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math/0008108 | Let MATH. For each MATH let MATH denote the corresponding NAME coordinate on MATH. In terms of NAME coordinates, MATH acts on a point of MATH with coordinates MATH by taking it to the point with coordinates MATH. Let MATH be NAME coordinates of MATH in MATH. By assumption, MATH for every MATH. Let MATH be defined by the following equations on NAME coordinates, MATH . It's clear that MATH. Let MATH with NAME coordinates MATH. Let MATH . Note that MATH. Set MATH. We divide the proof in five steps. CASE: Let MATH with NAME coordinates MATH. Then the following two conditions hold. MATH . Indeed, let's check REF first. If MATH then MATH by definition. Conversely, suppose first that MATH. Then MATH, and thus REF holds. Suppose now that MATH. Let MATH. Then MATH. Let MATH such that MATH and MATH. Using an induction argument, we need only show that MATH. Since MATH and MATH, there are MATH such that MATH and MATH. NAME find MATH such that MATH and MATH. If MATH or MATH, set MATH or MATH, respectively. Suppose that MATH and MATH. Then MATH, and hence there is MATH. Set MATH. Then MATH and MATH. Since MATH, REF imply that MATH. Hence, since MATH, REF imply that MATH as well. Thus REF holds. Let's now check REF . If MATH then MATH, and hence REF holds trivially. Assume that MATH, and fix MATH. Define MATH by letting MATH and MATH for each MATH, where MATH is chosen such that MATH and MATH. Observe that such MATH exists because REF holds. Moreover, MATH is independent of the choice of MATH, because if MATH is such that MATH and MATH, then MATH, and hence the claimed independence of MATH follows from REF . If MATH and MATH are distinct elements of MATH, then MATH for every MATH such that MATH and MATH. Indeed, if either MATH or MATH, then REF holds by definition of MATH or MATH. Suppose that MATH. If MATH, then MATH . If MATH, then MATH . Either way, REF holds. Now, REF follows from REF and an induction argument. REF is complete. CASE: Let MATH with NAME coordinates MATH. If REF hold, then MATH. Indeed, let MATH as in REF . Define MATH and MATH by letting for each MATH, MATH . Define a map MATH by letting MATH for each MATH and each MATH. For each MATH let MATH and MATH be NAME coordinates of MATH. Because of our choice of MATH, the power of MATH in the expression for MATH is minimum exactly among those MATH. Let MATH be the limit of MATH as MATH goes to REF, and MATH . NAME coordinates of MATH. Of course, MATH. Now, MATH if and only if MATH. Moreover, if MATH then MATH where the last equality holds by REF . So MATH, and hence MATH. REF is complete. CASE: If MATH is an ordered tripartition of MATH such that MATH, then MATH and MATH are NAME coordinates of MATH, where MATH . Indeed, let MATH be a basis of MATH in MATH. Write MATH with MATH . Since no NAME coordinate of MATH is zero, we may assume that the matrix MATH is the identity. Since MATH, a basis of MATH is given by the vectors MATH defined by MATH . In particular, MATH. It follows as well that MATH are NAME coordinates of MATH. REF is complete. CASE: If MATH, then there is an ordered tripartition MATH of MATH such that MATH, and such that MATH is in the orbit of MATH defined by REF . Indeed, let MATH with NAME coordinates MATH. Since MATH, we have MATH, and hence REF hold by REF . Since REF holds, we may replace MATH by a certain point in its orbit and assume that MATH for every MATH. Hence, using REF , it follows from REF that MATH, where MATH. If MATH set MATH. REF is complete in this case. If MATH then MATH, and thus MATH. In addition, MATH because MATH. Let MATH be any non - empty subset and put MATH. Now, MATH since MATH. Then MATH. REF is complete. CASE: If MATH is an ordered tripartition of MATH such that MATH, then MATH. Indeed, let MATH be NAME coordinates of MATH. By REF , MATH . If MATH then MATH and MATH. If MATH then MATH and MATH. In either REF holds. By REF as well, REF holds trivially with MATH. By REF , MATH, thus finishing REF show the lemma. |
math/0008108 | Let MATH . For each MATH let MATH denote the corresponding NAME coordinate on MATH. For each MATH let MATH denote the corresponding NAME coordinate on MATH. Let MATH. Let MATH be NAME coordinates of MATH in MATH and MATH be NAME coordinates of MATH in MATH. By assumption, MATH for every MATH and MATH for every MATH. Let MATH be defined by the following equations on NAME coordinates: MATH where MATH . Given MATH with NAME coordinates MATH and MATH, let MATH . Put MATH . Note that MATH and MATH. Set MATH and let MATH . We divide the proof in five steps. CASE: If MATH then MATH. Indeed, let MATH with NAME coordinates MATH and MATH. Since MATH, there is MATH such that MATH for every MATH and MATH for every MATH. Clearly, REF hold. As for REF , it holds if and only if MATH for all MATH and MATH satisfying REF . Since MATH, we need only show that MATH . But the three equalities above hold because REF holds. So MATH, finishing REF : Let MATH with NAME coordinates MATH and MATH. Then the following four conditions hold. MATH . Indeed, REF follow as in the proof of REF . Let's check REF now. Suppose there are MATH and MATH. Since MATH and MATH, there is MATH such that MATH and MATH. Since MATH and MATH, and since REF holds, there is MATH such that MATH and MATH. Since MATH, REF imply that MATH. However, MATH, reaching a contradiction. Analogously, if MATH then MATH. Suppose now that there are MATH and MATH. Since MATH and MATH, and since REF holds, there is MATH such that MATH and MATH. Since MATH and MATH, there is MATH such that MATH and MATH. Since MATH, it follows from REF that MATH. However, MATH, reaching a contradiction. Analogously, if MATH then MATH. So REF is proved. Let's check REF . As in the proof of REF , there are MATH and MATH such that MATH . If MATH and MATH are distinct elements of MATH, let MATH and MATH such that MATH and MATH. Since MATH, REF imply that MATH. So MATH, and hence REF holds. REF is complete. CASE: Let MATH with NAME coordinates MATH and MATH. If REF hold, then MATH. Indeed, for MATH, MATH, MATH and MATH, define the map MATH by letting MATH for each MATH and MATH for each MATH. NAME choose MATH, MATH, MATH and MATH as indicated below. Let MATH as in REF . For each MATH set MATH and MATH. For each MATH set MATH and MATH. For each MATH and each MATH set MATH and MATH, and put MATH . Choose the remaining MATH and MATH in such a way that MATH. As for the remaining MATH and MATH there are three cases to consider. CASE: If MATH set MATH and MATH if MATH and MATH and MATH if MATH. CASE: In REF does not hold, but MATH set MATH . CASE: In REF does not hold, but MATH set MATH and MATH as in REF , but exchanging MATH with MATH, MATH with MATH, and the MATH with the MATH. If REF hold, then REF holds as well. So the three cases above are independent. Moreover, one of the three cases occur. Indeed, if MATH and MATH, then REF occurs by REF . On the other hand, if MATH then REF holds, and if MATH, then REF holds. We chose MATH in such a way that it factors through MATH. Indeed, since MATH, it is enough to check that MATH for all MATH, what can easily be done in each of the three cases above. Let MATH for each MATH, and let MATH and MATH be NAME coordinates of MATH. Because of our choice of MATH, the power of MATH in the expression for MATH is minimum exactly among those MATH. Likewise, the power of MATH in the expression for MATH is minimum exactly among those MATH. Let MATH be the limit of MATH as MATH goes to REF. As in the proof of REF , we have MATH. Then MATH, finishing REF : Let MATH. Then there are ordered tripartitions MATH and MATH of MATH and MATH, respectively, such that REF hold, and such that MATH is in the orbit of MATH defined by REF . Indeed, let MATH and MATH be the NAME coordinates of MATH. By REF , MATH. Hence REF hold by REF . Since REF holds, we may replace MATH by a certain point in its orbit and assume that MATH for every MATH and MATH for every MATH. Let MATH and MATH. As in the proof of REF , we have MATH. There are four cases to consider. CASE: If MATH and MATH, set MATH and MATH. Clearly REF holds, whereas REF hold by REF . Since MATH, REF is complete for REF . CASE: If MATH and MATH, then set MATH for a certain non - empty subset MATH to be defined below. Set MATH. As in the proof of REF , we have MATH. Clearly REF holds. We need only choose MATH such that REF hold, and we'll do it according to whether REF or REF occurs. If REF or REF occurs, set MATH. If REF does not occur but REF does, set MATH . With any of the above choices REF hold for MATH and MATH. CASE: If MATH and MATH, proceed as in REF , exchanging MATH with MATH. CASE: If MATH and MATH, let MATH and MATH be non - empty subsets, and put MATH . As in REF , we need only choose MATH and MATH such that REF hold. If REF occurs, set MATH and MATH. If REF occurs but REF does not, set MATH and MATH. If REF occurs but REF does not, set MATH and MATH. With any of the above choices REF hold for MATH and MATH. REF is complete. CASE: If MATH, where MATH and MATH are ordered tripartitions of MATH and MATH such that REF hold, then MATH. Indeed, let MATH and MATH be NAME coordinates of MATH. As in the proof of REF , MATH . If MATH then MATH and MATH. If MATH then MATH and MATH. In either REF holds. Analogously, REF holds. Now, REF is simply a restatement of REF if MATH and MATH. In the remaining cases, REF can be obtained from REF as well. Finally, as in the proof of REF holds trivially with MATH. By REF , MATH, thus finishing REF show the lemma. |
math/0008108 | We take two preliminary steps. CASE: If MATH, then MATH and either MATH or MATH. Indeed, let MATH, and assume that MATH. Let MATH . Then MATH. Now, MATH is not a base point of MATH for any MATH by REF . So MATH is the minimum order of MATH at MATH for each MATH. Analogously, MATH is the minimum order of MATH at MATH for each MATH. So MATH. Assume now that MATH. Let MATH. We claim that MATH if and only if MATH. In fact, MATH by REF . So, if MATH then REF holds. Conversely, suppose REF holds. Since MATH, by NAME - NAME, MATH imposes independent conditions on MATH. So, since MATH by REF , it follows from REF that MATH, where MATH. We may thus assume that MATH. Suppose by contradiction that MATH. Replacing MATH by a larger subset, if necessary, we may assume MATH. Since MATH imposes independent conditions on MATH, there is MATH such that MATH for every MATH and MATH for MATH. So MATH by REF . Let MATH be the blow - up of MATH along MATH. By REF , there is an invertible sheaf MATH on MATH such that MATH and MATH, and such that MATH is the image of the restriction map MATH. Let MATH lifting MATH and MATH. So MATH. Since MATH and MATH, it follows from REF that MATH. So MATH, and hence MATH for every MATH. The contradiction proves our claim. Now, MATH. By analogy, REF holds if and only if MATH. So MATH. REF is complete. CASE: Assume that MATH and either MATH or MATH. Then MATH. Indeed, it follows from REF and the hypothesis that MATH if MATH. Now, if MATH then MATH by REF is complete. REF prove REF follows by analogy. Let's prove REF . By definition, MATH and MATH are the images of MATH in MATH and MATH, respectively, for each MATH. Hence, if MATH then MATH and MATH. Conversely, assume that MATH and MATH. By REF , we have MATH. In addition, MATH when MATH and MATH when MATH. So, if MATH and MATH then MATH by REF . Now, if MATH or MATH then MATH where the first and last equalities follow from REF is proved. Let's prove REF . Suppose that MATH. Then MATH as well. Thus MATH by REF . Analogously, MATH. By REF , MATH. REF is proved. Finally, let's prove REF . Suppose MATH for a certain MATH. Then the numerical data MATH associated with MATH and MATH is equal to that associated with MATH and MATH. Analogously, the numerical data MATH associated with MATH and MATH is equal to that associated with MATH and MATH. Thus MATH by REF . |
