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math/0005060
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Notice first that for MATH, MATH and MATH big enough, then the numbers that appear in the right hand side of the estimates in REF form an estrictly decreasing sequence. That is, MATH . Let us check the inclusion MATH, for example. Suppose first that MATH, then MATH . If MATH, the inclusion is obvious. Otherwise, MATH . Then MATH, and so MATH. Assume now MATH. Then, MATH . In this case there is not any cube MATH satisfying MATH and so, by our convention, MATH. That is, the inclusion holds in any case. The other inclusions are proved in a similar way.
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math/0005060
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Let us proof REF . The second statement is proved in an analogous way. Let MATH be as in REF . If MATH (in particular, MATH), then MATH (the latter inclusion holds provided MATH). Assume now MATH. If MATH, then MATH and the result is trivial. If MATH, we denote by MATH a cube centered at MATH with side length MATH. Then, MATH and so MATH. Thus MATH . Therefore, MATH and MATH.
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math/0005060
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CASE: Let MATH and MATH. If MATH, there exists some MATH with MATH and MATH. Then MATH and so MATH (as in REF ). CASE: Let MATH and let MATH be such that MATH. We know that MATH . So we are done if we see that MATH. As in REF , we have MATH . Thus MATH . CASE: Let us see the first inequality. If MATH and MATH belongs to some cube MATH with MATH, then MATH because otherwise, as in REF , we would get MATH . However, since we assume MATH, the cube MATH is bigger than MATH and contains MATH. So MATH, which is a contradiction. Since MATH and this cube is much bigger than MATH, if MATH is big enough we get MATH . As this holds for all MATH with MATH, we obtain MATH . This inequality also holds if MATH with MATH, since in this case MATH. We consider now the second inequality in REF . Let MATH be such that MATH . If MATH with MATH for some MATH, by REF we get MATH . Since this is satisfied for all MATH such that MATH, MATH . If MATH has been chosen big enough, then MATH and one has MATH . Thus MATH . If MATH and MATH with MATH, then by REF we also get MATH (in particular MATH). Then REF holds in this case too (with MATH). CASE: Suppose first that MATH. In this case we must show that MATH . Let MATH be such that MATH. We know that MATH . By the definition of MATH, it is enough to see that MATH. This follows from the inclusion MATH, which holds because MATH and then we can apply REF (in fact, a slight variant of REF ). Suppose now that MATH. It is enough to show that MATH . Let MATH be such that MATH. By definition we have MATH . We are going to see that MATH . Assume MATH. Then, since MATH (for MATH big enough), MATH . Notice that from the first inequality in REF we get MATH. In this situation we have MATH . This is not possible, since by REF we would have MATH and then we would get MATH. This would imply MATH and also MATH, and then MATH which is a contradiction because we are assuming that REF does not hold. So REF is true and MATH. Thus MATH . Since this holds for any MATH such that MATH, we get MATH .
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math/0005060
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Let us see REF first. So we assume that there exist some cube MATH containing MATH with MATH. Since MATH, we have MATH. In particular, MATH. By REF and the second inequality of REF we get MATH . So REF holds if we take MATH big enough. Consider now REF . By REF have MATH . Thus we can write MATH . Let us estimate the first integral on the right hand side of REF . Using the first inequality in REF we obtain MATH . Let us consider the last integral in REF (only in the case MATH). By REF we have MATH . From REF we get REF .
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math/0005060
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In the first two cases MATH for some MATH. In the latter case, by REF and our construction, there exists some MATH such that MATH. So in any case MATH for some MATH. Arguing as in REF , for MATH big enough, it is easily checked that MATH, and so MATH.
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math/0005060
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Assume that MATH and that either MATH, MATH or MATH on MATH, or MATH. By the preceding claim, MATH for some MATH. Then, MATH .
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math/0005060
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First we will prove the first statement. By REF , we know that MATH. Let MATH be such that MATH. We must have MATH. Otherwise, MATH and MATH, which is not possible. Since MATH, we have MATH. We know MATH because REF holds for MATH. By REF (for MATH and MATH) we get MATH . The term MATH is also bounded above by MATH because MATH and MATH are doubling, MATH, and it is easily checked that MATH. The estimates on MATH and MATH follow from from the definition of these functions and the estimate MATH for MATH.
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math/0005060
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Suppose that MATH is such that either MATH, MATH or MATH on MATH. By REF we have MATH for some MATH. By construction, the center of MATH belongs to some cube MATH with MATH. It is easily seen that MATH. Thus MATH . Since MATH and MATH are contained in a common cube of the generation MATH, by REF we get MATH and so MATH .
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math/0005060
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Consider MATH. We want to see that MATH is very close to MATH on this cube. By REF we have to deal with the cube MATH. Let us see that if MATH is such that MATH, then MATH. Notice that MATH. Now, by the definition of good cubes, there exists some MATH such that MATH, which implies MATH. For MATH big enough, we have MATH, and then MATH. So MATH. Let us estimate the term MATH . Recall that MATH . By the arguments above, if MATH and MATH, then MATH has been chosen for supporting MATH or MATH, that is, MATH. Then, MATH . By REF we obtain MATH (we have used that MATH, with MATH depending on MATH). Then we get MATH . For MATH, we have MATH . Let us estimate the first term on the right hand side. By REF we obtain MATH . On the other hand, by REF , the second term on the right hand side of REF is bounded above by MATH. Thus we have MATH if we choose MATH small enough.
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math/0005060
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If MATH, we have already seen that MATH. If MATH, then MATH (because MATH and MATH). By construction, we have MATH . If MATH on MATH, then MATH. Now we consider the case MATH such that MATH on MATH. By REF there exists some MATH with MATH. Recall that by REF , if MATH, we have MATH . So if MATH and MATH, we have MATH. Therefore, MATH. Then, MATH . Therefore, MATH. By REF we get MATH . Recall also that, by REF , MATH . From the definition of MATH and REF , we derive that MATH and MATH have the same sign. On the other hand, from REF we get MATH . So by the definition of MATH anb MATH we have MATH and by REF we obtain MATH (assuming MATH small enough). By REF and since MATH and MATH have the same sign, REF holds also in this case.
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math/0005060
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Take MATH such that MATH. By REF we get MATH for MATH. Therefore, MATH . By REF we have MATH . So we only have to estimate MATH. Take MATH with MATH. Since MATH, by REF we have MATH. Applying REF we get MATH . It is easily checked that MATH. Then we get MATH.
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math/0005060
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Consider first the case MATH. If MATH is such that MATH, then MATH and so MATH. Therefore, MATH . By REF , we get MATH on MATH. Assume now MATH. Let MATH. From the definition of MATH, we have MATH for any MATH, MATH, with MATH. Also, by REF we have MATH . Consider now MATH with MATH. Observe that MATH and MATH. We have MATH . By REF we obtain MATH. By REF , on MATH we have MATH and also MATH. Thus, MATH on any cube MATH containing MATH. Take now MATH with MATH. On this cube MATH, and then we have MATH . Again by REF , we get MATH on MATH. Thus, MATH on MATH. Going on, we will obtain MATH for all MATH on any cube MATH containing MATH.
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math/0005060
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Let MATH be fixed. We are going to estimate the sum MATH . Let MATH be such that MATH. Since MATH is a bad cube, there exists some MATH such that MATH. Then we have MATH. Since MATH and MATH, we get MATH, and so MATH. By the finite overlapping of the cubes MATH in MATH, we have MATH . Now, from the construction of MATH, it is easy to check that MATH. This fact and the bounded overlapping of the cubes MATH give MATH . Summing over MATH, as the supports of the functions MATH are disjoint for different MATH's, we obtain MATH .
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math/0005060
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The set MATH is open because MATH is lower semicontinuous. Since for MATH-a.e. MATH there exists a sequence of MATH-doubling cubes centered at MATH with side length tending to zero, it follows that for MATH-a.e. MATH such that MATH there exists some MATH-doubling cube MATH with MATH and so MATH. The existence of the functions MATH of REF is a standard known fact. REF follows from the other statements in the lemma. So the only question left is REF . Notice that, since MATH, we have MATH for each MATH. To construct the functions MATH we would like to start by the smallest cube MATH, and go on with the bigger cubes MATH following an order of non decreasing sizes. Since in general there does not exist a cube MATH with minimal side length in the family MATH, we will have to modify a little the argument. For each fixed MATH we will construct functions MATH, MATH, with MATH, satisfying REF . Finally, applying weak limits when MATH, we will get the functions MATH. The functions MATH that we will construct will be of the form MATH, with MATH and MATH. To avoid a complicate notation, suppose that the cubes MATH, MATH, satisify MATH (we can assume this because we are taking a finite number of cubes). We set MATH and MATH where the constant MATH is chosen so that MATH. Suppose that MATH (for some MATH) have been constructed, satisfy REF and MATH where MATH is some constant (which will be fixed below). Let MATH be the subfamily of cubes MATH, MATH, such that MATH. As MATH (because of the non decreasing sizes of MATH), we have MATH. Taking into account that for MATH by REF , and using that MATH is MATH-doubling and REF , we get MATH . Therefore, MATH . So we set MATH and then MATH . The constant MATH is chosen so that for MATH we have MATH. Then we obtain MATH (this calculation also applies to MATH). Thus, MATH . If we choose MATH, REF follows for the cubes MATH. Now it is easy to check that REF also holds. Indeed we have MATH . Finally, taking weak limits in the weak-MATH topology of MATH, one easily obtains the required functions MATH. The details are left to reader. A similar argument can be found in the proof of REF.
