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math/0002152
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Let MATH such that such that the functional MATH belongs to MATH. We have to show that MATH and MATH. So we will see that, for any cube MATH, MATH and for any two doubling cubes MATH, MATH . First we will show that REF holds for any doubling cube MATH. In this case the argument is almost the same as the one of CITE. We will repeat it for the sake of completeness. Without loss of generality we may assume that MATH . We consider an atomic block defined as follows: MATH where MATH is a constant such that MATH. By the definition of MATH, we have MATH . Since MATH is doubling, MATH . Now we have MATH . Therefore, MATH . Since MATH on MATH, we have MATH . From REF we get MATH . So REF holds in this case. Assume now that MATH is non doubling. We consider an atomic block MATH, with MATH and MATH where MATH is such that MATH. Let us estimate MATH. Since MATH is doubling and MATH, MATH . Since MATH, we have MATH . Thus MATH . As MATH, we also have MATH . Therefore, taking into account that MATH satisfies REF , and using REF , MATH . By REF we get MATH . That is, MATH which implies REF . Finally, we have to show that REF holds for doubling cubes MATH. We consider an atomic block MATH similar to the one defined above. We only change MATH by MATH in REF : MATH and MATH where MATH is such that MATH. Arguing as in REF (the difference is that now MATH and MATH are doubling, and we do not have MATH) we will obtain MATH . As in REF , we get MATH . By REF we have MATH . Thus MATH . Since MATH is doubling and satisfies REF , by REF we get MATH and we are done.
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math/0002152
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This lemma is an easy consequence of the NAME representation theorem. The same argument as the one of CITE works here.
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math/0002152
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Notice that, by REF , MATH for MATH. Now we repeat the arguments in CITE again. We consider the diagram MATH . The map MATH is an inclusion and MATH is the canonical injection of MATH in MATH (with the identification MATH for MATH). By REF , MATH is a closed subspace of MATH. By NAME 's closed range theorem (see CITE), MATH is also closed in MATH. Now it is easily checked that the NAME theorem and the fact that MATH imply MATH.
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math/0002152
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We assume MATH. The proof for MATH is similar. For some fixed MATH and all MATH, we will prove that there exists some MATH such that for any MATH we have the following good MATH inequality: MATH . It is well known that by this inequality one gets MATH if MATH. We denote MATH and MATH . For the moment we assume MATH. For each MATH, among the doubling cubes MATH that contain MATH and such that MATH, we consider one cube MATH which has `almost maximal' side lenght, in the sense that if some doubling cube MATH with side lenght MATH contains MATH, then MATH. It is easy to check that this maximal cube MATH exists, because MATH. Let MATH be the cube centered at MATH with side length MATH. We denote MATH. Then, assuming MATH small enough we have MATH, and then MATH. Indeed, by construction, we have MATH. Then, as MATH are doubling cubes containing MATH, MATH . Thus, for MATH, MATH . By NAME 's covering theorem there are MATH (depending on MATH) subfamilies MATH, MATH, of cubes MATH such that they cover MATH, they are centered at points MATH, and each subfamily MATH is disjoint. Therefore, at least one the subfamilies MATH satisfies MATH . Suppose, for example, that it is MATH. We will prove that for each cube MATH, MATH if MATH is chosen small enough. From this inequality one gets MATH . Then, MATH . Let us prove REF . Let MATH. If MATH is doubling and such that MATH, then MATH. Otherwise, MATH, and since MATH and MATH are doubling, we have MATH assuming MATH, and so MATH which contradicts the choice of MATH because MATH and MATH. So MATH, implies MATH . On the other hand, we also have MATH since MATH is doubling and its side length is MATH. Therefore, we get MATH and then, by the weak MATH boundedness of MATH and the fact that MATH is doubling, MATH . Thus, REF follows by choosing MATH, which implies REF , and as a consequence we obtain REF (under the assumption MATH). Suppose now that MATH. We consider the functions MATH, MATH, introduced in REF . Since for all functions MATH and all MATH we have MATH and MATH, operating as in REF we get MATH. On the other hand, MATH and so MATH. Therefore, MATH . Taking the limit as MATH, REF follows.
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math/0002152
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Assume MATH. The operator MATH is sublinear and it is bounded in MATH and MATH. By the NAME interpolation theorem, it is bounded on MATH, MATH. That is, MATH . We may assume that MATH has compact support. Then MATH and so MATH. Thus MATH, and so MATH. By REF , we have MATH . The proof for MATH is similar: Given MATH, we write MATH. It easily seen that the same argument as for MATH can be applied to the function MATH. On the other hand, MATH is bounded on MATH over constant functions. Indeed, since MATH is bounded from MATH into MATH, we get MATH . We also have MATH and so MATH.
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math/0002152
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We will assume MATH. CASE: Taking into account REF , for MATH-almost all MATH such that MATH, there exists a cube MATH satisfying MATH and such that if MATH is centered at MATH with MATH, then MATH . Now we can apply NAME 's covering theorem to get an almost disjoint subfamily of cubes MATH satisfying REF . CASE: Assume first that the family of cubes MATH is finite. Then we may suppose that this family of cubes is ordered in such a way that the sizes of the cubes MATH are non decreasing (that is, MATH). The functions MATH that we will construct will be of the form MATH, with MATH and MATH. We set MATH and MATH where the constant MATH is chosen so that MATH. Suppose that MATH have been constructed, satisfy REF and MATH where MATH is some constant (which will be fixed below). Let MATH be the subfamily of 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. Now it is easy to check that REF also holds. Indeed we have MATH and by REF , MATH . Thus we get REF . Suppose now that the collection of cubes MATH is not finite. For each fixed MATH we consider the family of cubes MATH. Then, as above, we construct functions MATH with MATH satisfying MATH and, if MATH, MATH . By REF there is a subsequence MATH which is convergent in the weak MATH topology of MATH and in the weak MATH topology of MATH to some function MATH. Now we can consider a subsequence MATH with MATH which is convergent also in the weak MATH topologies of MATH and MATH to some function MATH. In general, for each MATH we consider a subsequence MATH with MATH that converges in the weak MATH topologies of MATH and MATH to some function MATH. We have MATH and, by the weak MATH convergence in MATH and MATH, the functions MATH also satisfy REF . To get REF , notice that for each fixed MATH, by the weak MATH convergence in MATH, MATH and so REF follows. CASE: For each MATH, we consider the atomic block MATH, supported on the cube MATH. Since MATH, by REF we have MATH which implies REF .
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math/0002152
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For simplicity we assume MATH. The proof follows the same lines as the one of CITE. The functions MATH having compact support with MATH are dense in MATH, MATH. For such functions we will show that MATH . By REF , this implies MATH . Notice that if MATH has compact support and MATH, then MATH and MATH. Thus MATH, and then MATH. So the hypotheses of REF are satisfied. Given any function MATH, MATH, for MATH we take a family of almost disjoint cubes MATH as in the previous lemma, and a collection cubes MATH as in REF in the same lemma. Then we can write MATH . By REF , we have MATH, and by REF , MATH . Due to the boundedness of MATH from MATH into MATH, we have MATH . Therefore, MATH . Since MATH is of weak type MATH, we have MATH . On the other hand, as MATH is bounded from MATH into MATH, MATH . Thus MATH . So the sublinear operator MATH is of weak type MATH for all MATH. By the NAME interpolation theorem we get that MATH is bounded on MATH for all MATH. In particular, REF holds for a bounded function MATH with compact support and MATH.
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math/0002152
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We have already seen REF . Let us prove REF. Suppose that MATH, for example. Let MATH be a function supported on some cube MATH. Suppose first MATH (this is always the case if MATH). We have MATH . So it is enough to show that MATH . Let MATH be the point (or one of the points) in MATH which is closest to MATH. We denote MATH. We assume that MATH is a point such that some cube with side length MATH, MATH, is doubling. Otherwise, we take MATH in MATH such that satisfies this condition, and we interchange MATH with MATH. We denote by MATH a cube concentric with MATH with side length MATH. So MATH. Let MATH be the biggest doubling cube centered at MATH with side length MATH, MATH. Then MATH, with MATH, and one can easily check that MATH . Moreover, MATH and so, for MATH, MATH because MATH. Then we get MATH and from REF , we obtain REF . Suppose now that MATH. Since MATH is centered at some point of MATH, we may assume that MATH. Then MATH can be covered by a finite number of cubes MATH centered at points of MATH with side length MATH. It is quite easy to check that the number of cubes MATH is bounded above by some fixed constant MATH depending only on MATH. We set MATH . Since REF holds for the cubes MATH (which support the functions MATH), we have MATH . Now we are going to prove REF. Let MATH be supported on a cube MATH. Assume MATH and suppose first MATH. We consider the same construction as the one for REF. The cubes MATH, MATH and MATH are taken as above, and they satisfy MATH, MATH, MATH and MATH. Recall also that MATH is doubling. We take the atomic block (supported on MATH) MATH where MATH is a constant such that MATH. For MATH we have MATH . Then we have MATH . Since MATH is doubling, we have MATH . Therefore, MATH . If MATH, operating as in the implication MATH, we get that REF also holds.
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math/0002152
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Suppose that MATH. Let us see that this implies MATH. The estimates are similar to the ones that we used to show that CZO's bounded on MATH are also bounded from MATH into MATH. Assume, for example MATH. We have to show that if MATH, then MATH . We denote MATH. Then we write MATH . Since MATH is antisymmetric, we have MATH. On the other hand, since the NAME transform is bounded from MATH into MATH, we also have MATH . By the same argument, we get MATH . Also, it is easily seen that MATH for MATH, and so MATH . On the other hand, if MATH and MATH, we have MATH and so MATH. Finally, since MATH, MATH for MATH, and thus MATH. Therefore, MATH . So MATH, and thus we also have MATH, for any MATH. Now, some standard calculations show that REF is satisfied: MATH . Since MATH, we have MATH. Also, as usual, we have MATH . Finally, the last integral can be estimated using the boundedness of the NAME transform from MATH into MATH. Thus REF holds.
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math/0002152
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Let MATH be a cube concentric with MATH such that MATH, with MATH for some MATH. Then MATH for all MATH, with MATH depending on MATH and MATH. Observe also that if we take MATH so that MATH, then MATH and so MATH . Therefore, MATH . On the other hand, if MATH is big enough, then MATH. Indeed, MATH and so MATH, assuming MATH big enough. This implies MATH, and then, by construction, MATH. As a consequence, on the right hand side of REF , there is no overlapping in the terms MATH, and then MATH and REF follows.