math/0008108 | By REF , MATH for each MATH and each MATH. In other words, each MATH is homogeneous. Let MATH. Clearly, MATH if and only if MATH. So, by REF , MATH if and only if MATH, MATH . On the other hand, MATH and MATH for each MATH and MATH. So, as MATH runs over MATH, the associated numerical data MATH runs over a finite set. Therefore, the covering MATH of MATH is finite. Assume now that MATH. Let MATH. Set MATH and put MATH . Then MATH if MATH and MATH if MATH. By REF , MATH . Now, since MATH, it follows that MATH . Then MATH by REF . So MATH is convex. |
math/0008108 | To ease notation, let MATH . Let MATH consist of the points of the form MATH for all MATH satisfying MATH where MATH means the smallest non - zero number among the four listed. If REF holds, then MATH for every MATH, and hence MATH. We take three preliminary steps. CASE: For each MATH there are ordered tripartitions MATH of MATH and MATH of MATH such that REF below hold. MATH . Indeed, for each MATH let MATH . Since MATH, there is a unique MATH such that MATH . Let MATH. Then REF hold. By analogy, we construct MATH satisfying REF . REF is complete. CASE: If MATH and there are ordered tripartitions MATH of MATH and MATH of MATH satisfying REF for a certain MATH, then REF hold. Indeed, if MATH then MATH and MATH. Hence MATH . Assume that MATH. Let MATH and MATH be ordered tripartitions of MATH and MATH satisfying REF for a certain MATH. If there are MATH and MATH, then MATH and MATH by REF . Now, MATH because MATH. Using REF we get MATH reaching a contradiction. If there are MATH and MATH, then MATH and MATH by REF . Using REF as before, we get MATH, reaching a contradiction. So REF holds. By analogy, REF holds. REF is complete. CASE: Let MATH for MATH such that REF holds. Let MATH and MATH be ordered tripartitions of MATH and MATH such that REF hold. If REF holds, then so do REF . If REF holds, then so do REF . Indeed, define MATH by letting for each MATH, MATH . By REF , MATH . If REF holds, let MATH for (any) MATH, and define MATH by letting for each MATH, MATH . Then MATH . In addition, it follows from REF that MATH for every MATH, with MATH if and only if MATH. So MATH, proving REF . If REF holds, let MATH for (any) MATH, and define MATH by letting for each MATH, MATH . As before, MATH, proving REF . REF is complete. Let's prove that MATH satisfies REF . Indeed, let MATH for MATH such that REF holds. By REF , there are ordered tripartitions MATH of MATH and MATH of MATH such that REF hold. So REF hold by REF . If MATH then REF hold as well by REF . The uniqueness of the tripartitions follows from REF . Let MATH be an open neighbourhood of MATH. Let's prove REF . Assume MATH, and let MATH be an ordered tripartition of MATH satisfying REF . Define MATH by letting for each MATH, MATH . Let MATH, where MATH is such that MATH lies in MATH. Now, MATH for each MATH because MATH. Then REF follows from REF . So REF hold by REF . Analogously, REF holds. Let's prove REF . Assume that MATH. Let MATH and MATH be ordered tripartitions of MATH and MATH satisfying REF . Define MATH according to the following three cases. CASE: If MATH let for each MATH, MATH REF If REF does not hold, but MATH let for each MATH, MATH REF If REF does not hold, but MATH define MATH for each MATH as in REF , but now exchanging MATH with MATH. As in the proof of REF , it follows from REF that the three cases above are independent, and one of them occurs. Let MATH, where MATH is such that MATH lies in MATH. Now, MATH for each MATH and MATH for each MATH because MATH. Then REF hold in each of the three cases above. So REF hold by REF . |
math/0008108 | Let MATH. By REF , we may assume that MATH. Let MATH be the open neighborhood of MATH given by REF . To ease notation, let MATH . In addition, if MATH let MATH. Let MATH, MATH, MATH and MATH be the invertible sheaves given by REF . As in REF, let MATH denote the blow - up of MATH along MATH. Fix isomorphisms MATH and MATH for each MATH, and MATH and MATH for each MATH. If MATH and MATH, choose them such that MATH patch MATH and MATH to the sheaf MATH given by REF . Consider the corresponding evaluation maps, MATH . Let MATH and MATH. Let MATH and MATH. Let MATH . Consider the natural actions of the tori MATH and MATH on MATH and MATH, and their respective actions on MATH and MATH. Let MATH and MATH denote the orbits of MATH and MATH under these actions. Let MATH be an open neighborhood of MATH. We divide the proof in three steps. CASE: NAME show that the closure MATH satisfies MATH if MATH. In fact, assume that MATH. Then MATH. By REF , the map MATH induces a closed embedding MATH such that MATH. For each ordered tripartition MATH of MATH let MATH and denote by MATH the orbit of MATH in MATH under the action of MATH. By REF , all the NAME coordinates of MATH in MATH are non - zero. Thus, by REF , the closure MATH is the union of the orbits MATH obtained from all ordered tripartitions MATH of MATH such that MATH . Since MATH, REF is equivalent to REF . By REF , each ordered tripartition MATH of MATH satisfying REF satisfies REF for a certain MATH. Conversely, for each MATH there is an ordered tripartition MATH of MATH satisfying REF . Now, let MATH be an ordered tripartition of MATH satisfying REF , and let MATH such that REF hold. Let MATH . Let MATH and MATH be the following compositions, MATH where in each composition the first map is the natural injection, and the third map is the natural surjection. Equivalently, MATH and MATH are the evaluation maps corresponding to the isomorphisms MATH and MATH induced by MATH and MATH for each MATH. Let MATH and MATH. Since MATH by REF , from REF we get MATH. In addition, MATH. Let MATH, and denote by MATH the closed embedding induced by MATH, as in REF . Let MATH denote the orbit of MATH under the natural action of MATH. Since MATH, there is a bijection MATH taking a subspace MATH to MATH. Note that MATH . By REF , MATH . So, it follows from REF that MATH, and hence MATH. Thus, by REF , MATH finishing REF : NAME show that the theorem holds if MATH. In fact, assume that MATH. By REF , MATH and MATH for every MATH. If MATH as well, then MATH for every MATH, and hence the theorem holds. On the other hand, if MATH then the closure MATH satisfies MATH where the second equality holds by REF . Now, the case where MATH is analogous. REF is complete. CASE: NAME show that the theorem holds if MATH. Since MATH, by REF , the map MATH induces a closed embedding MATH such that MATH. Let MATH . Let MATH be the orbit of MATH under the induced action of MATH on MATH. By REF , we have MATH, where MATH. For each pair of ordered tripartitions MATH of MATH and MATH of MATH, define the subspaces MATH and MATH by REF . By REF , all the NAME coordinates of MATH in MATH and of MATH in MATH are non - zero. Thus, by REF , the closure MATH is the union of the orbits under MATH of the pairs MATH obtained from all ordered tripartitions MATH of MATH and MATH of MATH satisfying REF . As in REF is equivalent to REF . Since REF is equal to REF , by REF , for each pair consisting of an ordered tripartition MATH of MATH and an ordered tripartition MATH of MATH satisfying REF , there is a certain MATH satisfying REF . Conversely, for each MATH there are ordered tripartitions MATH of MATH and MATH of MATH satisfying REF . Now, let MATH be an ordered tripartition of MATH and MATH an ordered tripartition of MATH satisfying REF , and let MATH such that REF hold. Let MATH and MATH be given by REF , and put MATH . Let MATH and MATH be the evaluation maps corresponding to the isomorphisms MATH and MATH induced by MATH and MATH for each MATH. Let MATH and MATH be the evaluation maps corresponding to the isomorphisms MATH and MATH induced by MATH and MATH for each MATH. Let MATH and MATH. In addition, let MATH, and denote by MATH the closed embedding induced by MATH, as in REF . Let MATH and MATH. Then MATH and MATH, as in REF . Let MATH, and denote by MATH the closed embedding induced by MATH. Let MATH denote the orbit of MATH in MATH under the action of MATH. Let MATH denote the orbit of MATH under the natural action of MATH on MATH. Since MATH and MATH, there is a bijection MATH taking a pair MATH to MATH. Let MATH. Note that MATH . By REF , we have REF . In addition, by REF , MATH . We will apply REF to show that MATH. First, note that MATH. Indeed, MATH for every MATH by REF . Now, MATH is non - empty and contained in MATH, and MATH for every MATH. Hence MATH for every MATH. Analogously, MATH. So, REF applies. If MATH, we get MATH immediately. Assume MATH, and let MATH. Then MATH where the middle equality follows from REF . Analogously, MATH. Since MATH patch MATH and MATH to the sheaf MATH given by REF , it follows from REF that MATH patch MATH and MATH to the sheaf MATH where MATH is the pull - back of MATH to the blowup MATH of MATH along MATH. Using REF , we get MATH where MATH is the pull - back to MATH of the dualizing sheaf MATH of MATH. Applying REF , we get MATH. Since MATH, we get MATH from REF . Thus, by REF , MATH finishing REF . |
math/0008108 | By REF , the covering MATH of MATH is finite. Hence, it follows from REF that MATH is projective. Let's prove that MATH is connected. Let MATH. For each MATH let MATH . By REF , each MATH is an interval, and finitely many MATH suffice to get a covering MATH of MATH. Assume MATH. Then MATH and MATH. By REF , for each MATH either MATH or MATH; in either case MATH. Since each MATH is irreducible by REF , it follows that MATH and MATH lie in the same connected component of MATH. So MATH is connected. If MATH, then MATH for each MATH. Hence REF follows from REF . Assume now that either MATH or MATH. Let's show that MATH is of pure dimension MATH. Let MATH. Then MATH by REF . So, it's enough to show that either MATH or there is MATH such that MATH. To ease notation, let MATH . There are two cases to consider. CASE: Assume that there is MATH such that MATH . Let MATH satisfying REF , and put MATH . Then MATH. For each MATH such that MATH let MATH. We claim that MATH. Indeed, if MATH then either MATH or MATH, and hence MATH. On the other hand, since MATH or MATH, if MATH then MATH or MATH, and hence MATH. Therefore MATH if MATH. Now, if MATH then MATH. Moreover, it follows from REF that MATH if MATH and MATH if MATH. By REF , MATH if MATH. Thus MATH by REF , where MATH. On the other hand, either MATH or MATH. By REF , MATH. So MATH, finishing the proof. CASE: Assume now that there is no MATH satisfying REF . In addition, assume that MATH. We claim that MATH . Indeed, suppose that MATH. Since there is no MATH satisfying REF , we must have MATH. Then MATH by REF . So MATH because MATH. Now, REF holds for MATH, reaching a contradiction. Thus MATH. Analogously, MATH. Now, REF holds for MATH. Thus MATH. Finally, if MATH then MATH by REF . Thus MATH, and REF holds. Define MATH by letting MATH if MATH and MATH otherwise. Put MATH . Then MATH. Let MATH for each MATH such that MATH. Now, MATH for every MATH because MATH. So MATH, and hence MATH if MATH. Let MATH. Then either MATH for some MATH or MATH. Since MATH, in either case MATH by REF . On the other hand, MATH if MATH. Hence MATH as in REF . |
