATH has finite (or empty) support. We divide the proof of REF into two cases. CASE: MATH. We claim that MATH. The support of MATH, which is finite (or empty) by assumption, is given by MATH . Because there are no sinks, the source map MATH is surjective and MATH. Since MATH and MATH are finite this implies that MATH and by REF MATH is unital with MATH. Furthermore, since MATH we have MATH and using REF MATH . Now applying REF with MATH gives MATH or MATH . If MATH is empty the right hand side vanishes and REF holds. If MATH is nonempty, then MATH and for each MATH we have MATH. Hence MATH and we may use the definition of a NAME MATH-family to write MATH. Summing over all vertices in MATH and substituting above gives MATH which is REF . CASE: MATH. Once again let MATH be the support of MATH. CASE: Assume MATH is empty. Let MATH. Since there are no sinks, for each MATH there exists MATH such that MATH. Because MATH is empty, either MATH for some MATH or MATH for some MATH. Thus MATH and the left hand side of REF contains MATH . If we let MATH and MATH, then REF shows that MATH so REF holds. CASE: Suppose MATH is nonempty. Since MATH is nonempty and finite and since MATH has no sinks, it follows that MATH is nonempty and finite. Thus MATH. For convenience of notation, let MATH and MATH. Then MATH . Furthermore, since MATH is finite, we know that any vertex in MATH can emit only finitely many vertices. Thus MATH for all MATH. Applying this to the above equation and using REF gives MATH . Now we also have MATH and combining this with REF gives MATH so REF holds and we are done. |
math/0106161 | By REF and the universal property of MATH, there exists a homomorphism MATH with the property that MATH. Since this homomorphism is equivariant for the gauge actions, it follows from the NAME Uniqueness Theorem for NAME algebras CITE that MATH is injective. Furthermore, since MATH has no singular vertices, the MATH's generate MATH. Thus MATH is also surjective. |
math/0106161 | From REF we have MATH. |
math/0106161 | Induct on MATH: multiply the formula for MATH by MATH. |
math/0106161 | We shall first show that this is a NAME MATH-family. The projections MATH are mutually orthogonal because the MATH's have mutually orthogonal ranges, and are orthogonal to the MATH's because of the factor MATH. To see that the MATH's are mutually orthogonal suppose that MATH. Then, without loss of generality, we may assume that there exists MATH, and because MATH we see that MATH and MATH. Thus MATH and MATH is orthogonal to MATH. Furthermore, MATH and since MATH whenever MATH, we have MATH so the first NAME relation holds. Note that the elements MATH in MATH are all sinks in MATH. Thus to check the second NAME relation, we need only consider edges whose source is some MATH. If MATH is a subset of MATH and MATH, then MATH is a finite set and MATH. Furthermore, since MATH we see that each vertex in MATH is the source of finitely many edges. Hence MATH and MATH . Thus if we fix MATH, we have MATH . Now we have MATH which equals MATH by REF . Thus the MATH's and MATH's form a NAME MATH-family. REF also implies that we can recover MATH as MATH so the operators MATH and MATH generate MATH. For the last comment note that MATH implies MATH for some MATH, and hence MATH. |
math/0106161 | Applying the proposition to the canonical family MATH of MATH gives a NAME MATH-family MATH that generates MATH. Thus we have a homomorphism MATH whose image is MATH. If MATH is the gauge action on MATH and MATH is the gauge action on MATH, then we see that MATH for all MATH. Since each projection MATH is nonzero, it follows from the NAME Uniqueness Theorem for graph algebras CITE that MATH is injective. |
math/0106161 | Let MATH be a finite subset of MATH. Then MATH is isomorphic to the graph algebra MATH by REF , and this isomorphism is equivariant for the gauge actions. Furthermore, the projections in MATH corresponding to the vertices of MATH are all nonzero: MATH because MATH is, and MATH because of the existence of a MATH such that MATH. Applying the NAME Uniqueness Theorem for graph algebras CITE to the corresponding representation of MATH shows that MATH is faithful on MATH, and hence is isometric there. Thus MATH is isometric on the subalgebra generated by MATH. Since MATH has no singular vertices, this subalgebra is dense in MATH. Hence MATH is isometric on all of MATH. |
math/0106161 | If MATH has an exit, then either there is an edge of the form MATH with MATH, or there is an edge of the form MATH. In the first case, MATH and MATH so MATH is an exit for MATH. In the second case MATH and MATH. Hence there exists MATH such that MATH for which MATH. But then MATH and MATH so MATH is an exit for MATH. Conversely, suppose MATH has an exit. Since MATH has no sinks, there exists an edge MATH with MATH and MATH. If MATH, then MATH is an exit for MATH. If MATH, let MATH. Then MATH, so MATH. Since MATH we see that MATH is an exit for MATH. |
math/0106161 | We shall prove the theorem by showing that the representations MATH and MATH of MATH are faithful, and then MATH is the required isomorphism. Write MATH as the increasing union of finite subsets MATH, and let MATH be the MATH-subalgebra of MATH generated by MATH. By REF there are isomorphisms MATH that respect the generators. Since all the loops in MATH have exits by REF , the NAME Uniqueness Theorem for graph algebras CITE implies that MATH is an isomorphism, and hence is isometric. Thus MATH is isometric on the MATH-subalgebra MATH of MATH. But since MATH has no sinks or infinite emitters, MATH is generated by the MATH's and thus MATH is a dense MATH-subalgebra of MATH. Hence MATH is isometric on all of MATH, and in particular, it is an isomorphism. |
