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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 embedding. Next we determine the ratios of the constants MATH. Let MATH be a point of the form MATH. For such MATH, we have the decomposition MATH (see REF ). So we have the splitting MATH where MATH is same as in the proof of REF , and MATH is the theta function MATH of degree REF with the characteristic MATH . By the same way, we have MATH for MATH. We can check that MATH by REF . So we have MATH . By the same way, we see MATH and MATH . Moreover, we have MATH (see REF ). By the transformation REF , we have MATH for any pair of MATH - invariant characteristics MATH, and MATH. By explicit calculation of the above formula, we obtain MATH . Comparing these with REF , we have MATH since it holds MATH . Because the commutativity of diagram is established, by using REF , we see that MATH that is MATH. By the same calculation using REF - REF , we obtain MATH . These equalities gives the ratios in the assertion. |
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 system of MATH. Hence it holds for all MATH. |
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 and MATH, where MATH are the exceptional curves with respect to MATH, and MATH is a general line on MATH. For MATH, there is no divisor MATH on MATH such that MATH since MATH. This implies REF since MATH . |
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 this induces the equality MATH on MATH. Putting MATH, we obtain MATH . |
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. Finally, MATH which is equal to MATH by condition (cm), thereby proving the result. MATH . |
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 . Therefore NAME 's lemma implies that MATH is a quasi-isomorphism. Since MATH is also a quasi-isomorphism (see CITEEF), so is MATH . Finally, the long periodic exact sequences for both cyclic modules, together with the five lemma, give an isomorphism of the cyclic homologies. MATH . |
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 relations identify two cycles in the same orbit, and MATH . The complex is MATH-graded; therefore, to find MATH it is sufficient to consider elements of type MATH, in which MATH is identified with MATH and the MATH belong to MATH . We have MATH iff MATH and the result follows. MATH . |
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 relations identify two cycles in the same orbit, and MATH . The complex is MATH-graded; therefore, to find MATH it is sufficient to consider elements of type MATH, in which MATH is identified with MATH and the MATH belong to MATH . We have MATH iff MATH and the result follows. MATH . |
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 the result. |
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 a boundary simplex of MATH, and let MATH be the edge of MATH with endpoints MATH, MATH. If MATH then we can assume by ambient isotopy of MATH with support in MATH that MATH and MATH has no critical points in the interior of MATH. If MATH then let MATH be the edge of MATH with endpoints MATH. We can assume by ambient isotopy of MATH with support in MATH that MATH and that the critical points of MATH in the interior of MATH are in bijective correspondence to those in the interior of MATH. The latter case occurs at most MATH times for any edge of MATH, since the star of MATH in MATH contains MATH simplices. Thus MATH has MATH critical points in MATH. The MATH - skeleton of the dual cellular decomposition of MATH is contained in MATH. Thus MATH . Since MATH has MATH tetrahedra, MATH - simplices and at most MATH edges, MATH comprises MATH vertices. |
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 - NAME embedding having only critical points in the MATH vertices of MATH, that is, 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 thus the MATH vertices of MATH and at most one point in each of the MATH open edges of MATH. This yields the lemma. |
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 unique open MATH-simplex MATH of MATH. There is a MATH - NAME embedding MATH with the following properties. CASE: MATH. CASE: MATH and MATH has exactly four critical parameters in MATH with respect to MATH, with critical points in the vertices of MATH, and CASE: MATH for any MATH, and MATH has exactly one critical parameter in MATH with respect to MATH, namely with critical point in the barycenter of MATH. Compare the MATH - dimensional sketch in REF . By construction, MATH has at most one critical point in each open simplex of MATH, namely in its barycenter. Thus the critical points of MATH with respect to the MATH - skeleton of the dual cellular decomposition of MATH are its MATH vertices and at most three critical points in each of its MATH open edges. This yields the lemma. |
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 in MATH is a disc, and the intersection of any two of these discs is connected. A MATH - stratum of MATH in MATH is a connected component of MATH. Since MATH is not contained in the boundary of a single connected component of MATH, the closure of any component of MATH is a disc. Since MATH is not a union of two arcs that are each contained in the boundary of a connected component of MATH, the intersection of the closures of two connected components of MATH is connected. Therefore MATH is the MATH - skeleton (including MATH) of a simple cellular decomposition MATH of MATH, and the dual of MATH has no loops or multiple edges, thus, is a triangulation. Any vertex of MATH in MATH gives rise to two vertices of MATH. Thus MATH has MATH vertices. The MATH - skeleton of MATH contains MATH as a path of MATH edges, namely seven edges corresponding to edges of MATH (the thick dotted lines in REF ) and four edges in MATH (the thin dotted lines in the figure). |