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math/0002152
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Let MATH be two doubling cubes in MATH, with MATH. Let MATH be the first cube of the form MATH, MATH, such that MATH. Since MATH, we have MATH. Therefore, for the doubling cube MATH, we have MATH, with MATH depending on MATH, MATH, MATH and MATH. In general, given MATH, we denote by MATH the first cube of the form MATH, MATH, such that MATH, and we consider the cube MATH. Then, we have MATH, and also MATH. Then we obtain MATH where MATH is the first cube of the sequence MATH such that MATH. Since MATH, we also have MATH. By REF , if we set MATH, we get MATH .
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math/0002152
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For all MATH, we will show the pointwise inequality MATH where, for MATH, MATH is the non centered maximal operator MATH and MATH is defined as MATH . The operator MATH is bounded on MATH for MATH, and MATH is bounded on MATH for MATH because MATH is bounded on MATH (see CITE). Then the pointwise REF for MATH implies the MATH boundedness of MATH for MATH. If MATH is a bounded function we can apply REF because, by the MATH boundedness of MATH, it follows that MATH. On the other hand, by REF it is easily seen that we can assume that MATH is a bounded function. So REF implies that MATH is bounded on MATH, MATH. Let MATH a family of numbers satisfying MATH for any cube MATH, and MATH for all cubes MATH. For any cube MATH, we denote MATH . We will show that MATH for all MATH and MATH with MATH, and MATH for all cubes MATH with MATH. In the final part of the proof we will see that from the preceeding two inequalities one easily gets REF . To get REF for some fixed cube MATH and MATH with MATH, we write MATH in the following way: MATH where MATH and MATH. Let us estimate the term MATH: MATH . Now we are going to estimate the second term on the right hand side of REF . We take MATH. Then we have MATH . Notice that we have used that MATH which holds because MATH. Then we get MATH . By REF , to prove REF we only have to estimate the difference MATH. For MATH we have MATH . Taking the mean over MATH, we get MATH and so REF holds. Now we have to check the regularity REF for the numbers MATH. Consider two cubes MATH with MATH. We denote MATH. We write the difference MATH in the following way: MATH . Let us estimate MATH. For MATH we have MATH . So we derive MATH. Let us consider the term MATH. For MATH, it is easily seen that MATH . Thus MATH . Let us turn our attention to the term MATH. Operating as in REF , for any MATH, we get MATH . Taking the mean over MATH for MATH and over MATH for MATH, we obtain MATH . The term MATH is easy to estimate too. Some calculations very similar to the ones for MATH yield MATH. Finally, we have to deal with MATH. For MATH, we have MATH . We have MATH . Thus MATH . Taking the mean over MATH, we get MATH . So by the estimates on MATH and MATH, the regularity REF follows. Let us see how from REF one obtains REF . From REF , if MATH is a doubling cube and MATH, we have MATH . Also, for any cube MATH (non doubling, in general), MATH, and then by REF we get MATH . On the other hand, for all doubling cubes MATH with MATH such that MATH, where MATH is the constant in REF , by REF we have MATH . So by REF we get MATH for all doubling cubes MATH with MATH and, using REF again, we obtain MATH . From this estimate and REF , we get REF .
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math/0002154
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Let MATH. Then it is decomposed as MATH with MATH. Consequently MATH can be non-zero only for MATH. Hence MATH with MATH . Since MATH is a (non-zero) scalar we have found MATH. For such a MATH the condition MATH is equivalent to MATH. Sandwiching with MATH and MATH gives the desired intertwining relations for all MATH. Conversely, any MATH with MATH satisfying these relations intertwines MATH and MATH.
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math/0002154
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The preservation of addition and the multiplication is a straight-forward corollary of REF . The statement for the conjugates is derived in the same way as CITE.
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math/0002154
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Note that if MATH, MATH and MATH is an isometry, then MATH since MATH. Hence MATH for any MATH. With this it is easy to check that MATH is unitary. The first inclusion of REF together with CITE imply that MATH is an intertwiner from MATH to MATH. With that it is easy to check that MATH (compare the proof of CITE). Next, for MATH, MATH and MATH isometries, MATH, and MATH one has MATH, and hence we can compute MATH establishing REF . Finally, putting MATH so that consequently MATH and MATH, and choosing MATH gives the desired invariance properties of MATH with MATH.
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math/0002154
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Thanks to Izumi's result, REF , and due to the first inclusion in REF , all what we have to verify is the relation MATH for all MATH whenever MATH. For MATH there is some isometry MATH with some MATH. Then this is just MATH thanks to naturality.
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math/0002154
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It follows from MATH and the first inclusion in REF that MATH. The elements are clearly linearly independent as MATH are orthonormal isometries in MATH. Now the statement follows since MATH by NAME reciprocity.
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math/0002154
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By REF we have to show that MATH . So first we assume that MATH is in the right-hand side, and such a MATH satisfies MATH in particular for all MATH. So choose an isometry MATH with some MATH. Then, by REF , MATH whereas MATH . Equating these and multiplying by MATH from the left and MATH from the right we obtain, using again REF , MATH . Since this is supposed to hold for any MATH we may now take the sum over full orthonormal bases of MATH so that we find MATH for all MATH. Now recall that MATH is a linear combination MATH with MATH. But MATH . Therefore, using MATH implies MATH as well as orthonormality of the MATH's, we find that MATH for all MATH implies MATH for all MATH and all MATH. Taking the adjoint and applying the left inverse MATH yields MATH as the monodromy matrix MATH is obtained CITE from MATH. Hence we have MATH for all MATH. But MATH for all MATH if and only if MATH by CITE. Consequently MATH whenever MATH, so that indeed MATH. Conversely, if we start with MATH, that is, MATH then we find MATH for all MATH. Hence, if MATH and MATH is an isometry with some MATH, then MATH by REF . Thus MATH satisfies the desired intertwining relations.
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math/0002154
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Again by REF , we only need to show that for MATH the linear space of intertwiners MATH satisfying MATH for all MATH is given by MATH whenever MATH and vanishes otherwise. Thus suppose that MATH satisfies MATH for all MATH. Then in particular MATH whenever MATH is an isometry and MATH. Sandwiching this with MATH and MATH yields by use of REF MATH and since this is supposed for any subsector MATH of any MATH we can sum over orthonormal bases of MATH so that we arrive at MATH . If MATH then MATH with MATH some necessarily non-zero multiple of an isometry. Therefore we have found that MATH for all MATH implying MATH. Now note that if MATH then MATH so that our calculation also yields MATH. We conclude that the intertwiner space on the left-hand side of REF is zero unless MATH. But if MATH, then MATH as well as MATH, so that clearly MATH. Then the conclusion follows from REF .
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math/0002154
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Let MATH be an isometry, MATH. Then MATH where we used the BFE MATH.
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math/0002154
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Using REF , we can compute MATH and now the result follows by REF and since MATH.
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math/0002154
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Immediate from CITE.
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math/0002154
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Literally the same as the proof of REF , apart from the simplification in the second half that we now only have to consider MATH.
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math/0002154
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Using once again REF , we first show that if a non-zero MATH satisfies MATH for all MATH, then this implies MATH. So suppose we have such a MATH. Since MATH there will also be some MATH and an isometry MATH. Then the intertwining condition reads MATH whenever MATH and MATH is an isometry. Multiplication with MATH from the right yields MATH where we exploited MATH to apply REF . Now we can multiply by MATH from the right, and then we may use a summation over full orthonormal bases of MATH to obtain MATH for all MATH. Now note that a non-zero MATH is necessarily of the from MATH with MATH a non-zero multiple of an isometry. Hence we find MATH for all MATH, proving that the left-hand side of REF is zero unless MATH. On the other hand, if MATH then MATH. Hence, for an arbitrary MATH we find MATH, and therefore the naturality of the relative braiding of CITE gives us MATH for any MATH. By taking adjoints this reads MATH so that the desired intertwining relation is automatically fulfilled in particular for MATH. This completes the proof.
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math/0002154
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We compute MATH where we used the BFE for the relative braiding CITE, MATH.
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math/0002154
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Analogous to the proof of REF , this is reduced to REF by use of REF .
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math/0002164
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In local coordinates, the formula MATH becomes REF for the tensor MATH on MATH; thus REF are equivalent. The NAME rule for graded NAME algebras shows that MATH . Thus MATH is a differential on MATH if and only if MATH. The bracket MATH is skew-symmetric: MATH . As for the NAME rule, we have MATH . The anomalous term MATH vanishes for all MATH, MATH and MATH if and only if MATH is a differential.
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math/0002164
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By REF , a weak equivalence MATH of dg NAME algebras induces a bijection MATH; indeed, MATH is filtered by subspaces MATH, and similarly for MATH. It remains to prove that MATH induces bijections MATH for all MATH and MATH. Given MATH, define a dg NAME algebra MATH where MATH is placed in degree MATH. The construction MATH behaves like a based loop space of MATH at MATH, in the sense that MATH . To prove this, we must first show that these groupoids have the same objects, that is, that MATH if and only if MATH. If MATH, we see from REF that MATH . It follows, by induction on MATH, that MATH for all MATH, hence MATH. The remainder of the proof of REF is straightforward. That MATH is a bijection now follows on applying REF to the weak equivalence of filtered dg NAME algebras MATH . Finally, MATH is a bijection, since MATH.
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math/0002164
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It suffices to define MATH on the generators MATH of MATH over MATH. By the hypotheses on MATH, we have MATH so the definition of MATH is forced. By induction on MATH, we see that MATH . It follows that MATH, and hence that, for MATH, MATH . Thus MATH . The sum over MATH reduces to MATH, and the right-hand side to MATH.
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math/0002164
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From the formula MATH it follows that MATH and that MATH for MATH; the lemma follows easily from this formula.
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math/0002164
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We have MATH . Since MATH, this in turn equals MATH . The sum over MATH reduces to MATH, and the lemma follows.
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math/0002164
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It is clear that MATH is compatible with the differential on MATH: MATH . It is also easy to see that MATH is a morphism of graded NAME algebras, since MATH . It only remains to check that MATH is a weak equivalence; this is a variant on the ``exactness of the variational bicomplex." We learned the idea used in the following proof from NAME. Let MATH be a connected open subset of MATH, and let MATH. We must show that MATH if and only if MATH is a multiple of MATH. It is clear that this is so if MATH, since in that case, MATH. The operators MATH and MATH generate an action of the NAME algebra MATH on MATH, whose NAME subalgebra acts by the semisimple endomorphism MATH with kernel MATH. Suppose that MATH. Since MATH for MATH, we see that the irreducible MATH-module spanned by MATH is finite-dimensional. Since a finite-dimensional representation of MATH on which MATH has non-negative spectrum is trivial, we conclude that MATH; hence MATH lies in MATH, and as we have seen, is a multiple of MATH.