math/0106161 | Suppose that MATH satisfies REF . Let MATH be a loop in MATH with MATH for MATH. Then MATH is a loop in MATH. Let MATH be an exit for this loop, and without loss of generality assume that MATH and MATH. Since MATH, there exists an edge MATH in MATH from MATH to MATH. Since MATH we know that MATH and hence MATH is an exit for MATH. Conversely, suppose that MATH satisfies REF . Let MATH be a loop in MATH. Then MATH for all MATH and MATH. Hence there exists a loop MATH in MATH with MATH for all MATH. Let MATH be an exit for MATH in MATH, and without loss of generality assume MATH and MATH. Let MATH. Since MATH has entries in MATH we know that MATH. Hence MATH is an exit for MATH. |
math/0106161 | MATH satisfies REF MATH MATH satisfies REF MATH MATH satisfies REF . |
math/0106161 | Since we can approximate any MATH by a linear combination of MATH, a MATH argument shows that MATH is NAME for every MATH. We define MATH by MATH. Since MATH the map MATH is an adjointable operator on the NAME MATH-module MATH, and hence defines (left multiplication by) a multiplier MATH of MATH CITE. Taking adjoints shows that MATH for all MATH so MATH strictly. It is easy to check that MATH. |
math/0106161 | For the sake of simplicity we consider the case in which a single tail has been added to a sink MATH. As mentioned earlier, any element MATH may be (uniquely) written as MATH for some MATH and some finite set MATH. To extend MATH, we let MATH be the direct sum of infinitely many copies of MATH, define MATH to be the projection on the MATH summand, and let MATH be the partial isometry whose initial space is the MATH summand and whose final space is the MATH, with MATH taking the first summand of MATH onto MATH. Now for any MATH we write MATH (uniquely) as MATH and define MATH . One can check that MATH is a NAME MATH-family, and hence REF holds. For the same reasons, throwing away the extra elements of a NAME MATH-family gives a NAME MATH-family. The last statement in REF holds because MATH . For the first part of REF , just use REF to see that every representation of MATH factors through a representation of MATH. We still have to show that the image of MATH is a full corner. List the elements of MATH and the elements MATH. Define MATH. Then given any MATH and MATH we see that for large enough MATH we have MATH and MATH. Hence REF applies and the sequence MATH converges strictly to a projection MATH in MATH satisfying MATH . Thus the corner MATH is precisely MATH. To see that this corner is full suppose MATH is an ideal containing MATH. Then MATH contains MATH and MATH. Furthermore, if MATH is a vertex on the tail attached to MATH, then there is a unique path MATH with MATH and MATH, and MATH . Thus MATH contains MATH and hence is all of MATH. |
math/0106161 | For every MATH, define MATH. For every edge MATH with MATH not a singular vertex, define MATH. If MATH with MATH, a singular vertex, then MATH for some MATH and we define MATH. The fact that MATH is a NAME MATH-family follows immediately from the fact that MATH is a NAME MATH-family. |
math/0106161 | We prove the case where MATH has just one singular vertex MATH. If MATH is a sink, then the result follows from REF . Thus we need only consider when MATH is an infinite emitter. Given a NAME MATH-family MATH, and a nonnegative integer MATH we define MATH and MATH. Note that the MATH's are projections because the MATH's have orthogonal ranges. Furthermore, MATH for every MATH. Now, for every integer MATH, define MATH and set MATH . As mentioned previously any MATH may be written (uniquely) as MATH for some MATH and some finite set MATH of vertices on the added tail. For every MATH, define MATH to equal MATH on the MATH component of MATH and zero elsewhere. That is, MATH. Similarly, for every MATH with MATH, define MATH on the MATH component; MATH. For each vertex MATH on the added tail, define MATH to be the projection onto MATH; MATH. Note that, because the MATH's are non-decreasing, MATH for each MATH. Thus, for each edge MATH of the tail, we may define MATH to be the inclusion of MATH into MATH (where MATH is taken to mean MATH). More precisely, MATH where the MATH is in the MATH component. Finally, for each edge MATH and for each MATH, we have that MATH. Thus, we can define MATH as MATH where the nonzero term appears in the MATH component. Now for any MATH we can (uniquely) write MATH for some MATH and some finite subset MATH of vertices on the added tail. Thus we may define MATH. It then follows from calculations similar to those in CITE that MATH is a NAME MATH-family satisfying the bulleted points. |
math/0106161 | Again, for simplicity we assume MATH has only one singular vertex MATH. If MATH is a sink, then the claim follows from REF . Therefore, let us assume that this singular vertex is an infinite emitter. Let MATH denote the canonical set of generators for MATH and let MATH denote the NAME MATH-family in MATH constructed in REF . Define MATH. Also, list the elements of MATH and the elements of MATH. For each nonnegative integer MATH let MATH. It follows from REF that the sequence MATH converges to a projection MATH satisfying MATH . From these relations one can see that MATH. We shall now show that MATH. Since MATH is generated by a NAME MATH-family, it suffices to show that MATH satisfies the universal property for MATH. Let MATH be a NAME MATH-family on a NAME space MATH. Then by REF we can construct a NAME space MATH and a NAME MATH-family MATH on MATH such that MATH for every MATH, MATH for every MATH with MATH, and MATH for every edge MATH whose source is MATH. By the universal property of MATH we have a homomorphism MATH from MATH onto MATH that takes MATH to MATH and MATH to MATH. Now MATH for any MATH, so MATH. Let MATH and MATH. Then MATH and MATH. Finally, if MATH, then MATH for some MATH, MATH, and MATH. Thus MATH is a representation of MATH on MATH that takes generators of MATH to the corresponding elements of the given NAME MATH-family. Therefore MATH satisfies the universal property of MATH and MATH. Finally, we note that the corner MATH is full by an argument similar to the one given in REF . |