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 MATH correspond to the numbers in REF at the MATH - strata of MATH. The broken lines indicate MATH. One sees MATH points of MATH. Thus the MATH crossings of MATH contribute to MATH points in MATH. Similarly, REF show the parts of MATH corresponding to the sub-arcs of MATH in REF . We see MATH respectively MATH points of MATH. In REF , we show two of the four parts of MATH corresponding to REF , and obtain by symmetry REF points of MATH. The broken lines in the figures close up to three meridians (forming MATH), since the writhing number of MATH vanishes. We have MATH. The second claim of the lemma follows, since any MATH - cell of MATH in MATH meets MATH in at most two arcs. |
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 MATH then MATH. If MATH then MATH is connected by construction of MATH. According to REF , if MATH for MATH then MATH is connected. Finally, let MATH and MATH. Then MATH is connected by REF and since we have subdivided the MATH - cells of MATH that meet MATH. |
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. By induction, MATH bounds a disc in MATH and does not bound a disc in MATH. Thus by the general hypothesis, MATH. By maximality of MATH, either MATH is a critical point of MATH of type MATH, MATH or MATH, or MATH is a circle containing a critical point of MATH of type MATH. Since MATH by our general hypothesis, it follows by induction that the connected component of MATH containing MATH in its boundary is not a disc, for MATH. Thus the critical point in MATH is of type MATH. |
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 parameters in MATH, we can choose MATH so that MATH, by repeating this process. Thus we can assume that MATH. Let MATH be local coordinates around MATH as in REF . Let MATH and MATH such that MATH is a disc, and MATH (respectively, MATH) is the only critical parameter of MATH with respect to MATH in MATH (respectively, MATH). Define MATH. We have MATH. There is a compact disc MATH such that MATH is a copy of MATH, for all MATH. Let MATH be as in Construction REF. We change MATH into an embedding MATH in the following way, compare REF . CASE: For MATH, let MATH. CASE: When the parameter MATH increases from MATH to MATH, MATH pushes a finger along MATH towards MATH, so that MATH . CASE: For MATH, set MATH . CASE: In MATH, MATH induces an isotopy mod MATH with support in MATH, relating MATH with MATH. CASE: For MATH, let MATH. When we push a finger along MATH, critical points of MATH of type MATH or MATH occur at MATH. By Construction REF, these are also critical points of MATH of type MATH and MATH. Critical points of MATH in MATH do not occur. It follows that the critical points of MATH and MATH in MATH coincide, MATH, although the order of the corresponding critical parameters has changed. One sees that MATH has exactly two critical points with respect to MATH less than MATH, namely MATH and MATH. One iterates this construction and removes all critical points of type MATH of MATH. By the symmetric construction, one also removes all critical points of type MATH of MATH, and the lemma follows. |
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 MATH. We have CASE: MATH, if MATH is of type MATH or MATH, CASE: MATH, if MATH is of type MATH CASE: MATH, if MATH is of type MATH, and CASE: MATH, if MATH is of type MATH, Since MATH and MATH, and since MATH has at most MATH critical points of type MATH, the lemma follows. |
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 closure of any connected component of MATH is a ball, hence MATH. Thus the two contractions are allowed, that is, do not introduce multiple edges. Any edge MATH is adjacent to exactly two MATH - cells MATH of MATH that are different from MATH. By hypothesis on MATH, we have MATH. Thus we can further contract along the edges of MATH that connect the barycenter of MATH with the barycenters of MATH, whithout to introduce multiple edges. Any vertex MATH is endpoint of exactly two edges MATH of MATH that are not contained in MATH. Since MATH, we can further contract along the edges of MATH that connect MATH with the barycenters of MATH. These MATH contractions yield MATH. |
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 MATH (respectively, MATH) are the critical points of MATH that correspond to the smallest (respectively, biggest) critical parameter of MATH with respect to MATH, denoted by MATH (respectively, MATH). This is possible by introducing at most one critical point in each of the eight edge germs at MATH. Thus MATH. By a small isotopy, we factorize the critical points of MATH in MATH, except MATH, by critical points of type MATH, MATH, MATH and MATH. Any critical point of MATH in the interior of an edge of MATH factorizes by one critical point of type MATH and some other critical points in MATH. Let MATH be a vertex, corresponding to a critical parameter MATH of MATH. If a component of MATH intersects MATH in exactly one (respectively, two) arcs, then the critical point MATH factorizes by one (respectively, two) critical points of type MATH, one critical point of type MATH (with consistent signs), and some critical points outside MATH. Thus we obtain a MATH - NAME embedding MATH with MATH. Now, a MATH - NAME embedding MATH with the claimed properties is given by the restriction of MATH to MATH, for small MATH. |
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 parameters of MATH that MATH (respectively, MATH) contains a circle, in contradiction to the hypothesis on MATH. Thus MATH intersects the MATH - cells of MATH in arcs. Similarly one shows that MATH is a disjoint union of open discs, and MATH is a disjoint union of open MATH - balls. Thus the open MATH - strata of MATH are discs. Since both MATH and MATH intersect MATH and MATH is connected, MATH has an intrinsic vertex in MATH. Since MATH is not a critical parameter of MATH, MATH is simple. In conclusion, MATH is the MATH - skeleton of a simple cellular decomposition MATH of MATH. Let MATH be a MATH - cell of MATH. If MATH then it separates two MATH - cells of MATH, since MATH is dual to a triangulation. If MATH then it separates the MATH - cell of MATH corresponding to MATH from another MATH - cell of MATH. |