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math/0005060
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Given MATH, for each integer MATH, we consider the generalized NAME decomposition of MATH given in the preceding lemma, with MATH. We will adopt the convention that all the elements of that decomposition will carry the subscript MATH. Thus we write MATH, as in REF . We know that MATH is bounded and satisfies MATH (because MATH). We will show that MATH in MATH and MATH as MATH too. It is not difficult to check that MATH tends to MATH in MATH. Indeed, if we set MATH, then MATH as MATH, because MATH. Thus MATH and so MATH in MATH. Let us see that MATH as MATH. We denote MATH. Then we have MATH . The estimates for each term MATH are (in part) similar to the ones in REF for estimating MATH over atomic blocks. We write MATH . Taking into account that MATH, it is easily seen that MATH (the calculations are similar to the ones in REF ). Let us consider the last term on the right hand side of REF now. By REF we get MATH . We split the second integral on the right hand side of REF as follows: MATH . As in REF , we have MATH . Finally we have to deal with MATH. Consider MATH and MATH. Then MATH because MATH for some constant MATH. Indeed, for MATH we have MATH and MATH . Recall also that MATH and MATH. Then we get MATH for all MATH. Thus MATH. So REF holds and then MATH . When we gather the previous estimates, we obtain MATH . Taking into account the finite overlap of the cubes MATH (recall that they are NAME cubes covering MATH), we get MATH and we are done.
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math/0005063
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If MATH laxly centralizes MATH, let MATH, MATH, MATH, and MATH be given as in the definition of lax centrality. Let MATH be denoted by MATH. Since MATH, we have MATH, which implies that MATH. If MATH, we let MATH be the subalgebra of MATH of pairs related by MATH, and form the pushout square MATH where MATH is defined by MATH. Let MATH be given by MATH, MATH where MATH is given by MATH, and MATH where MATH is the homomorphism shown in the pushout square. Note that MATH is onto, because it is a pushout of the onto homomorphism MATH. In the notation of CITE, we have MATH. Let MATH. We have MATH, and we have MATH because that is the pushout of MATH along MATH, and MATH. To show that MATH, we must use the fact that MATH. Indeed, by CITE, we have MATH .
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math/0005063
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Let MATH, MATH, MATH, and MATH witness the fact that MATH laxly centralizes MATH. Since MATH, there is a natural one-one homomorphism MATH. It is assumed in the definition of lax centrality that MATH, and the same goes for MATH and MATH. Thus, there are MATH, MATH and onto homomorphisms MATH, MATH. Let MATH. Then the diagram MATH is a pullback square, where MATH and MATH. Clearly MATH is onto, and since MATH is one-one, MATH is one-one as well. Let MATH be the kernel of the homomorphism MATH given by MATH, and MATH the kernel of the homomorphism MATH given by MATH. Then MATH, because MATH is one-one. Also, MATH and MATH are easily seen to be onto. Thus, MATH, MATH. Let MATH. To show that MATH, MATH, MATH, and MATH witness the lax centrality of MATH by MATH, it suffices to prove that MATH and MATH. Suppose MATH. There is a MATH such that MATH. Let MATH, MATH be such that MATH and MATH. Then MATH. It follows that MATH. On the other hand, MATH. If MATH, then we have MATH, which implies that MATH. Thus, we have MATH. It follows that MATH. That MATH follows by similar arguments.
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math/0005063
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Let MATH, MATH, MATH, and MATH be given as in the definition of lax centrality, and such that MATH, MATH, and denote MATH by MATH. Let MATH, MATH be tuples of algebras in MATH, with MATH and MATH disjoint, such that MATH and MATH. Since MATH, we have embeddings MATH. Let MATH be a filter on MATH maximal with respect to the property that MATH implies MATH, where MATH is the kernel of the natural map from MATH to MATH. We have MATH, because MATH. Let MATH be an ultrafilter on MATH extending MATH and containing MATH. Such an ultrafilter exists, because MATH. We claim that if MATH, then MATH. For, if MATH, then MATH. On the other hand, if MATH, we have by the maximality of MATH that neither MATH nor its complement MATH can be adjoined to MATH. This implies that there is a MATH such that MATH and MATH. But MATH. By the remark following the definition of lax centrality, MATH laxly centralizes MATH. However, MATH, so MATH. Also, MATH, because MATH is a monolith. Thus, MATH laxly centralizes MATH. This holds for every MATH, so it holds for MATH for any particular MATH. We have MATH and laxly centralizing MATH, but MATH is maximal for that property. It follows that for any MATH, MATH, which verifies the claim. Let MATH be the restriction to MATH of the ultrafilter congruence on MATH. Since MATH for all MATH, we have MATH. Thus, MATH.
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math/0005064
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By the usual averaging over time modulations argument we may assume that MATH is a free solution to the wave equation. By time reversal symmetry we may assume MATH is a forward solution MATH in which case it suffices to show the scale-invariant estimate MATH . By the usual NAME arguments (for example, CITE) we may assume that MATH is restricted to a dyadic annulus; by scale invariance we may assume that MATH is supported on the annulus MATH. By NAME in MATH we have MATH . By NAME in MATH, polar co-ordinates, and the frequency localization we have MATH . By the frequency localization we may restrict MATH to the range MATH. The claim then follows from the embedding MATH and the NAME MATH restriction theorem MATH for the sphere MATH (see for example, CITE).
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math/0005064
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By REF the above estimates split into the spatial estimates MATH and the temporal estimates MATH . But these follow from NAME multiplication laws (REF , or CITE; alternatively, use fractional NAME, NAME embedding and NAME) and the hypothesis MATH. (In fact, one has some regularity to spare in all of these estimates, except for the first temporal estimate).
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math/0005070
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We prove REF . If the assertion does not hold, there exists an index MATH such that MATH. This implies that the self intersection number of each component of MATH is MATH, which is a contradiction (compare CITE). REF follows from REF as MATH are strictly positive.
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math/0005070
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Take a two dimensional cone MATH and assume that MATH is strictly positive. Let MATH be the canonical subdivision of MATH and let MATH be the vertices of MATH on this cone. Write MATH. Assume that MATH and MATH and MATH. Take MATH with MATH and put MATH and MATH. Then MATH and MATH are positive integers and MATH. This implies that MATH. Suppose that MATH and put MATH. Then by CITE the intersection number of (each component of) MATH is MATH. Then the negativity of the intersection number implies that MATH. Thus each component of MATH is a rational sphere of the first kind. This implies also that MATH and MATH. Put MATH. Then we can extend MATH to get a regular simplicial subdivision MATH such that its restriction to MATH is defined by the vertices MATH. Thus we get a toric resolution MATH. Changing MATH outside of MATH if necessary, we may assume by REF that MATH is a subdivision of MATH. Thus we get a canonical morphism MATH which factors MATH by MATH. By the definition, MATH is the composition of blowing-up at MATH intersection points of respective components of MATH and MATH in MATH. Note that MATH is line-admissible unless MATH is not strictly positive and MATH and MATH. This is the situation where MATH is the blowing up at the intersection of MATH and MATH. This does not occur if MATH satisfies MATH-condition. Now the assertion follows by the induction on the cardinality of MATH.
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math/0005070
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The first assertion follows by an inductive argument. Write MATH with positive rational numbers MATH. As MATH and MATH, MATH are positive integers. By the definition of MATH, we can write MATH for some MATH. The assertion for MATH holds and MATH. Assume that MATH with MATH. As MATH and MATH is the vertices of the canonical subdivision of MATH, there exists MATH, MATH, such that MATH . Thus MATH. The inequality MATH can be proved similarly by using the fact that MATH is the vertices of the canonical subdivision of the cone MATH from MATH (compare CITE). Now we show the second assertion. The inclusion MATH is obvious. Suppose that MATH is not contained in MATH. Suppose that MATH. Then we can write MATH for some positive integers MATH. If MATH, this gives a contradiction by comparing the MATH-th coefficient: MATH. Suppose that MATH. Write MATH as above. Then MATH. Thus we get MATH which contradicts to the assumption.
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math/0005070
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As MATH has an isolated singularity, MATH must contain a monomial of type MATH or MATH. Suppose that MATH. Let MATH be the vertex of MATH adjacent to MATH by an edge. It is clear that the non-compact face MATH which has MATH as a face and is unbounded to the direction of the MATH-axis has covector MATH. One can see that there exists no other non-compact face which is unbounded to the MATH-axis direction and bounded to MATH-direction. Let MATH be the compact face which has MATH as a boundary and let MATH be the corresponding covector. As MATH contains MATH, we need to have MATH. Now the last assertion follows from MATH.