math/0106161 | Let MATH be a desingularization of MATH. Then we may use REF to extend the MATH-families to MATH-families. Since every loop in MATH has an exit, it follows that every loop in MATH has an exit, and thus we may apply REF to get an isomorphism MATH that restricts to our desired isomorphism from MATH onto MATH. |
math/0106161 | Since MATH is an action of MATH on MATH, there exists a NAME space MATH, a faithful representation MATH, and a unitary representation MATH such that MATH . Let MATH and MATH. Also let MATH be a desingularization of MATH. For simplicity, we shall assume that MATH has only one singular vertex MATH. If MATH is a sink, then it follows from REF that there exists a NAME space MATH, and a NAME MATH-family MATH in MATH that restricts to MATH. We shall define a unitary representation MATH as follows: If MATH, then we define MATH. If MATH is in the MATH component of MATH, then we define MATH also in the MATH component of MATH. We define MATH on all of MATH by extending it linearly. Now let MATH, be the representation for which MATH and MATH, and let MATH be the gauge action on MATH. Then one can check that MATH . (To see this simply check the relation on the generators MATH and use the fact that MATH for all MATH.) Now if we define MATH by MATH, then we see that MATH for all MATH. Since MATH and since MATH has no sinks or infinite emitters, it follows from REF that MATH is faithful. Now, if MATH denotes the canonical inclusion of MATH into MATH, then we see that MATH (since each map agrees on the generators MATH). Because MATH and MATH are both injective, it follows that MATH is injective. If MATH is an infinite emitter, then an argument almost identical to the one above works. We obtain a faithful representation MATH and a unitary representation MATH as before, and we then extend this NAME space to MATH, where MATH as in REF . Similarly, we define MATH as follows: If MATH, then MATH. If MATH, then MATH. The rest of the argument follows much like the one above. If MATH has more than one sink or more than one infinite emitter, then we simply account for multiple tails. The previous argument will still work, we need only keep track of the multiple pieces added on when extending MATH to MATH. |
math/0106162 | Suppose MATH. Then MATH . Also, if MATH, then MATH . Finally, if MATH, MATH, and MATH, then MATH so MATH is hereditary. Furthermore, if MATH and MATH, then MATH and MATH which implies that MATH. Thus MATH is saturated. |
math/0106162 | Note that MATH is a saturated set containing MATH and therefore contains MATH. Thus MATH is contained in MATH. For inclusion in the other direction, notice that any nonzero product of the form MATH collapses to another of the form MATH and from an examination of the various possibilities and the hereditary property of MATH we deduce that MATH is an ideal. Since MATH contains the generators of MATH, it follows that MATH. The last two remarks follow easily. |
math/0106162 | If MATH is empty the claim holds vacuously. If MATH, then since MATH is simple we know that MATH and thus MATH. By REF there exist MATH, MATH, and MATH for MATH such that MATH . Furthermore, since MATH we may assume that MATH when MATH and MATH when MATH. (We remind the reader that if MATH, then MATH for some MATH and MATH.) Now define MATH. Since MATH we see that MATH is a projection. Furthermore, MATH and since MATH is a projection this implies that MATH. Therefore MATH and MATH. |
math/0106162 | Let MATH. Set MATH and MATH. We define an ultragraph MATH as follows: MATH . Note that if MATH, then MATH so MATH and since MATH is hereditary it follows that MATH and MATH. Thus MATH is well-defined. Let MATH be the canonical NAME MATH-family in MATH. For each MATH and MATH define MATH . Note that if MATH, then by REF MATH for some finite subsets MATH and some finite subset MATH. Thus if MATH then we see that MATH which is in MATH. Hence MATH is well-defined. We shall now show that MATH is a NAME MATH-family. Clearly, the MATH's have mutually orthogonal ranges since the MATH's do. Thus we simply need to verify the four properties of REF . CASE: We have that MATH, MATH, and MATH. CASE: If MATH, then MATH. On the other hand, if MATH, then MATH so MATH. CASE: If MATH, then MATH so MATH. On the other hand, if MATH, then MATH. CASE: Let MATH and MATH. If MATH, then MATH and MATH so MATH and MATH . If MATH, then since MATH we have that MATH. Also note that MATH implies MATH by REF . Thus the fact that MATH and the fact that MATH is saturated imply that there is at least one edge MATH with MATH. Hence MATH. Thus MATH . Now since MATH is a NAME MATH-family with MATH if and only if MATH, the universal property gives a homomorphism MATH whose kernel contains only those projections corresponding to vertices that are not in MATH. Since MATH is simple, the kernel of MATH is either MATH or MATH. Thus MATH is either MATH or MATH, and MATH is either MATH or MATH. Since MATH is a saturated hereditary subset, this implies that either MATH or MATH. |
math/0106162 | Let MATH and MATH be a loop in MATH with no exits. This implies that MATH for MATH, and MATH. In particular, the MATH's have ranges that are singleton sets. Define MATH and MATH. If MATH equals the (finite) collection of all subsets of MATH, then MATH is a hereditary subset of MATH. We shall show that MATH is NAME equivalent to MATH. Define MATH and let MATH be the graph MATH. We claim that MATH is generated by the NAME MATH-family MATH. Certainly this family lies in the corner. On the other hand, if MATH and MATH, then MATH unless both MATH and MATH have sources in MATH. Thus the claim is verified and the gauge-invariant uniqueness theorem for MATH-algebras of graphs CITE implies that MATH. To see that this is a full corner of MATH, suppose that MATH is an ideal in MATH containing MATH. Then MATH is an ideal of MATH and REF implies that MATH is a saturated hereditary subcollection containing MATH and hence containing MATH. But this implies that MATH contains the generators of MATH and hence is all of MATH. Therefore, MATH is NAME equivalent to MATH. Since MATH is a loop of length MATH, we see from CITE that MATH which is NAME equivalent to MATH. |