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 from MATH by deletion of one MATH - stratum with at most MATH vertices in the boundary. If MATH is of type MATH, then one obtains MATH from MATH by deletion of the two connected components of MATH. We estimate the number of intrinsic vertices in the boundary of these MATH - strata. Let MATH be the MATH - stratum of MATH that contains MATH. By minimality of MATH, there are at most MATH arcs in MATH, each connecting the two components of MATH. Thus by REF both components of MATH have at most MATH vertices in their boundary. |
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. Because MATH acts as the identity on MATH, MATH is an orientation preserving self homotopy equivalence of MATH, that is, MATH. Defining MATH in an analogous way using the second MATH factor of MATH, we have thus constructed the desired map given by MATH . It is clear from the construction that MATH restricted to MATH is just the inclusion MATH . Now we use the following theorem (see CITE). The space of orientation preserving self-homotopy equivalences on REF - sphere has the homotopy type of MATH, where MATH is the universal covering space for the component in the double loop space on MATH containing the constant based map. This proves that MATH is not homotopy equivalent to MATH but we have, using the NAME formula with field coefficients, MATH thus the map MATH induced by injection is injective for any field coefficients. |
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 or REF, and MATH if MATH. MATH is defined in a similar way. We know that the cup product of MATH and MATH does not vanish, so we have MATH. Hence MATH . Therefore we see that MATH must have a component in MATH. The only element like that is MATH, so MATH must involve this element. The result for MATH follows immediately from that for MATH and MATH by multiplication. |
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 . Thus MATH. |
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 can lift MATH to a cycle in MATH. This is represented by a map MATH. Now note that there is a map from the image of MATH to MATH given by MATH, where MATH and we can choose MATH close to MATH. In fact, we can choose MATH so close to MATH such that MATH also. Then using the MATH action, represented by MATH, we define a map to MATH, representing a cycle in MATH: for each MATH in the image of MATH we get a loop around MATH defined by MATH. Therefore the cycle MATH lifts to MATH which represents a cycle in MATH. |
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 is not a polynomial in the MATH. Take the largest such MATH. By the isomorphism in REF this would create an element in MATH that is not a polynomial in MATH and MATH. But this is impossible because this would give rise to a new generator in MATH corresponding to this new element in MATH and this contradicts the minimality of 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 the set MATH is linearly independent. Therefore it is an additive basis for the space spanned by elements of the form MATH. Note that MATH has a subalgebra isomorphic to MATH and an additive basis for this is MATH where MATH and MATH are equal to REF or REF, so an additive basis for MATH will contain all elements of this form. To prove the theorem in the case MATH we need to show that the set MATH where MATH and MATH is an additive basis of MATH. We start by proving that these elements generate additively MATH. Suppose we have an element MATH. From REF it is known that every element in MATH is a sum of elements of the form REF. Therefore MATH . It is also known that MATH is in MATH and if MATH is not in MATH we can write it as sum of elements in MATH. Thus MATH is a sum of elements in MATH. The next step is to show that the elements in MATH are linearly independent. We know that for a fixed degree MATH, the dimension of MATH is given by MATH because of the vector space isomorphism REF. But this is precisely the number of elements in MATH of degree MATH. So they must be linearly independent otherwise their span would not be the space MATH. This means that the set MATH defined above is an additive basis for MATH. Therefore MATH is an isomorphism. In the case MATH, MATH maps MATH to MATH and this is not an element in the form REF. However we can prove an analogous result to REF stating that any word in the MATH is a sum of elements of the form MATH. This is clear because MATH for all MATH and MATH. Repeating the steps for the case MATH and using isomorphism REF instead of isomorphism REF it follows easily that MATH is also an isomorphism. |
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 , assuming only these relations, that any word in these generators is a sum of elements of the form REF. Thus if we prove that this set of elements give an additive basis of MATH we prove that there are no more relations between the generators MATH, because the existence of another relation would give rise to one between the elements of the form REF and they would not be linearly independent. We will prove that by induction. The induction hypothesis is that up to dimension MATH elements of form REF are linearly independent, thus there are no relations between them up to dimension MATH. Suppose there was one of minimal degree MATH in MATH. The first step is to show that it would be between the MATH only. Assume it was given by a finite sum of the type MATH where MATH is a word on the MATH and MATH where MATH is an element in MATH and MATH equals MATH or MATH. Then from REF with MATH it follows that we must have MATH . We can group together the terms in which MATH is the same, thus we can write the relation as MATH where now MATH runs over a set of basis elements of MATH. This implies that we have a relation of the type MATH . Using REF with MATH we show that it is between the MATH, because MATH implies MATH . A relation in the MATH projects, under the map MATH, to one on the MATH in MATH. Using isomorphism REF this would give a relation in degree MATH between the MATH and MATH in MATH. But this contradicts the induction hypothesis because such relation implies one in MATH with MATH at most equal to MATH. |