math/0008108 | Let MATH, and denote by MATH the curve gotten from MATH by splitting the branches of MATH at each MATH and connecting them by a chain of MATH rational smooth curves; see REF Let MATH . Let MATH. By definition, MATH and MATH are the limit canonical aspects with foci on MATH and MATH of a certain regular smoothing MATH of MATH. Let MATH be the induced smoothing of MATH; see REF By REF , the limit NAME scheme MATH of MATH satisfies MATH where MATH and MATH are the ramification divisors of MATH and MATH. Now, since MATH we have MATH . Using the above expressions in REF , since MATH, we get MATH. So MATH is a limit NAME divisor. Conversely, if MATH is a smoothing of MATH, then there is MATH such that MATH induces a regular smoothing MATH of the curve MATH obtained from MATH by splitting the branches of MATH at each MATH and connecting them by a chain of MATH rational smooth curves; see REF Let MATH and MATH be as in REF . By REF there are vector subspaces MATH and MATH such that MATH and MATH are the limit canonical aspects of MATH with foci on MATH and MATH, respectively. Let MATH. By definition, MATH. Now, proceed as in the above paragraph to conclude that the limit NAME scheme MATH of MATH satisfies MATH. |
math/0008108 | Assume that MATH. Then we may choose MATH such that MATH. Define MATH by letting MATH for each MATH. Then MATH and MATH. Let MATH . Then MATH and MATH. For each MATH let MATH . Let MATH be the dense open subset of points MATH such that MATH and MATH. Then MATH for each MATH. We claim that there are MATH such that MATH for MATH. Indeed, let's prove the claim by induction on MATH. First, REF hold trivially for MATH. Now, let MATH and suppose that there are MATH such that REF hold. Let MATH. Then there is MATH such that MATH . By REF , MATH for each MATH. Now, if MATH then MATH, and hence MATH. However, MATH because MATH, reaching a contradiction. So MATH, and hence MATH. In addition, if MATH for a certain MATH, then MATH . Now, MATH because MATH. Moreover, MATH. So MATH . Since MATH, it follows that MATH contradicting REF . Hence MATH for any MATH. The induction proof of our claim is complete. For each MATH let MATH. The following conditions on MATH, MATH define an open subset MATH. By our claim, MATH. Let MATH. Since MATH for each MATH, also MATH. As shown in REF imply that MATH. Since MATH, and since MATH and MATH, it follows that MATH for each MATH. So MATH are ramification points of MATH, and hence of MATH. Since MATH it follows from REF that MATH sit on the support of a limit NAME divisor on MATH. Since MATH, there must be a MATH - dimensional family of limit NAME divisors. |
math/0008108 | If MATH then MATH is simply a point by REF , thus proving REF in this case. We may assume from now on that MATH. (The case where MATH is completely analogous.) Given MATH, the number of MATH such MATH is MATH. Thus the number of MATH such that MATH is MATH. If MATH then MATH if and only if MATH. Indeed, MATH because MATH. In addition, if MATH then the number of MATH such that MATH is at least MATH, and hence MATH because MATH. Let MATH be the number of strata MATH of MATH with MATH. By REF , MATH if and only if MATH and MATH. Now, given MATH satisfying MATH, there is MATH such that MATH and MATH. (For instance, pick MATH given by MATH for each MATH.) If also MATH satisfies MATH and MATH then there is MATH such that MATH, and hence MATH by REF . It follows that sending MATH to MATH gives a REF correspondence between the set of strata MATH of MATH with MATH and the set of MATH with MATH. Thus MATH. If MATH then MATH, thus proving REF in this case. Moreover, MATH if and only if MATH, thus proving REF in this case. Assume from now on that MATH. By analogy, MATH, where MATH is the number of strata MATH of MATH with MATH. Fix MATH. Let MATH such that MATH . Then there is MATH such that MATH if and only if MATH . Now, MATH satisfy REF if and only if there is MATH such that MATH . So, there is a REF correspondence between the set of pairs MATH satisfying REF for a certain MATH and the set of MATH such that MATH. The number of MATH such that MATH is MATH. Let MATH be the number of strata MATH of MATH such that MATH . Then MATH. Now, by REF holds if and only if MATH . It follows from REF that sending MATH to MATH gives a REF correspondence between the set of strata MATH of MATH satisfying REF and the set of pairs MATH satisfying REF for certain MATH and REF for a certain MATH. Thus MATH . Since MATH, we proved REF . Without loss of generality, assume from now on that MATH. Clearly MATH, and hence MATH. So, MATH only if MATH, and hence only if MATH. It follows that MATH is irreducible only if MATH. Assume from now on that MATH. Then MATH for each MATH. It follows from REF that MATH if and only if MATH. Hence MATH, thus proving REF . Now, if MATH then MATH by REF . |
math/0008109 | Recall that MATH is the odd automorphism given by the matrix REF . Consider first MATH, that is, the MATH-invariants in MATH. Letting MATH and MATH, it is clear that this space consists of elements of the form MATH and MATH, and hence is isomorphic to MATH. Therefore, since MATH is a subgroup (since all elements are now even) of MATH generated by the MATH copies of MATH's, we have MATH . |
math/0008109 | By NAME - NAME duality REF we have MATH as MATH module, where the summation is over strict partitions MATH with length MATH. Therefore combined with REF this gives us MATH . Now by REF and the fact that irreducible MATH-modules are self-contragredient we have MATH where the summation is over all strict partitions of length MATH. |
math/0008109 | We will establish REF together. The argument we will present follows closely the one used in CITE to derive the NAME duality from the MATH-duality. The MATH-duality says that MATH, where the summation is over all strict MATH with MATH. Observe that the space MATH may be identified with the zero-weight space MATH. Putting these together we have MATH where MATH runs over all strict partitions MATH. So to recover REF it suffices to show that MATH is isomorphic to the irreducible MATH-module MATH, which is the contents of REF. First note that the normalizer of MATH acts on the MATH-weight spaces by permutation. Since the determinant character is invariant under permutation of weights, we see that MATH is invariant under the action of MATH. Now the induced right action (see REF) of the operators MATH (see REF ) when acting on MATH satisfy the commutation relations of the MATH's in MATH, for MATH. This action of the MATH's combined with the action of MATH then gives a right action of MATH on MATH. Set MATH in the remainder of the proof. (However, it is convenient to continue making the distinction between MATH and MATH.) Consider MATH . Recall that MATH, MATH are the standard coordinates of MATH. It is not difficult to see that the space MATH inside MATH, which is MATH, may be identified with the space spanned by vectors of the form MATH, where MATH denotes either MATH or MATH and MATH is a permutation of MATH. Thus this space is in bijection with the space MATH. On this space the normalizer of MATH acts, and so we obtain an action of the symmetric group MATH. We also have an action of MATH (see REF ), which gives rise to the action of MATH in MATH, for MATH. This action combined with that of MATH gives a left action of MATH on MATH. Furthermore the induced right action of MATH above and this action commute in the usual sense according to REF. Also our left action of MATH permutes the upper indices of the vector MATH, whereas the left action of the MATH's changes the parity of MATH. Thus this is the left regular action of MATH. Our right action of MATH, on the other hand, permutes the lower indices of MATH, whereas our right action of MATH changes the parity of MATH. Thus our right action is the right regular representation of MATH. From the general theory of finite supergroup CITE, the left-hand side of REF , which is isomorphic to MATH under the left and right actions of MATH, is equal to the summation MATH over all strict partitions of MATH. Decomposing MATH in the right-hand side of REF into a direct sum of irreducible MATH-modules, we see that this is only possible when each MATH itself is irreducible as a MATH-module. Comparing with REF , we see that this irreducible module is isomorphic to MATH. |
math/0008112 | Let MATH, for MATH. Clearly, if MATH, then MATH for each MATH. A simple induction argument shows that it suffices to prove the proposition when MATH, which we shall assume for the remainder of the proof of REF . The Proposition will be a consequence of the following lemma. Let MATH, MATH be as in REF with MATH. If MATH, then there exists a formal mapping MATH such that MATH, where MATH, and MATH . We now show that REF follows from REF , whose proof will be given below. Set MATH and MATH. Observe that the formal mapping MATH, where MATH satisfies MATH. Hence, since MATH, it follows that MATH. Also, by REF , MATH, which proves REF . |
math/0008112 | Let MATH, and hence MATH. Without loss of generality, we may assume that MATH, MATH, are linearly independent over MATH and hence MATH, MATH, are also linearly independent over MATH. We write MATH, with MATH, and for example, MATH for the MATH . Jacobian matrix of MATH with respect to MATH. Consider the MATH-linear system MATH with MATH, MATH, and MATH. We take MATH, MATH and solve REF for MATH, which is possible since the rank of MATH is MATH. Moreover, if we denote by MATH, MATH, indices such that MATH with MATH, then the solution MATH is of the form MATH where MATH. Write MATH where MATH is given by REF . Observe that the coefficients of MATH are in MATH, MATH, and MATH. We define MATH, MATH, to be the formal flow of MATH, that is, the solution of the initial value problem MATH where MATH denotes the vector of coefficients of MATH. The fact that MATH implies that MATH. Since MATH, it is easy to verify that the identity REF holds. This completes the proof of REF and hence that of REF . |
