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math/0009200 | According to the case by case calculation, we have MATH for MATH. Since MATH satisfies the cocycle condition, we obtain the result. |
math/0009200 | By direct calculations, we have MATH for the twelve MATH. According to REF , the equality REF shows that MATH is a character on MATH. So we obtain the result for any MATH. |
math/0009200 | Note that REF asserts MATH where the constant depends on the path of integration. So the numerator is not identically zero, and it is same for the denominator. By the assumption we have MATH . By using REF we can check that MATH for MATH. This implies single-valuedness of MATH. |
math/0009200 | By REF , the zero divisor of the i-th component of MATH coincides with that of the i-th component of MATH via the isomorphism MATH. So we can write as MATH with certain constants MATH. It shows the diagram in question is commutative. Since MATH is an embedding and MATH is an isomorphism, we see that MATH is an embeddin... |
math/0009200 | The map MATH is essentially anti-canonical map (see REF). Hence it follows from REF . |
math/0009200 | By REF , we have MATH for MATH. We must show MATH for MATH. Let MATH be the point MATH . Then MATH is fixed by MATH and MATH. Moreover, MATH for MATH and MATH. So we have MATH by REF . Since MATH (see REF ), we obtain REF for MATH and MATH. By the same way, we see that REF holds for any member MATH of the generator sys... |
math/0009200 | By REF , a function MATH defines the meromorphic function MATH on MATH. So, by REF , we have the isomorphism of MATH - vector space: MATH . Hence we have REF . Next let us recall that MATH is the blow up of MATH at REF points. We denote this blow up by MATH. Then the NAME - NAME group MATH has the free generator MATH a... |
math/0009200 | By REF , the restriction of meromorphic function MATH on MATH is of order REF. In fact, MATH, so the numerator vanishes at only MATH with order REF, the denominator vanishes at only MATH with order REF, and MATH (see REF). Hence the map MATH is an isomorphism. Moreover, by REF , we have the equality MATH on MATH, and t... |
math/0009200 | REF is a direct consequence of REF . REF is obtained by the same argument as the proof of REF . |
math/0009213 | Most of the relations which should be satisfied by a cyclic module are easy to check; the main difficulty arises for the relation MATH . We will use repeatedly the following property of MATH . Let us first compute the square of MATH . Take MATH . Suppose by induction that: MATH . Then MATH this proves the induction. Fi... |
math/0009213 | It is clear that MATH is a simplicial map; it also commutes with the cyclic operation, due to the fact that MATH is central in MATH . At the simplicial module level, there is a factorization of MATH as MATH where MATH and MATH . The module MATH is induced by the inclusion map MATH from the trivial MATH-module MATH . Th... |
math/0009213 | Let MATH be a generator for MATH and let MATH be its norm. The homology of MATH with coefficients in MATH is given as follows (see CITE): MATH . REF yields the result. MATH . |
math/0009213 | We shall use the following resolution (see for instance CITE): There is a MATH-bimodule projective resolution of MATH given by MATH NAME by MATH over MATH yields the following complex: MATH . The space MATH is generated by the cycles in MATH subjected to the relations given by the image of MATH . Since MATH the relatio... |
math/0009213 | In the first place, MATH is equal to MATH so that we know the homologies of MATH (see REF ). Next, we have MATH . Then, using REF , we get the following formula: MATH . In particular, MATH . An induction on MATH yields the result. MATH . |
math/0009214 | We shall use the following resolution (see for instance CITE): There is a MATH-bimodule projective resolution of MATH given by MATH NAME by MATH over MATH yields the following complex: MATH . The space MATH is generated by the cycles in MATH subjected to the relations given by the image of MATH . Since MATH the relatio... |
math/0009214 | In the first place, MATH is equal to MATH so that we know the homologies of MATH (see REF ). Next, we have MATH . Then, using REF , we get the following formula: MATH . In particular, MATH . An induction on MATH yields the result. MATH . |
math/0009214 | We consider only the MATH because the other cases may be obtained by translating the quiver along the cylinder on which it lies. We then determine the radical of MATH and the latter's projective cover, through their representations on the quiver. Iterating this process until the radical obtained is projective yields th... |
math/0009214 | The MATH are known, owing to NAME 's lemma. The other MATH are obtained from the resolutions of REF , applying the functor MATH . |