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. Thus MATH has a subalgebra MATH. Composing these inclusions of MATH and MATH as subalgebras of MATH with cup product multiplication in MATH we get a map MATH . MATH is an algebra homomorphism because MATH is commutative and it is compatible with filtrations (the obvious one on MATH and the filtration MATH on MATH coming from the fibration on the left in REF). The degeneration of the spectral sequence at the MATH - term implies that MATH is an algebra isomorphism. This proves isomorphism REF in the case MATH. For the case MATH note that the map MATH is surjective, so there are MATH and MATH in MATH such that MATH and MATH generate the ring MATH, where MATH is the generator of the cohomology of MATH and MATH is such that MATH. MATH is the generator of the cohomology of the MATH factor. Now we need to prove that MATH in MATH in order to claim that the subalgebra of MATH generated by MATH and MATH is isomorphic to MATH. MATH in MATH . Using isomorphisms REF we can show that the rank of MATH is REF. But in MATH the cycles MATH are linearly independent. We will show that MATH evaluated on all these classes is REF. The only one at which is not obviously REF is MATH. Let the map MATH represent REF - cycle MATH. Then MATH and this vanishes because MATH is a spherical class, that is, MATH. Again composing these inclusions of MATH and MATH as subalgebras of MATH with cup product multiplication we get a map MATH which is an algebra isomorphism. |
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 in REF , the isomorphism MATH is the composite of the restriction MATH with integration over the fiber of the projection MATH of the fibration REF. Since integration over the fiber kills the elements of MATH, for each MATH there is MATH such that MATH where MATH, MATH and MATH. Therefore we have MATH, because MATH (proved in REF ). Thus MATH. Since MATH is an isomorphism we get MATH. From knowing that all generators in MATH have multiplicative order REF it follows that MATH must have an infinite number of generators, just by comparing the dimensions MATH for each MATH. Recall that REF implies that MATH contains a free noncommutative ring on REF generators that projects to non-zero elements in MATH and the dimensions MATH increase as MATH increases. |
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 is given by MATH where MATH or MATH. Thus comparing REF and an additive basis for MATH we conclude that the homology groups of MATH and MATH are the same. We just need to show that MATH induces those isomorphisms. The elements MATH, MATH and MATH are the images of the generators of MATH and MATH under the injective maps MATH and MATH induced by inclusions MATH and MATH. On the other hand the restriction of MATH to MATH is given by the map MATH and we know that MATH maps REF - dimensional generator of MATH to the element in MATH obtained as the NAME product of MATH and MATH. This proves that MATH induces an isomorphism in homology with MATH and MATH coefficients for all primes MATH with MATH. If MATH then an additive basis for MATH is given by the set of elements of the form REF. It follows from REF that the homology groups MATH and MATH are isomorphic. The elements MATH are the images of the generators of MATH and MATH are the images of the generators of MATH. In this case the map MATH stated in REF takes the generators of MATH to the elements MATH which are the three generators of the free noncommutative subalgebra of MATH. Therefore we get another isomorphism in homology. |
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 MATH and using REF we also have MATH . Secondly, since by taking inverses, REF yields MATH so we also get MATH . Finally, we notice that MATH and using REF we also have MATH . If we define the matrix MATH, then REF, and REF give MATH, while REF, and REF give MATH, so MATH is indeed unitary. (Here MATH denotes the unit in MATH.) Let us observe now that MATH . These equalities prove exactly that MATH . |
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. Define the sequence MATH . By REF , one can find two sequences MATH and MATH in MATH, and a sequence of unitaries MATH, such that MATH . Define MATH, MATH, and MATH, MATH, so that we have MATH . Fix now a complex number MATH. On the one hand, we have MATH . This gives, for every MATH, the equalities MATH . Notice however that, using the unitary invariance (REF for MATH unitary), together with REF, gives, for every MATH, the equalities MATH . On the other hand, if we define MATH (which is obviously a unitary in MATH), then for all MATH we have MATH . This gives MATH . Using the unitary invariance, combined with real scalar homogeneity (REF with MATH), the equality REF gives MATH which forces MATH . Combining REF with REF gives MATH . It is obvious that, by construction, we have MATH, which means that MATH. Using the norm continuity of the quasi-trace (see CITE), combined with REF gives the desired result. |
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 MATH, and MATH. Let us work in the AW*-algebra MATH. Obviously MATH is again an AW*-factor of type MATH, so it carries its quasi-trace MATH. By the uniqueness of the quasi-trace, it is obvious that MATH . Notice that MATH, so in particular MATH belongs to MATH. In MATH, we have MATH, and MATH, which means that MATH. (Here MATH denotes the dimension function in MATH). By REF , we get MATH which combined with REF yields MATH which obviously proves REF. General Case. One knows (see CITE) that MATH is also an AW*-factor of type MATH. Moreover, if we denote by MATH the quasi-trace of MATH, we have (as above) MATH . Define the idempotent MATH, and the projection MATH. Using REF we have MATH. If we denote by MATH the dimension function of MATH, then by REF we also have MATH, which gives MATH. Using the obvious equality MATH, by the particular case above applied to MATH, we get MATH . |
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 MATH, we get MATH . |
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 from the proof of REF , we have MATH . On the other hand, we have MATH so we get MATH . Finally, using the preceding Corollary, we get MATH . |