math/0008112 | We shall prove the proposition by induction on MATH. If MATH, then the proposition holds with either MATH or MATH, depending on whether or not all vector fields in MATH vanish at MATH. Now assume that the proposition holds for MATH; we shall prove it for MATH. If all the MATH vanish at MATH, then again MATH is the only formal manifold satisfying the conclusion of the proposition. If not, without loss of generality, and after making a formal change of coordinates, we may assume that the vector field MATH is in MATH. Consider the NAME subalgebra MATH of all formal vector fields in MATH of the form MATH. It is easy to see that any MATH may be written uniquely in the form MATH with MATH and MATH. We write MATH and consider the NAME algebra MATH of formal vector fields in the indeterminates MATH obtained from MATH by replacing MATH by MATH. By the inductive hypothesis, there is a formal manifold MATH to which all the vector fields in MATH are tangent and such that MATH. Let MATH, MATH, be generators of the manifold ideal MATH. Let MATH be the formal manifold whose manifold ideal is generated by the MATH, MATH, regarded as elements of MATH (independent of MATH). Then all the vector fields in MATH are tangent to MATH and MATH. We shall first show that any MATH is tangent to MATH. For MATH and MATH as before, we shall use the notation MATH. Since any MATH can be written in the form MATH with MATH, and since the MATH are independent of MATH, it suffices to show that any vector field in MATH is tangent to MATH. For this we expand MATH as a power series in MATH and obtain MATH where MATH is the vector field in MATH given by MATH . Since MATH it is tangent to MATH and, hence, we have MATH for some MATH. Hence, MATH is tangent to MATH by REF . It remains to show that MATH. Since all the vector fields in MATH are tangent to MATH, we have MATH. The opposite inequality follows from the assumption that MATH and the fact that MATH. To finish the proof of REF , we must show that the formal manifold MATH provided by REF is unique, and if MATH is as in the statement of REF then MATH. Let MATH be generators of the manifold ideal MATH. Since MATH is tangent to MATH, it follows that MATH for MATH and MATH. Let MATH be the ideal in MATH generated by MATH. Note that MATH, MATH, and hence MATH is a manifold ideal of the same codimension as MATH. Since MATH is tangent to MATH, MATH is tangent to MATH. By the induction hypothesis, the manifold ideal MATH is unique and MATH. Thus, MATH implies that there are MATH such that MATH . Now, since MATH we conclude, by substituting REF , that MATH for MATH. Hence, MATH. This also proves the uniqueness of MATH, since if the dimensions of MATH and MATH were the same, then necessarily MATH. This completes the proof of REF . |
math/0008112 | It suffices to prove the proposition using the special choice of coordinates MATH, MATH made above. Thus, we may assume that MATH, MATH. Since MATH is a formal manifold of codimension MATH and MATH, there are MATH such that MATH is generated by MATH. By using the special form of the MATH (that is, substituting MATH for MATH), we may assume that MATH is independent of MATH, that is, MATH, MATH. Since the MATH are tangent to MATH and MATH, MATH and MATH, we obtain MATH where MATH and the MATH are MATH-matrices whose entries are in MATH. Let MATH be a fundamental system of solutions for the system of differential equations given by REF with MATH. Denote by MATH the MATH-matrix in which the MATH are columns. Since MATH is a solution of the system REF (in particular with MATH) there exists MATH, where MATH, such that MATH. Since MATH is invertible, we have MATH and, hence, each component of MATH is in MATH. Moreover, MATH generate MATH. Proceeding this way to eliminate MATH, we finally obtain MATH such that MATH, MATH, satisfy the conclusion of REF . |
math/0008112 | By REF , it follows that the iterated NAME mapping MATH is of the form MATH where MATH . In particular, we obtain MATH . The identity REF is an immediate consequence of REF . This completes the proof of REF . |
math/0008112 | By iterating REF , we obtain, for any multi-index MATH, MATH . In particular, applying REF with MATH, we conclude that, for each multi-index MATH, MATH . Since MATH, MATH are tangent to MATH, we deduce that MATH for all MATH and all multi-indices MATH. Hence, MATH, which proves that MATH and MATH. Assume that MATH and MATH, for all MATH. We shall prove it for MATH. By applying REF with MATH, we conclude, using also REF , that for MATH, MATH where the last equality follows from the inductive hypothesis and the fact that the MATH are tangent to MATH. Hence, MATH is independent of MATH. Consequently, MATH which is REF by the inductive hypothesis. A similar argument, using instead the identity MATH which follows from REF , shows that MATH for MATH. This completes the proof of REF . To prove REF let MATH be defined as in REF. By using REF , and REF we first observe that MATH . In light of REF , and REF , the conclusion REF is an immediate consequence of REF below with MATH and MATH. |
math/0008112 | The implication MATH is clear. The opposite implication follows by an inductive argument based on the facts that MATH is a homomorphism, and MATH and MATH are derivations. For instance, we have, for MATH, MATH . A similar identity holds for MATH. The details are left to the reader. The equivalence MATH follows by simply writing REF in terms of the components MATH of MATH. The last statement of REF is a direct consequence of REF . |
math/0008112 | A simple computation (which is left to the reader) shows that REF holds with MATH and MATH . |
math/0008112 | Among MATH and all their repeated commutators, we can pick out, by the assumption in the lemma, MATH such that MATH evaluated at MATH span MATH. By REF , we may assume that there are MATH and MATH, MATH for MATH, such that MATH . Let us write MATH . By assumption the vectors MATH evaluated at MATH span MATH. It is not difficult to see that this fact implies that MATH are linearly independent over MATH (and hence span MATH) since they are linearly independent over MATH when evaluated at MATH. On the other hand, by REF , these vectors are all in the image of the Jacobian MATH, considered as a linear map from MATH to MATH. Hence MATH and the lemma is proved. |
math/0008112 | Let MATH be such that MATH form a basis for MATH. By multiplying the MATH by suitable power series (clearing the denominators), we obtain MATH and MATH still form a basis for MATH since MATH is linear over MATH. |
math/0008112 | The main step in the proof is to show that there are formal vector fields MATH on MATH and MATH, MATH for all MATH, such that MATH . To prove this, we write MATH and denote by MATH the homomorphism defined by MATH for MATH. By abuse of notation, we also denote by MATH the mapping given by applying MATH to each component. By REF , for any MATH, we have MATH, where MATH . Hence, we conclude that MATH, and therefore by REF MATH . On the other hand, observe that if MATH, with MATH linearly independent over MATH, then MATH are linearly independent over MATH. Therefore we have MATH . Hence, by using REF , and the assumption that MATH, we obtain MATH . We therefore conclude again using REF that MATH . Now we write MATH and MATH . By REF and the second equation of REF , we have MATH, and hence, by REF , MATH. By clearing denominators and again applying REF ( MATH ), it follows that there are MATH and MATH, MATH, such that REF holds. In view of the first equation of REF , the conclusion of REF for MATH now follows by applying REF with MATH. |
math/0008112 | We first note that the conclusion of the theorem is independent of the choice of the NAME variety mapping MATH. This can be seen by using the notation of REF and observing that MATH, where MATH is a parametrization of the formal manifold MATH . Hence it suffices to prove REF using the special choice of NAME variety mapping MATH introduced in REF. Let MATH be the formal manifold whose ideal is generated by MATH and MATH, for MATH. The conclusion in REF is now a consequence of the definition of MATH in REF and of REF. |
math/0008114 | CASE: Suppose MATH and MATH. Then MATH as MATH implies MATH as MATH and that there exists a MATH such that MATH. Let MATH be given by MATH. Then it is not difficult to see that MATH is an isomorphism. Therefore MATH is isomorphic to MATH by REF . And by the same argument MATH is isomorphic to the groupoid MATH-algebra of MATH. CASE: Since MATH is the suspension of MATH, for every MATH there exist unique MATH and MATH such that MATH. Define MATH, and let MATH be the completion of MATH. Then by the Theorem in CITE MATH is a MATH imprimitivity bimodule. For completeness, we write down the module structures and the inner products. Module structures. Suppose that MATH, MATH and MATH. Then MATH give that MATH is a left MATH and right MATH bimodule with MATH for every MATH, MATH and MATH. Inner products. Define MATH and MATH by MATH . |
math/0008114 | Suppose that MATH is the edge set of the MATH-th coordinate space of MATH. Then by REF MATH . For MATH, MATH with MATH and the canonical projection to the MATH-th coordinate space MATH, define MATH and MATH by MATH . Then every MATH is a unitary element in MATH, and MATH is generated by MATH. If we denote MATH as MATH, then by REF MATH . |
math/0008114 | We have the following NAME exact sequence. MATH . We consider MATH as the automorphism MATH given by the NAME isomorphism of NAME. Define MATH by MATH for MATH. Then the induced automorphism MATH is the required isomorphism. For MATH, let MATH be the induced unitary element as in the proof of REF . Then MATH is homotopic to MATH. Hence if we denote MATH as MATH, then MATH is given by MATH and the induced automorphism MATH is the dimension group automorphism MATH of the adjacency matrix MATH. Therefore MATH is the inverse of MATH, and MATH is the same as MATH. Since MATH is isomorphic to MATH, MATH is trivially the identity map. Thus the six-term exact sequence is divided into the following two short exact sequences; MATH and MATH . Therefore we conclude that MATH . |
math/0008127 | Let MATH denote the unitary group of MATH, endowed with NAME measure MATH. Let MATH . For each MATH, we get MATH so MATH for all MATH. Furthermore, MATH . Lastly, for MATH, set MATH . Assume now that MATH, that is, MATH for all MATH. Then MATH . Since whenever MATH, MATH and any MATH is a linear combination of unitaries from MATH, we get that MATH . Since for any MATH, MATH and MATH, we get that MATH . We also have MATH. Hence MATH is in the center of MATH; moreover, MATH and MATH. We claim that MATH is invertible. Writing MATH as a direct sum of matrix algebras, we find that MATH, where MATH, and MATH is related to the value of MATH on the minimal projection of MATH corresponding to the MATH-th matrix summand in the direct sum decomposition of MATH. To show that MATH is invertible, is it sufficient to show that MATH is a strictly positive scalar for all MATH. Repeating the argument above with MATH replaced by MATH, we find that MATH is in the center of MATH and is non - negative. Furthermore, MATH since the subset of MATH with MATH has non - zero measure. Now let MATH. Then MATH; for all MATH, MATH; and for all MATH, MATH as desired. |
math/0008127 | Let MATH be as above. Consider the NAME - NAME MATH - algebra MATH, generated by the operators MATH and MATH. Then MATH. The automorphism MATH is identified with the NAME automorphism MATH: indeed we have MATH, MATH and MATH. Choose MATH as in REF . Let MATH. Then MATH. Let MATH. It follows that MATH. It is easily seen that the map MATH defines an isomorphism of MATH with the MATH bimodule MATH defined in REF. Hence by the results of that section, MATH. Moreover, since MATH we get that MATH. |
math/0008127 | Fix MATH. By the lemma in CITE we can find a MATH such that for every pair of elements MATH in the unit ball of MATH such that MATH we have the implication MATH . Let MATH. So assume that MATH is a finite set such that each MATH, MATH and MATH, MATH and MATH is an element of MATH such that MATH for MATH. By our choice of MATH we then have MATH for all MATH. We also claim that MATH for all MATH. To see this we first note that since MATH we have that MATH. Thus MATH for all MATH with MATH. Similarly, MATH and hence MATH . Since MATH some routine functional calculus shows that MATH and MATH. Combining these inequalities, the inequalities in the previous paragraphs and a standard interpolation argument we get MATH . These four inequalities will be needed at the end of the proof. Assume that MATH and MATH. By NAME 's extension theorem we may assume that MATH is defined on all of MATH (and takes values in MATH). Now choose MATH such that MATH, MATH, and MATH . Using the techniques in the proof of REF we may replace the (not necessarily unital) maps MATH and MATH with unital maps MATH and MATH such that MATH. (Here we use the facts that MATH and MATH.) Similarly, choose MATH such that MATH, and MATH . By NAME 's extension theorem we may assume that MATH is defined on all of MATH. Define MATH by MATH for all MATH and MATH by MATH for all MATH. Since we have arranged that MATH is unital (and MATH is unital by assumption) we see that MATH. Since MATH is a contractive completely positive map, this shows that MATH is a positive operator of norm less than or equal to one. Since it is clear that MATH is a completely positive map this, in turn, implies that MATH is also a contractive map (see CITE). Hence MATH and MATH . Thus we only have to check that MATH, for MATH. But, letting MATH we have MATH . Hence for any MATH we have the desired inequality. |