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math/0002164
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We must show that if MATH is a symplectic manifold, the inclusion MATH of sheaves of dg NAME algebras is a weak equivalence. By the NAME theorem (in its original sense!), it suffices to consider a convex subset MATH of MATH with its standard symplectic structure, and NAME tensor MATH . Let MATH be the differential associated to the NAME element MATH of MATH; it is given by the formula MATH . Clearly, the dg NAME algebra MATH is a resolution of MATH. The complex MATH is isomorphic to the cone of the morphism MATH where MATH is the NAME complex of the jet-space MATH and MATH is its quotient by the constant functions. To see this, one identifies MATH with MATH and MATH with MATH. REF now follows from the NAME theorem for MATH.
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math/0002164
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Lifts MATH of MATH are characterized by the equations MATH and MATH . Let MATH be a section of MATH such that MATH; there are no obstructions to the existence of MATH, because MATH is NAME. Since MATH, we see that there is a section MATH of MATH such that MATH; again, there are no obstructions to the existence of MATH. Taking the bracket of this equation with MATH, we see that MATH . But MATH vanishes by the NAME rule, while MATH. By REF , we conclude that MATH.
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math/0002164
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CASE: MATH is a morphism of complexes (that is, MATH): Let MATH be an element of MATH. It is obvious that MATH since MATH and MATH both vanish. As for MATH, we have MATH and MATH . CASE: MATH preserves the NAME bracket: If MATH, we have MATH . On the other hand, MATH . From these formulas, we see that MATH. Finally, it is clear that MATH, as they must, since MATH and MATH commute in MATH.
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math/0002164
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We must show that MATH vanishes. Rewritten using the operators MATH and MATH, and taking into account that the operators MATH and MATH graded commute with MATH and MATH, and that MATH, we see that this equals MATH which vanishes, since MATH and MATH.
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math/0002164
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We have MATH . The formula follows, since MATH.
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math/0002164
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There is short exact sequence of complexes MATH and hence, for MATH, a long exact sequence MATH . There is an isomorphism between the complex MATH and the NAME complex MATH, obtained by mapping MATH to MATH and MATH to the basis vector MATH of MATH. Likewise, the complex MATH is isomorphic to the reduced NAME complex MATH. The NAME lemma for MATH shows that MATH induces isomorphisms between the groups MATH and MATH and the group MATH. The composition of MATH with the boundary map MATH vanishes: if MATH, we have MATH . We conclude that there is a short exact sequence MATH and hence that MATH is indeed an isomorphism onto the cohomology of MATH.
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math/0002169
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Obviously, REF follows from REF . For MATH the statement was proved by NAME and NAME REF . Now suppose that MATH and REF is minimal. Since MATH is globally generated, NAME 's theorem ensures that a generic map MATH is injective. Then, in the following commutative diagram columns and rows are exact and MATH is locally free: MATH . The minimality of the middle row yields the minimality of the last row. Using induction on MATH, we may assume that the last row is weakly admissible. Then, the middle row is also weakly admissible.
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math/0002169
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We first prove that if MATH is semistable, then MATH. In fact, if MATH then MATH split as MATH and by weakly admissibility MATH then we have MATH which contradicts the semistability of MATH. Now suppose that MATH is semistable and MATH for some MATH with MATH and let MATH be the largest of such MATH. Since MATH the minor MATH of MATH, obtained by cutting off the last MATH rows and MATH columns, remains of maximal rank so MATH is a vector bundle. A surjective morphism MATH is defined by the diagram MATH where the first two vertical map are the natural projections. Observe that MATH have the same rank MATH as MATH and MATH . Then, MATH must be semistable otherwise any torsionless quotient sheaf destabilizing it would also destabilize MATH. By induction on MATH, we may assume that the second row of REF is strongly admissible. In particular, we have MATH that gives a contradiction. Finally, if MATH is stable (respectively, semistable), then MATH but, from the exact sequence MATH we have MATH then MATH (respectively, MATH).
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math/0002169
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Let MATH be the minimal resolution of MATH where MATH . The stability of MATH ensures the vanishing MATH so that MATH. Then, from REF we easily find the following data: MATH and from REF we have MATH . Now, by splitting MATH as MATH with MATH, the above formula becomes MATH . Since MATH and MATH so MATH . Finally, distributing the direct sums appearing in the definition of MATH and MATH REF becomes MATH .
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math/0002169
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It can be verified by direct computation from REF that, if MATH has natural pair, then the codimension of MATH is zero. Conversely, let MATH, MATH be two non-negative integers. Since all finite difference MATH are non decreasing functions of MATH, then MATH and by the previous lemma MATH . If MATH, we have MATH and MATH, since MATH implies MATH. This forces MATH to be a natural pair.
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math/0002169
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This is a verification; we outline the main steps of the computation. In the first place, one must ensure that the natural pair MATH is actually associated to vector bundles in MATH. This amount to show that from REF the pair MATH has the appropriate NAME classes and that REF hold. From REF , any pair MATH such that MATH is a natural pair of the form MATH. From resolutions REF we find that MATH, MATH must satisfy REF . Then, it remains to verify that MATH is uniquely determined from MATH, MATH, MATH and satisfy REF . By substitution, the inequalities MATH turn into MATH . Since the intervals MATH are disjoint for MATH varying in MATH, then REF give the only suitable value for MATH.
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math/0002169
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Moduli space of stable rank MATH vector bundles on MATH are smooth. By the previous proposition they can have only one connected component.
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math/0002169
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Let MATH be the admissible pair associated to a vector bundle MATH in MATH. Then, one has MATH if and only if MATH. By semicontinuity of cohomology groups and REF , it is enough to restrict ourselves to the case where MATH is general. So, by REF one has MATH if and only if MATH and the condition MATH is equivalent to MATH by REF .
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math/0002170
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Let MATH. We prove that MATH for any MATH. Then we conclude that MATH and hence MATH. Since MATH, we obtain MATH. Now REF and the assertion of the Proposition for MATH imply that MATH . Hence, MATH is the symmetrizer of the algebra MATH. Let MATH. We prove that MATH for any MATH. For this we use the relations of the algebra MATH and REF . Observe that in REF only MATH occurs. Therefore, we can use that MATH for all MATH. Let now MATH. Then MATH . The case MATH can be proven similarly.
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math/0002170
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Let MATH. Using the relations REF - REF we obtain MATH . The other cases can be shown similarly.
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math/0002180
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This is a direct corollary of the general category-theoretic REF , under `NAME below. Alternatively, a direct argument can be used, as follows. Given a colax symmetric monoidal functor MATH we must define a functor MATH. On objects, take MATH. To define MATH on morphisms, first note the following: CASE: for each MATH, there is a map MATH in MATH given by MATH CASE: for each map MATH in MATH, there is a corresponding map MATH in MATH . Now if MATH is a map in MATH, let MATH be the composite MATH where MATH is the second component of MATH. This defines a functor MATH. Conversely, take a functor MATH. Define a functor MATH by MATH on objects; if MATH is a morphism in MATH then define MATH where the map MATH in MATH is given by MATH . To define MATH, obviously MATH is the unique map MATH. For the MATH's, first define maps MATH in MATH (for MATH) by MATH . Then define MATH to be MATH . After performing all the checks we see that the two processes are mutually inverse, and that they can be extended to apply to transformations too. Thus we obtain the required isomorphism of categories. MATH .
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math/0002180
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REF is immediate from the second half of the proof of REF . An easy induction, again using this half of the proof, shows that MATH equals MATH so by the comments in REF , we have REF . MATH .
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math/0002180
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CASE: From the definition of MATH in the proof of REF , it is easy to show that MATH, and similarly MATH. Hence MATH . Also, MATH is the unique map from MATH to MATH. CASE: Similarly, an easy induction on MATH shows that MATH where MATH is the map defined in REF . But by the comments in REF , MATH is a homotopy monoid if and only if each MATH is an equivalence. MATH .
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math/0002180
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For each MATH, let MATH denote the standard MATH-simplex MATH with its MATH vertices collapsed to a single point, and this point declared the basepoint. Informally, MATH can be described as follows: CASE: MATH, for example, MATH is a single point, MATH, and MATH looks like MATH CASE: MATH is defined on morphisms by the standard face and degeneracy maps of simplices CASE: MATH is defined by face maps, for example, MATH is the evident inclusion MATH which is a homotopy equivalence. Formally, it's easiest to employ the description of homotopy monoids given in REF . So, first consider the usual functor from MATH to MATH, mapping MATH to MATH and defined on morphisms by face and degeneracy maps. Since all the face and degeneracy maps take vertices to vertices, this functor induces another functor MATH with MATH. We may also view MATH as a functor MATH and MATH are respectively coproduct and initial object in MATH, so they are product and terminal object in MATH. So by REF , MATH corresponds to a colax monoidal functor MATH with MATH. To see that MATH is a homotopy monoid we must first of all check that the unique map MATH is a homotopy equivalence, and then check that the map MATH induced by the two maps MATH in MATH is also a homotopy equivalence (see REF ). The first check is trivial, and for the second it is easy to construct a homotopy inverse. (From the conceptual angle, note however that there is no canonical choice of a homotopy inverse: see the picture of MATH above, for instance.) MATH .