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 contact form MATH, and it is easy to conclude that MATH is elliptic. |
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. We explain this for MATH (the argument for MATH is analogous). In fact, the required volume preserving contactomorphism MATH homotopic to MATH is given by the following explicit formula: MATH . This completes the proof. |
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 map on the finite path space MATH. In particular, for MATH define MATH. It is easy to check that MATH is injective, that it preserves source and range, and that it is onto the set MATH. We define MATH similarly. In particular, if MATH, define MATH. If MATH is a finite path whose range is a singular vertex MATH, we define MATH, where MATH is the tail in MATH added to MATH. To show that MATH is a bijection, we construct an inverse MATH. Notice that every MATH either returns to MATH infinitely often or it ends up in one of the added infinite tails. More precisely, MATH has one of two forms: either MATH or MATH, where each MATH is either an edge of MATH or a MATH. We define MATH by MATH and we define MATH . It is easy to check that MATH and MATH are inverses, and we have established REF . To prove REF , let MATH and MATH in MATH. Then there exists a finite path MATH in MATH such that MATH and MATH for some MATH lying on the path MATH. Note that the vertices of MATH that are on the path MATH are exactly the same as the vertices on the path MATH. Hence MATH must also be a vertex on the path MATH. Now, because MATH preserves source and range, MATH is a path that starts at MATH and ends at MATH, which is a vertex on MATH. Thus MATH. For the converse let MATH and MATH, and suppose that MATH in MATH. Then there exists a finite path MATH in MATH with MATH and MATH for some vertex MATH on the path MATH. Notice that if MATH is a vertex on one of the added infinite tails, then MATH must have passed through MATH, and so must have MATH. Thus we may assume MATH. Now MATH is a finite path in MATH that starts and ends in MATH, so it can be pulled back to a path MATH with source MATH and range MATH. Since MATH lies on the path MATH, it lies on the path MATH, and thus MATH is a path from MATH to some vertex of MATH. Hence MATH in MATH. |
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 this, note that if MATH is a sink in MATH, then it cannot be a part of a loop in MATH; and if MATH is an infinite-emitter in MATH, then MATH is the source of two edges, which would necessarily create an exit for any loop. Since none of the singular vertices of MATH appear in MATH, it follows that MATH is a loop in MATH with no exits. This establishes REF . Now suppose MATH is the base of exactly one simple loop MATH in MATH. Then MATH is a simple loop in MATH. If there were another simple loop MATH in MATH based at MATH, then MATH would be simple loop in MATH based at MATH that is different from MATH. Thus if MATH satisfies REF , then MATH satisfies REF . Now suppose MATH satisfies REF . Let MATH be the base of a simple loop MATH in MATH. If MATH, then MATH is a simple loop in MATH based at MATH. Since MATH satisfies REF , there is a simple loop MATH in MATH different from MATH. Certainly, MATH is a simple loop in MATH and, because MATH is injective, MATH must be different from MATH. Now suppose MATH is on an added infinite tail; that is, MATH for some MATH. Then MATH must have the form MATH for some MATH. Now, MATH is a simple loop in MATH based at MATH and hence MATH is a simple loop in MATH based at MATH. Since MATH satisfies REF , there must be another simple loop MATH in MATH based at MATH. Now MATH will be a simple loop in MATH based at MATH. If MATH is not a vertex on MATH, then MATH will be another simple loop based at MATH that is different from MATH. On the other hand, if MATH is a vertex of MATH, then MATH has the form MATH, where MATH. Since MATH is a simple loop based at MATH, we know that MATH for MATH. Hence MATH is not a vertex on the path MATH. Therefore MATH is a simple loop based at MATH. Furthermore, it is different from the loop MATH, because if they were equal then we would have MATH, which contradicts the fact that MATH and MATH are distinct. Thus MATH satisfies REF . |
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 that there is a path MATH from MATH to MATH. But then MATH is a path from MATH to MATH in MATH. Hence MATH. Now assume MATH is cofinal and for every singular vertex MATH we have MATH. If MATH has a sink MATH, then since MATH is cofinal it follows that MATH. Furthermore, since MATH it must be the case that MATH is the only sink in MATH. Hence MATH is obtained from MATH by adding a single tail at MATH. Now if MATH, then since MATH we must have that MATH eventually ends up in the tail. If MATH, then either MATH is in the tail or MATH. Since MATH this implies that in either case MATH. Hence MATH is cofinal. Now assume that MATH has no sinks. Let MATH and MATH. We must show that MATH in MATH. We will first show that it suffices to prove this for the case when MATH and MATH. If MATH, a vertex in one of the added infinite tails, then because MATH has no sinks, MATH must be the source of some edge MATH with MATH and we see that MATH in MATH implies MATH in MATH. Likewise, if MATH, a vertex in the infinite tail added to MATH, then MATH in MATH implies MATH in MATH. Thus we may replace MATH by MATH. Hence we may assume that MATH and MATH. Since MATH is a finite path in MATH whose source is in MATH, REF implies that MATH, where MATH is either an infinite path in MATH or a finite path in MATH ending at a singular vertex. If MATH is an infinite path, then cofinality of MATH implies that MATH and REF implies that MATH. If MATH is a finite path ending at a singular vertex, then MATH by assumption and so MATH. Thus MATH is cofinal. |
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-family. |