math/0008127 | By CITE it suffices to show the inequality MATH. Let MATH be given and MATH be any finite set containing the unit of MATH and such that MATH, MATH. Let MATH be an approximate unit such that MATH, for all MATH. Since we can manufacture a quasicentral approximate unit out of the convex hull of MATH CITE, we may further assume that MATH is quasicentral in MATH (and still fixed by MATH). Choose MATH according to the previous lemma and take MATH large enough that MATH, MATH for MATH. Since MATH for all MATH it is clear that MATH for MATH and all MATH. Hence letting MATH, for some MATH, the previous lemma implies that MATH is bounded above by MATH which is bounded above by MATH . This inequality implies the result. |
math/0008127 | Denote by MATH the range of MATH. Define recursively MATH . Let MATH be the cardinality of MATH. Then MATH has dimension at most MATH times the dimension of MATH, so that MATH. Let MATH be the orthogonal projection onto MATH; then MATH are clearly an increasing sequence, and MATH for all MATH. Let MATH . Then MATH for all MATH. Note that the rank of MATH is the same as that of MATH, which is bounded by MATH . Set MATH, MATH. Since MATH and thus MATH, if MATH then we have MATH for all MATH. Since MATH is self-adjoint, also MATH if MATH. Hence MATH if MATH. Let MATH be the orthocomplement of MATH in MATH. Then in the decomposition MATH, each MATH has the form MATH that is, it is a ``block tri - diagonal" matrix. Hence using the convention MATH we have the identities MATH . Now MATH so that for any MATH, MATH . Similarly, MATH . Hence MATH . The last inequality is due to the fact that for a fixed MATH, the operator MATH is a block-diagonal operator, with the blocks having orthogonal ranges, so that MATH. Choose the smallest integer MATH with MATH, and set MATH. Then MATH; by construction, MATH and the rank of MATH is bounded by MATH. |
math/0008127 | We have a short exact sequence MATH with splitting MATH given by MATH. Let MATH, MATH and MATH be self-adjoint elements, each of norm at most MATH, so that each MATH has finite rank. Let MATH be the sum of the ranks of MATH. Let MATH . Fixing MATH and a positive integer MATH, let MATH . Then the sum of the ranks of MATH is at most MATH, so there exists a projection MATH of rank MATH, so that MATH for all MATH and MATH. Consider the collection MATH of at most MATH self-adjoint operators on MATH. Let MATH be as in REF for the given value of MATH and the short exact sequence REF. By applying REF with these choices of MATH, MATH, MATH, and MATH, we find a positive, finite - rank MATH, so that: CASE: MATH, hence MATH for all MATH and MATH CASE: MATH for all MATH and MATH CASE: The rank of MATH is at most MATH. Let MATH. Then MATH satisfies the hypotheses of REF . Indeed, MATH and MATH if MATH, and MATH, etc. Hence by REF , as long as MATH we have MATH . Note that MATH . Hence setting MATH we have MATH . On the other hand, MATH . Since the rank of MATH is at most MATH it follows that MATH so that MATH . Since MATH can be written as the closure of the linear span of elements of the form appearing in MATH, the statement of the theorem follows. |
math/0008127 | REF follows from REF and the results of CITE. Hence it is sufficient to prove REF ; for that one only needs to prove that MATH, in view of the behavior of entropy with respect to inductive limits. We now proceed by induction on MATH. Since MATH, the statement is true for MATH. Applying REF to MATH gives MATH, which is equal to MATH by the induction hypothesis. |
math/0008127 | Let MATH be the GNS NAME space associated to MATH, and let MATH be the cyclic vector associated to MATH. Let MATH be the NAME algebra on two generators CITE. Without loss of generality, by replacing MATH with MATH and MATH with MATH we may assume that the GNS representation MATH satisfies MATH. Let MATH be the unitary induced on MATH by MATH. We shall covariantly identify MATH as the crossed product by an certain endomorphism MATH of the algebra MATH described in REF , taken with the automorphism MATH. By REF , we then have MATH, which, in view of our identification, is the same as MATH, hence proving the Proposition. The remainder of the proof is essentially a special case of the techniques used in CITE where more general NAME - NAME algebras were shown to have a crossed product structure. Consider now the NAME space MATH and the representation MATH given by MATH. Consider the isometry MATH defined by MATH where MATH denotes the image of the unit of MATH in MATH, and MATH. Denote by MATH the unitary implementing MATH. Denote by MATH the unitary MATH acting on the NAME space MATH. One sees that MATH, and MATH. MATH. This actually follows from REF, since MATH is isomorphic to the NAME - NAME algebra associated to the MATH bimodule MATH. For the reader's convenience, we give a proof. Let MATH be the vector - state on MATH, associated to the vector MATH. Then MATH, and one can easily verify that: CASE: MATH for all MATH, MATH, MATH; and REF MATH, for all MATH. It follows from CITE that MATH and MATH are free in MATH. Since MATH is (obviously) isomorphic to MATH, MATH, and since MATH is injective and MATH, the claim is proved. From now on, write MATH. Denote by MATH the closed linear span MATH . (each monomial above has exactly MATH terms equal to MATH and the same number of terms equal to MATH). Note that because of the relation MATH, MATH, each MATH is a MATH - subalgebra of MATH. The action of MATH on a vector MATH can be described as follows: MATH (here we identify MATH with MATH). Denote by MATH the MATH - algebra MATH. Then MATH and hence MATH determines an endomorphism of MATH. It can be easily seen that MATH is isomorphic to the crossed product of MATH by this endomorphism (compare the discussion after REF for the definition). Moreover, MATH leaves MATH invariant and commutes with the endomorphism MATH CITE. It remains to show that MATH, where MATH and MATH are as in REF . We proceed by induction. In the case that MATH, MATH, MATH. Let MATH . Then MATH is invariant under the action of MATH on MATH. Consider the isomorphism MATH . For each MATH, MATH has the form MATH. It follows that the representation of MATH on MATH is faithful. Denote by MATH the ideal MATH . Write MATH . Then MATH; hence the representation of MATH obtained by restricting its action to the space MATH is faithful. The action of MATH on MATH can be explicitly written as MATH . Denote by MATH the compact operator given by MATH . Then the map MATH is a MATH - algebra isomorphism of MATH with MATH. For all MATH, the subspaces MATH are invariant under the action of MATH. Denote by MATH the representation of MATH obtained by restricting its action on MATH to the space MATH. The image MATH lies inside MATH. By our assumption, MATH. This implies that MATH. We claim that MATH is faithful. Indeed, assume it's not, and some MATH is annihilated by MATH. Write MATH, where MATH and MATH. Since MATH, and MATH, it follows that MATH and MATH. But for MATH, MATH. Proceeding inductively, we see that MATH is faithful. We have therefore proved that MATH which means that MATH. Since MATH is given by MATH, it follows that the dynamical system MATH. |
math/0008127 | Since MATH is free from MATH and commutes with MATH, we see that for all MATH, MATH. Now let MATH be such that MATH. Then MATH because MATH and MATH are free with respect to MATH. But then MATH and MATH are free with respect to MATH, since for all MATH, MATH, and any element in MATH has the form MATH, for some MATH. The proof that MATH are free with respect to MATH proceeds along similar lines. |
math/0008127 | We will denote by MATH the defining map from MATH onto a dense subspace of MATH, and similarly for the other MATH spaces. We will also need the isomorphism MATH given by MATH. CASE: Let MATH. Choose MATH so that MATH, that is, MATH. There exists and element MATH such that MATH that is, MATH . Thus MATH. CASE: Given MATH take MATH such that MATH, hence MATH. A counter example to REF is provided by taking MATH, MATH, MATH, MATH and MATH. A counter example to REF is provided by taking MATH, MATH and MATH as in the above example, MATH and MATH. |
math/0008127 | Consider in MATH the unitary MATH. Consider the subalgebras MATH . Since MATH is free from MATH with amalgamation over MATH, it is easily seen from the definition of freeness and REF that the algebras MATH and MATH are also free with amalgamation over MATH. Furthermore, the restriction of MATH to MATH is MATH, and hence the GNS representation associated to each MATH is faithful. It follows from the embedding result (see REF) that MATH and MATH together generate the reduced free product MATH. It is easily seen that the algebra MATH is generated by MATH and MATH, and is isomorphic to MATH via the map MATH . It is clear that MATH intertwines the automorphisms MATH and MATH. |
math/0008127 | Since MATH has trivial MATH - theory, it follows from NAME 's exact sequence for free products (see REF) that MATH is zero. Once MATH is known to be simple and purely infinite, it will follow from NAME 's fundamental results CITE, that there exists a partial isometry MATH so that MATH and MATH. Set MATH. Then MATH, in such a way that MATH corresponds to the partial isometry MATH, and therefore REF is satisfied. To see that REF is satisfied, note that MATH is the linear span of MATH; hence it is sufficient to check that MATH and MATH all lie in MATH. This is clearly true of MATH and MATH, since MATH and MATH lie in MATH. Furthermore, since MATH, and the projections MATH and MATH are orthogonal, it follows that MATH and MATH are both zero. Since MATH and MATH lie in the center of MATH and MATH, it follows that MATH, and hence MATH, so that REF is satisfied as well. Note that MATH, where MATH. We shall now apply REF to show that MATH is simple and purely infinite; the following five observations show that the hypotheses of this theorem are satisfied. CASE: MATH and MATH are faithful states. CASE: Let MATH; then MATH is a partial isometry belonging to the spectral subspace of the state MATH corresponding to MATH, that is, MATH for all MATH. CASE: Let MATH and MATH; then MATH and we have MATH, MATH. CASE: Consider MATH . By REF , MATH and MATH are free with respect to MATH, so MATH and it is well known (compare REF) that MATH with MATH . Therefore, MATH is equivalent in MATH to a subprojection of MATH and MATH contains the diffuse abelian subalgebra MATH. CASE: The centralizer of MATH contains a diffuse abelian subalgebra; hence by REF, MATH is simple and thus MATH is full in MATH. The above facts allow us to apply REF, and we conclude that MATH is simple and purely infinite. |
