paper stringlengths 9 16 | proof stringlengths 0 131k |
|---|---|
math/0107059 | The proof is clear by construction. |
math/0107059 | Assume MATH and MATH. Then MATH and MATH imply MATH and MATH with strict containment by REF. By REF we conclude MATH and MATH. This gives MATH as desired. |
math/0107059 | Let MATH, then there are MATH and MATH such that MATH and MATH . Thus MATH. This implies MATH or MATH by REF. |
math/0107059 | Suppose under the assumptions of the lemma we had MATH, then there was a MATH with MATH such that MATH. Pick any MATH such that MATH, and let MATH be a translate of MATH such that MATH. Clearly MATH is an element of MATH. The multi-tile MATH cannot be in the tree MATH because of MATH, but its geometry would qualify it to be in the tree, namely, MATH and MATH for some MATH. Thus it must have been selected for a tree MATH at a previous stage of the selection process. However, the geometry of MATH qualifies it to be in the same tree MATH, namely, MATH and MATH because MATH for some MATH and MATH by REF and the fact that MATH and MATH intersect. This gives a contradiction to the maximality of MATH. |
math/0107059 | Since MATH is a finite union of intervals, there are two types of points in MATH; those that are the left endpoint of connected components in MATH, and those that are right endpoints. Clearly it suffices to prove the bound for left endpoints only. For each MATH and each left endpoint MATH of MATH consider the interval MATH. We claim that these intervals are pairwise disjoint. This claim implies the conclusion of the lemma because these intervals are contained in MATH. To prove the claim, assume MATH and MATH have nonempty intersection. If MATH, we necessarily have MATH because then both MATH and MATH are endpoints of dyadic intervals of length MATH. Thus we can assume MATH. Then MATH is contained in the interior of the dyadic interval MATH of length MATH with right endpoint MATH. However, the interior of MATH is disjoint from MATH and MATH is contained in MATH, a contradiction to REF . This proves the claim. |
math/0107059 | To prove the first statement, let MATH and MATH be two neighboring intervals of MATH and assume MATH is larger. Let MATH be the dyadic interval which contains MATH and has MATH times the length of MATH. We have to prove MATH. However, MATH is contained in MATH, und thus does not contain any interval MATH with MATH. By maximality of MATH we have MATH. This proves the first statement of the lemma. To prove the second statement, let MATH be any MATH- dyadic interval of length MATH and observe the dichotomy that either MATH is contained in an interval of MATH, or MATH is partitioned into intervals of MATH which are strictly smaller than MATH. The latter is necessarily the case if MATH is nonempty or if MATH for some MATH. This proves the second statement. |
math/0107059 | There is a dyadic interval MATH of same length as MATH which is contained in MATH and MATH. By definition of MATH, MATH contains a MATH for some multi-tile MATH. This together with REF proves the lemma. |
math/0107059 | Most of the proof is exactly as in the proof of REF , the only new statement is the one on the distance between two neighboring intervals. Let MATH be such an interval. It suffices to show every different interval MATH with MATH has distance at least MATH from MATH. The case MATH then follows by symmetry. The case MATH is easy, since the distance between MATH and MATH is a multiple of MATH. Thus consider MATH. Let MATH be the dyadic interval of length MATH which contains MATH and let MATH be the dyadic interval of length MATH whose left endpoint is equal to the right endpoint of MATH. Clearly MATH is disjoint from MATH, whereas MATH is contained in MATH. The crucial observation is that MATH cannot be contained in MATH, because two neighboring intervals in MATH differ by at most a factor MATH. Thus the distance from MATH to any point in MATH is larger than MATH, which proves the claim. |
math/0107059 | This follows easily from the fact that lacunarity depends only on the frequency intervals. |
math/0107059 | If MATH, then MATH. Since MATH this contradicts REF. |
math/0107059 | Assume to get a contradiction that MATH. By size comparison we then necessarily have MATH. By REF we have MATH. Hence REF implies MATH, which implies MATH (We made a point of not using REF to conclude this). Hence the geometry of the multi-tile MATH qualifies it to be in the tree MATH. But it is not, because different selected trees have no multi-tiles in common, so by maximality of MATH it must be an element of a tree that was selected prior to MATH. This contradicts the assumed order of selection of MATH and MATH. |