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 have orthogonal ranges. Furthermore, MATH for every MATH. Now for every integer MATH define MATH and set MATH . For every MATH define MATH acting on the MATH component of MATH and zero elsewhere. That is, MATH. Similarly, for every MATH with MATH define MATH on the MATH component. For each vertex MATH on the infinite tail define MATH to be the projection onto MATH. That is, MATH. Now note that because the MATH's are non-decreasing, MATH for each MATH. Thus for each edge of the form MATH we can 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 MATH. To see this recall that MATH, and thus MATH. Therefore we can define MATH by MATH where the nonzero term appears in the MATH component. We will now check that the collection MATH is a NAME MATH-family. It follows immediately from definitions and the NAME relations on MATH that MATH for every MATH that is not of the form MATH or MATH, and that MATH for every MATH not on the infinite tail. Furthermore, it is easy to check using the definitions that the MATH's are mutually orthogonal and that MATH for every edge MATH on the infinite tail. Now note that for every MATH, MATH . Finally, let MATH be a vertex on the infinite tail. The edges emanating from MATH are MATH and MATH, and we have MATH where the nonzero term is in the MATH component. Also MATH where the nonzero term is again in the MATH component. We then have the following: MATH . Thus MATH and we have established that MATH is a NAME MATH-family. It is easy to verify that the bulleted points in the statement of the lemma are satisfied. |
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 generated by a NAME MATH-family, in order to show that MATH it suffices to prove that MATH satisfies the universal property of 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 in MATH whose source is MATH. Now by the universal property of MATH, we have a homomorphism MATH from MATH onto MATH that takes MATH to MATH and MATH to MATH. For any MATH we have MATH, so MATH. Let MATH. If MATH, then MATH and MATH. Finally, if MATH then MATH for some MATH, and MATH so that 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. We now show that MATH. Just as in CITE, we have that MATH converges strictly in MATH to a projection MATH and that for any MATH with MATH, MATH . Therefore the generators of MATH are contained in MATH and MATH. To show the reverse inclusion, let MATH and MATH be finite paths in MATH with MATH. We need to show that MATH. If either MATH or MATH does not start in MATH, then MATH by the above formula. Hence we may as well assume that both MATH and MATH start in MATH. Now, if MATH as well, there will exist unique MATH with MATH and MATH. In this case, MATH and MATH, so MATH . On the other hand, if MATH then MATH for some MATH. We shall prove that MATH by induction on MATH. Suppose that MATH for any paths MATH and MATH with MATH. Then if MATH we shall write MATH for finite paths MATH and MATH with MATH. Now there are precisely two edges, MATH and MATH with source MATH. Thus MATH which is in MATH. Hence MATH. Finally, we note that MATH is full by an argument identical to the one given in CITE. |
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. Let MATH and MATH. Then MATH . Now since MATH we see that MATH and hence MATH. Consequently, MATH. Similarly, whenever MATH, then MATH. Now, by definition, every MATH with MATH has MATH. Therefore the above equation shows us that MATH which is an element of MATH by the previous paragraph. Hence MATH. |
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 MATH. Now having chosen MATH, we do one of two things: if MATH for all MATH, we stop. If there exists MATH such that MATH, then we choose a path MATH with MATH and MATH. We then continue in this manner to produce a path MATH, which may be either finite or infinite. Note that since MATH has either a finite or countable number of elements, we may choose MATH in such a way that MATH for all MATH. Now if MATH is an infinite path we define MATH. On the other hand, if MATH is a finite path then one of two things must occur. Either MATH is a singular vertex or there is an edge MATH with MATH and MATH. Continuing in this way, we see that having chosen MATH, either MATH is a singular vertex or there exists MATH with MATH and MATH. Using this process we may extend MATH to a path MATH that is either infinite or is finite and ends at a singular vertex. Now since every vertex on MATH is an element of MATH we certainly have MATH. Also, for every element MATH there exists a MATH such that MATH so we have MATH. |
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 a maximal tail in MATH. Thus by REF we have MATH for some MATH. Now MATH is either an infinite path in MATH or a finite path in MATH ending at a singular vertex. In either case MATH is a maximal tail in MATH. Furthermore, MATH . Therefore MATH is a maximal tail. Now if MATH, then we are in the case described in REF and the claim holds. Let us therefore suppose that there exists MATH. If we define MATH to be the vertices on the tail added to MATH, then we see that MATH implies that MATH. Now for each vertex MATH with MATH there are two edges, MATH and MATH, with source MATH. Since MATH and MATH, it must be the case that MATH has the form MATH for some finite path MATH in MATH. Consequently, MATH is a finite path in MATH ending at MATH, and MATH. Now let MATH. Note that if MATH and MATH, then MATH since MATH is hereditary. Because MATH it follows that we must have MATH. Furthermore, since MATH there exists MATH with MATH and MATH. But then MATH and MATH and hence MATH. Thus MATH, and by REF is a breaking vertex. All that remains is to show that MATH . Let us suppose that MATH. If MATH, then MATH. But because the MATH's for MATH can only reach elements of MATH and MATH, the only way to have MATH for all MATH is if we have MATH. Hence MATH is the only element of MATH and MATH. Thus we have established all of the claims in REF . For the converse let MATH be a maximal tail. Consider the following two cases. CASE: MATH . We shall show that MATH is a maximal tail in MATH. Since MATH is a saturated hereditary subset of MATH, the set MATH certainly satisfies REF in the definition of maximal tail. We shall prove that REF also holds. Let MATH. If it is the case that MATH, then we