math/0008127 | Consider the free product conditional expectation MATH. The GNS representation associated to MATH is faithful, by definition. Let MATH. By REF , MATH also gives rise to a faithful GNS representation. By REF, MATH and MATH are free with respect to MATH. It follows from the assumptions and the embedding result (see REF) that the MATH - algebra generated by MATH and MATH in the GNS representation of MATH associated to MATH is isomorphic to MATH. The desired isomorphism follows. |
math/0008127 | Using the embedding result (see REF), we can reduce to the case that MATH and MATH, by replacing MATH with the algebra generated by MATH and MATH. We have the following sequence of covariant inclusions, justified below: MATH . Inclusion REF is implied by REF , together with the embedding result (see REF). Isomorphism REF is implied by REF . |
math/0008127 | Because of the embedding result (see REF) and the behavior of entropy with respect to inductive limits, it suffices to prove the statement when MATH is a set with two elements. By monotonicity of MATH, the inequality MATH in REF is clear. Let MATH be the NAME algebra generated by the unilateral shift MATH and denote by MATH the vacuum expectation on MATH. CITE showed that MATH is a semicircular element with its spectrum an interval. Hence by functional calculus MATH contains a unitary MATH such that MATH, for all MATH. It is not difficult to see that MATH contains no nonzero compact operator, and thus MATH is faithful. By REF , the amalgamated free product dynamical system MATH can be covariantly embedded into MATH, where MATH and where MATH . So by monotonicity again, MATH. Using REF we have MATH . |
math/0008127 | Consider the reduced group MATH - algebra MATH of the nonabelian free group on MATH generators, where MATH is the cardinality of MATH. Let MATH be the unitary generators of MATH corresponding to the free generators of MATH. Let MATH be the automorphism such that MATH. Then from CITE and CITE, MATH, Let MATH where MATH is the canonical tracial state on MATH, and let MATH . Then by REF , MATH. Let MATH. Then MATH for all MATH and MATH and, by REF , the family MATH is free with respect to MATH. It is not difficult to see, though somewhat tedious to write down in detail, that the inclusion representation of MATH on MATH is a multiple of the GNS representation, MATH, of MATH on MATH. Indeed, one chooses vectors MATH such that MATH spans a dense subset of MATH and so that the representation of MATH on MATH is equivalent to MATH, for all MATH. Hence there is an isomorphism MATH sending the copy of MATH in MATH to MATH. We see that the automorphisms MATH and MATH are conjugate via MATH. Hence, by monotonicity of MATH we have MATH. |
math/0008127 | If MATH has no cycles then MATH is (conjugate to) a free permutation so MATH by REF . In general, by making a cycle decomposition of MATH and using REF , we see that in order to prove REF we may without loss of generality assume that MATH itself is a cyclic permutation, MATH, of a finite set MATH. However, then MATH, where each MATH is conjugate to MATH. Hence again applying REF we have MATH. |
math/0008127 | Assume without loss of generality that MATH. Given MATH we can find a MATH - invariant state MATH such that MATH. Let MATH be the canonical conditional expectation and define a state MATH. Then one checks that MATH and MATH. But under these conditions, CNT - entropy is also monotone (compare CITE) and so MATH. Since MATH was arbitrary, we are done. |
math/0008127 | Since MATH is faithful, by REF the GNS representation of MATH is faithful. Consider the MATH - algebra MATH, where MATH is the free shift and let MATH be the implementing unitary. It is fairly easy to see (compare the proof of CITE) that there exists a MATH - isomorphism MATH such that MATH. Hence MATH. But REF above and our main result imply that MATH. |
math/0008127 | Replacing MATH with MATH, if necessary, we may assume that there exists a NAME unitary in the centralizer of MATH. In this setting, CITE ensures that MATH is a simple MATH - algebra, where MATH is the NAME algebra and MATH is the vacuum state. Moreover, this free product is nuclear since it is isomorphic to a NAME - NAME algebra over MATH. Thus NAME 's absorption theorem for MATH (that is, MATH, for any simple, separable, unital, nuclear MATH - algebra MATH, compare CITE) together with the previous proposition implies the result. |
math/0008127 | By adding a unit to MATH and replacing MATH by its crossed product by MATH, we may assume that MATH is unital and MATH is inner. We may also assume that there exists a MATH - invariant state MATH with faithful GNS representation. Indeed, it follows from REF that the covariant embedding MATH, where we regard MATH and MATH is a unitary which implements MATH, is entropy preserving. Now any vector state arising from the second copy of MATH will be MATH - invariant and have faithful GNS representation since it is a cyclic vector. So, we assume that MATH is unital, MATH is inner and MATH is MATH - invariant with faithful GNS representation. Consider the free product MATH . Here MATH is the average of NAME measure on the second copy of MATH and the tensor product of MATH, NAME measure and an arbitrary faithful state on MATH, and MATH is the vacuum state on the NAME algebra MATH. Since there is a NAME unitary in the centralizer of MATH (coming from the copies of MATH), CITE implies that MATH is simple. Moreover, MATH is nuclear being isomorphic to a NAME - NAME algebra over MATH (compare CITE, CITE). Since MATH is MATH - equivalent to zero, MATH satisfies the Universal Coefficient Theorem. Since MATH is type I it also satisfies the UCT and hence, by a result of CITE so does MATH. Moreover, it follows from NAME 's six term exact sequence for MATH - theory (see REF) that MATH has the MATH - theory of MATH. Hence MATH is a simple, unital, purely infinite, nuclear MATH - algebra which satisfies the UCT and has the MATH - theory of MATH. So, by the classification results of NAME - NAME (CITE, CITE), MATH. By our main result we have that the automorphism MATH has the same entropy as MATH. Hence defining MATH we get the desired entropy preserving covariant embedding. |
math/0008130 | Let MATH and MATH be as in the statement and MATH. Because MATH is surjective, it follows that MATH are linearly independent at MATH. Each function MATH is the defining function of a hyperface because MATH must map faces of codimension MATH to faces of codimension MATH. |
math/0008130 | We may assume that MATH is dense in MATH. Fix MATH and we let MATH if MATH, for some MATH and MATH. We need to show that MATH is well-defined and bounded. Thus, we need to prove that MATH, if MATH and MATH, for some MATH and MATH. We will show that, for each MATH, there exists a constant MATH such that MATH . Let MATH, and let MATH . Then MATH is in MATH, and it follows from REF that we can find MATH such that MATH. Let MATH. Using again REF , we obtain, for MATH that MATH and hence MATH. A standard argument using the asymptotic completeness of the algebra of pseudodifferential operators shows that we can assume that MATH has order MATH. Let then MATH which gives, MATH . The desired representation of MATH on MATH is obtained by extending MATH by continuity to MATH. To extend MATH further to MATH, we proceed similarly: we want MATH for MATH and MATH. Let MATH and MATH be as in REF . We need to prove that MATH if MATH. Now, because MATH is dense in MATH, we can find MATH in MATH the norm closure of MATH and MATH such that MATH. Choose an approximate unit MATH of the MATH-algebra MATH, then MATH (in the sense of generalized sequences). We can replace then the generalized sequence (net) MATH by a subsequence, call it MATH such that MATH, as MATH. By density, we may assume MATH, for some MATH. Consequently, MATH, as MATH. Then MATH because MATH. |
math/0008130 | This is just the summary of the above discussion. |
math/0008130 | For brevity, let MATH be the space on the right-hand side in REF. Also, let MATH. Then we get for all MATH that is, MATH, and MATH. On the other hand, for MATH, there exists MATH such that for all MATH hence, MATH which gives the second equality in REF. By CITE, it remains to show MATH . Because of MATH, MATH is elliptic; by the usual symbolic argument we get MATH satisfying MATH. Furthermore, for MATH another definition chase yields as before MATH thus, MATH because of MATH and REF. Since we have MATH, this completes the proof. |
math/0008130 | Let MATH, MATH. We shall prove first that, for small MATH, there exists MATH such that MATH and MATH, for all non-degenerate representations MATH on MATH. Because the family MATH, MATH, extends to a first-order differential operator on MATH, the adiabatic groupoid of MATH, we obtain that MATH induces an element in MATH, which explains the choice of the power MATH. To be precise, let MATH, MATH, and MATH be the identity elements. If MATH, MATH, denotes the evaluation map as in REF (so that, in particular MATH, MATH), then we have MATH for MATH, and MATH. Thus, the family MATH leads to an element MATH. Choose a quantization map MATH for MATH as in CITE, and denote by MATH the metric on MATH, so that the principal symbol of MATH is MATH. Then the function MATH is an order two symbol on MATH, see REF , and MATH satisfies MATH. From the results CITE, we know that the function MATH is continuous at MATH (in fact everywhere, but that is all that is needed), and hence MATH will be invertible in MATH for MATH small. We define then MATH and a straight-forward computation gives MATH, for MATH, a dense subspace of MATH, and for MATH but small. Since MATH is (essentially) self-adjoint, we obtain that MATH is the inverse of (the closure of) MATH. This means MATH, for MATH but small, according to our conventions. Let now MATH and MATH be arbitrary. Then there exists a continuous function MATH with MATH such that MATH, and we obtain MATH by the composition property of the functional calculus for continuous functions, for MATH small enough. Because of MATH this completes the proof. |
math/0008130 | The proof is the same as that of the boundedness of operators of order zero, using NAME 's trick CITE. Let us briefly recall the details. It suffices to show MATH, for some constant MATH independent of MATH. Choose MATH with MATH. This is possible because MATH is defined and continuous on the sphere bundle MATH of MATH, a compact space. Let MATH be smooth with MATH (this is defined only outside the zero section), and let MATH be an operator with principal symbol MATH. Then MATH with MATH of order MATH. By replacing MATH with MATH such that MATH has order MATH and MATH, we obtain that the order of the operator MATH is less than MATH. Continuing in this way, we may assume that MATH has order MATH, so in particular MATH is bounded. Then MATH for MATH. |
math/0008130 | The statement follows from the boundedness of MATH, if MATH, so assume that MATH. Then the proof is the same as that of the previous lemma if in the proof of that lemma we replace MATH with MATH and take MATH. |
math/0008130 | The first statement follows from the previous lemma. Fix MATH such that MATH. REF gives MATH for some MATH and all MATH . Consequently, there is MATH with MATH for all MATH. Since MATH is essentially self-adjoint by REF and MATH, its range MATH is dense by CITE. By REF, we obtain for MATH which completes the proof. |