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math/0002180
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Take a homotopy monoidal category MATH in MATH, and construct from it a monoidal category as follows. CASE: MATH. CASE: What we want to define is a functor MATH what we actually have are functors where MATH is the unique map MATH in MATH. So for each MATH and MATH, choose (arbitrarily) a pseudo-inverse MATH to MATH, and define MATH as the composite MATH . CASE: The next piece of data we need is a natural isomorphism between MATH and MATH. To see why such an isomorphism should exist, consider what would happen if the MATH's were genuine inverses to the MATH's. Then the MATH's would satisfy the same coherence and naturality axioms as the MATH's (with the arrows reversed), and this would guarantee that all sensible diagrams built up out of MATH's commuted. Hence MATH would be strictly associative. As it is, MATH is only inverse to MATH up to isomorphism, and correspondingly MATH is associative up to isomorphism. In practice, let us choose (at random) natural isomorphisms MATH for each MATH and MATH. Then a natural isomorphism MATH can be built up from the MATH's and MATH's. The exact formula for MATH is rather complicated, and only included for the record. Most readers will therefore want to skip the next paragraph and ignore REF . In order to define MATH, consider the diagram at the top of REF . The composites around the outside are MATH and MATH, so we must find natural isomorphisms inside each of the four inner squares. The bottom-right square, in which MATH are the two surjections MATH in MATH, is genuinely commutative. In each of the other three squares, imagine taking each arrow labelled by a MATH, reversing its direction, and changing the MATH to a MATH. The imaginary square would then be genuinely commutative in each case, which means that the actual square is commutative up to isomorphism. This is the thought behind the formula for MATH given in the rest of REF . (For the usage of MATH, see page REF.) CASE: We must now check that the associativity isomorphism just defined satisfies the famous pentagon coherence axiom. This asserts the commutativity of a certain diagram built up from components of MATH, that is, built up from MATH's and MATH's. However, this diagram does not commute, which is perhaps unsurprising since MATH and MATH were chosen independently. But all is not lost: for recall the result that if MATH is an equivalence of categories, then MATH can be exchanged for another natural isomorphism MATH so that MATH is both an adjunction and an equivalence (see CITE). So when we chose the natural isomorphisms MATH and MATH above, we could have done it so that MATH was an adjunction. Assume that we did so. Then this being an adjunction says that certain basic diagrams involving MATH and MATH commute (namely, the diagrams for the triangle identities CITE): and that is enough to ensure that the pentagon commutes. Any reader who followed the construction of MATH will see that the pentagon involves REF terms of the form MATH or MATH. Checking that it commutes is therefore an appreciable task, but in the absence of higher technology there is no alternative. CASE: So far we have only mentioned binary tensor, and not units. To construct the unit object MATH of MATH, choose a pseudo-inverse MATH to the equivalence of categories MATH (in other words, pick an object of MATH), and define MATH as (the image of) the composite MATH . CASE: We need left and right unit isomorphisms MATH natural in MATH. To define them, choose natural isomorphisms MATH in such a way that MATH is an adjunction (MATH left adjoint to MATH); this is possible by the result referred to under `NAME above. (In fact, it's not strictly necessary to use that general result, since the involvement of the category MATH makes the situation trivial; but the argument from general principles is conceptually cleaner.) Then MATH and MATH can be built up from MATH's, MATH's, MATH and MATH. For the record only, MATH is defined in REF , which can be explained in the same way that REF was. CASE: The final check is that the triangle axiom holds; this is the `other' coherence axiom for monoidal categories, along with the pentagon. It is built up out of MATH, MATH and MATH, hence out of MATH's, MATH's, MATH and MATH, and commutes for the same reason that the pentagon commutes. MATH .
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math/0002180
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Let MATH be a monoidal transformation. Let MATH and MATH be the (arbitrarily-chosen) functors used in the construction of MATH, and MATH the natural transformations, and similarly MATH etc.for MATH. We now construct a monoidal functor from MATH to MATH. The functor part is MATH. For the rest of the structure we need isomorphisms MATH (natural in MATH) and MATH. The first can be extracted from the diagram in which the right-hand square commutes and the left-hand square commutes up to isomorphism. The second arises similarly from the diagram Once these coherence isomorphisms have been written down explicitly, it is just a matter of checking the axioms. MATH .
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math/0002180
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Let MATH be the choices for MATH, and MATH those for MATH. Then MATH and MATH are isomorphic objects of MATH, so their images in MATH under the functor of REF are also isomorphic. In other words, MATH in MATH. MATH .
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math/0002180
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There is a REF-category MATH whose objects and REF-cells are the same as those of REF category MATH, and whose REF-cells are homotopy classes of homotopies. An equivalence in this REF-category is just a homotopy equivalence. (All homotopies mentioned here must respect basepoints.) Moreover, cartesian product MATH and the one-point space MATH make MATH into a monoidal REF-category. So by REF , a homotopy semigroup MATH in MATH gives rise to a weak semigroup in MATH with underlying space MATH. Now we only have to see that a weak semigroup in the monoidal REF-category MATH gives rise to a MATH-space. (In fact they are more or less the same thing, as the argument reveals.) I shall only do this informally. A weak semigroup in MATH consists of a based space MATH, a multiplication MATH with MATH, and a REF-cell MATH between the two maps MATH such that MATH satisfies the pentagon axiom. Now a REF-cell in MATH is a homotopy class of homotopies, so we may pick a representative MATH of MATH. The pentagon axiom then says that a certain pair of homotopies MATH (built up from MATH's) belong to the same homotopy class. Choose a homotopy MATH between MATH and MATH: then MATH is essentially a map MATH where MATH is the solid pentagon (as in CITE), and the data MATH thus describes a MATH-space. MATH .
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math/0002180
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First we show that MATH is a MATH-fold homotopy comonoid, that is, a homotopy MATH-algebra in MATH. Let MATH be the colax monoidal functor of REF , exhibiting MATH as a homotopy comonoid. Also let MATH be the MATH-fold smash product, MATH . Observe that since MATH distributes over MATH, MATH naturally becomes a multi-monoidal functor MATH observe moreover that MATH preserves homotopy equivalences. Assembling all of this, we get a composite MATH and this is a colax multi-monoidal functor MATH in which the components of MATH are equivalences. Thus MATH defines a MATH-fold homotopy comonoid, and MATH as required. To finish the proof we simply use the observation made in REF that MATH defines a homotopy-preserving monoidal functor MATH . Composing this with MATH yields a MATH-fold homotopy monoid MATH whose value at MATH is MATH. MATH .
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math/0002180
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Simply apply the axioms for a class of equivalences REF to the commuting diagrams in the definition of monoidal transformation REF . MATH .
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math/0002183
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Since MATH admits a finite MATH-invariant measure, by REF , we have that MATH. Let MATH . Then MATH, MATH, and MATH. Therefore MATH admits a finite MATH-invariant measure, and hence MATH is closed. Hence replacing MATH by MATH, we may assume that MATH is normal in MATH. Now without loss of generality we can replace MATH by MATH and MATH by MATH, and assume that MATH is a discrete subgroup of MATH. Moreover, MATH.
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math/0002183
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Take any MATH. By REF , there exists a compact set MATH such that the set MATH is infinite. Therefore there exists MATH such that MATH. Therefore there exists MATH such that MATH. Put MATH. Then MATH. Put MATH . Then MATH. Suppose there exists a closed subgroup MATH of MATH containing MATH such that MATH is closed. Then by REF , we have that MATH is a subgroup of MATH. Note that for MATH, if MATH then MATH. Hence MATH for all MATH. Therefore MATH is closed in MATH, and hence in MATH. Thus MATH, and MATH. This shows that replacing MATH by MATH, we may assume that MATH. This completes the proof of the claim.
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math/0002183
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Let MATH be such a subgroup. Then the claim follows from replacement of MATH by MATH.
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math/0002183
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Let MATH denote a locally finite MATH-invariant NAME measure on MATH. Since MATH is a proper map, the projected measure MATH on MATH is a locally finite and MATH-invariant. Now the claim follows from the uniqueness, up to constant multiple, of the locally finite MATH-invariant measures on MATH.
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math/0002183
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Since the fibers of MATH have finite invariant measures, we have that MATH has a finite MATH-invariant measure. Therefore MATH has a finite MATH-invariant measure. Again MATH has a finite MATH-invariant measure. Now the claim follows from REF .
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math/0002183
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For any MATH, there exists a closed connected subgroup MATH of MATH such that MATH has a finite MATH-invariant measure, MATH, and MATH. Therefore MATH. Now MATH. Hence MATH has a finite MATH-invariant measure. Therefore MATH is closed. Let MATH be the one-parameter unipotent subgroup of MATH such that MATH. Since MATH, we have that MATH and MATH . Now MATH. This shows that MATH . Case of MATH-rank-MATH : In this case by REF , the group MATH is the NAME closure of the subgroup generated by MATH for all MATH. Therefore by REF , MATH is a normal subgroup of MATH. Hence the claim holds in this case. Case of MATH-rank-MATH : Put MATH. Then MATH. Now suppose that the unipotent radical of MATH is nontrivial. Since MATH, MATH is contained in a unique minimal parabolic subgroup, say MATH of MATH. Let MATH denote the unipotent radical of MATH. Then MATH and MATH. Hence MATH for all MATH. Let MATH. Since MATH we have that MATH. Therefore MATH. Hence MATH is compact. Therefore MATH, and hence MATH which is a contradiction. Thus MATH is a reductive subgroup of MATH. Since MATH, we obtain that the solvable radical of MATH is compact. This completes the proof of the claim.
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math/0002183
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Due to REF , without loss of generality we can pass to the quotient MATH and assume that MATH. Now arguing as in the proof of REF , we conclude that MATH contains a neighbourhood of MATH in MATH. Now since MATH contains a neighbourhood of MATH in MATH, the claim follows. MATH .
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math/0002185
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We may assume that MATH. Since MATH is exact, it follows from NAME 's theorem CITE that the inclusion map of MATH into MATH is nuclear. Thus, there are MATH and unital completely positive maps MATH and MATH such that MATH . Let MATH be the corresponding linear functional defined as in REF . Since MATH is weak-MATH dense in MATH, we can approximate MATH by linear functionals in MATH in the weak-MATH-topology. It follows that we can approximate MATH in the point-norm topology by completely positive maps MATH such that MATH is in MATH. Thus, for arbitrary MATH, we may find such MATH with MATH . Let MATH. Since MATH and MATH, MATH is invertible. Thus, we can define a unital completely positive map MATH by MATH for MATH. Taking MATH sufficiently small, we have MATH . Finally, put MATH and we are done.
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math/0002185
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CASE: We follow the proof of REF. We give ourselves a finite subset MATH and MATH. By the assumption, there is MATH satisfying the conditions in REF for MATH, MATH and MATH. If MATH for MATH, then we put MATH. We define MATH by MATH for MATH. It is not hard to check that MATH has the desired properties. CASE: Let MATH be an increasing net of finite subsets of MATH containing the unit MATH. By the assumption, there are finite subsets MATH and a net of positive definite kernels MATH such that MATH and that MATH . We may assume that MATH for all MATH. Let MATH be the NAME multiplier associated with the positive definite kernel MATH. Then, MATH's are completely positive contractions and it can be seen that MATH and that MATH . Let MATH be the restriction map from MATH onto MATH, that is, MATH for MATH. For MATH and MATH, we define a complete contraction MATH by MATH for MATH. Then, we have MATH for MATH and MATH. To prove that MATH is nuclear, we take a unital MATH-algebra MATH. It suffices to show that MATH. Let MATH be the canonical quotient map. Since MATH is nuclear, we observe that MATH is a well-defined contraction. Since MATH's are completely positive contractions, so are MATH (see REF) and we have MATH . Combining this with the factorization MATH we see that MATH. CASE: This is obvious.
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math/0002186
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For MATH denote by MATH the ring of germs at MATH of real-analytic functions on MATH. For MATH near MATH, we can think of an element MATH in MATH as a MATH-tuple MATH satisfying the condition in REF . Hence the subsheaf MATH coincides with the sheaf of relations MATH . Since the sheaf MATH is coherent by REF, it follows that MATH is locally finitely generated over MATH which proves the lemma.