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math/0005070
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Assume that MATH with MATH. CASE: If MATH, we have MATH. As MATH, MATH divides MATH. Thus MATH if and only if MATH and MATH. In this case, MATH and MATH. Assume that MATH. By the definition of MATH, MATH is a non-compact face with dimension REF. In particular, MATH. By REF , we have MATH. CASE: Suppose that MATH. Then MATH. This implies MATH. Put MATH with MATH and MATH. Then by the above equality, we have MATH. Put MATH. By the definition, MATH divides the minors of MATH which are MATH. Thus MATH for MATH. Thus MATH is an integral covector for MATH. It is clear that MATH. Assume that MATH for some MATH. By the monotonity of the coefficients REF , we have MATH for MATH. Thus MATH and MATH for MATH.
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math/0005070
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Denote by MATH. Suppose that MATH and MATH. Then MATH by REF and MATH. By the assumption, we have MATH. As the continuous fraction MATH is given by MATH (MATH copies of REF), we get MATH and the assertion follows immediately.
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math/0005070
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This follows from REF .
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math/0005070
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Let MATH be the primitive covectors in MATH inserted by the canonical subdivision from MATH. If MATH, the assertion is immediate from REF , as MATH. We assume that MATH. If MATH, the assertion is obvious. Assume that MATH. By REF , MATH is given by MATH for MATH. Thus MATH. If MATH, MATH is monotone increasing by REF and we see that MATH and the assertion follows immediately. Assume that MATH. Then MATH is monotone decreasing for MATH. Thus MATH if and only if MATH. If this is the case, MATH is the unique covector in common. Thus the assertion follows from these observations.
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math/0005070
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The last assertion follows from by REF as MATH.
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math/0005070
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From the congruence MATH modulo MATH, it is clear that MATH. Thus MATH. The equality REF results from MATH . Thus MATH and MATH. The last congruence equation is equivalent to MATH. Assume that MATH. As MATH, there exists positive integer MATH, MATH, such that MATH modulo MATH. Put MATH. We see that MATH. Take MATH which satisfies the congruence MATH modulo MATH. Then MATH takes the form MATH with MATH and thus MATH. For the positivity of MATH, we need to have MATH. The integrity of MATH implies MATH . As MATH can move MATH and MATH or MATH, this congruence equation has a positive solution MATH. Then put MATH for such a solution MATH. This gives a covector MATH.
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math/0005070
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The first statement is a conclusion of CITE. REF follows from the NAME criterion and CITE.
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math/0005070
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We first check when the central exceptional divisor MATH is rational by using REF (see also CITE). If this is the case, we compute the number of arms from MATH. If this number is less than REF, we show that MATH. Recall that the number of arms in the resolution graph is the sum of MATH for non-regular cones MATH. CASE: Let MATH. Put MATH. Then MATH with MATH. Note that MATH, MATH and MATH. By CITE MATH is rational if and only if REF MATH or REF MATH. If REF holds, then MATH. We have MATH, MATH and MATH. If MATH, MATH gives MATH arms. Hence, in any case the resolution graph of MATH has at least three arms centered at MATH. In REF , we have MATH, MATH and MATH. If MATH, we have at least three arms in the resolution graph. Suppose that MATH. Then the number of arms at MATH is MATH, unless MATH and MATH. In this case, the resolution graph has two similar arms and MATH is normally smooth with MATH. NAME of other cases: CASE: Let MATH. Then MATH with MATH and MATH. The dual NAME diagram MATH has REF arms MATH, MATH, MATH where MATH and MATH. The central divisor MATH is rational if and only if MATH. If MATH is rational, then MATH and MATH and MATH. Hence, the resolution graph has at least three arms. CASE: Let MATH. Then MATH with MATH and MATH. The dual NAME diagram MATH has REF arms MATH, MATH, MATH where MATH and MATH. The divisor MATH is rational if and only if MATH which is equivalent to MATH. We have MATH, MATH and MATH. As MATH has MATH-copies of arms, MATH has at least three arms. CASE: Let MATH. Then MATH with MATH. The dual NAME diagram MATH has REF arms MATH, MATH, MATH where MATH, MATH and MATH. The divisor MATH is rational if and only if MATH. In this case, we have MATH, MATH and MATH. Thus MATH has three arms. CASE: Let MATH. Then MATH with MATH. The dual NAME diagram MATH has REF arms MATH, MATH, MATH, MATH where MATH and MATH. By the weighted homogenuity, we have the equality MATH which implies that MATH. Hence MATH. By CITE, MATH is rational if and only if either REF MATH, or REF MATH. By symmetry, we may assume that the first case REF . Then MATH, MATH, MATH, MATH. Thus the resolution graph has at least three arms. CASE: Let MATH. Then MATH with MATH . By the weighted homogenuity, we must have MATH which implies that MATH and MATH. Thus MATH. The dual NAME diagram MATH has REF arms MATH, MATH, MATH and MATH where MATH, MATH and MATH. The divisor MATH is rational if and only if MATH for some MATH and MATH. Then MATH and MATH, MATH, MATH and MATH. Thus the MATH has at least REF arms.
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math/0005070
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This is a summary of the following three lemmas.
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math/0005070
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We mainly use REF . Let MATH. The equation is MATH . Hence, we have the following conclusions. CASE: MATH if and only if there exists an integer MATH such that MATH and MATH. This is equivalent to MATH. And in this case MATH. CASE: MATH if and only if there exists an integer MATH such that MATH. And in this case MATH. CASE: MATH if and only if there exists an integer MATH such that MATH. And in this case MATH. REF is obvious now. CASE: One can see this by comparing the three sets MATH. In case MATH, we have MATH and MATH. Hence, MATH. In case MATH, we have MATH for MATH. Hence, MATH. In case MATH, we have MATH and MATH. Hence, MATH.
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math/0005079
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CASE: Let MATH be a MATH-extension of MATH. Since MATH is irreducible, so is MATH. Set MATH and MATH. Note that MATH, or MATH according as MATH is of real, complex, or quaternionic type, respectively. Since MATH, it follows from the NAME reciprocity that MATH where MATH denotes the multiplicity of MATH in MATH. On the other hand, since MATH the multiplicity MATH must be either MATH or MATH. If MATH (that is, MATH), then MATH for some MATH-module MATH which is not isomorphic to MATH. Since MATH, MATH is isomorphic to MATH, so that MATH is also a MATH-extension of MATH. Since MATH appears in MATH once, MATH. Therefore the identity REF applied to MATH implies MATH. This together with the equality MATH implies that the two mutually non-isomorphic MATH-extensions MATH and MATH of MATH are of the same type as MATH. If MATH, then MATH. Since any MATH-extension of MATH is contained in MATH as a direct summand, MATH is the unique MATH-extension of MATH. The type of MATH can be read from the equality MATH. This equality in particular implies that MATH must be even, in other words, MATH is not of real type when MATH. These prove REF except the last statement. The last statement can be seen as follows. If MATH has two MATH-extensions MATH and MATH, then MATH is a nontrivial real MATH-module of dimension one with trivial MATH-action, and MATH is isomorphic to MATH, proving the last statement. CASE: Suppose MATH has no MATH-extension. It is known that MATH is MATH-extendible if MATH is of quaternionic type, see the remark following REF. If MATH is reducible, then each direct summand of MATH is a MATH-extension of MATH which contradicts the assumption that MATH has no MATH-extension. Therefore, MATH is irreducible. Noting that MATH, it follows from the NAME reciprocity that MATH which implies the statement on the type of MATH. The last statement in REF follows again from the NAME reciprocity.
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math/0005079
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For MATH and MATH, denote by MATH, MATH, and MATH the set of generators of MATH according to MATH, MATH, and MATH, respectively. Note that the dimension of MATH is twice that of MATH and MATH. Then it is easy to find the generators and relations of MATH according to MATH as in REF .
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math/0005079
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See CITE for the former statement in REF . To see the uniqueness in REF , we note that if MATH and MATH are MATH-extensions of MATH, then MATH. Since MATH acts trivially on MATH and MATH is of odd order, MATH must be a trivial MATH-module. Therefore MATH is also isomorphic to MATH, which means that MATH and MATH are isomorphic as MATH-modules. This proves the uniqueness of the MATH-extension of MATH. It remains to prove REF . Let MATH be a normal subgroup of MATH which contains MATH and MATH is a normal cyclic subgroup of MATH of order MATH. By REF above, MATH has a MATH-extension, say MATH. Then MATH is a MATH-extension of MATH.