math/0009216 | Let MATH be the barycentric subdivision of MATH. All regular neighborhoods occuring in this proof are to understand with respect to MATH. For any simplex MATH of MATH, choose a vertex MATH of MATH. Let MATH be the barycenter of MATH. By ambient isotopy of MATH with support in MATH, we can assume that MATH. Let MATH be ... |
math/0009216 | Let MATH be the dual cellular decomposition of MATH. Since MATH is polytopal, MATH is polytopal as well. Thus MATH has a NAME diagram MATH. We choose coordinates MATH for MATH so that no edge of MATH is parallel to the MATH - plane. Then a sweep-out of MATH by planes parallel to the MATH - plane gives rise to a MATH - ... |
math/0009216 | Let MATH be the barycentric subdivision of MATH. Let MATH be the MATH - skeleton of the dual cellular decomposition of MATH. Since MATH has a diagram, MATH also has a diagram. A sweep-out by Euclidean planes yields a MATH - NAME embedding MATH with MATH as set of critical points. The critical points of MATH in MATH are... |
math/0009216 | Let MATH be the barycentric subdivision of MATH. Since MATH or its dual is shellable, also MATH is shellable. Thus there is a shelling order MATH on the open tetrahedra of MATH such that MATH is a compact MATH - ball for all MATH. Let MATH. If MATH contains exactly MATH open MATH - simplices of MATH, then there is a un... |
math/0009216 | Since MATH is the MATH - skeleton of a cellular decomposition MATH dual to a triangulation, any connected component of MATH is a ball, the intersection of any two of these compact balls is connected, and the MATH - strata of MATH in the interior of MATH are discs. We show that the closure of any MATH - stratum of MATH ... |
math/0009216 | We construct MATH according to REF shows the two annuli contained in MATH that correspond to the two sub-arcs of MATH shown in REF ; the annuli are cut along the dotted lines (left and right side of the rectangles in the figure). The figure shows the pattern of MATH, where the numbers in REF at the edges of MATH in MAT... |
math/0009216 | By REF , the closure of each MATH - stratum of MATH is a disc and the closure of any connected component of MATH is a ball. Thus MATH is the MATH - skeleton of a simple cellular decomposition of MATH. Let MATH be the closures of two connected components of MATH. It remains to show that MATH is connected. If MATH and MA... |
math/0009216 | For the first part of the lemma, let MATH be small enough such that there is no critical parameter of MATH with respect to MATH in MATH. Then, MATH contains a small circle MATH around MATH. For MATH, one obtains MATH from MATH by isotopy induced by MATH. For the second part of the lemma, note that MATH bounds a disc in... |
math/0009216 | Let MATH be a critical parameter of MATH of type MATH, and let MATH be the corresponding critical point. We use the notations of Construction REF. If MATH then we replace MATH by the critical point of MATH of type MATH that corresponds to the critical parameter MATH. Since there are only finitely many critical paramete... |
math/0009216 | For any MATH, define MATH. The homeomorphism type of MATH changes only at critical parameters of MATH with respect to MATH. Let MATH be a critical parameter of MATH with respect to MATH. We choose MATH so that MATH is the only critical parameter of MATH in the interval MATH. Denote by MATH the NAME characteristic of MA... |
math/0009216 | By hypothesis on MATH, it is the MATH - skeleton of a simple cellular decomposition MATH of MATH, and MATH is a MATH - cell contained in the boundary of two different MATH - cells MATH of MATH. We contract MATH along the edges that connect the barycenter of MATH with the barycenters of MATH. By hypothesis on MATH, the ... |
math/0009216 | By definition, there is a MATH - NAME embedding MATH with MATH and MATH. Let MATH be the open MATH - cells of MATH that contain MATH. Pick two different vertices MATH in MATH. We change MATH into a MATH - NAME embedding MATH by pushing a finger from MATH towards MATH and and from MATH towards MATH, so that MATH and MAT... |
math/0009216 | Let MATH be a MATH - cell of MATH and assume that some component MATH of MATH is a circle. It bounds a disc MATH. Let a collar of MATH in MATH be contained in MATH (respectively, in MATH). Since MATH has no critical parameters of type MATH (respectively, MATH), it follows by induction on the number of critical paramete... |
math/0009216 | Let MATH. One observes that MATH comprises at most four points more than MATH, namely the points of MATH. The claim follows by induction, with MATH. |
math/0009216 | Let MATH be not of type MATH. Then any simple arc in MATH with boundary in MATH is parallel in MATH to a sub-arc of MATH. Thus MATH by the minimality of MATH. If MATH is of type MATH, then MATH is isotopic to MATH. If MATH is of type MATH or MATH, then we apply the analysis in the proof of REF and see that one obtains ... |