math/0106162 | If MATH contained a loop with no exits, then REF would imply that MATH contains an ideal NAME equivalent to MATH. Hence MATH could not be simple. |
math/0106162 | Suppose that MATH is simple. Then REF implies that every loop in MATH must have an exit. Furthermore, if MATH is a saturated hereditary subcollection of MATH, then it follows from REF that either MATH or MATH. Conversely, suppose that MATH satisfies the two properties above. If MATH is an ideal in MATH, then REF tells us that MATH is a saturated hereditary subcollection of MATH. Hence MATH equals either MATH or MATH. If MATH, then clearly MATH. On the other hand, if MATH, then since every loop in MATH has an exit we may use the NAME Uniqueness Theorem to conclude that the projection MATH is injective, and thus MATH. |
math/0106162 | To see that MATH, first note that MATH. Also whenever MATH, then because MATH is saturated we have that MATH for any finite subset MATH, and hence MATH. Thus by induction we have MATH for all MATH. To see that MATH we shall show that MATH is a saturated hereditary subcollection. We shall begin by proving inductively that each MATH is hereditary. For the base case, we have by hypothesis that MATH is hereditary. Now assume that MATH is hereditary and consider MATH. If MATH, then by the definition of MATH either MATH or MATH. In either case, MATH. To see that MATH is closed under subsets, Let MATH and MATH be typical elements of MATH. Then MATH which is in MATH because MATH is closed under unions. Finally, suppose that MATH is a typical element of MATH and that MATH with MATH. Then MATH and since MATH is hereditary MATH. Also, MATH so MATH is a finite subset of MATH. Thus MATH and MATH is hereditary. Since MATH is the union of hereditary sets, it follows that MATH itself is hereditary. To see that MATH is also saturated, let MATH be a vertex with MATH and MATH. Since MATH and since there are only finitely many edges with source MATH, we see that there exists MATH such that MATH. Thus MATH and MATH. Hence MATH is saturated. Therefore MATH and to prove the claim, we let MATH. Then MATH for some MATH and MATH for some MATH and some finite subset MATH. Similarly, MATH for some MATH and some finite subset MATH. Continuing inductively we see that MATH where the MATH's are all finite sets. |
math/0106162 | Let MATH be a desingularization of MATH CITE. Then since the class of NAME is closed under stable isomorphism CITE, and since MATH has loops if and only if MATH has loops, we see that it suffices to prove the claim for ultragraphs with no singular vertices. Suppose MATH has no singular vertices. If MATH has no loops, then write MATH as the increasing union of finite subsets MATH, and let MATH be the MATH-subalgebra of MATH generated by MATH. By CITE there are isomorphisms MATH. Since MATH has no loops, it follows from CITE that each MATH has no loops. Since MATH is a finite graph with no loops, MATH is a finite-dimensional MATH-algebra CITE. Because MATH has no singular vertices, the MATH's are dense in MATH and MATH. Thus MATH is the direct limit of finite-dimensional MATH-algebras, and consequently MATH is an NAME. Conversely, suppose that MATH has a loop MATH. CASE: MATH has an exit. Because MATH has no sinks, we may assume without loss of generality that there exists an edge MATH with MATH and MATH. Now MATH and so MATH is an infinite projection. Since a projection in an NAME is equivalent to one in a finite-dimensional subalgebra it cannot be infinite. Hence MATH is not AF. CASE: MATH has no exits. Since MATH has no sinks, it follows from REF that MATH contains an ideal NAME equivalent to MATH which is not AF. Hence MATH cannot be AF. |
math/0106162 | Let MATH be a vertex in MATH. Then MATH. Choose any vertex MATH with MATH. By hypothesis, MATH connects to a loop MATH in MATH. Without loss of generality we may assume that there exists a path MATH in MATH with MATH and MATH. Now since MATH for MATH and MATH, we see that there exists a loop in MATH with vertices MATH. Furthermore, since MATH, MATH for MATH, and MATH we see that there is a path in MATH from MATH to this loop. |
math/0106162 | If MATH contains a loop without an exit, then REF tells us that MATH contains an ideal NAME equivalent to a commutative MATH-algebra. Since ideals are hereditary subalgebras this implies that MATH is not purely infinite. Now suppose that every loop in MATH contains an exit, but that there is a vertex MATH that does not connect to a loop. Let MATH and MATH. Note that MATH implies MATH, and thus MATH and MATH restrict in such a way that we may form the ultragraph MATH. Let MATH be the generating NAME MATH-family. Then REF implies that MATH. This combined with the fact that MATH implies MATH shows that MATH is a NAME MATH-family. Since MATH does not connect to a loop, we see that MATH has no loops and hence MATH satisfies REF . Thus by the NAME uniqueness theorem CITE, we see that MATH is isomorphic to the subalgebra MATH. We shall show that this subalgebra is hereditary. Let MATH and MATH. Then for any MATH and MATH we see from a consideration of cases that MATH will have the form MATH for some MATH and MATH. Since these elements span dense subsets in MATH and MATH, we see that for all MATH and MATH we have MATH. It follows from CITE that MATH is hereditary. But now, since MATH does not contain any loops, REF implies that MATH is AF. Hence MATH cannot be purely infinite. Conversely, suppose that MATH is an ultragraph in which every loop has an exit and every vertex connects to a loop. Let MATH be a desingularization of MATH CITE. Then MATH satisfies REF if and only if MATH does, and also every vertex in MATH connects to a loop if and only if every vertex in MATH connects to a loop. Since MATH is isomorphic to a full corner of MATH and because pure infiniteness is preserved by passing to corners, it therefore suffices to prove the converse for ultragraphs with no singular vertices. Let us therefore assume that MATH has no singular vertices. If MATH is the edge matrix of MATH, then it follows from CITE that MATH. Now since MATH satisfies REF , it follows from CITE that MATH satisfies REF . Also, since every vertex in MATH connects to a loop, it follows from REF that every vertex in MATH connects to a loop. Therefore, CITE implies that MATH is purely infinite. |