math/0107059 | By the previous lemma we only need to prove that MATH has been selected prior to MATH. However, REF together with MATH implies MATH, which in turn implies that MATH has been selected prior to MATH. |
math/0107059 | As before, we can use REF to conclude that if MATH, for MATH, then MATH has been selected prior to MATH. Thus we may assume MATH and MATH have been selected in this order. Assume to get a contradiction that MATH, then we have by dyadicity MATH. Let MATH be a multi-tile in MATH for which MATH is minimal. Let MATH be the tree so that MATH. If there is another multi-tile MATH for which MATH, then we conclude by REF that MATH. Hence MATH and MATH are in the same tree MATH. Now let MATH for some MATH with MATH. We aim to show MATH, which will prove the lemma, because then one of the trees MATH and MATH has to be empty, a contradiction. Observe that by REF we have MATH. By a similar argument for MATH we have MATH. By assumption on the size of tree tops we have MATH. If MATH, then this implies MATH, which in turn implies MATH. We may thus assume MATH. Then we can merely conclude MATH and MATH. By a similar argument with MATH in place of MATH we have MATH and MATH. But MATH for some index MATH. Hence MATH. However, for some possibly different index MATH we have MATH. Hence MATH and MATH have non-empty intersection, and thus by REF we have MATH. However, we have seen before MATH and we have MATH. Hence MATH, which proves the claim. |
math/0107059 | We may assume that the MATH are real and positive. By estimating MATH by a weighted sum of MATH for dyadic MATH (as before MATH has the same center as MATH but MATH times its length), it suffices by the triangle inequality to show Let MATH be as above, MATH be real and positive, and let MATH. Then there exists an absolute constant MATH such that MATH . Fix MATH. Let MATH be the best constant MATH for which REF obtains for all subsets of MATH (in place of MATH itself) and all MATH; our objective is to show that MATH is bounded uniformly in MATH and MATH. By construction there exists a subset MATH of MATH and MATH such that MATH . Squaring this we obtain MATH . One only has a contribution if MATH or MATH. By symmetry and positivity we thus have MATH . Moving the MATH summation inside the inner product, and using NAME and the definition of MATH, we may estimate the left-hand side by MATH . On the other hand, by disjointness we have MATH. Inserting this into the previous estimates we obtain MATH as desired. |
math/0107059 | We may write MATH . By the triangle inequality it suffices to prove for each MATH and an analoguous estimate for the term MATH, which is clearly an easy variant of the above. Since MATH for all MATH and the intervals MATH are pairwise disjoint, we have MATH . The claim now follows by REF . |
math/0107059 | We can write MATH where MATH is a suitable bump function adapted to MATH. From the decay of the kernel of MATH we thus have the pointwise estimate MATH and the claim easily follows. |
math/0107059 | We can truncate MATH so that the kernel is supported on the interval MATH; from REF it is easy to see that this does not affect the MATH boundedness of MATH or REF. The claim then follows by partitioning space into intervals of length MATH and applying the MATH boundedness hypothesis to each interval separately. |
math/0107059 | We first consider REF. The first inequality is just by definition. By the remark just after REF , we only have to prove the second inequality for MATH. Then this inequality follows from REF since MATH contains the singleton tree MATH with top data MATH where MATH is defined by MATH and MATH is the MATH-dyadic interval of length MATH which contains MATH. It is easily verified that these top data indeed turn MATH into a tree. Now we consider REF. Observe first of all that MATH because both intervals are MATH-dyadic. By translating MATH, we may as well assume MATH. Namely, we have to translate MATH by at most ten times its length, so that MATH stays the same up to some bounded factor. We consider the two cases MATH and MATH. Assume first MATH. Observe that by MATH-non-lacunarity, MATH is contained in MATH. Hence MATH is contained in MATH. Pick MATH so that MATH is an endpoint of MATH. Then MATH by definition of the right hand side, because the multiplier MATH is supported in MATH and satisfies REF, possibly with a constant. Now the claim follows from REF, which proves REF in the case MATH. Now assume MATH. We consider again a singleton tree MATH with top data MATH, MATH so that MATH is an endpoint of MATH. Again, by MATH-non-lacunarity of MATH we see that these top data indeed turn MATH into a tree. Then the multiplier MATH satisfies REF with respect to this singleton tree as one can easily see, possibly with some constant, and REF follows by definition of the tree size. |