math/0009219 | First note that for MATH we have MATH, that is, MATH. Due to the fact that MATH is a holomorphic section MATH . By the isometry MATH . If we compare this two expressions and take the definition of the coherent vectors we obtain MATH . |
math/0009219 | Take any MATH with MATH. Denote by MATH the operator of pointwise multiplication of the sections with the function MATH. MATH . Using the isometry REF we can rewrite the last expression and obtain MATH . |
math/0009219 | Let MATH. For the scalar product we calculate MATH . |
math/0009220 | As in CITE we define a map MATH where MATH is the space of all orientation preserving self-homotopy equivalences of MATH. Given MATH we define a self map of MATH, denoted by MATH, via the composite MATH where MATH, respectively MATH, denote inclusion into, respectively projection onto, the first MATH factor of MATH. Be... |
math/0009220 | It is clear that MATH includes terms MATH, just by looking at the cell structure. Note that the cup product is defined using the diagonal map MATH. If MATH generates MATH then MATH. Now we need to define the duals MATH and MATH of MATH and MATH respectively. Let MATH be the element in MATH such that MATH, MATH if MATH ... |
math/0009220 | Although in this section we are working with MATH coefficients we will prove a stronger result by showing that the statement is true also over MATH. Note that MATH and we show that MATH. For example, in the case when MATH consider MATH, where MATH and MATH represent the cycles MATH and MATH respectively. Then MATH . Th... |
math/0009220 | We know that MATH commutes with all other elements and we have equations MATH . We also know that MATH, MATH commutes with MATH for MATH and MATH. These facts together with the two equations imply that we can always bring any copy of MATH to the right of MATH and the MATH, adding, if necessary, words on the MATH. |
math/0009220 | Note that since MATH is a codimension REF submanifold of MATH, there is a circle bundle MATH where MATH is a neighborhood of MATH in MATH: MATH . Therefore for any map representing a cycle MATH we can obtain a cycle in MATH by lifting the map to MATH using the fibration REF. More precisely, using the section MATH we ca... |
math/0009220 | The existence of injections MATH with MATH or REF imply that we have elements MATH in MATH. Let MATH be the subring generated by MATH. Suppose there is an element of minimal degree MATH in MATH. From isomorphism REF we can conclude that such an element would be mapped to a sum of elements MATH with MATH. For some MATH,... |
math/0009220 | Consider the elements of the form MATH, with MATH in MATH. If they are not linearly independent, choose a maximal linearly independent subset MATH. It follows from REF that this is a basis for MATH. Now consider the image in MATH of MATH. This is given by MATH with MATH. These are elements of the form MATH, MATH and th... |
math/0009220 | We already know from REF that the generators of the NAME ring are MATH. Therefore it is sufficient to prove that the only relations between them are the ones in MATH, the commutativity of MATH with MATH and MATH plus the ones on MATH coming from the definition of an exterior algebra MATH. We also know from REF , assumi... |
math/0009220 | The proof is based in the argument used by NAME in CITE with some necessary changes. MATH has subalgebras MATH. From REF we know that MATH is homotopy equivalent to MATH where MATH denotes the universal covering space of MATH. Therefore we have a map MATH. The composite of MATH with the previous map gives us a map MATH... |
math/0009220 | Note that REF shows that MATH is the tensor product of the algebras MATH and MATH. Since MATH we can conclude that MATH is a commutative algebra with two generators of multiplicative order REF. Next we show that MATH has an infinite number of generators and all have order REF. This proves the proposition. As we stated ... |
math/0009220 | The map MATH restricted to MATH or the second MATH factor is just the inclusion in MATH. Moreover, using the NAME formula for homology with coefficients in a field F, we get MATH . Let MATH be MATH or MATH with MATH. Note that in this case MATH has only a generator in dimension REF. Therefore an additive basis for MATH... |
math/0009221 | Consider the function defined by MATH . It is obvious that MATH is a homeomorphism. Since MATH is invertible, the spectrum of MATH is contained in MATH, so by functional calculus there exists an invertible positive element MATH, with MATH, such that MATH, which means MATH . Define the elements MATH . First, we have MAT... |
math/0009221 | Denote, for simplicity, the projection MATH by MATH. The assumption is that MATH. Then we have a MATH-isomorphism MATH, such that MATH. It is obvious that, since MATH, there exists an element MATH such that MATH. By REF , we can find a sequence MATH of invertible elements in MATH (the unit in MATH is MATH), with MATH. ... |