math/0106162 | Since MATH is simple, it follows from REF that MATH is cofinal and satisfies REF . If MATH has no loops, then MATH is AF by REF . If MATH has a loop, then every vertex connects to that loop due to cofinality, and MATH is purely infinite by REF . |
math/0106162 | Let MATH be the index set of MATH, and let MATH denote the subring of MATH generated by the rows MATH of MATH and the point masses MATH. If we let MATH, then CITE implies that MATH and MATH. Now MATH . Let us examine MATH. When MATH, then MATH . If MATH, then MATH is eventually zero, and the above equations reduce to MATH and MATH for MATH. Hence MATH is generated by MATH and MATH and MATH has rank REF. Thus MATH. Next we shall show that MATH maps onto MATH. Since MATH is a ring generated by MATH we see that MATH equals the collection of all sums of products of the MATH's and MATH's. But for the matrix MATH above, any product of the MATH's and MATH's may be written as a sum of MATH and the MATH's. Hence MATH. But, MATH and MATH, so MATH maps onto MATH. Hence MATH. |
math/0106162 | Let MATH be the generating NAME MATH-family in MATH. We shall first show that MATH is NAME equivalent to MATH. Note that since MATH, MATH restricts to a NAME MATH-family. Now MATH by REF . If we let MATH, then MATH and MATH is generated by MATH. Since MATH is a transitive ultragraph that is not a single loop, we see that MATH satisfies REF . It then follows from the NAME Uniqueness Theorem CITE that MATH. All that remains to show is that MATH is a full corner of MATH. Suppose that MATH is an ideal in MATH containing MATH. Since MATH for all MATH we see that MATH. But then MATH contains the generators of MATH and MATH. Hence the corner is full and MATH is NAME equivalent to MATH. We shall now show that MATH. To do this we shall first show that MATH. If it was the case that MATH, then we could find a linear combination such that MATH . Also since MATH we may assume that MATH and MATH. Let MATH be the (necessarily finite) set of edges that are the initial edge of a MATH. Because MATH is an infinite emitter, it follows that MATH is a nonzero projection. Hence MATH which is a contradiction. Therefore MATH, and MATH is generated by the projection MATH. Consequently, MATH. |
math/0106162 | Recall that a character for MATH is a nonzero homomorphism MATH. We shall show that there is a unique character on MATH. Let MATH be a generating NAME MATH-family. Since MATH by REF we see that the projection MATH is a character. We shall now show that this character is unique. Let MATH be a character. Set MATH. Then MATH is a nonzero ideal and MATH is a saturated hereditary subcollection. Since MATH is transitive, we see that MATH satisfies REF . Therefore, the NAME Uniqueness Theorem CITE implies that MATH contains one of the MATH's, and MATH is nonempty. Because MATH is nonempty and MATH is transitive, it follows that MATH. Now since MATH is nonzero, we cannot also have MATH in MATH. Therefore, MATH, and this implies that MATH for all MATH and MATH for all MATH. Since MATH is generated by MATH, and MATH we see that MATH is completely determined by its value on MATH. Because MATH is a projection, MATH. Thus MATH is unique. Now if MATH was an NAME algebra, then MATH would be generated by an NAME family MATH. Let MATH be the gauge action on this NAME algebra. Because there is a unique character MATH on MATH, we see that MATH for all MATH. Also, since MATH is nonzero, MATH for some MATH. Thus MATH for all MATH which is a contradiction. |
math/0106162 | Let MATH. Then MATH is a saturated hereditary subcollection of MATH. The short exact sequence MATH induces the following cyclic six term exact sequence for MATH-theory: MATH . It follows from REF that MATH. Thus MATH and MATH. Also, REF tells us that MATH is NAME equivalent to MATH, and it then follows from REF that MATH and MATH. Thus the above exact sequence becomes MATH and MATH. Now we see that MATH and thus MATH is unital. Therefore, if MATH were the MATH-algebra of a graph, then this graph would have to have a finite number of vertices. It then follows from CITE that there exists an exact sequence MATH for some finite sets MATH and MATH. Hence MATH, which is a contradiction. |
math/0106162 | Using the language of CITE, let MATH be a universal NAME representation of MATH in MATH which is NAME covariant (that is, coisometric on MATH). We shall show that MATH is a NAME MATH-family in MATH. Since MATH is a homomorphism, we trivially have MATH and MATH. Because MATH is a NAME representation we have MATH. Also MATH so MATH. Finally, if MATH is the source of finitely many vertices, then MATH and MATH. It then follows from the fact that MATH is NAME covariant that MATH. Hence MATH is a NAME MATH-family and the universal property of MATH gives a homomorphism MATH with MATH and MATH. Let MATH and MATH be the inclusion maps. Then MATH is a NAME representation. To see that MATH is also NAME covariant, let MATH with MATH. Then MATH and hence MATH. But then MATH . Since MATH is a NAME representation which is NAME covariant, the universal property of MATH CITE implies that there is a homomorphism MATH which commutes with MATH. But then MATH and MATH so MATH and MATH are inverses for each other. |
math/0106162 | Since MATH we have MATH. If MATH, then for any MATH there exists a finite linear combination with MATH. Thus MATH for all MATH. Hence MATH and the claim is proven. |
math/0106162 | Induct on MATH. Multiply the formula for MATH by MATH. |