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math/0005079
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The proof of CITE holds in the real category with slight modification. For reader's convenience we shall give the argument when MATH is finite. The case when MATH is infinite is easy since the action of MATH on MATH is transitive, see CITE for details. We first note that if there exists an equivariant isomorphism MATH, then it must satisfy the equivariance condition MATH for any MATH where MATH. Suppose MATH. Then MATH, MATH is finite cyclic, say, of order MATH, and since REF is excluded, MATH is not of real type. Choose an element MATH such that MATH is the rotation through the angle MATH. By the assumption we have a MATH-linear isomorphism MATH. Set MATH, which is also a MATH-linear isomorphism. Then we connect MATH and MATH continuously in the set of MATH-linear isomorphisms of the fiber MATH-module along the arc of MATH joining MATH and MATH. This is possible because the set of MATH-linear isomorphisms of the fiber MATH-module is homeomorphic to a general linear group over MATH or MATH depending on the type of MATH (remember that MATH is not of real type), and it is arcwise connected. Thus we have a bundle isomorphism between MATH and MATH restricted to the arc of MATH joining MATH and MATH. We extend this isomorphism to an entire isomorphism over MATH using the equivariance condition MATH. When MATH, we choose a MATH-linear isomorphism MATH and a MATH-linear isomorphism MATH. Similarly to the above, we connect MATH and MATH as MATH-linear isomorphisms along the arc of MATH joining MATH and MATH, and then extend it to an isomorphism over MATH using the equivariance condition. But it is not always possible to connect MATH and MATH when MATH is of real type because the set of MATH-linear isomorphisms of the fiber MATH-module, which is homeomorphic to MATH, is not arcwise connected. In this case, however, we have another assumption that MATH is MATH-extendible for MATH or MATH since REF is excluded. By REF MATH has two MATH-extensions, say MATH and MATH. Thus the character of MATH as a MATH-module is of the form MATH for some nonnegative integers MATH and MATH with MATH, so that the set of MATH-linear isomorphisms between MATH and MATH is homeomorphic to MATH. Since the inclusion map from MATH to MATH induces a surjection on the MATH level, it is possible to choose a MATH-linear isomorphism MATH so that MATH and MATH can be connected in the set of MATH-linear isomorphisms of the fiber MATH-module.
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math/0005079
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There is an element MATH in MATH with MATH as the character of the fiber MATH-module by REF , and we have the semi-group isomorphisms MATH where the former isomorphism is given by sending MATH to MATH and the latter is given by taking orbit spaces by the MATH-action. In fact, the map sending MATH to MATH is the inverse of the former isomorphism, where MATH is viewed as a MATH-vector bundle through the quotient map from MATH to MATH (see REF for details), and the latter is an isomorphism because the action of MATH on MATH is free. As is well known, MATH is generated by the trivial line bundle MATH and the NAME line bundle MATH with relation MATH. Therefore, if we denote by MATH the two generators of MATH corresponding to MATH and MATH in MATH through the above isomorphism, then the lemma follows except the last statement. To see the last statement, we note that MATH is a nontrivial real MATH-line bundle over MATH with trivial fiber MATH-module and that MATH proving the last statement.
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math/0005079
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CASE: We note that MATH acts on MATH by conjugation. Since the minimal polynomial of any element in MATH has distinct root it is diagonalizable. So two elements in MATH are in the same orbit if and only if they have the same eigenvalues which are MATH because MATH. This implies that MATH has exactly MATH connected components because there are MATH possibilities of the MATH eigenvalues. CASE: MATH acts transitively on MATH, and the isotropy subgroup at an element of MATH is isomorphic to MATH; so MATH is homeomorphic to a homogeneous space MATH which has two connected components (see CITE for more details). CASE: As observed in REF above, MATH elements MATH respectively lie in the MATH different connected components of MATH. Through the inclusion map from MATH to MATH, they respectively are mapped to MATH where MATH is the MATH matrix MATH. Since MATH and MATH are conjugate by MATH whose determinant is negative, the MATH elements above in MATH are in a same connected component of MATH if and only if the number of MATH's as entries are congruent modulo MATH. This implies REF .
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math/0005079
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It follows from REF that MATH and MATH. Therefore the lemma follows from REF .
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math/0005079
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Let MATH be the set of (not necessarily invariant) complex structures on the fiber MATH. The collection MATH of MATH over MATH forms a MATH-fiber bundle over MATH, the MATH-action on MATH being induced from that on MATH. Then a MATH-invariant complex structure on MATH can be viewed as a continuous MATH-equivariant cross section of the MATH-fiber bundle. The image of the cross section lies in MATH because MATH acts trivially on MATH. In order to construct a continuous MATH-equivariant cross section of MATH, we choose a pair of points from MATH and MATH (that is, one point from each), which can be connected by a continuous cross section of MATH restricted to the arc MATH in MATH joining MATH and MATH. Not all pairs of those points are connected by such a cross section as observed later. But, once we find such a cross section, we can extend it to an entire MATH-equivariant cross section using the equivariance as is done in the proof of REF . On the other hand, we know in CITE that isomorphism classes of complex MATH-vector bundles over MATH are distinguished by the complex fiber MATH- and MATH-modules. Therefore, the number MATH of inequivalent MATH-invariant complex structures on MATH is equal to the number of pairs of connected components in MATH and MATH which are connected through MATH. Suppose MATH is even. Denote by MATH and MATH, for MATH and MATH, the connected components of MATH containing MATH and MATH connected components of MATH, respectively. If MATH and MATH are connected through MATH, then so are MATH and MATH. Counting the number of choices of pairs of connected components in MATH and MATH which are connected through MATH, one has MATH . On the other hand, if MATH and MATH are connected through MATH, then so are MATH and MATH and one has MATH . For MATH odd, a similar argument proves that MATH .
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math/0005079
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Since MATH is of real type, the character of MATH is also MATH; so we may view MATH as a complex irreducible character of MATH. We have proved in CITE that the semi-group MATH of isomorphism classes of complex MATH-vector bundles over MATH with multiples of MATH as the character of fiber MATH-modules is generated by four elements MATH with relation MATH, where MATH are complex MATH-vector bundles over MATH with MATH as the fiber MATH-module such that the fiber MATH-modules (respectively, MATH-modules) of MATH and MATH, where MATH, MATH, MATH, and MATH denote MATH or MATH, agree if and only if MATH (respectively, MATH). In fact, the two non-isomorphic fiber MATH-modules (respectively, MATH-modules) of MATH are complex conjugate to each other, so the complex conjugate (or dual) bundles of MATH and MATH are respectively MATH and MATH. Let MATH be the realification map. It is surjective by REF . Since any complex MATH-vector bundle is isomorphic to its complex conjugate bundle as real MATH-vector bundles, MATH and MATH. Therefore the relation MATH on MATH reduces to MATH on MATH. It follows that for each fixed fiber dimension there are at most two elements in MATH. We claim that there is no other relation. It suffices to show that there are exactly two elements in MATH for a fixed fiber dimension. If there is only one element for a fixed fiber dimension, say MATH, then the unique bundle must have MATH inequivalent MATH-invariant complex structures because the number of elements in MATH of (real) fiber dimension MATH is exactly MATH CITE. This contradicts REF . It remains to show that the two generators MATH and MATH are related by tensoring with a nontrivial MATH-line bundle with trivial fiber MATH-module. The fiber MATH-modules of MATH and MATH at MATH are isomorphic but the fiber MATH-modules of them at MATH are not, more precisely, they are related through the tensor product with the nontrivial real MATH-dimensional MATH-module defined by MATH, see REF . Therefore MATH is obtained from MATH by tensoring with a real MATH-line bundle with trivial fiber MATH-module, whose fiber at MATH is the trivial MATH-module and the fiber at MATH is the nontrivial MATH-module. The existence of such line bundle is guaranteed by REF .
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math/0005079
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The map MATH is surjective by REF and injective except for Cases A and B by REF . In both Cases A and B the target of MATH is a semi-group generated by one element by REF while the domain of MATH is generated by two elements with the relation as in REF . This implies that MATH is two to one. Finally, we note that tensoring elements in MATH with a nontrivial MATH-line bundle with trivial MATH-module does not change the fiber MATH-modules (respectively, fiber MATH- and MATH-modules) in Case A (respectively, Case B). This implies the last statement in the theorem.
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math/0005079
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REF - REF follow from REF , and REF follows from REF . We now prove the last statement in the theorem. After setting MATH and MATH, it is obvious that the inverse images of the pairs MATH in the remark after REF by the semi-group homomorphism MATH in REF are real MATH-line bundles with trivial fiber MATH-module. Moreover, MATH preserves the two tensor product operations, one on MATH with real MATH-line bundles and the other on MATH by the pairs MATH. Therefore, REF implies that, except for Cases A and B, the generators of MATH are related through tensor product with real MATH-line bundles with trivial fiber MATH-module. The same argument also holds for MATH by REF . For Cases A and B, the statement follows from the last statement of REF .
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math/0005079
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The proof is elementary and left to the reader.
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math/0005079
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CASE: Since MATH acts freely on MATH, every real MATH-line bundle over MATH with trivial fiber MATH-module is the pull-back of a real line bundle over MATH by the quotient map MATH. Suppose MATH is even. Then MATH is trivial, so pullback line bundles by MATH have trivial first NAME classes, which means that the underlying line bundles over MATH are trivial. According to CITE, an equivariant line bundle is trivial if and only if its underlying bundle is trivial. Thus, MATH are both trivial when MATH is even. If MATH is odd, then MATH above is an isomorphism. Therefore, exactly one of MATH has trivial first NAME class. This together with the result in CITE mentioned above shows that exactly one of MATH is trivial equivariantly. CASE: Set MATH. Since MATH acts freely on MATH, MATH are pullback of real MATH-line bundles over MATH by the quotient map MATH. Here MATH is of order two and acts on the circle MATH as reflection, so we may think of MATH as MATH. According to REF (or REF ) there are four real MATH-line bundles over MATH. Since the map MATH is an isomorphism in this case, they are distinguished by their fiber MATH-modules over the points MATH. More precisely, there are two possibilities for the fiber MATH-modules at MATH and MATH respectively since there are two real one-dimensional MATH-modules (the trivial one and the nontrivial one), and hence altogether there are four real MATH-line bundles over MATH. Moreover, MATH-line bundles are trivial if and only if the fiber MATH-modules at MATH are isomorphic (see also CITE). If MATH is even, then all pullback line bundles by MATH are trivial as discussed in REF ; so MATH are all trivial. If MATH is odd, then the pullback by MATH preserves the triviality of line bundles because MATH is an isomorphism. Since there are exactly two trivial MATH-line bundles over MATH, two of MATH are trivial and the other two are nontrivial.