math/0009221 | The proof will be carried on in two steps. Particular Case: Assume MATH. Denote, for simplicity MATH by MATH, and MATH by MATH. By the Parallelogram NAME (see CITE) we have MATH . Since MATH, we get MATH . Using the intermediate value property for MATH, we can find a projection MATH such that MATH. Put MATH. We have MA... |
math/0009221 | Put MATH. If we define MATH, then MATH is invertible, and MATH. This computation shows that it is enough to prove REF in the case when MATH. Let MATH. Arguing as above, there exists some MATH such that MATH. So if we put MATH, we have MATH, with both MATH and MATH self-adjoint idempotents. Since this obviously forces M... |
math/0009221 | Let MATH. As in the proof of the preceding Corollary, there exists an element MATH, such that MATH. Put MATH, MATH, so that MATH. But now, we also have MATH, MATH, which means in particular that MATH. So, if we work in the AW*-factor (again of type MATH) MATH, we will have MATH. On the one hand, using the notations fro... |
math/0009227 | Take a contact form MATH on MATH. Assume that MATH is a coorientation-preserving conservative contactomorphism (the coorientation-reversing case can be treated similary), so MATH preserves a volume form MATH for some continuous positive function MATH. A simple calculation shows that then MATH preserves the continuous c... |
math/0009227 | Indeed, each such MATH represents the action in cohomology of a diffeomorphism of the form MATH, where MATH is the canonical lift to MATH of a periodic automorphism of MATH, MATH . Obviously, MATH is homotopic to a volume preserving contactomorphism of MATH. It suffices to show that the same is valid for MATH and MATH.... |
math/0009228 | First define a map MATH. If MATH, then MATH will have one of two forms: either MATH is not a singular vertex, in which case MATH, or else MATH is a singular vertex, in which case MATH for some MATH. We define MATH by MATH where MATH is the path described in REF . Since MATH preserves source and range, it extends to a m... |
math/0009228 | If MATH is a loop in MATH with no exits, then all the vertices on MATH emit exactly one edge. Hence none of these vertices are singular vertices, and MATH is a loop in MATH with no exits. If MATH is a loop in MATH with no exits, then we claim that none of the singular vertices of MATH can appear in the loop. To see thi... |
math/0009228 | Assume MATH is cofinal and fix MATH. Suppose MATH. Because MATH is cofinal, MATH in MATH. Thus by REF , MATH in MATH. Now let MATH be any singular vertex. Then MATH is the infinite tail MATH added to MATH. By cofinality of MATH, MATH connects to MATH, and since any path that connects to MATH connects to MATH, we know t... |
math/0009228 | For every vertex MATH in MATH, define MATH. For every edge MATH in MATH with MATH not a singular vertex, define MATH. If MATH is an edge in 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-f... |
math/0009228 | We prove the case where MATH has just one singular vertex MATH. If MATH is a sink, then the result follows from CITE. Therefore let us assume that MATH is an infinite-emitter. Given a NAME MATH-family MATH we define MATH and MATH for each positive integer MATH. Note that the MATH's are projections because the MATH's ha... |
math/0009228 | Again for simplicity we assume that MATH has only one singular vertex MATH. 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 and MATH. To prove the proposition, we will show that MATH is a full corner in MATH. Since MATH is genera... |
math/0009228 | Let MATH be a desingularization of MATH. Use REF to construct MATH-families from the given MATH-families. Then apply CITE to get an isomorphism between the MATH-algebras generated by the MATH-families that will restrict to an isomorphism between MATH and MATH. |
math/0009228 | This follows from CITE and the fact that the class of NAME is closed under stable isomorphism (see CITE). |
math/0009228 | By CITE and the fact that pure infiniteness is preserved by passing to corners, every vertex connects to a loop and every loop has an exit implies pure infiniteness. For the converse we note that the proof given in CITE works for arbitrary graphs. |
math/0009228 | Letting MATH denote a desingularization of MATH, we have MATH . |
math/0009228 | That MATH is immediate from REF . We show the reverse inclusion by showing that the generators of MATH are in MATH. Letting MATH denote the NAME MATH-family defined in the proof of REF , the generators for MATH are MATH. Clearly for MATH, we have MATH, so all that remains to show is that for every MATH we have MATH. Le... |