math/0106162 | For convenience of notation, let MATH. We shall prove the claim by induction on MATH. For MATH the equality holds easily. Therefore, assume the equality is true for MATH and we shall prove it for MATH. MATH where this last line follows from REF . The final claim follows from the fact that the terms MATH and MATH are orthogonal when MATH. |
math/0106162 | Let MATH denote the right hand side of the above equation. If MATH is not an infinite emitter, then MATH and MATH. Hence MATH. To see the reverse inclusion let MATH and MATH. Choose MATH. Since MATH, REF implies that there exists a finite linear combination MATH with the MATH's mutually orthogonal and MATH. Define MATH, and let MATH and MATH. Let MATH. Because MATH, we know from REF that MATH. Thus for every MATH we may find a finite linear combination MATH such that MATH. Now since MATH, there exists an edge MATH such that MATH and MATH is not equal to any of the MATH's. Since the MATH's are mutually orthogonal, we have that MATH for all MATH. Hence MATH . Since this inequality holds for all MATH we have MATH. Thus MATH . Now for every MATH we have that MATH. Since MATH has no sinks, this implies that MATH is the union of a finite number of vertices that emit finitely many edges. Since MATH is finite CITE implies that MATH. Furthermore, since MATH is finite, it is an element of MATH and the equality MATH shows that MATH. Since the MATH's all emit finitely many edges and since MATH was arbitrary, the above shows that MATH and hence MATH. |
math/0106162 | Let MATH denote the point mass at MATH. Then the map MATH extends to an isomorphism from MATH onto MATH. |
math/0106163 | As MATH is word-hyperbolic, the universal cover MATH has a compactification to MATH, and the action by deck transformations extends to MATH. The action of MATH has two fixed points MATH and MATH, and MATH acts properly discontinuously on MATH with quotient MATH. The result follows as MATH is the interior of this manifold. |
math/0106163 | We simply take the image in MATH of a curve MATH in MATH that is a core of the solid torus constructed above. It is easy to ensure that the image is embedded. |
math/0106163 | Any infinite cyclic subgroup MATH yields a geodesic line in MATH by the Flat Torus Theorem (see CITE) and so it is a quasi-convex subgroup (see REF for details). The main theorem of CITE implies that MATH is a missing boundary manifold, hence the interior of a solid torus. In the semi-hyperbolic case an infinite cyclic group lies in the center of its centralizer (which is a finite extension of the infinite cyclic group) and the latter is both semi-hyperbolic and a quasi-convex subgroup of MATH. |
math/0106163 | Consider the cover MATH of MATH and pick base-points MATH of MATH and MATH of MATH. As MATH is homeomorphic to MATH, there is a curve MATH and a solid torus MATH embedded in MATH, with MATH based at MATH, so that MATH and MATH are as required. We shall find a finite cover MATH of MATH which is covered by MATH so that the image of MATH embeds in MATH. As MATH is compact, there is a finite collection MATH of inverse images of MATH in MATH such that if each of these map to distinct points in an intermediate cover MATH between MATH and MATH, then MATH embeds in this cover. Choose paths joining MATH to each point in MATH and let MATH be the set of elements in MATH that are represented by the images of these paths. Let MATH. By the above theorem, there exists a subgroup MATH separating MATH from MATH. Let MATH be the corresponding cover. It is easy to see that the elements of MATH map to different points in MATH, and hence MATH embeds in MATH. |
math/0106163 | As MATH and MATH are homotopic, there exists an annulus MATH with boundary components MATH and MATH. On passing to a cover with a sufficiently thick tube MATH around MATH, the annulus MATH is contained in the tube MATH. Hence MATH has a lift that is contained in MATH. |
math/0106163 | Consider MATH. The inverse image of the solid torus MATH in MATH contains an unknotted cylinder MATH that contains a unique component MATH of the inverse image of MATH. The group MATH is the kernel of the map MATH that maps the longitude of the torus MATH (hence those of MATH and MATH), to MATH and the meridian to MATH. As MATH is an unknot (because MATH is a geodesic) this kernel is MATH, hence MATH. This implies that MATH is isotopic to MATH. |
math/0106163 | As MATH and MATH are homotopic, there is an annulus MATH bounding MATH and MATH. By passing to a cover MATH with a sufficiently thick tube MATH around MATH, we may ensure that MATH is contained in MATH. The Core lemma now implies that MATH and MATH are isotopic. |
math/0106163 | Let MATH be a topological geodesic corresponding to a primitive class in MATH. Then the geodesics MATH and MATH are homotopic hence there exists a finite cover MATH such that MATH and MATH lift to isotopic curves. Since MATH is homotopic to identity there is no obstruction in lifting it to a homotopy equivalence MATH of the finite covering MATH. One can further assume (by means of some isotopy on MATH) that MATH (pointwise). The proof of the core lemma shows that MATH is atoroidal hence hyperbolic (since NAME). Furthermore the restriction MATH is a homeomorphism. Therefore MATH is homotopic to an isometry of MATH (by NAME rigidity) and hence isotopic to an isometry (by NAME 's theorem). Let MATH be this isometry. Since the isometry group of a hyperbolic manifold of finite volume is finite it follows that MATH is of finite order. Further MATH has an extension MATH (by identity) to all of MATH, by asking MATH to keep pointwise MATH. In particular MATH is a periodic homeomorphism of MATH. Let consider the lift MATH of MATH to the universal covering MATH (which is also the universal covering of MATH). The action of MATH extends continuously to the compactification (over the boundary sphere) to a homeomorphism of the ball MATH. The action by deck transformations extends to one by