math/0008130 | By the previous corollary, both MATH and MATH extend to bounded operators. |
math/0008130 | We notice that if MATH satisfies the assumptions of the lemma, then MATH satisfies them as well. We can assume then that MATH. We shall check only that MATH. The relation MATH can be proved in the same way or follows from the first one by taking adjoints. Let MATH be with MATH, MATH be a sequence converging to MATH, and define MATH. Then we have MATH, thus, MATH first defined on the dense subspace MATH, has a unique bounded extension with MATH because of MATH and MATH . Since MATH is the limit of MATH, we obtain the formula for the principal symbol as well. |
math/0008130 | From REF we know that MATH is in MATH. By REF , applied to MATH and MATH, we have MATH, thus, MATH. Taking square roots completes the proof. |
math/0008130 | Assume first that MATH. Then the theorem follows from REF . For arbitrary MATH, MATH, and hence we get MATH by using functional calculus with continuous functions. A look at REF completes the proof. |
math/0008130 | If we change the metric on the compact space MATH, we obtain a new NAME operator, and MATH will be replaced by a different operator MATH. However, by REF , MATH and MATH are bounded for all even integer MATH. By interpolation, they are bounded for all MATH. This proves the independence of the NAME space on the choice of a metric on MATH. The last claim follows from REF if MATH is an integer. Let MATH. Then MATH. Using this fact and applying the NAME principle to MATH, with MATH, we obtain the desired result for all MATH. |
math/0008130 | We have MATH, by REF , because MATH and MATH. The identity for the principal symbol follows from the corresponding one in REF . |
math/0008130 | By construction, MATH is, up to similarity, the restriction of MATH to one of the fibers MATH, with MATH. |
math/0008130 | Using the deformation retract MATH, we define (up to isomorphism) MATH. Then MATH, the isomorphism being uniquely determined up to homotopy. Moreover, we have a (non-canonical) isomorphism MATH of vector bundles, which allows us to define a MATH-module structure on MATH. By replacing MATH with an isomorphic bundle, we can assume then that MATH, as NAME modules. Choose an admissible connection on MATH. REF then gives that MATH. |
math/0008130 | Let MATH, as above. CASE: Choose MATH such that MATH and define MATH. Then MATH. Similarly, we find a right inverse for MATH up to compact operators. Thus, MATH is NAME. CASE: The operator MATH is the product of the bounded operator MATH and of the compact operator MATH. CASE: For MATH in a MATH-algebra MATH and MATH a bounded MATH-representation of MATH, the spectrum of MATH is contained in the spectrum of MATH (we do not exclude the case where they are equal). If MATH, this gives REF , by taking MATH or MATH. If MATH, MATH, is self-adjoint, elliptic, then we use the result we have just proved for MATH, the NAME transform of MATH. |
math/0008130 | An injective representation MATH of MATH-algebras preserves the spectrum, and in particular, MATH is invertible if, and only if, MATH is invertible. Denote by MATH the algebra of bounded operators on MATH. The morphism MATH induced by MATH is also injective. Fix MATH. Then MATH is invertible if, and only if, MATH is invertible. By replacing MATH with MATH, we obtain MATH. We see then that MATH is invertible modulo MATH if, and only if, MATH is invertible modulo compact operators. This gives MATH. We thus obtain REF if we take MATH or MATH, the NAME transform of MATH. CASE: By definitions, MATH is NAME if, and only if, MATH defines a NAME operator MATH. Then MATH . For REF , a similar reasoning holds: MATH . |
math/0008130 | Again, REF follow from REF . The assumption MATH implies MATH. Because the groupoid obtained by reducing MATH to MATH is amenable, the representation MATH, MATH is injective on MATH. This gives MATH, MATH, for all MATH. |
math/0008130 | The representation MATH given by the restrictions MATH and the homogeneous principal symbol is injective. This gives REF . For MATH note that we have MATH for the NAME transform MATH of MATH, and MATH. To obtain REF from REF as above, it is enough to observe that the operator MATH, (with MATH elliptic of order MATH, fixed) belongs to MATH if, and only if, MATH for all MATH, and that MATH if, and only if, MATH. |
math/0008130 | It is enough to prove that the representation MATH is injective on MATH, because we can recover the principal symbol of a pseudodifferential operator from its action on functions, as explained in the previous section. The groupoid MATH is amenable because the composition series of REF are associated to the groupoids MATH, which are amenable groupoids. It is then enough to prove that each representation of the form MATH is contained in the vector representation. Let MATH be an interior point. By considering a small open subset of MATH, we can reduce the problem to the case when the manifold MATH is of the form MATH. Then the result is reduced to the case MATH using REF . But for this case MATH is isomorphic to the crossed product algebra MATH and the vector representation corresponds to the natural representation on MATH in which MATH acts by multiplication and MATH acts by translation. This representation is injective (it is actually often used to define this crossed product algebra). From this the result follows. |
math/0008130 | We are going to apply REF , with MATH ranging through the set of hyperfaces of MATH; this is possible because of REF . Furthermore, note that by the definition of the groupoid structure on MATH in REF , the boundary hyperfaces MATH of MATH are closed, invariant submanifolds with MATH. For each boundary hyperface MATH of MATH, we have by REF MATH where MATH is missing if MATH. Since MATH is essentially self-adjoint and elliptic, REF completes the proof. |
math/0008130 | Let MATH be a minimal face of MATH (that is, not containing any other face of MATH). Then MATH is a compact manifold without corners and hence the NAME operator on MATH contains MATH in its spectrum. The above theorem then shows that MATH. On the other hand, MATH is positive, and hence MATH. This completes the proof. |
math/0008131 | We shall write MATH, for simplicity, where the filtration is as defined in REF . The filtration of the complex computing the NAME homology of MATH then gives rise to a spectral sequence with MATH, by standard homological algebra. By the definition of the NAME complex of MATH, MATH . This completes the proof for NAME homology. The proof for cyclic homology is similar . |
math/0008131 | We shall use MATH . Denote MATH for simplicity. The projective limits give rise to a MATH exact sequence MATH for every fixed MATH (see REF from the Appendix). From the assumption that MATH for MATH and MATH, we know that MATH becomes stationary for MATH and MATH. This shows that the MATH term above vanishes for MATH, and hence MATH if MATH. It also gives that the natural morphism MATH induces an isomorphism of the MATH-terms of the corresponding spectral sequences, for MATH and MATH. Using then the fact that homology and inductive limits commute we obtain MATH, where for the last isomorphism we have used that the spectral sequence associated to MATH is convergent because MATH. |
math/0008131 | Each of the algebras MATH is a topologically filtered algebra in its own if we let MATH (so the filtration of MATH depends only on MATH and MATH). For the algebras MATH, the NAME complex is defined as for any topologically filtered algebra, except that there is no need to take an additional direct limit with respect to MATH. The NAME complex of MATH is then given by MATH . Because taking homology is compatible with inductive limits, we obtain that MATH . Denote by MATH the spectral sequence associated by REF to the topologically filtered algebra MATH. Again because homology is compatible with inductive limits, MATH. The result then follows from REF . |
math/0008131 | The assumption that MATH is independent of MATH has as a consequence that we need not take inductive limits with respect to MATH in the definition of the NAME complex of MATH. Denote MATH for simplicity. The NAME complex of MATH is complete, in the sense that MATH . The projective limits give rise to a MATH exact sequence MATH as MATH, for every fixed MATH (see REF ). The spectral sequence MATH associated to the complex MATH is convergent because MATH. Consequently, the homology groups of MATH are endowed with a filtration MATH such that MATH . Assume now that MATH, the assumption that MATH for all MATH gives that MATH for all MATH or MATH. Let MATH, MATH, and MATH, MATH. Then REF gives that the natural map MATH gives an isomorphism MATH and induces the zero map MATH. Using REF from the Appendix, we obtain that MATH and MATH for any fixed MATH. Because MATH, the result follows. |
math/0008131 | Each of the algebras MATH is a topologically filtered algebra in its own, if we let MATH (so the filtration really depends only on MATH and MATH). We shall write MATH instead of MATH, and hence MATH. The NAME complex of MATH is then given by MATH . Because taking homology is compatible with inductive limits, we obtain that MATH . Denote by MATH the spectral sequence associated to the natural filtration of the NAME complex the algebra MATH. Because homology is compatible with inductive limits, MATH. The result then follows from REF . |
math/0008131 | This is completely analogous to the previous results, so we will be sketchy. The natural filtration on the complex MATH induces a spectral sequence whose MATH term is the MATH-homology of MATH. This spectral sequence is proved to be convergent as in any of the above two theorems. |
math/0008131 | This statement is local because we are dealing with germs. So we can assume that MATH and MATH. The complex whose cohomology we have to compute is then the projective tensor product of the corresponding complexes for MATH and MATH an appropriate number of times. By the NAME Lemma, the complex corresponding to MATH has cohomology only in dimension MATH. The cohomology of germs at MATH of NAME forms on MATH is seen to be generated by MATH in dimension MATH and by MATH in dimension MATH (MATH). The cohomology of a tensor product of these complexes is isomorphic to the tensor product of their cohomologies as graded vector spaces (by the topological NAME theorem). This proves the theorem. |
math/0008131 | Consider for each face MATH of MATH the usual NAME complex MATH whose homology is, by (a variant of) NAME 's theorem, the absolute cohomology of MATH. Denote by MATH the hyperfaces containing MATH and by MATH their defining functions. Fix a local product structure in a neighborhood of the face MATH and choose a smooth cutoff function MATH with support in that neighborhood and equal to MATH in a smaller neighborhood of MATH. The map MATH where the local product decomposition near MATH is used to lift MATH to a smooth form on MATH, is a chain map. The NAME complex is a resolution of the constant sheaf on a manifold with corners (no factors of MATH are allowed). From this and the previous lemma we obtain that the cochain map MATH, where the sum is take over all faces of MATH, gives an isomorphism in cohomology. This proves the proposition. |
math/0008131 | That the cohomology of the first complex is isomorphic to MATH is of course well known. The second isomorphism follows from the NAME spectral sequence applied to the map MATH and the above proposition. |