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math/0005079
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Recall from the last statement in REF that all generators are related through tensor product with the real MATH-line bundles MATH and MATH according as MATH and MATH, respectively. These line bundles are all trivial if MATH is even by REF . So the existence of one trivial generator implies triviality of the other generators, and this finishes the proof in case that MATH is even. In the following we assume that MATH is odd. Denote by MATH a real irreducible MATH-module with MATH as its character. Recall that the fiber MATH-module of a generator is MATH if both MATH and MATH are nonzero, and MATH otherwise. In case MATH, we choose elements MATH and MATH in MATH such that MATH is the rotation through the angle MATH and MATH is the reflection about the MATH-axis. Then MATH (respectively, MATH) is generated by MATH and MATH (respectively, MATH). CASE: It suffices to show that the fiber MATH-module of a generator extends to a MATH-module. In case that MATH, the fiber MATH-module of MATH is MATH and it is MATH-extendible by REF . The other case is that either MATH or MATH and in this case the fiber MATH-module of MATH is MATH which is MATH-extendible by REF . CASE: In this case MATH by REF and the fiber MATH-modules of generators are MATH which is MATH-extendible by REF . So there is at least one trivial generator, say MATH. Since MATH or MATH, the tensor product of MATH with MATH has different fiber MATH-module from that of MATH at the point MATH such that MATH. So we get the other generator MATH. Since MATH is trivial by REF , so is MATH. CASE: In this case MATH by REF and the fiber MATH-modules of generators are MATH because either MATH or MATH is zero. Set MATH. Then MATH has a MATH-extension, say MATH, by REF . Note that the fiber modules of MATH at MATH and MATH are isomorphic to MATH and MATH, respectively. Thus MATH is isomorphic to the product bundle MATH by REF . We next consider triviality of the generators MATH. It suffices to show that at least one generator, say MATH, is trivial. Then so is the other generator MATH. We assume that MATH. The other case MATH can be proved similarly. MATH is of real type. Since MATH is not of quaternionic type by REF , it suffices to prove that MATH is not of complex type. Suppose that MATH is of complex type. Then there is a complex MATH-module MATH such that MATH and MATH as complex MATH-modules. We note that the realifications of MATH and MATH are MATH, and since MATH is MATH-invariant, MATH or MATH for MATH where MATH and MATH denote the characters of MATH and MATH respectively. Since MATH and MATH by assumption, MATH has two MATH-extensions of complex type by REF but no MATH-extension. It follows that MATH is MATH-extendible but not MATH-extendible, so MATH is MATH-invariant but not MATH-invariant. Namely, MATH and MATH, so that MATH. Therefore MATH because MATH is odd. On the other hand, since MATH is an element of MATH, MATH. Therefore MATH, but this contradicts that MATH. Thus MATH must be of real type. Since MATH is of real type by the claim above, MATH is irreducible and its character is MATH-invariant. It follows that there is a trivial complex MATH-vector bundle MATH over MATH with MATH as the fiber MATH-module, see CITE. Since MATH, there are two MATH-extensions of MATH, say MATH and MATH. Their complexifications MATH and MATH are non-isomorphic because MATH. Moreover these modules are both MATH-extensions of MATH. Thus the fiber MATH-module, say MATH, of MATH at MATH must be either MATH or MATH. It follows that the realification of MATH is either MATH or MATH. Therefore the realification of MATH, which is trivial, is isomorphic to one of MATH. CASE: In this case MATH by REF and the fiber MATH-modules of the generators are MATH. By a similar argument to REF there are two trivial generators, MATH and MATH. It suffices to show that the other two generators MATH and MATH are nontrivial. Consider the following isomorphisms MATH . Note that MATH is isomorphic to the product bundle MATH, where MATH, or MATH according to the type of MATH. It follows that MATH. Both MATH and MATH are nontrivial for all MATH. Note that the fiber MATH-module of MATH at MATH is the trivial MATH-module MATH while the fiber MATH-module at MATH is the nontrivial MATH-module MATH by the remark after REF . Then MATH (respectively, MATH) acts on MATH (respectively, MATH) as multiplication by MATH (respectively, MATH). Recall that MATH acts on both MATH and MATH trivially, that is, as multiplication by MATH. Assume that MATH is trivial. Then there exists a MATH-module MATH such that MATH and MATH. Thus MATH and MATH act on MATH as multiplication by MATH and MATH, respectively. Hence MATH acts on MATH as multiplication by MATH, and since MATH is odd, MATH also acts on MATH as multiplication by MATH. But this contradicts that MATH acts trivially on MATH. In the same way we can prove that MATH is also nontrivial. Since MATH is trivial, MATH must be nontrivial by REF and the claim above. Replacing MATH by MATH we can similarly prove that MATH is nontrivial. CASE: Case A: In this case MATH are MATH in REF . Since MATH is odd, MATH has a MATH-extension by REF . So we may assume that one generator, say MATH, is trivial. Then the following isomorphisms MATH imply that MATH is nontrivial since MATH is nontrivial by REF . CASE: In this case MATH are MATH in REF . Remember that MATH and MATH, see the proof of REF . Since MATH is trivial by Theorem C in CITE, MATH is also trivial. In the following we shall prove that MATH is nontrivial. Assume that MATH is trivial, that is, it is isomorphic to the product bundle MATH for some MATH-extension MATH of the fiber MATH-module MATH. MATH is of real type. If MATH is not of real type, then we may view MATH as the realification of a complex product bundle MATH, but this contradicts that MATH is the realification of the nontrivial bundles MATH and MATH in MATH. Denote by MATH the character of MATH. Every fiber MATH-module of MATH, which is MATH, is isomorphic to MATH and it is irreducible of complex type by REF . It is well known in representation theory that the character of MATH is zero on MATH. Thus MATH is always zero on MATH, where MATH. It follows that we have MATH so MATH. This implies that MATH is either irreducible of complex type or reducible with different direct summands of real type. In the sequel we show that neither case occurs. It is easy to see that the latter case does not occur because if it does, then each summand of MATH is a MATH-extension of MATH which contradicts the uniqueness of the MATH-extension of MATH by REF . Now, suppose MATH is irreducible and of complex type. We claim that the set MATH of MATH-invariant complex structures on MATH is not empty. Then it contradicts that MATH is of real type. Since MATH is irreducible and of complex type, there exists a MATH-invariant complex structure on MATH, that is, MATH. This means that each connected component of MATH is invariant under the action of MATH because MATH has two connected components. On the other hand, since the order MATH of MATH is odd and the number of connected components of MATH is two, each connected component is also invariant under the MATH-action. Therefore, it is invariant under the MATH-action because MATH and MATH generate MATH. Now we note that each connected component of MATH is homeomorphic to MATH (see CITE) and that any smooth action of a finite group on MATH is linear, so the MATH-action on MATH has a fixed point, that is, MATH.
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math/0005082
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By the first resolvent identity one has MATH . Therefore MATH and, by duality (here MATH is considered as an element of MATH), MATH . This ends the proof.
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math/0005082
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By REF one has MATH and so MATH solves REF ; by linearity MATH is also a solution. As regard REF let us at first note that MATH and MATH . Therefore one has MATH which immediately implies that MATH satisfies REF .
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math/0005082
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By REF MATH necessarily differs from (the restriction to MATH of) MATH by a MATH-independent, densely defined operator MATH. Being densely defined, MATH has an adjoint and MATH. Therefore, being MATH injective, MATH is closable and so, being MATH bounded, MATH is closable. Denoting by MATH the closure of MATH, the closure of MATH is given by MATH, which satisfies REF by REF .
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math/0005082
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We have already proven that, under our hypotheses, MATH is a pseudo-resolvent, that is, MATH . We proceed now as in the proof of [REF ]. By [REF] MATH, being a pseudo-resolvent, is the resolvent of a closed operator if and only if it is injective. Since MATH would imply MATH by REF we have MATH (see REF ) and so MATH. Since, as we have seen before, REF implies, when MATH, MATH one has MATH and so MATH . This gives the denseness of MATH. Indeed MATH, which is equivalent to MATH for all MATH, implies MATH. Let us now define, on the dense domain MATH, the closed operator MATH which, by the resolvent identity REF , is independent of MATH; it is self-adjoint since MATH . To conclude, the uniqueness of the decomposition MATH is an immediate consequence of REF .