homeomorphisms of the compactification obtained by adding the boundary of the group MATH, because this is word-hyperbolic. In our case the boundary is the sphere at infinity. But MATH is homotopic to identity, hence the action induced on the boundary is trivial. This shows that MATH is the identity on the boundary sphere. A periodic homeomorphism of MATH which restricts to identity on the boundary is the identity. We will use REF CITE improved by NAME (see CITE) in its variant stated in (CITE,REF ). It states that a compact NAME group acting effectively on a connected topological manifold has a nowhere dense fixed point set. One considers the finite cyclic group action induced by our periodic homeomorphism on the ball MATH. This action extends to the sphere MATH by the identity outside the upper hemisphere. Then the fixed point set contains a REF-ball and the previous result shows that the action cannot be effective, and hence MATH must be the identity map. In particular MATH is isotopic to the identity. |
math/0106164 | Let MATH and MATH be meridians corresponding to the components of MATH. CASE: Let MATH and MATH the corresponding cyclic branched coverings of MATH. If MATH and MATH are the associated monodromy maps, we have MATH, MATH, MATH and MATH. Therefore, MATH and the two coverings are equivalent. CASE: The link MATH is equivalent to the mirror image of MATH. Let MATH be an orientation reversing homeomorphism of MATH sending MATH to MATH. Since MATH preserve both the orientations of the two components of MATH, we have MATH and MATH, where MATH and MATH are the generator meridians associated to MATH. Therefore, MATH and the two coverings are equivalent. CASE: The link MATH is equivalent to the link MATH with the opposite orientation in the second component. Let MATH be the identity map on MATH, then MATH and MATH. Therefore, MATH and the two coverings are equivalent. CASE: Let MATH be the homeomorphism of MATH realizing the equivalence. Without loss of generality we can assume that MATH and MATH. Then MATH and MATH. Therefore, MATH and the two coverings are equivalent. |
math/0106164 | CASE: See REF of the previous proposition. CASE: Assume MATH. Since MATH is odd, we have MATH. Therefore, MATH is equivalent to MATH and from the previous proposition we obtain MATH. Now, let MATH. Since MATH is odd, we have MATH. Therefore, MATH is equivalent to MATH and from the previous proposition we obtain MATH. |
math/0106164 | CASE: The MATH-fold strictly-cyclic covering of MATH is REF-fold covering of the torus knot/link of type MATH CITE, which can be presented as the closure of a MATH-string braid. So it has bridge number MATH and then MATH CITE. CASE: The MATH-fold strictly-cyclic covering of MATH is REF-fold covering of MATH, which can be presented as the closure of a MATH-string braid. So it has bridge number MATH and then MATH CITE. |
math/0106164 | The fundamental group MATH of the orbifold MATH admits the presentation MATH where MATH and MATH are the homotopy classes of two meridians MATH and MATH around the components MATH and MATH of the link, and MATH is the relation deriving from the standard presentation of the group of MATH. From the definition of a singly-cyclic covering we have MATH for the epimorphism MATH defined by MATH and MATH. Consider the subgroup MATH such that MATH, where MATH. Let MATH be the epimorphism defined by MATH and MATH. This epimorphism induces the MATH-fold cyclic covering MATH such that the axis of the cyclic group action is the component MATH of MATH corresponding to the meridian MATH. Therefore, the underlying space of the orbifold MATH is MATH and the singular set is MATH. Obviously, MATH is a trivial knot in MATH with singularity index MATH, and MATH is a MATH-periodic knot/link which also has singularity index MATH. Since MATH is equivalent to MATH with MATH, then we have MATH. Using MATH and MATH, the diagram of coverings is commutative. This completes our proof. |
math/0106165 | Let MATH be the natural map. We have MATH iff MATH, so for any MATH, MATH is either REF or the order of MATH. The result will follow once we show that, for MATH, the following statements are equivalent: CASE: MATH and MATH are in the same orbit; CASE: MATH; CASE: MATH. Now MATH, so MATH, from which it follows that REF implies REF . If MATH, then MATH for some MATH, which gives MATH. Thus REF implies REF , and trivially REF implies REF . |
math/0106165 | Note first that for MATH and MATH we have MATH. We prove that MATH by induction on MATH. For MATH we have MATH, so since MATH is a cycle, MATH. Suppose then that the result is true for some MATH, and let MATH and MATH. We compute MATH and MATH . Hence MATH iff MATH . Now, for MATH and MATH or REF, MATH is an element of MATH with last entry MATH, so MATH. Further, MATH . (The last step here uses the quandle condition.) It follows that MATH proving REF , and with it the lemma. |
math/0106165 | We show that the chain map MATH sending MATH to MATH is a projection onto the subcomplex MATH. We must prove the following two statements. CASE: If MATH then MATH. CASE: If MATH then MATH. For MATH, MATH, so REF is true in this case. Let MATH REF and MATH, and suppose that MATH. Then either MATH or MATH, and it follows that MATH (using induction in the first case). Thus REF is proved. As for REF , this is clear for MATH, so suppose that it holds for some MATH and take MATH and MATH. Then MATH . Since MATH is in MATH, so is MATH, and REF follows. |
math/0106165 | Let MATH have MATH for some MATH with MATH. Since MATH for MATH or REF (and, as for any MATH, MATH), we have MATH . Fix MATH and MATH, and set MATH. If MATH, the first sum in REF is empty. If MATH and MATH, MATH, so MATH. For MATH, MATH, so again MATH, and it follows that MATH. |