math/0008131 | We shall prove this lemma for MATH, the other case being completely similar, and actually even simpler. The subspaces MATH are defined in the following way. Fix an increasing, countable exhaustion MATH by compact submanifolds. For any vector bundle MATH, we denote by MATH the space of classical symbols supported above some compact set in the base MATH. We then choose a quantization map MATH as in CITE, whose main property is that it induces a bijection MATH for all MATH (including MATH or MATH). Denote by MATH the natural projection and by MATH the set of symbols with support in MATH, then an increasing triple filtration for MATH is defined, for MATH first, by MATH . Choose now a radial completion of MATH, which is then a diffeomorphism of MATH onto the interior of the ball bundle MATH. This identifies MATH with the subset MATH of those smooth functions on MATH with support above the compact set MATH. We use this identification to define the topology on MATH, which in turn gives MATH the induced topology. To define MATH in general, let MATH be the space of NAME symbols of order MATH, with support in MATH, and only rational singularities with total order MATH in each defining function of a hyperface of MATH outside MATH (we do not count negative orders). Then we define MATH as the image of MATH, with the induced topology. This definition is such that MATH, if MATH. The topology on this space is defined similarly. In this way, MATH becomes a closed subset of MATH whenever MATH, MATH, and MATH. Then we endow MATH with the strict inductive limit topology. REF - REF of a topologically filtered algebra are then satisfied. |
math/0008131 | We know from REF that MATH. Using again the NAME, we obtain that MATH is isomorphic to MATH and the operator MATH is the projection which sends MATH to REF, is the inclusion MATH and the identity on the other factors. The above computation of MATH then gives the result. |
math/0008131 | We shall use REF . For MATH, REF gives MATH . Moreover, the assumptions of REF are satisfied, by REF , and this gives MATH . This gives the result. |
math/0008131 | Whenever the periodicity operator MATH of the exact sequence REF is surjective, we have MATH the projective limit being taken with respect to the operator MATH. The conclusion then follows from REF . |
math/0008131 | The proof is essentially the same as for the corresponding result for the algebra MATH, so we will be brief. The filtration on MATH is induced from the filtration on MATH. This and the specific form of the MATH-terms then give MATH . Moreover, the differential MATH is the same as that for MATH if MATH, but is trivial for MATH. Thus, if MATH, we get as above that MATH . Consequently, MATH . |
math/0008131 | We first prove the analogue of REF in our new settings: namely, the MATH-term of the cyclic spectral sequence associated to the topologically filtered algebra MATH by REF is given by MATH if MATH, and otherwise by MATH . Moreover, the periodicity morphism MATH vanishes if MATH and is the natural projection if MATH. Indeed, REF together with a relative version of the HKR isomorphism give that MATH . Moreover, it follows that the operator MATH is the projection just above. Now we can compute each term and use the homotopy invariance of the relative NAME cohomology to see that the MATH-homogeneous relative cohomology groups MATH vanish for MATH and that for MATH we have MATH . On the other hand, we have MATH if MATH, and, for MATH, we have MATH . The first assertion is a direct consequence of the above discussion and REF . The computation of MATH then follows as in the proof of REF . |
math/0008131 | The MATH-unitality follows from REF . For the second part, we proceed essentially as in REF . Choose an anti-symmetric tensor in the last MATH-variables MATH with MATH. We denote by MATH the quantization of MATH. Let MATH be the total degree. Because MATH where the dots represent terms of order at most MATH, the quantity MATH is of total order at most MATH and hence, modulo terms of order MATH, MATH is easily checked to be exactly MATH. |
math/0008131 | The MATH-unitality follows from REF , as above. To prove the rest of this proposition, we can either use the same method as the one used to prove REF , or we can argue that this proposition actually follows from REF . |
math/0008131 | The computation of MATH proceeds exactly as in the case of manifolds with boundary. From REF we know that the MATH term of the spectral sequence associated with the order filtration on the NAME complex is given by MATH . In our case, MATH which is the main reason why this result is true, and explains the thinking behind REF . Thus the NAME structure on MATH can be related to the natural NAME structure on the cotangent bundle MATH. In particular, the restriction of the given NAME structure to the interior of MATH is a symplectic structure. We now recall the definition of the symplectic NAME operator MATH. Let MATH be the two form on the interior of MATH that defines the symplectic form. The form MATH has only NAME singularities at the hyperfaces due to our assumptions on MATH. This then defines the MATH-operator by the formula CITE MATH . Then MATH and MATH on the interior of MATH, by CITE, and hence also on MATH because the forms in the latter space are determined by their restriction to the interior of MATH. Now, keeping in mind that by construction the operator MATH maps MATH-homogeneous MATH-forms into MATH-homogeneous MATH-forms, we get MATH . Using REF and the relative homotopy invariance of the relative NAME cohomology, we see that the only non-zero homogeneous component corresponds to MATH, and then that it is given by MATH . From the fact that MATH, if MATH, we see that MATH, for all MATH, and hence we deduce the result. The convergence of the spectral sequence to MATH follows from REF . By replacing MATH with MATH we obtain the stated formula for MATH. |
math/0008131 | By a standard argument using the 'SBI'-exact sequence, it is enough to prove this statement for NAME homology. Consider then the spectral sequences associated to the two algebras and the order filtration on their NAME complexes by REF . The induced map is then an isomorphism at the MATH-term, by REF , because MATH . |
math/0008131 | Only the last statement is not similar to some other proofs in the previous sections. Choose an open neighborhood MATH of MATH in MATH such that MATH is a deformation retract of MATH. Also, let MATH be a smooth function supported in MATH which is one in some smaller neighborhood of MATH. This data gives rise to a lifting of any smooth function on MATH to a function on MATH with support in MATH, and hence also to a linear lift MATH. The computation of MATH identifies the boundary map in the NAME cohomology spectral sequence and gives the result. |
math/0008131 | Let MATH be the dimension of MATH, as before. The dual of REF holds true for NAME cohomology. This implies that the inclusion MATH is isomorphic (in the sense of the above Proposition) to the dual of the map MATH. But this last map is checked to be an isomorphism. |
math/0008131 | We will proceed by induction on MATH to show that the morphism MATH is zero for any MATH. Let MATH denote the dimension of the manifold MATH, as above. The statement of the theorem is obviously true for MATH. The proof of REF , show that the groups MATH are generated by elements of order MATH with respect to the degree filtration, and that all cycles of order less that MATH are boundaries. This is a direct consequence of the computation of the MATH terms of the spectral sequences associated to the degree filtration. Assuming now the statement to be true for all values less than MATH, we obtain from the `SBI' exact sequence that the groups MATH are isomorphic to MATH. This shows that the groups MATH are generated by elements of order at most MATH in the degree filtration. It follows that they map to elements of order less than MATH in MATH, and hence they vanish by the above remark. The last statement follows directly from the vanishing of MATH in the NAME 's exact sequence. |
math/0008133 | Consider the complex REF, which computes the continuous cohomology of MATH, and let MATH be the map MATH . As in REF , the map MATH descends to the quotient to induce an isomorphism MATH of complexes, that is MATH, which establishes the isomorphism MATH, as desired. |
math/0008133 | CASE: First we check that MATH and MATH are injective. Indeed, assume that MATH, for some MATH. Then, if MATH, we have MATH and hence MATH. Consequently, we have MATH, with MATH, as desired. The same argument shows that if MATH and MATH have a point in common, then the standard subgroups MATH and MATH are conjugated in MATH. The injectivity of MATH follows from the injectivity of MATH, indeed, if MATH, let MATH as above, and conclude that MATH, by the uniqueness of the NAME decomposition. As above, this implies that MATH. Since the differential MATH is a linear isomorphism onto its image (that is, it is injective) and MATH is injective, it follows that MATH is a local homeomorphism onto its image (for the locally compact topologies), and that its image is an analytic submanifold (CITE, p. REF ). The set MATH is an algebraic variety on which MATH acts with orbits of the same dimension, and hence MATH is proper. This proves that MATH is a homeomorphism. Using an inverse for MATH, we obtain that MATH is also a homeomorphism. To prove now REF , consider a standard subgroup MATH, and let MATH be the dimension of MATH. Then MATH on MATH, and MATH contains MATH as an open component. It follows that, if MATH, then MATH. This shows that MATH is a union of sets of the form MATH. This must then be a disjoint union because the sets MATH are either equal or disjoint, as proved above. Now, if MATH has semisimple part MATH, then MATH, for some standard subgroup MATH, and hence MATH. The sets MATH are open in the induced topology because the map MATH is continuous. |
math/0008133 | There exists a (not natural) isomorphism MATH of vector spaces. By the above lemma, the inclusion of MATH of MATH-modules induces natural isomorphisms MATH because the functor MATH is compatible with inductive limits and with direct sums. The naturality of these isomorphisms and the Five Lemma show that MATH . This is enough to complete the proof. |
math/0008133 | It follows from the definition of MATH that, if MATH, then MATH vanishes in a neighborhood of MATH. Conversely, if MATH is in MATH, then we can find some polynomial MATH, with MATH, such that MATH is bounded from below on the support of MATH by, say, MATH, then MATH. The second isomorphism follows from the first isomorphism using REF . |
math/0008133 | Let MATH be a standard subgroup of MATH. Recall first that MATH is a finite group that acts freely on MATH, which gives a MATH - equivariant isomorphism MATH . Let MATH be a smooth MATH - module. The NAME - NAME spectral sequence applied to the module MATH and the normal subgroup MATH gives natural isomorphisms MATH . Combining these two isomorphisms, we obtain MATH . The result then follows from REF , which implies directly that MATH . The proof is now complete. |
math/0008133 | The result follows from Luna's Lemma. For MATH-adic groups, Luna's Lemma is proved in CITE, page REF, Properties ``C" and ``D." |
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