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math/0005082
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Writing MATH by REF one has MATH . Thus by REF there follows MATH and so, since MATH implies MATH, one has MATH . Injectivity of MATH and MATH for any MATH then follows by injectivity of MATH (see REF ), REF , injectivity of MATH, and the definitions of MATH. Being MATH densely defined, one has MATH and so injectivity of MATH give denseness of the range of MATH. Being MATH closed, its domain is a NAME space with respect to the graph norm and we can apply the open mapping theorem to the continuous map MATH. Thus to conclude the proof we need to prove that the range of MATH is closed. By [REF] MATH if and only if the range of MATH is closed; by the closed range theorem (see for example, [REF]) the range of MATH is closed if and only if the range of MATH is closed, and this is equivalent to the range of MATH being closed. Therefore, since MATH when either MATH if MATH is surjective, or when MATH if MATH has a dense range, one has MATH and so, since MATH is closed, it has a closed range by [REF].
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math/0005082
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By our hypotheses MATH necessarily differs from (the restriction to MATH of) the bounded sesquilinear form associated to MATH by a MATH-independent NAME form MATH. Therefore MATH is a semi-bounded, densely defined, closable NAME form. If MATH denotes the unique semi-bounded self-adjoint operator corresponding to the closure of MATH (see [REF] for the existence of MATH), then the operator MATH gives the thesis.
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math/0005083
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Suppose the conditions hold. The cone MATH yields a toric open embedding MATH, hence MATH is quasiaffine. To see that the map MATH is surjective, consider an affine chart MATH, where MATH is a maximal cone. Since MATH induces a bijection of maximal cones, there is a MATH such that MATH. Moreover, MATH was assumed to have a finite cokernel, so MATH is surjective. Since MATH is equivariant,this implies MATH. To check that the map MATH is affine, keep on considering MATH. It is easy to see that the inverse image of MATH is MATH . Using the bijection MATH we see that MATH contains no element of MATH. Consequently, the only cones of MATH mapped by MATH into MATH are the faces of MATH. By the above formula, this means MATH. So we see that MATH is affine. It remains to show that the strict transform is bijective. As to this, recall first that the invariant prime divisors of MATH are precisely the closures of the MATH-orbits MATH where MATH. We calculate the strict transform of a MATH-stable prime divisor MATH corresponding to a ray MATH. Since MATH is bijective, there is a unique ray MATH with MATH. It follows from REF that MATH is a multiple of the MATH-invariant prime divisor MATH corresponding to MATH. Note that MATH. To calculate the multiplicity of MATH in MATH, it suffices to determine the pullback of MATH via MATH. On the affine chart MATH, every invariant NAME divisor is principal, and if MATH is the primitive lattice vector in MATH then the assignment MATH induces a natural isomorphism MATH. Since we have MATH the pullback MATH corresponds to the map MATH. By REF , this map is an isomorphism and hence MATH. Again using bijectivity of MATH, you conclude that the strict transform is bijective. Thus the conditions are sufficient. Using similar arguments, you see that the conditions are also necessary.
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math/0005083
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Suppose MATH is a quotient presentation. Given an invariant affine open subset MATH, the preimage MATH is affine as well. There is an effective invariant principal divisor MATH with support MATH, because MATH is quasiaffine. So MATH is an effective NAME divisor with support MATH. By construction, MATH lies in the image of MATH. Conversely, suppose that MATH is a triangle. Set for short MATH. Let MATH and MATH denote the dual lattices of MATH and MATH respectively. Dualizing the triangle, we obtain a sequence MATH . For each prime divisor MATH, let MATH denote the dual base vector. For every invariant open set MATH we have the submonoid MATH generated by the MATH, where MATH is a prime divisor. Let MATH be the cone generated by MATH. For example, MATH. We claim that MATH is a face provided that MATH is an affine invariant open subset. Indeed by assumption, there is a MATH such that MATH is an effective NAME divisor with support MATH. So for each prime divisor MATH, we have MATH with equality if and only if MATH. So MATH is a supporting hyperplane for MATH cutting out MATH and the claim is verified. In particular, since MATH is a face of MATH, this cone is strictly convex. For later use, let us also calculate MATH. If MATH denotes the primitive lattice vector in the ray MATH corresponding to the divisor MATH, we have MATH . That implies MATH. So in particular, MATH is a primitive lattice vector, and MATH induces a bijection between the rays of MATH and MATH. Let MATH be the fan in MATH generated by the faces MATH, where MATH ranges over all invariant affine open subsets. By construction, this defines a quasiaffine toric variety MATH. It remains to construct the quotient presentation MATH. First, we do this locally over an invariant affine open subset MATH. Let MATH be the corresponding cone, and let MATH be the affine open subset defined by MATH. Clearly, the map MATH has a finite cokernel, since MATH was is assumed to be injective. We have MATH. Moreover, it follows from we saw above that the map MATH induces a bijection between the sets of primitive lattice vectors generating the rays of MATH and MATH. Therefore the induced map MATH gives a bijection between the primitive lattice vectors generating the rays in MATH and MATH. By REF , the associated toric morphism MATH is a quotient presentation. To obtain the desired quotient presentation MATH, we glue the local patches. Let MATH be two affine charts. The intersection MATH is affine, and the rays of MATH are in bijection with the invariant prime divisors in MATH. On the other hand, the rays of MATH are the images of the duals to the prime divisors in MATH. This implies that MATH.
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math/0005083
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Suppose MATH is a quotient presentation and consider two invariant affine charts MATH, MATH of MATH. Since MATH is an affine toric morphism, the preimages MATH are invariant affine charts of MATH. The restriction of MATH defines a quotient presentation MATH. Since MATH is separated, the intersection MATH is even affine. REF implies that the image MATH is again an affine toric variety. Conversely, let MATH be of affine intersection. Choose a splitting MATH, where MATH is the kernel of the canonical map MATH. It suffices to show that the canonical factorization MATH is a triangle. Let MATH be an invariant affine open subset. We have to check that the complement MATH is a NAME divisor. For each invariant affine open subset MATH, the intersection MATH is affine, so MATH is a NAME divisor. Hence MATH is a NAME divisor.
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math/0005083
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First, suppose that the quotient presentation MATH is defined by an inclusion of rings MATH. The weight module MATH is given by the homogeneous component MATH of degree MATH. Let MATH and MATH be the primitive lattice vectors generating the rays in MATH and MATH, respectively. Choose MATH representing MATH. Note that the MATH-invariant NAME divisor MATH on MATH is given by the function MATH . The reflexive fractional ideal MATH over the ring MATH corresponding to MATH is generated by the monomials MATH with MATH as functions on MATH. Obviously, the map MATH induces the desired bijection MATH. This is compatible with localization, hence globalizes.
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math/0005083
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The problem is local, so we can assume that MATH is given by an inclusion of rings MATH. By REF , the ring of invariants MATH is nothing but MATH.
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math/0005083
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Suppose that MATH maps to MATH. According to REF , the homogeneous components in MATH are invertible. NAME easily check that the multiplication maps MATH are bijective. So by REF , the quotient presentation MATH is a principal homogeneous MATH-space. Hence the condition is sufficient. Reversing the arguments, you see that the condition is necessary as well.
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math/0005083
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First, we check sufficiency. Let MATH be the preimage of the subgroup MATH. The group scheme MATH is finite, so its action on MATH is automatically closed. Consequently, the quotient MATH is a geometric quotient. NAME directly see that MATH is quasiaffine. Consider the induced toric morphism MATH. The strict transforms in MATH are injective, and their composition is bijective. So the map on the right is bijective, hence MATH is another quotient presentation. By construction, its triangle MATH factors through MATH. According to REF , it is a geometric quotient. So MATH is the composition of two geometric quotients, hence a geometric quotient. The condition is also necessary. Suppose MATH is a geometric quotient, that means the fibers MATH are precisely the MATH-orbits. By definition, MATH acts freely on MATH. By semicontinuity of the fiber dimension, the stabilizers MATH for MATH must be finite. Note that the stabilizers are constant along the MATH-orbits. Hence the stabilizers generate a finite subgroup MATH. Set MATH. As above, we obtain a quotient presentation MATH. By construction, MATH is a free geometric quotient for the action of MATH. Now REF , ensures that MATH is a principal homogeneous MATH-space. By REF , the triangle MATH factors through MATH. This implies that MATH factors through the group of invariant MATH-Cartier divisors.
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math/0005083
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By definition, we have MATH. Since MATH is an open embedding, MATH holds for each quasicoherent MATH-module MATH. This gives MATH. Because MATH, we have MATH. Consequently, MATH.
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math/0005083
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Choose MATH such that the homogeneous elements MATH define effective NAME divisors with support MATH. Then MATH has the same radical as the irrelevant ideal MATH. Suppose that MATH. Then the restrictions MATH are zero as well. Note that the preimage MATH is affine, with global section ring MATH. The NAME subring MATH defines a factorization MATH . According to REF , the map on the right is a principal bundle for the action of the diagonalizable group scheme MATH. Setting MATH, we conclude that the MATH-module MATH is zero as well. Hence MATH for some integer MATH, because MATH is of finite type. Consequently MATH with MATH. This shows that the condition is necessary. The converse is similiar.
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math/0005083
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First, we claim that MATH factors through the open subset MATH. For MATH choose MATH such that MATH is a unit in MATH. Then MATH is invertible on a MATH-saturated neighbourhood of MATH. Clearly, this neighbourhood is mapped into MATH. According to REF , the projection MATH is a categorical quotient for the MATH-action defined by the MATH-grading on MATH (here we use the assumption MATH). The composition MATH is MATH-invariant. So the universal property of categorical quotients gives a commutative diagram MATH which defines the desired morphism MATH.