math/0106165 | We let MATH and MATH be the inclusions. Define MATH by MATH for MATH. (For MATH the groups involved are REF.) A straightforward computation shows that MATH is a chain map. Since MATH, MATH induces MATH. Now MATH is injective, MATH is generated by MATH and MATH, and MATH. Hence there is a short exact sequence MATH where MATH and MATH. By REF , there is an isomorphism of chain complexes MATH such that, for MATH, MATH. Then MATH is isomorphic to the cokernel of MATH . But, for MATH, MATH, so this cokernel is isomorphic to MATH, and we are done. |
math/0106165 | A basis for MATH consists of all elements of MATH of the form MATH, and these are all cycles. The group MATH is generated by all elements of MATH of one of the forms MATH and MATH, and we have MATH . It follows that MATH is free abelian, with a basis consisting of the equivalence classes of triples MATH under the equivalence relation MATH generated by MATH . Given MATH, let MATH be the element of MATH such that MATH. Then MATH and MATH, so MATH. It follows that MATH iff MATH and MATH, so the set of equivalence classes of MATH may be identified with MATH. |
math/0106165 | As MATH runs over MATH, MATH runs over MATH, taking on each value MATH times. Thus MATH . The automorphism MATH given by MATH is in particular a bijection and carries MATH to itself, so the restriction MATH is also a bijection. Hence the sum MATH . |
math/0106165 | Since MATH, we have MATH . |
math/0106165 | For MATH, we have MATH and MATH . For MATH we have MATH and MATH . Applying REF MATH times, the sums agree as required. |
math/0106165 | For MATH, MATH and MATH . Applying REF MATH times, the sums agree as required. |
math/0106165 | MATH . |
math/0106165 | MATH by MATH applications of REF . |
math/0106165 | MATH and MATH . |
math/0106165 | MATH and MATH by MATH applications of REF . But these sums agree as the set MATH is the image of MATH under the bijection MATH. |
math/0106165 | We need to show that MATH . For MATH, we have MATH, while MATH as required. For MATH, we have MATH and MATH . By REF , the first sum adds to zero, and by REF we have MATH as required. For MATH, MATH which, by REF as above yields MATH . Now, MATH . The first sum is zero by REF , and applying REF we get MATH . Reindexing this sum by replacing MATH with MATH, we have MATH so that MATH as required. |
math/0106165 | We deal first with the case MATH. There is a chain map MATH induced by the projection of MATH onto its orbit rack. In REF, it is proved that for MATH, the element MATH of MATH is a cycle. Since the boundary maps in MATH are all zero, this means that we can define a chain map MATH by setting MATH . This is almost the same as the chain map used in the proof of REF. Then, for MATH, MATH . Hence, by REF , the induced map MATH is multiplication by MATH. It follows, since MATH is free abelian, that the torsion subgroup of MATH is equal to MATH and is annihilated by MATH, and that MATH. Since the reverse inequality was proved in CITE, the proof in the case of rack homology is complete. When MATH is a quandle, the other two cases follow from the case just proved, REF , and REF. |
math/0106165 | We identify MATH with MATH under the operation MATH. There is a short exact sequence of abelian groups MATH where MATH and MATH. Note that for MATH, MATH. Let MATH be a function such that MATH, and define MATH by MATH. The function MATH sending MATH to MATH is a bijection. Now define MATH by MATH; because MATH is coprime to MATH, MATH is also a bijection. We have, for MATH, MATH and MATH . Hence MATH . On the other hand, MATH so MATH is the desired isomorphism. |
math/0106165 | Here the underlying sets of both racks are naturally identified with MATH. We use MATH for the rack operation in MATH, and MATH for that in MATH. Thus, for MATH, MATH . Define functions MATH and MATH from MATH to itself by MATH and MATH. Since MATH, MATH is an involution. Since MATH, we have that MATH iff MATH, which in turn is equivalent to MATH. Since MATH only takes on the values REF and MATH, this implies that MATH. Hence MATH . On the other hand, MATH so MATH is the desired isomorphism. |
math/0106167 | We must show that MATH. All elements in the left hand side are of form MATH . It would be better to divide these elements into five parts: CASE: MATH, or MATH. CASE: MATH, and MATH . CASE: MATH, and MATH . CASE: MATH, and MATH . CASE: MATH, and MATH . For part MATH, let MATH (we leave to the reader the rest of this case). We have MATH . It is obvious that if we define MATH, then MATH is also a shuffle and the result is in MATH. For case MATH, let MATH, and MATH, where MATH, and MATH. Now let MATH. Then it is easy to check that MATH is also a shuffle and we have MATH and MATH. On the other hand we have MATH and MATH. So elements of case MATH cancel the elements of case MATH. Now let us do the case MATH. We assume MATH, where MATH. We have MATH where MATH . It is easy to show that the permutation MATH is a MATH-shuffle. Similarly one can do case MATH and then by counting the proof is finished. |
math/0106167 | The existence of MATH is part of the generalized NAME theorem CITE. We just prove that MATH. Let us calculate the action of MATH on a typical element MATH. For every MATH with MATH, we have MATH, so we should only check the identity MATH. For MATH for simplicity let MATH, and MATH. Then it is not difficult to verify that MATH for every MATH, and MATH. |
math/0106167 | Let MATH. As in the proof of REF , we have MATH . Let MATH. In MATH because we are working with normalized chains, all parts are zero except for MATH or MATH. We denote the first part by MATH and the second part by MATH. We show that MATH and MATH. For MATH and MATH let MATH . The reader can easily check that MATH for all MATH and all MATH. For the rest of the elements in MATH, we have MATH for all MATH and all MATH, where MATH . Now, we have MATH. We have shown that MATH . Next, we compute MATH. For MATH and MATH let MATH . For MATH and all MATH we have MATH, and if we denote MATH then MATH for all MATH and all MATH. Finally, we have MATH for MATH. We have shown that MATH. The proposition is proved. |
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