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math/0002186
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Define MATH, MATH, where MATH and MATH are the integer valued functions defined by REF respectively. Given MATH, choose MATH and MATH as in REF . Now consider the set of vector valued real-analytic functions MATH (as in REF ) defined in MATH. For each subset of MATH functions in this set, we take all possible MATH minors extracted from their components. Then by REF , the set MATH is given by the vanishing of all such minors. Since MATH is arbitrary, MATH is a real-analytic subvariety. To show that MATH is also real-analytic, we repeat the above argument for the set of vector valued real-analytic functions MATH (as in REF ). Both subsets MATH are proper by the choices of MATH and MATH.
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math/0002186
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We may assume MATH. Since MATH is CR at MATH, there is a neighborhood of MATH in MATH such that the piece of MATH in that neighborhood is contained as a generic submanifold in a complex submanifold of MATH (called the intrinsic complexification of MATH) of complex dimension MATH (see for example, CITE). By a suitable choice of holomorphic coordinates MATH with MATH as in the proposition, we may assume that the intrinsic complexification is given by MATH near MATH. Then MATH is a generic submanifold of the subspace MATH. Hence in the rest of the proof it suffices to assume that MATH is generic and MATH. We may therefore find holomorphic coordinates MATH, with MATH being the codimension of MATH in MATH and MATH, such that if MATH is a local defining function of MATH near MATH, then MATH is an invertible MATH matrix. By the implicit function theorem, we can write MATH near MATH in the form MATH where MATH is a MATH-valued holomorphic function defined in a neighborhood of MATH in MATH and vanishing at MATH. We now apply the definition of minimum degeneracy given in REF to the (complex valued) defining function of MATH given by MATH . It can be easily checked that the identity REF holds with MATH replaced by MATH (even though here MATH is complex valued). Consider the basis of MATH vector fields on MATH given by MATH where, as above, MATH. Observe that since MATH is independent of MATH, for MATH and MATH, MATH . Since MATH is of minimum degeneracy at MATH, it follows that for MATH in a neighborhood of MATH in MATH, MATH . By a standard complexification argument (see for example, REF), we conclude that for MATH and MATH near the origin, we also have MATH . Hence there exists an integer MATH such that for MATH in a neighborhood of MATH, MATH . In particular, if MATH is MATH times the number of multi-indices MATH with MATH, the map MATH given by MATH is of constant rank equal to MATH for MATH in a neighborhood of MATH in MATH. By the implicit function theorem, there exists a holomorphic change of coordinates MATH with MATH such that MATH. It follows that MATH is independent of MATH for all MATH, MATH, and hence, by the choice of MATH, for all MATH. Therefore, if we write the complexification of REF in the form MATH we conclude that MATH is independent of MATH. Hence the (complex valued) function given by MATH is independent of MATH, and MATH is given by MATH. Thus all vector fields MATH, MATH, are tangent to MATH and hence so are the vector fields MATH. After a linear change of the coordinates MATH we can write MATH near MATH in the form MATH where MATH is a real-analytic, real vector valued function. Hence the submanifold MATH given by MATH satisfies the required assumptions.
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math/0002186
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Since MATH is generic, and hence CR, we can choose a frame MATH of real-analytic MATH vector fields on MATH near MATH, spanning the space of all MATH tangent vectors to MATH at every point near MATH. (Here MATH, where MATH is the codimension of MATH.) We write MATH, where MATH, MATH, are real valued vector fields. We prove first that REF implies REF . By the condition that MATH is of minimum orbit codimension MATH at MATH, it follows that the collection of the vector fields MATH, MATH, generates a NAME algebra, whose dimension at every point near MATH is MATH. Therefore, by the (real) NAME theorem, we conclude that there exist MATH real-analytic real valued functions MATH with independent differentials, defined in a neighborhood of MATH, vanishing at MATH and such that MATH (that is, MATH is a CR function) for all MATH and MATH. Moreover, the local orbits of the MATH, MATH, are all of the form MATH with MATH and MATH sufficiently small. By REF (CITE, see also REF ), the functions MATH extend holomorphically to a full neighborhood of MATH in MATH. This proves that REF implies REF . For the proof that REF implies REF , we observe that since MATH extends holomorphically, we have MATH for all MATH and MATH. By the reality of MATH it follows that MATH for all MATH and MATH. Hence the set MATH is the CR orbit of MATH at MATH and is of dimension MATH, which proves REF . Since the implication MATH is trivial, the proof of the proposition is complete.
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math/0002186
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We take normal coordinates MATH vanishing at MATH (see for example, REF), that is, we assume that MATH is given by MATH near MATH, where MATH is a germ at MATH in MATH of a holomorphic MATH-valued function satisfying MATH . We may choose a frame MATH spanning the space of all MATH vector fields on MATH of the form MATH for MATH. In particular, MATH. Let MATH be the functions given by REF . Since, for MATH, the functions MATH are real and extend holomorphically, we conclude that MATH. We denote again the by MATH the extended functions. By the choice of the coordinates, MATH, MATH, MATH. By using the independence of the differentials of MATH and reordering the components MATH if necessary, we may assume that MATH . We make the following change of holomorphic coordinates in MATH near MATH: MATH . Note that on MATH, we have MATH for MATH. The reader can check that the new coordinates MATH are again normal for MATH. Indeed, MATH is given by MATH where MATH satisfies the analog of REF , with MATH for MATH. The desired coordinates are obtained by taking MATH that is, MATH and MATH with MATH, MATH. We take MATH for MATH. By the properties of the functions MATH, the submanifold MATH given by REF , with MATH close to MATH, is of finite type if MATH is a sufficiently small neighborhood of MATH in MATH. This completes the proof of REF .
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math/0002186
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Denote by MATH the dimension of the span in MATH of the MATH, MATH. By the assumption, there exists a MATH minor MATH, extracted from the components of the MATH, which does not vanish identically. Note that MATH. Let MATH, MATH, be such that MATH. Then for MATH and MATH as in the lemma, it follows that MATH, where MATH is the corresponding minor with MATH replaced by MATH. On the other hand, the dimension of the span of the MATH is also MATH. Since MATH is a MATH minor that does not vanish identically, the proof of the lemma is complete.
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math/0002186
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We write MATH. Let MATH be a local defining function for MATH. Then MATH is a local defining function for MATH in a neighborhood of MATH in MATH. By the definition of MATH-equivalence, we obtain MATH for any local parametrization MATH of MATH at MATH. By the second identity in REF , the holomorphic function MATH vanishes of order MATH at MATH on the submanifold MATH which is generic in MATH. This implies the first statement of the lemma. Since MATH is invertible and by the first statement of the lemma we have MATH, the map MATH must be invertible at MATH. Hence the first identity in REF implies the second statement of the lemma.
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math/0002186
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We first observe that every MATH-equivalence between MATH and MATH induces a linear isomorphism between MATH and MATH. Since MATH and MATH are MATH-equivalent for every MATH, this implies MATH. To complete the proof of REF , we argue by contradiction. We assume that MATH is CR at MATH but that MATH is not. If MATH is a local defining function for MATH and MATH is a local parametrization of MATH at MATH, we set MATH, MATH. Since MATH is assumed not to be CR at MATH, the collection of functions MATH satisfies the assumptions of REF . Let MATH be the integer given by the lemma. We take MATH and let MATH be a MATH-equivalence between MATH and MATH. If we set MATH, then the identity map is a MATH-equivalence between MATH and MATH. Hence, by REF , there exist a local parametrization MATH of MATH at MATH and a local defining function MATH for MATH near MATH such that MATH and MATH. We apply REF for the collection MATH defined above and MATH and conclude that MATH is not CR at MATH. Thus we have reached a contradiction, since MATH and MATH are biholomorphically equivalent. This completes the proof of REF . To prove REF , suppose that MATH and MATH are CR at MATH. Since MATH and MATH are CR and MATH by REF , we may assume that MATH and MATH, both contained in MATH with MATH, MATH and MATH and MATH generic in MATH (compare beginning of proof of REF ). By REF , MATH and MATH are also MATH-equivalent for every MATH. We observe that MATH is of minimum degeneracy at MATH if and only if MATH is of minimum degeneracy at MATH and the degeneracies of MATH and MATH at MATH are the same. Therefore, by replacing MATH by MATH and MATH by MATH, we may assume that MATH and MATH are generic (that is, MATH) in the rest of the proof. We show first that MATH. From REF of these numbers, there exists MATH such that MATH where, for MATH, MATH with MATH being a defining function for MATH near MATH, and MATH is the corresponding subspace for MATH. We may choose holomorphic coordinates MATH vanishing at MATH such that the MATH matrix MATH is invertible. In these coordinates we take a basis of MATH vector fields on MATH in the form MATH where we have used matrix notation so that MATH is viewed as a MATH matrix. We now choose a local parametrization MATH of MATH at MATH, and put MATH . We then choose MATH and MATH to be a MATH-equivalence between MATH and MATH which exists by the assumptions of the proposition. Replacing MATH by MATH we may assume without loss of generality that MATH is the identity map of MATH. By REF , we can find a local parametrization MATH of MATH at MATH satisfying MATH and a local defining function MATH of MATH near MATH satisfying MATH. Denote by MATH, MATH, the local basis of MATH vector fields on MATH given by the analog of REF with MATH replaced by MATH. (Observe that MATH coincides with MATH and hence is invertible). By the choice of MATH, we have MATH in a neighborhood of MATH in MATH, where MATH is a vector field whose coefficients vanish of order MATH at MATH. We put MATH . Then it follows from the construction that MATH and, in particular, MATH for all MATH and MATH as in REF . Hence, by making use of REF , we have MATH, which proves the first part of REF . To prove the second part of REF , assume that MATH is of minimum degeneracy at MATH and that MATH is not. We shall reach a contradiction by again making use of REF . From the definition of minimum degeneracy there exists an integer MATH such that MATH for MATH, and MATH near MATH. Hence the collection of real-analytic functions given by REF with MATH replaced by MATH satisfies the assumption of REF . We let MATH be the integer given by that lemma and choose MATH to be a MATH-equivalence with MATH satisfying MATH. As before, we may assume that MATH is the identity. Using again REF , we obtain the analogue of REF , with MATH replaced by MATH. We conclude by REF that the dimension of the span of the MATH given by REF , with MATH replaced by MATH, is not constant in any neighborhood of MATH. This contradicts the first part of REF and proves the second part of REF . The proof of REF is quite similar to that of REF , and the details are left to the reader. The proof of REF is complete.