must also have MATH, and hence there exists MATH such that MATH and MATH in MATH. But then MATH and MATH and MATH in MATH. On the other hand, if one of the MATH's is not in MATH, then it must be on an infinite tail MATH. Because MATH and MATH, we must have MATH for some MATH. Thus we can replace MATH with MATH and reduce to the case when MATH. Hence MATH also satisfies REF and is a maximal tail. Consequently, MATH is a primitive ideal by CITE, and MATH is a primitive ideal by REF . CASE: MATH for some breaking vertex MATH and MATH. As in Case I, it suffices to show that MATH satisfies REF in the definition of maximal tail. To see this, let MATH. If MATH, then we must have MATH and MATH. If MATH, then MATH must be on one of the added tails in MATH. Since MATH we must have that MATH is an element on MATH. In either case we see that MATH can reach an element of MATH in MATH. Consequently, MATH and MATH clearly satisfies REF . |
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. Thus we have proven REF . For REF , let MATH be a breaking vertex and set MATH and MATH. Then MATH. An argument similar to the one above shows that MATH. |
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 to MATH. Now suppose that MATH. Fix MATH. Note that either MATH in MATH or MATH is on the infinite tail added to MATH in MATH. Because MATH, there are infinitely many edges in MATH from MATH to vertices that connect to MATH. Thus no matter how far out on the tail MATH happens to be, there must be an edge in MATH whose source is a vertex further out on the tail than MATH and whose range is a vertex that connects to a vertex MATH for some MATH. Since MATH we must have MATH and thus MATH. Now assume that MATH. Then every vertex on the infinite tail attached to MATH connects to a vertex in MATH. In fact it is true that every vertex on the infinite tail attached to MATH connects to a vertex in MATH, which implies that every vertex on the infinite tail connects to a vertex in MATH. But this implies that there must be infinitely many edges from MATH to vertices that connect to MATH. Thus MATH. |
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 we derive that MATH for all MATH, and hence MATH. The other inclusion is immediate. The proposition is then a consequence of the previous lemma. |
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 represented by a word in the MATH's having only positive exponents. We assume that the degree of the word in the variable MATH is minimal among all linear combinations of words (with positive exponents) representing MATH. CASE: If this degree is less or equal to REF then there is nothing to prove. CASE: If the degree is REF then either MATH, MATH so using the induction hypothesis we are done, or else MATH, where MATH. Therefore MATH where MATH by the induction hypothesis and MATH. CASE: If MATH then MATH can be reduced to MATH. CASE: If MATH then MATH hence the degree of MATH can be lowered by one, which contradicts our minimality assumption. CASE: If MATH then MATH. One derives that: MATH hence we reduced the problem to the case when MATH is a word of type MATH. CASE: If the degree of MATH is at least REF we will contradict the minimality assumption. In fact, in this situation MATH will contain either a sub-word MATH, with MATH and MATH, or else a sub-word MATH, with MATH. CASE: In the first case using the induction we can write MATH, with MATH. CASE: Furthermore, if MATH then MATH, and hence the degree of MATH can be lowered by one. CASE: If MATH then MATH, and again its degree can be reduced by one unit. CASE: If MATH then either MATH or MATH is equal REF. Assume that MATH. We can therefore write: MATH contradicting again the minimality of the degree of MATH. CASE: In the second case we can write also MATH, MATH with MATH. CASE: If MATH or MATH equals REF then, after some obvious commutations the word MATH contains the sub-word MATH which can be replaced by MATH and hence diminishing its degree. CASE: If MATH then MATH. We use the homogeneity to replace MATH by a sum of elements of type MATH. Each term of the expression of MATH which comes from a factor which has the exponent MATH, has diminished its degree. The remaining terms are MATH, so they contains a sub-word MATH whose degree we already know that it can be reduced as above. This proves our claim. Eventually recall that the NAME traces MATH on MATH are multiplicative, and hence they satisfy: MATH. Therefore there is a unique extension of MATH from MATH to MATH. This ends the proof of our proposition. |
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 admissible functionals implies that: MATH . By the recurrence hypothesis one knows that: MATH and since: MATH one derives our claim. |
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 edges. Therefore one can find a sequence of open pentagon configurations lining on the path which joins MATH to MATH. We apply then REF iteratively, whenever we see one such o.p.c., or one o.p.c. appears at the next stage, as in the figure below: When this process stops, we find two semi-oriented paths MATH and MATH having the same endpoint MATH. So MATH and MATH. From minimality both these paths must be unoriented, and thus MATH and MATH are weakly equivalent. |
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 able to apply the induction hypothesis. CASE: By using REF several times one can also suppose that no exponents greater than REF occur in MATH. CASE: If the degree of MATH is REF then MATH with MATH. The induction hypothesis applied to MATH implies that MATH with MATH. Then several transforms of type REF will do the job. CASE: Consider now that the degree in MATH is at least REF. Then MATH contains a sub-word which has either the form MATH with MATH, or else one of the type MATH. The second case reduce to the first one as above. In the first case assume that MATH for some MATH. Then several applications of REF lead us to consider the sub-word MATH. CASE: If MATH we use two times REF and we are done. CASE: Otherwise use either REF and then REF if MATH or else both REF and then REF , if MATH. This proves that every vertex descends to MATH. But these vertices have not outgoing edges, as can be easily seen. When we use the unoriented edges some new vertices have to be added. But it is easy to see that these new vertices do not have outgoing edges either. Since any vertex has a semi-oriented path ending in MATH our claim follows. |