math/0107059 | Fix MATH. It suffices to show that MATH for all trees MATH in MATH. Fix MATH. We have to estimate both summands in REF. The second summand is immediate since MATH is a universally bounded operator in MATH. We consider the first summand in REF. By frequency modulation invariance we may assume that MATH. From REF it suffices to show that MATH whenever MATH is supported on MATH and obeys REF with MATH. First suppose that MATH vanishes on MATH. Then a simple computation following REF shows that MATH for all MATH, and the claim follows by summing in MATH. Thus we may assume that MATH is supported in MATH. It then suffices to show that MATH . By NAME 's inequality, we may estimate the left-hand side by MATH for a suitable choice of signs MATH. But then the claim follows since the expression inside the norm is simply a pseudo-differential operator of order REF in the symbol class MATH applied to MATH, with bounds uniform in MATH, MATH, MATH, and MATH (see CITE). |
math/0107059 | Fix MATH, MATH, MATH. We shall exploit scale invariance to reduce to the case MATH. By a frequency modulation leaving MATH and MATH invariant, we may as well assume MATH. By the triangle inequality it suffices to prove estimate REF where the summation goes over the subset MATH instead of MATH for any subset MATH. Fix such MATH and let MATH denote the set of non-lacunary indices, that is, MATH. We may assume MATH is non-empty and thus a tree with top data MATH. By REF we have that MATH is non-empty. Changing notation we may as well assume that all multi-tiles in MATH have lacunarity type MATH and thus MATH. Recall that MATH also satisfies the crucial nesting property of REF . Let MATH denote the set MATH; its smallest element is greater or equal MATH. Recall that by REF the map MATH is one-to-one. From REF we may estimate the left-hand side of REF as MATH where the spatial cutoff MATH and NAME cutoff MATH are defined as MATH and MATH for MATH. In particular, MATH has NAME support in the region MATH. We remark that our notation is sloppy here, the operator MATH also depends on the parameter MATH. We will always write MATH in combination with a function MATH, and the omitted index is always the one of the function MATH. Let MATH be chosen so that MATH for MATH and MATH; this can be done by the hypotheses on MATH stated in REF . We observe the simple bounds For all MATH, MATH, and intervals MATH of length MATH we have MATH . By interpolation and REF it suffices to prove the bounds MATH and (in the MATH case only) MATH . The second estimate is immediate from the boundedness of the MATH, while the third follows from the first and REF since MATH and thus MATH. Thus it suffices to prove the first inequality. Fix MATH, MATH, MATH. From REF we see that there exists MATH with MATH such that we have the pointwise estimate MATH on MATH. It thus suffices to show that MATH . This however was observed in REF. From this NAME and NAME we have MATH for all MATH and MATH. Summing in MATH, we see that we may bound each summand in REF by the right-hand side of REF. Thus the main difficulty is to obtain summability in MATH. In the non-degenerate case MATH this summability is obtained by noting there must be at least two lacunary indices in MATH, estimating those indices in MATH in space (and MATH in MATH), and taking all other indices in MATH. See for example, CITE, CITE, CITE and the discussion in the introduction of CITE. This however is not feasible in the general case for two reasons. Firstly, there might only be one lacunary index; and secondly, one can only get good MATH bounds for the MATH index when MATH. Thus we will have to invoke REF as outlined before. We shall need to replace MATH in REF by a product of cutoff functions. If MATH was a characteristic function, we could simply write it as a power of itself. However, it is an approximate characteristic function. From REF we have: MATH . The right hand side can be handled by REF . Let MATH, then MATH . By the triangle inequality, we can estimate this sum by MATH . By NAME and NAME REF, this is bounded by MATH . On the other hand, from REF and the lower bound in REF we have MATH . Summing this in MATH and evaluating the integral, we estimate the previous by MATH and the claim