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math/0002186
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Observe that MATH and MATH are generic submanifolds of MATH. We write MATH. Let MATH be a local defining function for MATH. Then MATH is a local defining function for MATH in a neighborhood of MATH in MATH. By the definition of MATH-equivalence, we obtain MATH for any local parametrization MATH of MATH at MATH. We choose MATH in the form MATH where MATH is a local parametrization of MATH at MATH. Similarly, we choose a local defining function MATH for MATH near MATH in MATH and put MATH. Since MATH is a MATH-equivalence between MATH and MATH, the identity map is a MATH-equivalence between MATH and MATH with MATH. We let MATH, MATH, be the MATH vector fields defined in a neighborhood of MATH in MATH given by REF (after reordering coordinates in the form MATH with MATH invertible). Similarly we define MATH by an analogue of REF with MATH replaced by MATH, where MATH is the defining function of MATH given by REF for the identity map so that MATH. (We may take the same decomposition MATH since MATH). Hence MATH with MATH a MATH vector field in a neighborhood of MATH in MATH whose coefficients vanish of order MATH at MATH. Observe that the vector fields MATH are tangent to MATH. By REF , there exists a MATH real-analytic matrix valued function MATH such that MATH . Differentiating REF with respect to MATH and applying MATH for MATH, we obtain MATH where MATH and MATH are real-analytic functions in a neighborhood of MATH in MATH, valued in MATH and in MATH matrices respectively. Using the relation MATH and the definition of MATH given above, we conclude MATH . We now choose a local parametrization MATH of MATH at MATH given by REF , that is, MATH, with MATH given by REF . Since the MATH are tangent to MATH, we conclude from REF that MATH . By the choices of MATH and MATH, we have the decompositions in MATH . We multiply both sides of REF on the right by the MATH constant matrix MATH with MATH being the MATH identity matrix. We conclude that MATH . We now use the assumption that MATH is MATH-nondegenerate. By this assumption, we can choose multi-indices MATH and integers MATH, with MATH, MATH, such that the MATH matrix given by MATH is invertible for MATH near MATH. Since MATH by REF and MATH, we conclude that MATH. Since MATH is generic in MATH, REF follows. From REF it follows in particular that MATH. Since MATH is invertible, we conclude that MATH is also invertible, and REF follows from REF by taking MATH. This completes the proof of REF .
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math/0002186
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Set MATH, where MATH are defined by REF respectively. Let MATH, MATH a real-analytic submanifold of MATH and MATH. We may assume that MATH and MATH have the same dimension, since otherwise there is nothing to prove. Let MATH be fixed. If, for some integer MATH, MATH and MATH are not MATH-equivalent, then we can take MATH to satisfy the conclusion of REF . Assume for the rest of the proof that MATH and MATH are MATH-equivalent for all MATH. Without loss of generality we may assume MATH. We shall make use of REF . Since MATH is CR, of minimum degeneracy, and of minimum orbit codimension at MATH, MATH is also CR, of minimum degeneracy, and of minimum orbit codimension at MATH. Furthermore, in the notation of REF , we have MATH, MATH. Hence we may apply REF to both MATH and MATH with the same integers MATH to obtain the decompositions MATH where both decompositions are understood in the sense of germs at MATH in MATH. Since MATH is finitely nondegenerate at MATH, there exists an integer MATH such that MATH is MATH-nondegenerate at MATH. Assume first that MATH and MATH are generic at MATH, that is, MATH. Then, for every MATH, the conclusions of REF hold. By conclusion REF of that lemma, for every MATH-equivalence MATH between MATH and MATH, the map MATH is a MATH-equivalence between MATH and MATH. Furthermore, MATH and MATH satisfy the assumptions of REF . By REF , there exists a biholomorphic equivalence MATH between MATH and MATH with MATH. As we mentioned in the beginning of REF, without loss of generality, we can assume that MATH is convergent. Then we may define the germ MATH of a biholomorphism at the origin as follows: MATH . It is then a consequence of REF that MATH satisfies the conclusion of REF if MATH. We now return to the general case in which MATH and MATH are not necessarily generic, and let MATH be a MATH-equivalence corresponding to the decomposition given by REF with MATH. By REF the mapping MATH is a MATH-equivalence between the generic submanifolds MATH and MATH and MATH. It follows from the generic case, treated above, that there exists a biholomorphic equivalence MATH between MATH and MATH such that MATH. We write MATH corresponding to the product MATH. We may now define MATH by MATH and conclude that MATH satisfies the desired conclusion of the theorem. This completes the proof of REF (assuming REF ).
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math/0002186
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For MATH as above and MATH, define a germ MATH at MATH of a holomorphic function on MATH by MATH . We use the consequence of the chain rule that any partial derivative of a composition of two holomorphic maps can be written as a polynomial expression in the partial derivatives of the components. Then it follows from the assumptions of the lemma that MATH is in the ring MATH. It is straightforward to see that, if MATH is given by MATH then REF holds and the map MATH satisfies the conclusion of the lemma.
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math/0002186
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For convenience we use the notation MATH so that the equation of MATH near MATH is given by MATH, or equivalently, by MATH. We first differentiate the identity REF in MATH. Using the chain rule we obtain the identity in matrix notation MATH where MATH. (Observe that MATH in REF depends on the map MATH). The invertibility of MATH implies the invertibility of MATH and hence of MATH for MATH near the origin (since MATH by REF ). Hence we conclude for MATH sufficiently small, MATH . Our next goal will be to express the right-hand side of REF and then its derivatives in terms of functions in MATH that vanish on certain vector subspaces. For this we introduce the notation MATH . We have the following lemma. With the notation above there exists a MATH matrix MATH, independent of MATH, with entries in MATH such that, for MATH in a neighborhood of MATH in MATH, MATH and MATH vanishes on the subspace MATH defined by REF . For simplicity we drop the argument MATH in MATH, MATH, MATH, MATH and MATH. We have MATH . The first factor in the right-hand side of REF can be expressed as a matrix valued polynomial in the entries of MATH and MATH with holomorphic coefficients in MATH. We now think of the entries of MATH as variables in MATH and those of MATH as part of the variables in MATH and write MATH with MATH independent of the variable in the first factor MATH and having entries in MATH. Since MATH, MATH vanishes on the subspace MATH defined by REF with MATH. By the standard formula for the inverse of a matrix, the third factor in the right-hand side of REF can be also written in the form MATH, where MATH is a matrix valued polynomial (with entries in MATH) depending only on part of the variables in MATH and independent of the variables in MATH and MATH. The second factor in the right-hand side of REF can also be written in the form MATH with the entries of MATH in MATH. This can be shown by using the chain rule in addition to the arguments used for the first and third factors. The proof of the lemma is completed by taking MATH and using the fact that MATH is a ring. For the sequel we shall need the following lemma, which is proved by repeated use of the chain rule, making use of the identities REF , MATH, and induction on MATH. The details are left to the reader. Let MATH and MATH be as in REF . Then for every MATH with MATH and every MATH, there exists MATH such that the following holds. For any MATH-equivalence MATH between MATH and MATH with MATH, MATH . If in addition MATH (that is, the differentiation in REF is taken with respect to MATH only) and if MATH vanishes on the subspace MATH defined by REF , then MATH vanishes on the subspace MATH. We now return to the proof of REF . By making use of REF we obtain the identity MATH where MATH as before and MATH is given by REF . We claim that for every MATH with MATH, there exists MATH, independent of MATH, vanishing on the subspace MATH, and such that the following identity holds for MATH in a neighborhood of MATH in MATH: MATH . Indeed, for MATH, REF follows directly from REF and for MATH, REF is a reformulation of REF . For MATH we prove the claim by induction on MATH. Assume that REF holds for some MATH. By differentiating REF with respect to MATH we obtain in matrix notation the identity MATH . By REF we have MATH where MATH is a MATH matrix with entries in MATH, vanishing on the subspace MATH. Since, as in the proof of REF , each entry of the matrix MATH can be written in the form MATH with MATH in the ring MATH, the identity REF for MATH replaced by any multiindex MATH with MATH follows from REF by observing that the ring MATH has a natural embedding into MATH. This completes the proof of the claim. We now use the condition that MATH is MATH-nondegenerate which is equivalent to MATH (see for example, CITEEF). From this, together with REF , we conclude that we can select a subsystem of MATH scalar identities from REF from which MATH can be solved uniquely by the implicit function theorem. We obtain MATH where MATH is a germ of a holomorphic map MATH, with MATH . Observe that the germ MATH depends only on MATH but not on MATH. We claim that there exists MATH, independent of MATH, such that MATH . In order to prove the claim we use the notation MATH and MATH, and for MATH, we denote by MATH the natural projection from MATH onto MATH. We define MATH by MATH . To show that MATH is in MATH, we must differentiate the right hand side of REF with respect to MATH and evaluate at MATH. By using the chain rule and the fact that each MATH is in MATH and vanishes when MATH, it is easy to check that for any multiindex MATH, MATH is a polynomial in MATH. This proves REF . We now differentiate the identity REF with respect to MATH. By using REF again, we find MATH, independent of MATH, such that MATH . For any MATH we decompose MATH and set for MATH as above and MATH, MATH . Clearly MATH is in MATH and is independent of MATH and MATH. Since for any MATH-equivalence MATH we have MATH for MATH with MATH by REF , it follows from REF that MATH. Hence REF implies MATH where MATH is a germ at MATH of a holomorphic map from MATH to MATH depending on MATH and whose restriction to MATH vanishes at MATH of order MATH. By taking complex conjugates of REF for MATH, and using the fact that MATH is equivalent to MATH, we obtain REF with MATH satisfying the conclusion of REF .
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math/0002186
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We prove the theorem by induction on MATH. We start first with the case MATH and assume that MATH is a MATH-equivalence between MATH and MATH with MATH. By conjugating REF with MATH replaced by MATH we obtain MATH with MATH and MATH as in REF . If we observe that MATH, we conclude that the second term on the right hand side of REF vanishes at MATH of order MATH when MATH. Our next goal will be to substitute REF into REF and to apply REF . For this, we define polynomials MATH and MATH, for MATH and MATH, to be the determinants of the parts of the jets MATH and MATH obtained from the first MATH rows and first MATH columns of the linear terms of MATH and MATH respectively (that is, corresponding to MATH and to MATH for MATH and MATH respectively). We also set MATH, MATH and for MATH, MATH . (Observe that MATH by the definition of MATH and MATH above.) Then MATH satisfies the assumptions of REF , in particular, REF holds since by REF we have MATH and the right hand side of REF is nonvanishing whenever MATH is a MATH-equivalence with MATH. From substituting REF into REF we obtain the identity MATH where the restriction of MATH to MATH vanishes of order MATH at the origin. Then for MATH, REF is a consequence of REF with MATH being the polynomial MATH given by the lemma. The required REF follows from REF and from the explicit formula for MATH in the lemma. Now we assume that REF hold for some fixed MATH and any MATH and shall prove them for MATH and any MATH. We replace the terms MATH and MATH by using REF with MATH and MATH replaced by MATH. We obtain MATH with the restriction of MATH to MATH vanishing of order MATH. Similarly to the preceeding proof of REF for MATH, the desired conclusion of the theorem follows by making use of REF .