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math/0005083
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Let MATH. Then MATH is invertible on a neighbourhood of the fibre of MATH over MATH. Looking at the commutative REF , we see that MATH is invertible at some point of the fibre of MATH over MATH. Since MATH is saturated, this means MATH. The reverse inclusion is clear by definition.
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math/0005083
|
Suppose that MATH are two base-point-free MATH-algebras, which define two morphisms MATH, with MATH. First, assume that MATH. Let MATH be saturated. Using REF , we infer MATH. To check the second condition for equivalence, note that MATH are NAME polynomial algebras. So the map MATH induces the desired isomorphism. Conversely, assume that the base-point-free MATH-algebras are equivalent. Let MATH be saturated, and let MATH be its degree. Consider the partial quotients MATH . Then the isomorphism MATH induces the identity on MATH. Thus the morphism MATH induces both, MATH.
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math/0005083
|
In REF , we already saw that the assignment is functorial in MATH. By REF , it is well-defined on equivalence classes and gives an injection from the set of equivalence classes to the set of morphisms. It remains to check that the identity morphism MATH arises from a base-point-free MATH-algebra. Indeed: you easily check that MATH, together with the adjunction map MATH is a base-point-free MATH-algebra defining the identity on MATH.
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math/0005083
|
Let MATH be a base-point-free MATH-algebra on the scheme MATH defining a morphism MATH. Set MATH. The map MATH defines a homomorphism MATH. Clearly, it suffices to show that this map is bijective. The problem is local, so we may assume that MATH is affine, hence each weight module MATH is trivial and MATH holds. According to REF , for each MATH-homogeneous unit MATH, the image MATH is a global unit. Since each weight module MATH is generated by such a homogeneous unit, we infer that MATH is bijective.
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math/0005083
|
By assumption, MATH has invertible homogeneous components. By the preceding Proposition, each base-point-free MATH-algebra MATH is isomorphic to the preimage MATH.
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math/0005091
|
It will be shown that the bijection between MATH and MATH given by MATH induces an isomorphism of posets MATH. To establish this, it suffices to show that to each codimension MATH flat MATH there corresponds a codimension MATH flat MATH. Write MATH, where MATH, and let MATH. The flat MATH may be realized as the set of solutions of the system of REF, where MATH is MATH and MATH. Then, MATH has codimension MATH in MATH if and only if MATH if and only if MATH if and only if MATH has codimension MATH in MATH.
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math/0005091
|
For MATH, define MATH by MATH where MATH and MATH are in MATH. It is then readily checked that MATH is the identity map. Furthermore, if MATH is another hyperplane of MATH, the composition MATH is given by MATH so is null-homotopic. Consequently, the classes MATH form a basis. Finally, using stratified NAME theory, one can show that the relative homology group MATH vanishes for MATH, see CITE. It follows that the natural inclusion MATH induces an isomorphism MATH. So the classes MATH form a basis for MATH as asserted.
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math/0005091
|
Denote points in MATH by MATH, where MATH and MATH satisfies MATH for each MATH. Similarly, denote points in MATH by MATH, where MATH and MATH. In this notation, we have MATH and MATH. Let MATH be the total space of the pullback of MATH along the map MATH. It is then readily checked that the map MATH defined by MATH is an equivalence of bundles. Since the bundle MATH admits a cross-section, so does the pullback MATH. Furthermore, the structure group of the latter is the pure braid group MATH. Consequently, the action of the fundamental group of the base MATH on that of the fiber MATH is by pure braid automorphisms. As such, this action is by conjugation (see for instance CITE or CITE), hence is trivial in homology.
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math/0005091
|
By the previous result, the bundle MATH is equivalent to the pullback of MATH along the map MATH of REF. An analogous result for the complements of the subspace arrangements MATH and MATH is established next. For MATH, view MATH as MATH and MATH as MATH. Denote points in the configuration space MATH by MATH, where MATH and MATH for each MATH. Define MATH by MATH where MATH for each MATH. It is readily checked that the restriction of MATH to MATH takes values in the configuration space MATH. Let MATH be the pullback of the bundle MATH along this restriction, with total space MATH consisting of all points MATH for which MATH. Since the hyperplane arrangement MATH is strictly linearly fibered over MATH, the complement of the subspace arrangement MATH may be realized as MATH . Define MATH by MATH. The map MATH is a homeomorphism. Moreover, the following diagram commutes. MATH . It follows that MATH is a bundle which is equivalent to the pullback of the bundle of configuration spaces MATH along the map MATH, and therefore has a cross-section.
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math/0005091
|
For each hyperplane MATH of MATH, let MATH denote the composition of MATH and the natural inclusion MATH. It will be shown that the composition MATH induces the identity in homology if MATH, and induces the trivial homomorphism if MATH, thereby establishing the result. For MATH, write MATH as in REF. Then MATH is given by MATH, where MATH, see REF. Since the restriction of MATH to MATH takes values in MATH, the restriction of MATH to MATH takes values in MATH. From REF, the map MATH is given by MATH, where MATH, and MATH is the point MATH. Let MATH, and define MATH by the equation MATH . Then, a calculation yields MATH and MATH where, as before, MATH. Recall that MATH was chosen sufficiently small so as to insure that the point MATH, where MATH, lies in MATH for all MATH, MATH. Since MATH, it follows that MATH for all MATH. In other words, MATH for all distinct MATH and MATH. If MATH, then MATH since MATH is generic in MATH. Thus, MATH, and the point MATH lies in the configuration space MATH for all MATH, including MATH. It follows that MATH is trivial in homology in this instance. If, on the other hand, MATH, then MATH and thus MATH is necessarily non-zero. In this instance, MATH, where MATH is given by MATH, which clearly induces the identity in homology.
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math/0005091
|
The proof is by induction on MATH. In the case MATH, MATH is an arrangement of MATH points in MATH, the fundamental group of the complement is MATH, the free group on MATH generators, and it is well known that MATH is isomorphic to the free NAME algebra MATH, see for instance CITE. In general, assume that the fiber-type arrangement MATH is strictly linearly fibered over MATH and that MATH as above. Then there is a split, short exact sequence of groups MATH, and by REF , the action of MATH on MATH is by pure braid automorphisms. As such, this action is by conjugation, hence is trivial on MATH. By REF , the descending central series quotients of MATH are free abelian, and there is a short exact sequence of NAME algebras MATH which splits as a sequence of abelian groups. The result follows by induction.
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math/0005091
|
From the exact sequence of NAME algebras REF noted above, it follows that MATH is isomorphic to the semidirect product of MATH by MATH determined by the NAME homomorphism MATH given by MATH for MATH. It suffices to show that the homomorphism MATH factors as asserted. From REF , and the results of NAME and NAME stated in REF , there is REF algebras with split exact rows MATH . Via the splittings, view MATH and MATH as NAME subalgebras of MATH and MATH respectively. Then for MATH and MATH, we have MATH in MATH. Thus MATH in MATH and MATH.
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math/0005091
|
By REF , one has MATH, where the sum is over all MATH and MATH for which MATH. The result follows.
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math/0005091
|
The proof is by induction on MATH. In the case MATH, there is nothing to show since MATH is a free group on MATH generators, MATH is isomorphic to the free NAME algebra MATH, and there are no codimension two flats in MATH. In general, assume that MATH is strictly linearly fibered over MATH and that MATH as before. Then MATH has a defining polynomial of the form MATH, see REF. View MATH as a subarrangement of MATH, where MATH. Then the set MATH generates MATH, where the generators MATH correspond to the hyperplanes MATH of MATH, and to the generators MATH of the free NAME algebra MATH under the additive isomorphism MATH. By REF , MATH is isomorphic to an extension of MATH by MATH. Consequently, the NAME bracket relations in MATH consist of those of MATH, and those arising from the extension. By induction, the NAME bracket relations in MATH are given by MATH for codimension two flats MATH contained only in hyperplanes MATH. So it remains to analyze those relations in MATH arising from the extension. These are given implicitly in REF . Recall from REF that MATH if MATH, and is zero otherwise. Thus the results of REF may be recorded as MATH where MATH and MATH is the NAME delta. Note that the expression on the right lies in MATH. Under the above identifications, these relations take the form MATH . Now one can check that MATH if and only if MATH is a codimension two flat in MATH if and only if MATH. Using this observation, the relation REF may be expressed as MATH . A calculation then shows that this relation is equivalent to MATH, where MATH is the codimension two flat in MATH contained in MATH and MATH. Since this relation holds for all MATH for which MATH, it follows that MATH as well.