math/0009228 | It is straightforward to see that if MATH, then MATH is a maximal tail CITE. Conversely, suppose that MATH is a maximal tail. We shall create a path in MATH inductively. Begin with an element MATH. If there exists an element MATH for which MATH, then we may use REF of maximal tails to choose a path MATH with MATH and M... |
math/0009228 | It follows from REF that any ideal in MATH has the form MATH for some saturated hereditary set MATH and some MATH. Let MATH be a desingularization of MATH. It follows from REF that MATH is primitive if and only if MATH is primitive. Now suppose that MATH, and hence MATH, is primitive. It follows from CITE that MATH is ... |
math/0009228 | To prove REF , let MATH and MATH. Then using REF we have MATH. We shall show that MATH. To begin, if MATH then MATH where the last step follows from REF . On the other hand, suppose MATH. Then since MATH every vertex MATH must connect to some vertex MATH. So we may replace MATH with MATH and repeat the above argument. ... |
math/0009228 | If MATH is a maximal tail, then from REF we have MATH for some MATH. Similarly, for each MATH we may write MATH for some MATH. Now MATH . So the claim holds when MATH is a maximal tail. Now let us consider the case when MATH is a breaking vertex. It follows from REF that MATH, where MATH is the path on the tail added t... |
math/0009228 | Since MATH is a bijection, we may use MATH to pull the topology defined on MATH in CITE back to a topology on MATH. Specifically, if MATH then MATH for some MATH and we define MATH. But from REF we see that this is equivalent to defining MATH. Now with this topology MATH, and consequently MATH, is a homeomorphism. |
math/0009231 | CASE: The first equation is the dimension formula CITE. The second equation follows from MATH, which is clear from the definition CITE. REF The results are known or trivial for MATH. Now use the transversal slice at MATH CITE to reduce a general case to this case. |
math/0009233 | This follows by a direct computation, making use of the following identities CITE: MATH . |
math/0009233 | In fact, it suffices to show that the ideal MATH is a vector space of dimension MATH. Let MATH be the span of MATH, where: MATH . There is an isomorphism of vector spaces MATH. Remark first that the following identities hold true in MATH: MATH . Then, by direct computation, we obtain that: MATH . From these relations w... |
math/0009233 | One uses the identity MATH for proving the multiplicativity for MATH, and then continue by recurrence for all MATH. |
math/0009233 | Define recursively the modules MATH as follows: MATH . The natural projection MATH is surjective. For MATH it is clear. For MATH we know that MATH, from the exact form of the relations MATH, generating the ideal MATH. We shall use a recurrence on MATH and assume that the claim holds true for MATH. Consider now MATH rep... |
math/0009233 | For MATH this is equivalent to the multiplicativity of the admissible functional. We will use a recurrence on MATH and assume that the claim holds true for MATH. By REF one can reduce the element MATH in MATH to a (non-necessarily unique) normal form MATH, where MATH, MATH and MATH. The multiplicativity of the admissib... |
math/0009233 | Consider two minimal elements MATH and MATH which lie in MATH. Then there exists some path MATH joining them. Since MATH is minimal the closest oriented edge - if it exists - must be in-going; and the same is true for MATH. If this path is not unoriented, then the minimality implies that there are at least two oriented... |
math/0009233 | The proof is similar to that of Pentagon Lemma. |
math/0009233 | We use an induction on MATH. For MATH the claim is obvious. Let now MATH be a word in the MATH's having only positive exponents. CASE: If its degree in MATH is zero or one, then we apply the induction hypothesis and we are done. CASE: If the degree in MATH is REF and MATH contains the sub-word MATH, then again we are a... |
math/0009233 | CASE: First, any o.p.c. can be decomposed into irreducible ones. Further, if each irreducible component satisfies REF then their composition verifies, too. CASE: The reduction transformations acting on different monomials of a linear combination commute with each other. CASE: Obvious. CASE: The simplification transform... |
math/0009233 | By using REF there are only finitely many words MATH on the top line, to check. Furthermore MATH, where MATH. The number of cases to study can be easily reduced, since: CASE: If MATH is the empty word, then REF holds; CASE: By homogeneity it suffices to consider MATH; CASE: Let MATH denote the reversed word associated ... |
math/0009233 | The existence in the first case is equivalent to MATH and MATH, so it is symmetric. In the second case also it is equivalent to MATH for all MATH, so it is again symmetric. The equivalence is trivial. |