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math/0002186
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Suppose that MATH is a formal equivalence. If MATH is convergent, it is clear that MATH is also convergent. Conversely, if MATH is convergent, then the first term on the right hand side of REF is a convergent power series in MATH by composition. Since MATH is a MATH-equivalence for every MATH, the remainder term is MATH, and hence MATH is also convergent by REF .
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math/0002186
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We use REF for MATH and substitute MATH for MATH, where MATH is given by REF . The corollary easily follows by taking MATH and MATH.
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math/0002186
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We shall take MATH to be the NAME number of MATH at MATH. In the notation of REF we take MATH, MATH and set MATH where MATH is defined by REF . Here MATH and MATH. Observe that MATH is a linear automorphism of MATH. It follows from REF that MATH. Furthermore it follows from REF that REF also holds. Hence we can apply REF . Let MATH be given by the lemma, so that REF holds. By REF , we obtain MATH with MATH. By a simple change of MATH and MATH, we obtain from REF the equivalent identity MATH for all MATH such that MATH and both MATH and MATH are sufficiently small. Here MATH and MATH are independent of MATH, the components of MATH are in the ring MATH and MATH is a germ at MATH in MATH, depending on MATH and vanishing of order MATH at MATH. Observe that the left-hand side of REF is independent of the parameter MATH, whereas the right-hand side contains this parameter. We choose MATH such that the function MATH does not vanish identically for MATH in a neighborhood of MATH in MATH, and put MATH in REF . For convenience we consider a holomorphic change of variable MATH near the origin in MATH, where MATH is determined by the identity MATH for an appropriate integer MATH. By a further simple change of MATH and MATH, we conclude from REF that the identity MATH holds for all MATH such that MATH and both MATH and MATH are sufficiently small. Again MATH is independent of MATH and its components are in the ring MATH and MATH is a germ at MATH in MATH, depending on MATH and vanishing of order MATH at MATH. We next expand both sides of REF in NAME series in MATH and equate the constant terms. The required properties of those terms are established in the following lemma. Let MATH and MATH be finite-dimensional vector spaces with fixed linear coordinates MATH and MATH respectively, and MATH be in the ring MATH with MATH. For a fixed integer MATH, consider the NAME series expansion MATH . Then MATH and, for every MATH, MATH is in the ring MATH. In addition, if MATH for some integer MATH, then MATH for all MATH such that MATH. We expand MATH in power series of the form MATH with MATH, where MATH, MATH, MATH. Then MATH is a polynomial in MATH satisfying the estimates REF . Since MATH we conclude that MATH and MATH for every MATH. Now assume that MATH. This means that MATH holds whenever MATH. For fixed MATH, this inequality together with MATH implies, in particular, that MATH in the last sum of REF or, equivalently, MATH in that sum. This completes the proof of the lemma. We now complete the proof of REF . We expand the right-hand side of REF in NAME series in MATH. Since the left-hand side is independent of MATH, we equate it to the constant term of the NAME series. The required conclusion of REF with MATH follows by applying REF for MATH to MATH with MATH, MATH and to MATH with MATH, MATH. The proof of REF is now complete.
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math/0002186
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We continue to work with normal coordinates near the origin MATH for MATH. We fix a local parametrization of MATH at MATH of the form MATH . Let MATH be a defining function for MATH near MATH and MATH be given by REF . For MATH, MATH and MATH, we consider the functions MATH . It follows from the properties of MATH, the chain rule and REF that the MATH are in the ring MATH. If follows from the definition that we can think of this ring as a subring of the following formal power series ring with polynomial coefficients MATH where MATH are complex coordinates in MATH . It is a standard fact from commutative algebra that any formal power series ring with coefficients in a NAME ring is again NAME; in particular, the ring REF is NAME. Hence there exists an integer MATH such that the subset MATH generates the same ideal in the ring REF as all the MATH, MATH, MATH. By the identity REF and the definition of MATH-equivalence we have for MATH and MATH, MATH . Hence we proved REF with the collection MATH, MATH, being the set of functions given by REF and MATH, MATH. This completes the proof of REF . We shall now prove REF . By the choice of the set REF , every germ MATH can be written in the form MATH where MATH are in the ring given by REF . Since MATH, the germ MATH can be substituted for MATH in each MATH to obtain a formal power series in MATH. From REF on MATH we obtain the following identities of convergent power series in MATH: MATH . In view of REF we conclude that MATH for MATH near the origin. This completes the proof of REF and hence that of REF .
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math/0002186
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Let MATH, MATH, be given. It follows from the definition of the ring MATH that MATH can be viewed as an element of MATH. By REF , given MATH, there exists MATH such that if MATH and MATH are as in the corollary, there exist MATH, MATH satisfying REF . We may now apply REF with MATH (and hence MATH) to conclude that there exists MATH with MATH, MATH satisfying the conclusion of the corollary.
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math/0002187
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If MATH has nonzero square, then, since MATH, MATH. But MATH is a free MATH module generated by MATH. So if MATH has no MATH-torsion, then MATH must be MATH. And then, of course, MATH, as well. Thus MATH. Since MATH, the sequence collapses. Now suppose MATH. Then for each generator MATH, there is a MATH which is linearly independent of MATH in MATH, and such that MATH. Since the action of MATH is homologically trivial, MATH is a free MATH-module on generators corresponding to those of MATH, so in fact MATH and MATH must be independent in MATH, as well. But MATH. This is only possible if MATH. MATH is generated by products of two-dimensional classes, so MATH, as well. It follows that MATH. The same argument shows that MATH. The conclusion about MATH follows immediately from tom CITE.
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math/0002187
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Suppose first that MATH acts on MATH. By REF , MATH is a free MATH module on MATH generators. Recall that MATH, where MATH and MATH generate MATH. Since MATH is a polynomial ring, it contains no zero-divisors, so it makes sense to localize at the set MATH consisting of all of the nonzero elements. We check easily that MATH. Now, each proper subgroup MATH is the kernel of some nonzero homomorphism MATH, and this MATH, viewed as an element of MATH, restricts trivially to MATH. So the MATH-singular set MATH contains only those points fixed by all of MATH. By the Localization Theorem, MATH is nonempty. Consideration of the isotropy representation of MATH at a fixed point MATH shows that there must be MATH such that MATH fixes a two-dimensional subspace MATH, while MATH acts by MATH. But MATH forms part of a MATH-sphere MATH fixed by MATH. If MATH reverses orientation on MATH, it also acts by MATH on MATH, contradicting homological triviality. If MATH, the same argument shows that MATH contains MATH points, but of course no contradiction ensues from the isotropy representation of MATH. If MATH is odd, a similar argument applies, except in the case where MATH and MATH. To ensure that MATH is central in MATH, we replace the one-dimensional generators of MATH, with their two-dimensional images under the NAME map, which generate the polynomial part of MATH. Finally, suppose MATH and MATH, with generators MATH, and MATH. The NAME fixed-point theorem implies that MATH. Thus MATH either contains at least one MATH-sphere, or consists of exactly four isolated points. In the first case, MATH acts effectively on the sphere, which is impossible. In the second case, the action of MATH on MATH must fix at least one point MATH. But MATH cannot act freely on the linking sphere to MATH, so some other element MATH fixes a MATH-sphere, and the argument proceeds as before.
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math/0002187
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If each component of MATH contains at least three spheres, then each sphere intersects its neighbor geometrically once, and the claim follows. If some component contains exactly two spheres, then each intersects the other twice. One of them might, a priori, represent a multiple of two in MATH. But REF cited above implies that it must be nontrivial in MATH.
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math/0002187
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Recall that MATH while for MATH odd, MATH . Let MATH denote the projection MATH. The NAME spectral sequence of the pair MATH has MATH . On the other hand, MATH is a free MATH-space, so MATH is canonically isomorphic to MATH. Since MATH is a relative manifold pair, NAME duality gives a commutative diagram: MATH . But MATH is generated by meridians to the spheres in MATH, and each of these is a MATH-fold cover of its image in MATH. Thus MATH is multiplication by MATH. Since the left-hand edge homomorphism MATH of the NAME spectral sequence is induced by the fiber inclusion MATH, we can conclude that MATH has exponent MATH. In other words, no MATH summand of MATH supports more than one non-zero differential. In rank counting arguments we can therefore treat its integral rank as though it were a MATH rank. Notice also that, since MATH, each nonzero class in MATH with MATH must be mortal. Consider the terms of MATH indicated in REF : Now, elements of MATH can only be killed by MATH, while MATH can be killed either by MATH or MATH. By the above observations, we have: MATH so MATH, as claimed.
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math/0002187
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From the homology spectral sequence of the covering MATH, we obtain a short exact sequence MATH. As we have already seen, MATH is multiplication by MATH in each factor. It follows that MATH is MATH-torsion-free. So in fact all classes in MATH with MATH are mortal. In particular, MATH must vanish, so MATH, so MATH. Suppose for a contradiction that MATH. The NAME spectral sequence then takes the following form (each entry with MATH is MATH for some MATH, so to save space, we simply indicate its rank): There are generators MATH and MATH such that MATH and MATH. By the multiplicative properties of the spectral sequence, this kills the entire row MATH, except MATH. Now, MATH, so there is some MATH such that MATH. Since MATH must also perish, there is MATH, independent of MATH, so that MATH. But then MATH, since MATH and MATH were already killed by MATH. Now MATH has rank MATH, and MATH. But MATH has rank MATH, so MATH must have rank MATH. This is a contradiction, so MATH.
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math/0002187
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For convenience, we continue to assume the action is not pseudofree, or fixed-point-free in the case of MATH. As above, let MATH denote the singular set with two homologically redundant spheres removed. For simplicity of notation, we assume MATH is connected; if it isn't, our argument will carry through on each piece. It follows from the work of CITE (see also CITE) that each MATH has an equivariant normal bundle. Thus MATH has a regular neighborhood MATH which is homeomorphic to the manifold obtained by plumbing together disk bundles MATH, MATH, over MATH according to the graph MATH . The boundary of such a plumbed manifold is a lens space MATH, and MATH is given by the determinant of the intersection matrix. In our case, the matrix is unimodular, so MATH is in fact a three-sphere. Thus MATH is homeomorphic to a connected sum of copies of MATH and MATH. But MATH is also a connected summand of MATH which carries all of its homology. By CITE, MATH.