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math/0005091
|
Given a fibration MATH with a section MATH, there is a homotopy equivalence MATH given by the composite: MATH where MATH is the loop space multiplication and such that the inclusions of MATH and MATH into MATH are maps of MATH-spaces. NAME, if the spaces involved have torsion free homology then MATH. By REF and NAME, one obtains MATH upon passing to the NAME algebra of primitives. This result is a topological analogue of REF as the underlying NAME algebra structure is ``twisted." Now apply these considerations to the fiber bundle MATH. The fiber in this case is MATH. REF follows by induction, and then REF by the NAME theorem. By the NAME Theorem, MATH is isomorphic to MATH, a tensor algebra on MATH generators of degree MATH. Thus the module of primitive elements is generated as a NAME algebra by the primitive elements in degree MATH which are in the image of the NAME map. Since the NAME map takes values in the the module of primitive elements, that module is generated as a NAME algebra by those spherical classes given by the homology classes of degree MATH. Next notice that the homology groups here are torsion free. Hence the NAME map factors through MATH. Furthermore, the homotopy groups of a loop space modulo torsion give a graded NAME algebra where the NAME bracket is induced by the classical NAME product, and the NAME map is a morphism of graded NAME algebras. Thus the induced map MATH is an epimorphism of NAME algebras. Since all spaces are simply connected, and are of finite type, the homotopy groups modulo torsion are finitely generated free abelian groups in any fixed degree. By REF and NAME concerning rational homotopy groups, the induced map MATH is also a monomorphism. The result follows.
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math/0005091
|
The realization of the bundle MATH as the pullback of the bundle of configuration spaces MATH along the map MATH from REF yields REF algebras MATH with exact rows, and, on the level of primitives, REF algebras MATH where MATH is induced by MATH. Since the underlying bundles admit cross-sections, the rows in the above diagrams are split exact. Via these splittings, view MATH and MATH as NAME subalgebras of MATH and MATH respectively. From the above considerations, it follows that the NAME algebra MATH is isomorphic to the semidirect product of MATH by MATH determined by the NAME homomorphism MATH given by MATH for MATH. Moreover, for MATH, we have MATH in MATH. Thus MATH in MATH and MATH.
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math/0005091
|
By REF , one has MATH, where the sum is over all MATH and MATH for which MATH. Since the homology suspension MATH is an isomorphism and MATH is the map in loop space homology induced by MATH, one has MATH, where the sum is over all MATH and MATH for which MATH. The result follows.
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math/0005091
|
REF follows from the fact that the homology of MATH is abelian while the homology of MATH is generated by NAME brackets of weight at least MATH in homological degrees greater than MATH. REF follows from the remarks at the beginning of this section. REF follows at once from the fact that the result holds in case MATH, which was established in REF . In case MATH, the NAME operation MATH is precisely the natural NAME bracket in the homology of a MATH-fold loop space, MATH. These NAME bracket relations are recorded in REF . As shown in CITE, a further property of the operation MATH is that MATH, where MATH denotes the homology suspension. Thus by induction on MATH, the asserted NAME bracket relations are satisfied modulo elements in the kernel of the suspension. Furthermore, MATH is primitive in case the classes MATH and MATH are primitive. In characteristic zero, and in case MATH is greater than MATH, the homology suspension induces an isomorphism on the module of primitives. Thus the asserted NAME bracket relations are satisfied.
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math/0005094
|
The above REF is the statement for MATH. We proceed by induction on MATH. Assume REF for some MATH. Then MATH . Since MATH the above integral can be expressed as a sum of non-negative terms.
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math/0005094
|
According to CITE MATH . We want to write the divisor in the form MATH for MATH. This gives MATH . Now MATH . We use the restriction property of MATH with respect to MATH. On MATH the class MATH is invariant under the action of MATH exchanging the punctures. So under the natural map MATH it descends to the restriction of the class MATH on the ambient space MATH. Now MATH . In a similar way, we get MATH and for MATH with an extra factor MATH for MATH, if MATH is even. Also we use MATH.
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math/0005098
|
Let MATH be a common log resolution of MATH and MATH, with MATH . REF implies that MATH, and hence MATH (that is, the difference MATH is effective). Therefore MATH and the statement follows.
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math/0005098
|
Since the relative canonical bundle MATH is effective, it follows from the definition that MATH for any ideal MATH. Then using REF we find that MATH . Taking MATH, this gives REF . For REF , fix MATH and use the subadditivity relation REF to deduce: MATH as asserted.
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math/0005098
|
It suffices by REF to show that MATH . But membership in the ideal on the right is tested locally at a general point of each irreducible component of MATH. So we can assume after shrinking MATH that MATH is smooth and irreducible, of codimension MATH, and in this REF is clear. For then MATH for all MATH, and MATH is resolved by taking MATH to be the blow-up of MATH along MATH. Writing MATH for the corresponding exceptional divisor, one has MATH . Consequently MATH as asserted.
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math/0005098
|
We can assume without loss of generality that MATH is MATH-exceptional and that MATH, so that MATH. Applying REF to the graded family MATH REF , it suffices to prove that MATH. We suppose to this end that we've fixed a large integer MATH such that the multiplier ideal MATH in question is computed on a log resolution MATH of MATH dominating MATH. Then MATH gives rise to a prime divisor on MATH on MATH - viz. the proper transform of MATH - with MATH and one has MATH for every MATH. Let MATH be the effective NAME divisor on MATH defined in the usual way by writing MATH. Since MATH, we see that MATH appears with coefficient MATH in MATH. Consequently MATH and therefore MATH as required.
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math/0005098
|
The symbolic powers MATH again form a graded family of ideals, so REF will apply as soon as we establish that MATH. Referring to the primary decomposition REF , it is enough to show that MATH . For a given index MATH, inclusion in MATH is tested at a generic point of MATH. So having fixed MATH we are free to replace MATH by any open subset meeting MATH. Therefore, by definition of the symbolic powers, we may assume after localizing that MATH. But in this case MATH, and MATH thanks to a variant of REF (compare CITE).
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math/0005100
|
The hereditary orders in MATH containing MATH form a partially ordered set which we will denote by MATH. If MATH lies minimally over MATH then one proves exactly as in CITE that MATH. Let MATH be a maximal order lying over MATH. We deduce that MATH, where MATH is the length of a maximal chain in MATH, starting in MATH and ending in MATH. A local computation shows that MATH is given by REF , which finishes the proof of the computation of MATH. If MATH is algebraically closed then by NAME 's theorem CITE one has that MATH where MATH is a vector bundle of rank MATH on MATH. Hence by NAME theory MATH.
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math/0005100
|
By the previous proposition it suffices to prove this for MATH. Let MATH be the regular projective curve associated to the function field of MATH CITE. Then MATH is a regular compactification of MATH. In particular MATH is a finite number of points, whence by the localization sequence MATH is finitely generated if and only if MATH is finitely generated. Hence we may assume that MATH is projective. By CITE one has MATH where MATH denotes the MATH-points of the Jacobian of MATH. It is well-known that MATH is not finitely generated if MATH (for example because in that case MATH is non-trivial and divisible CITE).
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math/0005100
|
It is clear that the categories in REF. and REF. have finitely generated (free abelian) NAME groups, and that MATH is not finitely generated if MATH is derived equivalent to a hereditary category MATH with all objects of finite length and an infinite number of nonisomorphic simple objects. In view of REF the proof is completed by using REF .
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math/0005100
|
For the convenience of the reader we repeat the proof. We have to show that MATH is generated by MATH and MATH. Let MATH be the full subcategory of MATH consisting of objects isomorphic to finite direct sums of the form MATH. Then MATH is a strict triangulated subcategory of MATH. This can be deduced from the fact that the formation of triangles in MATH is compatible with direct sums CITE. Sending MATH to MATH defines a right adjoint to the inclusion MATH. This yields a semi-orthogonal decomposition of MATH given by MATH. In particular MATH is generated by MATH and MATH. Now we repeat this construction with MATH. So if MATH is the full subcategory of MATH consisting of objects isomorphic to finite direct sums of the form MATH then we have that MATH is generated by MATH and MATH. Continuing this procedure we find that MATH is generated by MATH and MATH. This finishes the proof.
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math/0005100
|
The inclusions MATH define a map MATH. An inverse to this map is given by sending MATH to MATH (see REF for notations).
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math/0005100
|
Using the same method as in REF we find inductively using REF that MATH. Now it is easy to see that sending MATH to MATH defines an isomorphism MATH. This proves what we want.
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math/0005100
|
Let MATH and assume that MATH is maximal such that MATH. Then there is a triangle MATH . Applying MATH yields injections MATH. Since for MATH the non-trivial homology of MATH occurs in degrees MATH and since the projective dimension of MATH is less than or equal to REF it follows that MATH for MATH. In particular MATH for MATH. Since trivially MATH for MATH it follows that MATH. But then it follows from REF that MATH. Repeating this procedure with MATH replaced by MATH eventually yields that the homology of MATH is in MATH.
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math/0005100
|
If MATH is a map in MATH then one has to show that MATH, MATH. Since the complex represented by MATH clearly lies in MATH, this follows from the previous lemma.
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math/0005100
|
By REF one has MATH .
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math/0005100
|
By REF we have to show MATH. By REF it follows that MATH is equal to MATH. So it is sufficient to show that MATH. By REF one has MATH. Since MATH is an abelian subcategory of MATH, it also has finite NAME dimension. In particular if MATH there is a quotient category MATH of MATH which has finite length. Selecting a simple object in MATH yields a rank function on MATH which is non-trivial. Hence MATH. This yields a contradiction.
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math/0005100
|
Since one has MATH it follows that MATH . It is now easy to see that MATH which yields the result.
|
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