math/0009233 | We call the strands which come or arrive to the reduction blocks essential strands. CASE: If the essential arcs coming from the top block are disjoint from those arriving in the bottom block then MATH, MATH, where the first block is contained in MATH and the second one in MATH. These two reductions commute with each ot... |
math/0009233 | There are no restrictions arising from the above identification of two parallel strands because their labels are the same. This means that any permutation involving one of the two strands is also allowable for the second one, as well. Thus, we can always get such a normal form for the respective interactive configurati... |
math/0009233 | The only thing one needs to know is that: It suffices to consider the words MATH as above with MATH and MATH. REF shows that any admissible functional MATH on MATH satisfies: MATH . In the same way one shows that in the simplification process the minimal element in MATH associated to the word MATH must be the product o... |
math/0009233 | A trace MATH defined on MATH should satisfy the following identities: MATH . These are equivalent to: MATH . Remark that these are also sufficient conditions for an admissible functional be actually a trace on MATH. Moreover, the equations above can be expressed in the following algebraic form: MATH . The solutions of ... |
math/0009233 | We will prove that the commutativity constraints are verified by induction on MATH. The claim is true for MATH, and now we suppose that it holds for all algebras MATH, for MATH. In order to prove the claim for MATH it suffices to consider MATH and MATH belonging to some system of generators of MATH, as a module. In par... |
math/0009233 | Since MATH, we can suppose that MATH is a positive braid. All the elements in MATH associated to MATH are polynomials in the variables MATH of degree at most MATH. The substitutions MATH and MATH imply that, if MATH and MATH are representatives of the trace of MATH, then MATH, where MATH is a polynomial in MATH. It fol... |
math/0009233 | We will show that MATH are MATH-polynomials. The other case is analogous. Suppose first that MATH, where MATH is the set of positive braids and MATH is such that MATH. Then MATH where MATH denotes the length of MATH. In the process of computing the value of the trace on the word MATH we make two types of reductions: ei... |
math/0009233 | Let MATH (respectively MATH) denote the image of MATH (and respectively MATH) after the substitutions MATH, MATH and MATH for MATH. It is easy to check that MATH. By some more involved computations we verified that MATH. Since MATH the claim follows. |
math/0009233 | The proof is analogous to the proof of REF . |
math/0009236 | We should check that MATH: MATH . |
math/0009236 | We only check those identities that involve the cyclic operator MATH and leave the rest to the reader. CASE: MATH . MATH . CASE: MATH. For MATH, this is obvious. For MATH we have MATH . CASE: MATH . MATH . CASE: MATH . This is obvious. CASE: MATH . Let MATH. Then, MATH . Thus, MATH . This last identity completes our pr... |
math/0009236 | First we show that MATH commutes with cyclic operators. We have MATH . Since MATH we obtain MATH . On the other hand MATH . Thus, MATH commutes with cyclic operators. Next we show that MATH commutes with coface operators, that is, MATH . We check this only for MATH and leave the rest to the reader. We have MATH and MAT... |
math/0009236 | We construct a cocylindrical module MATH and show that the diagonal MATH of MATH is isomorphic to the cocyclic module MATH. We can then apply the cyclic NAME theorem to derive our spectral sequence. Let MATH . We define the horizontal and vertical cosimplicial and cyclic operators by MATH . One can check that MATH is a... |
math/0009236 | It is not difficult to see that REF defines an associative product on MATH. We check the MATH-module algebra condition. MATH . |
math/0009236 | By REF we can see that MATH . |
math/0009236 | We see that the maps MATH, where MATH define a presimplicial homotopy between MATH and MATH, so that MATH is an isomorphism on NAME cohomology and hence on (periodic)cyclic cohomology. In the case of MATH we see that since MATH the map MATH defines the homotopy MATH between MATH and MATH. So, MATH is MATH on NAME cohom... |
math/0009236 | We first prove that MATH is equivariant, that is, if MATH is an equivariant cochain then MATH: MATH . Next we check that MATH is a cyclic map. First we check that MATH commutes with the cyclic operators: MATH . Now since MATH is a right MATH-comodule algebra, we have MATH and since MATH we can see that MATH . Therefore... |
math/0009236 | As is shown in REF , NAME invariance is a formal consequence of two facts: inner automorphisms induce the identity map on cohomology and a generalized trace map exists. In our case, these are established in REF . |
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