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 transformations for MATH and MATH commute with each other. |
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 to MATH. If REF holds for MATH, then it also holds for MATH; CASE: Several cases, as MATH, can be easily tested at hand. The nontrivial situations are those when a REF -move (and then a REF -move) can be applied. It suffices therefore to check the case of MATH, since MATH is its reversed and the remaining MATH REF are consequences of these two. Then we have the situation depicted in the diagram: MATH where MATH are those from REF . If we apply REF whenever it is possible on MATH, then after a long computation we find a common minimal element for MATH and MATH. |
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 other. CASE: If there is an essential strand labeled MATH which intersects some essential strand of the other block, then it will intersect all of them. In particular MATH commutes with all letters of the reduction block. Moreover, a simple verification shows that, if MATH commutes with all letters of the monomial from the left hand side of one formula among REF , then it will commute with the elements from the right hand side of the same formula. This shows that the commutations depicted in the diagram can be realized after the first reduction transformation (of the upper block). This implies our claim. |
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 configuration. |
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 of the minimal elements associated to the two words MATH and MATH. This proves the claim. This shows that the cases left unverified can be reduced to those which we claimed above. |
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 these equations are those claimed above. |
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 particular we will choose the set of generators MATH from REF. For MATH, MATH the claim is obvious. It remains to check whenever MATH holds true. There are three cases to consider: CASE: MATH; CASE: MATH, with MATH; CASE: MATH, with MATH. which will be discussed in combination with the following six sub-cases: CASE: MATH, and MATH, CASE: MATH, and MATH, MATH, CASE: MATH, and MATH, MATH, CASE: MATH, MATH, MATH, MATH, CASE: MATH, MATH, MATH, MATH, CASE: MATH, MATH, MATH, MATH. REF are trivially verified by an immediate calculation. Furthermore one obtains: CASE: MATH CASE: MATH. CASE: MATH . CASE: MATH CASE: MATH . For the remaining cases, we need also to know more precisely the form of MATH. Specifically, let us write MATH with MATH and MATH. We can show by a direct computation that the equalities hold also for REF . Moreover, using Maple we have found that in the cases (REF, ii) and (REF, iii), for MATH, there are two additional identities, which are not consequences of those in the previous lemma. The difference MATH is expressed as a linear combination with polynomial coefficients in MATH and MATH. For arbitrary elements MATH as above the three traces above seem to be unrelated. A sufficient condition for the commutativity to hold is that the coefficients in front of these terms vanish. We derive therefore the following obstructions: CASE: MATH CASE: MATH . Furthermore the vanishing of MATH and MATH insures the commutativity of the admissible functional. |
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 follows that: MATH where we put: MATH . |
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: either one uses the relations from MATH, or else one replaces MATH by MATH (respectively MATH by MATH). In the first alternative the word MATH is replaced by MATH, where the MATH are words from MATH, the coefficients MATH are MATH-polynomials, and MATH. In the second case the word MATH is replaced by MATH where MATH and MATH. When we substitute for MATH and MATH their values as functions on MATH and MATH one finds that: MATH where MATH and MATH are MATH-polynomials. In particular: MATH . Now MATH where MATH are MATH-polynomials. Thus, it follows that: MATH where MATH are MATH-polynomials. Now, remark that the same reasoning holds true for non necessarily positive MATH, by getting rid of the negative exponents in MATH by making use of the cubic relation. The only difference is that one has to take into account the normalization factor in front of the trace. The claim follows. |
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 proof of the theorem. |
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 MATH . Since MATH and MATH the result is MATH . The proof of compatibility of MATH with codegeneracies is similar and we leave it to the reader. The theorem is proved. |
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 cocylindrical module. The proof is very long, but is totally similar to the proof of REF and is left to the reader. Next we show that the diagonal of MATH, MATH, is isomorphic with the cocyclic module MATH. To this end, we define the maps MATH and MATH by MATH . By a rather long computation one can verify that MATH is a morphism of cocyclic modules and MATH. Now, we can apply the generalized cyclic NAME theorem to derive our spectral sequence. Again the argument is similar to that used in CITE and hence omitted. |
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 cohomology and since MATH, in the normalized form MATH and MATH defines a homotopy between MATH and MATH. Thus, MATH is MATH on cyclic and periodic cyclic cohomology. |
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 commutes with coface operators MATH. Since MATH commutes with all coface operators and it is easy to check that it also commutes with codegeneracy operators. This proves the proposition. |
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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