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math/0002187
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We begin with a slight variant of the plumbing construction of the previous section: Let MATH, and label the spheres consecutively around MATH as MATH. Let MATH denote the ``north pole" of MATH. Choose orientations for each of the MATH, and let MATH denote the corresponding fundamental class. Finally, choose an orientation for MATH and let MATH denote the sign of the intersection at MATH. (When we considered MATH earlier, we implicitly chose orientations to make each MATH; here, because the spheres are arranged in a closed loop, this might not be possible.) With these conventions, MATH is obtained by plumbing together MATH-bundles MATH over MATH, each with NAME class MATH, according to the orientations given by the MATH. The plumbing graph is a circle, which we parameterize as MATH with the endpoints identified. MATH can be thought of as a torus fiber bundle over the plumbing graph with a fiber-preserving, free MATH action. It is not a priori a principal bundle, but if the MATH action on the fibers extends to a torus action, it will become one. With appropriate smoothing around the plumbing points, the torus action will extend over MATH, establishing the first part of the proposition. The MATH-bundle over MATH can be assembled by gluing copies of MATH via attaching maps MATH which incorporate the clutching functions for the MATH, the coordinate switches at each plumbing point, and the orientations MATH. (See REF , which is intended to invite comparison with the diagrams in CITE.) The maps are determined up to isotopy by their MATH representations. The clutching functions take the form MATH; the coordinate switches are of course MATH, and the orientation changes, MATH. Together, such matrices generate MATH, the structure group of the bundle. The MATH action has well-defined rotation numbers in the fiber over MATH, so it extends to a torus action in that fiber. The gluing maps in MATH define a trivialization of the bundle over MATH which is equivariant with respect to the MATH action. Using them, the torus action extends along all of the fibers. The structure of the torus bundle is thus determined by the total gluing function MATH; with slight abuse of notation, we may write MATH. A compatibility condition is imposed by the existence of the MATH action - namely, that the gluing map MATH must commute with order MATH rotations in each factor of MATH. We may analyze this requirement by lifting to the universal cover MATH. A rotation MATH lifts to a translation MATH. The requirement that MATH means that the line spanned by each MATH must be CASE: Normalized by MATH, if MATH. Since the total space of the bundle is the boundary of MATH, it is orientable, so MATH is one of MATH or MATH CASE: Centralized by MATH, if MATH, which implies MATH. In the latter case, MATH clearly commutes with the entire torus action, so MATH supports the structure of a principal bundle. NAME when MATH, MATH must respect the base-fiber splitting of the bundle MATH over MATH, so MATH is ruled out. We proceed to rule out MATH, also. If such a bundle is realized on the singular set of a MATH action, let MATH be a small meridional loop around MATH in MATH. Then MATH is homologous to MATH, and so MATH in MATH. But MATH is torsion-free, as we saw in REF. It is generated by any pair of meridians to neighboring two-spheres in MATH. This finishes the proof that MATH is a trivial principal MATH-bundle, and hence also the proof of the first claim. Our proof of the second claim is constructive, based on NAME and NAME 's model in the case of torus actions. The orbit space MATH is an annulus. Its outer boundary component MATH consists of MATH fixed points separated by MATH arcs whose stabilizers are copies of MATH. Its inner boundary MATH consists entirely of principal orbits. Adjoin a disk MATH to MATH. Because the torus bundle is trivial over MATH, there is no obstruction to lifting this adjunction to a MATH-equivariant gluing of MATH to MATH. The resulting manifold, denoted MATH, is simply connected and has the same intersection form as MATH, so it is homeomorphic to MATH. Finally, REF, NAME, and NAME CITE shows that MATH and MATH need not be MATH-equivariantly homeomorphic. They begin with a linear MATH action on MATH, and equivariantly connect sum a MATH-orbit of counterexamples to the NAME conjecture in MATH around one of the singular MATH-spheres. The resulting space is still homeomorphic to MATH, but the complement of the singular set has nonabelian fundamental group. In the linear example, MATH has the homotopy type of a torus.
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math/0002188
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By NAME duality if MATH is a root of MATH then MATH is also a root of MATH. Hence if we write MATH it follows that MATH . Let us set for brevity MATH. From REF we get: MATH hence MATH . By REF we have MATH which yields MATH as desired.
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math/0002188
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Given any MATH let MATH be the geodesic with initial condition MATH. Let MATH be the parallel transport along MATH from MATH to MATH. Let MATH be an orthonormal basis of MATH by eigenvectors of the symmetric transformation MATH. Let MATH be the parallel vector field along MATH with initial condition MATH. Given MATH we can write MATH for some smooth functions MATH. Then, MATH and since MATH, we have MATH . Now, from the NAME equation, we get that MATH. Therefore MATH that is, MATH . Since MATH is orthogonal we get MATH and therefore, MATH .
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math/0002189
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Take REF and write MATH as: MATH . The integral converges if: MATH, and in this case, it converges to the constant MATH, independent of MATH. Absorbing this into the main constant, the result for MATH is created. In the case MATH, REF is MATH . Thus again, the error is bounded by a constant, this time, dependent on MATH. Observe that MATH is immediately able to be replaced with MATH, demonstrating the continuity of the formulae. Lastly, bound MATH as: MATH . As MATH, the integral converges to MATH, again, another constant independent of MATH, which is absorbed into MATH. The second result is thus achieved, by observing that MATH.
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math/0002190
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Given equations are already satisfied on the disk for the vector fields MATH, MATH. Further at the points of the disk we define transversal to this disk vector fields MATH, MATH, MATH, in such a way that all the union of MATH vectors forms a basis at each point and also that REF is satisfied. Upon constructing the needed vector fields at the points of the disk we extend them to a neighborhood with the help of REF . Obtained structure MATH coincides with the structure MATH on the disk and does not necessarily do so outside. MATH .
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math/0002190
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Since the commutators of vector fields along MATH are determined by their MATH-prolongations outside MATH, we write the general form for a MATH-prolongation of the vector field MATH: MATH where MATH is the submodule of the module of vector fields, consisting of the vector fields vanishing on the submanifold MATH to the second order. If on the disk MATH the decomposition of the additional vector fields is written as MATH then the equations MATH with MATH have the following form MATH . This system decomposes (by MATH) on MATH determinate systems of MATH linear equations with MATH unknowns. The matrix MATH of each system is nondegenerate, hence the system possesses a solution. Thus the field MATH is constructed. Let us rectify it: MATH. We can assume that on the disk MATH the tangent vector fields have the original form MATH. In new coordinates the coefficients of the decomposition REF do not depend on MATH. So one can search for the prolongation of the field MATH in the form similar to REF , but with no dependence on MATH. Continuing the process we get some coordinates MATH, in which the disk MATH belongs to the subspace MATH and such that on this disk MATH . Now we prolong the vector fields to a neighborhood via REF . MATH .
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math/0002190
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Perturb the almost complex structure MATH in a neighborhood of the disk MATH so that it coincides with the standard integrable structure MATH near the boundary of this neighborhood: for every MATH there exists such an almost complex structure MATH that MATH in a small neighborhood of the disk MATH and MATH in a neighborhood of the boundary of the manifold MATH. Further one can suppose that MATH, where MATH can be equipped with an almost complex structure MATH, which coincides with MATH in MATH and which equals the standard integrable structure MATH in the complement. Let us supply the manifold MATH with the symplectic structure MATH, where MATH is the standard volume form, and also MATH and MATH. Decreasing if necessary the size of the neighborhood of the disk MATH we can suppose that the almost complex structure MATH is tamed by the symplectic structure MATH, that is, MATH for MATH. Denote by MATH the homology class of the sphere MATH. The disk MATH can be extended to the entire rational pseudoholomorphic curve MATH, which lies in the class MATH. Let us consider the space MATH of entire pseudoholomorphic curves MATH of the class MATH. Since the class MATH cannot be decomposed into a sum of homology classes MATH, MATH, with MATH, then NAME compactness theorem (REF or REF) implies the compactness of the space MATH, where MATH is the complex automorphisms group of the sphere MATH, MATH. Moreover for almost complex structure MATH of general position the space MATH is a smooth manifold of dimension MATH REF; CITE sec. CASE: Consider the space of nonparametrized pseudoholomorphic curves MATH. This space is a compact manifold of dimension MATH. Let us consider the evaluation map MATH, which is defined by the formula MATH for MATH, MATH. We suppose the group MATH acts on MATH by conjugation, MATH, MATH, whence the correctness of the definition for MATH. Since MATH-curves foliate the manifold MATH outside a small neighborhood of the image MATH (because there MATH), the map MATH has degree MATH, MATH. Therefore through every point, close to the curve MATH, some pseudoholomorphic curve MATH passes, which is homologous to the curve MATH. To eliminate the general position condition for MATH we take a sequence MATH of the general position almost complex structures, which tends to MATH in MATH-topology, and use the compactness theorem CITE B. REF Intersecting the obtained set of pseudoholomorphic spheres MATH with a small neighborhood MATH of the disk MATH, we get the desired set of the disks MATH in a neighborhood MATH. Smoothness of these disks follows from the standard elliptic regularity REF; CITE sec. CASE: MATH .
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math/0002190
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For a small neighborhood MATH of the point MATH the estimates of REFa of the paper CITE imply existence of a number MATH, dependent only on the almost complex structure MATH and the neighborhood MATH, such that for every MATH, MATH, MATH, and MATH there exists a pseudoholomorphic disk MATH such that MATH, MATH. Setting MATH we have MATH for all (now not necessarily unit) vectors MATH for which MATH. Since a compact set can be covered by a finite number of neighborhoods MATH, the first statement of the proposition is proved. The second part is a reformulation of the nonlinear NAME lemma CITE: if an almost complex structure MATH on a compact manifold is tamed by an exact symplectic structure MATH, then the derivative at zero of any pseudoholomorphic disk MATH, passing through a fixed point at the manifold, is bounded by a non-depending on the disk constant: MATH. MATH .
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math/0002190
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The inequality MATH is equivalent to the statement of REF because MATH, where the lower bound is considered over all mappings MATH, such that MATH .
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math/0002190
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The inequality MATH is evident because MATH, where the lower bound is taken over all pseudoholomorphic mappings MATH, MATH, and the norm is count with respect to the NAME metric. Let us prove the reverse. We follow the NAME 's proof CITE. Let MATH be a smooth curve from a point MATH to a point MATH such that MATH. Due to upper semicontinuity there exists a continuous on MATH function MATH, such that MATH and MATH that is, for sufficiently dense partition MATH we have MATH . Consider arbitrary pseudoholomorphic curve MATH, which satisfies the conditions MATH and MATH. Define for small MATH the curve MATH. Since MATH, REF imply that for small MATH it holds: MATH . Thus for sufficiently dense partition MATH . Since MATH is arbitrary constant, the theorem is proved. MATH .
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