follows from REF . Instead of estimating REF, it suffices by the above Lemma to to show that MATH . To do this we shall now introduce the crucial tree projection operators. Let the notation and assumptions be as above. Then for each MATH there exists a function MATH with the following properties. CASE: (Control by size) We have MATH . CASE: (MATH approximates MATH on MATH: lacunary case) For all MATH and MATH, we have MATH where MATH is a suitable NAME projection to the frequency region MATH. CASE: (MATH approximates MATH on MATH: non-lacunary case) For all MATH, MATH, and all intervals MATH of length MATH, we have the bounds MATH where MATH is the function MATH and the sets MATH have been defined in REF . Also, for all MATH, MATH, and intervals MATH of length MATH we have MATH . One can construct the MATH to be linear operators, but we shall not use this. Roughly speaking, MATH is the projection of MATH to the region MATH of phase space. Although such a description can easily be made rigorous in a NAME model, and is not too difficult in a lacunary NAME model, it is substantially more delicate in the NAME setting in the non-lacunary case. Note that, the function MATH can be controlled by REF . We prove this rather technical Proposition in REF. For now, we see why this proposition, combined with REF , gives REF. To exploit the fact that MATH approximates MATH on MATH, we use REF and the triangle inequality to estimate REF by the main term MATH plus MATH error terms of the form MATH where MATH. Let us first control the contribution of a term REF. Fix MATH. By the triangle inequality we may estimate REF by MATH . By NAME, we may estimate the previous by MATH . The first factor we estimate using REF. The second group of factors we estimate using REF . For the last group we use REF. Combining all these estimates and using REF, we see that we may estimate REF by MATH which of course sums to MATH . Expanding out MATH, we may estimate this by MATH computing the integral, we thus obtain MATH and the claim follows from REF . It remains to estimate REF. By repeating the proof of REF (but using REF in place of REF when MATH) it suffices to estimate MATH which we rewrite using REF as MATH . By REF , we may estimate this by MATH and the claim follows from REF. Observe that in applying REF we have used MATH, which had been a consequence of REF . This concludes the proof of REF , except for the proof of REF which shall be done in the next section. |
math/0107059 | From REF we may write MATH . Here MATH is a NAME projection whose multiplier is supported in MATH, is constant MATH for MATH, and vanishes on MATH. REF then follows by using repeated integration by parts to obtain pointwise bounds on MATH. There exists a MATH-dyadic interval MATH of length MATH whose left endpoint coincides with MATH, because MATH and hence MATH is a union of dyadic intervals of length MATH by REF . By REF there is a tile MATH with MATH. The bounds REF then follow from REF. |
math/0107059 | We shall prove REF for the sign +; the other sign follows by applying the frequency reflection MATH and conjugating MATH. The idea is to remove maximal trees in a greedy selection process from MATH until REF is obeyed for the remainder set. This procedure shall be given by iteration. If REF holds, we terminate the iteration. If REF does not hold, then there exists a tree for which MATH . Since the set of all possible trees MATH obeying REF is compact, we may select MATH so that MATH is maximal. By retaining the top data but adding further multi-tiles if necessary, we may assume that MATH is maximal in the sense of REF . We then add this tree MATH to MATH. Then, we remove all the multi-tiles in MATH from MATH. We then repeat this iteration until REF holds. Since MATH is finite, this procedure halts in finite time and yields a collection MATH of mutually distinct convex trees. Note that trees with a larger value of MATH will be selected before trees with a smaller value of MATH. REF holds by construction, so it only remains to show REF. As usual, we shall use the MATH method to prove this orthogonality estimate. One may think of this Lemma as a phase space version of REF , which was set entirely in physical space, and we shall need REF in the proof of this estimate. Write MATH. Observe that MATH for all MATH by construction. It suffices to prove REF separately for the set of all MATH which satisfy MATH and the set of all MATH which satisfy MATH for appropriate MATH. We first consider the set of trees which satisfy REF. For simplicity of notation we may assume this set is equal to MATH. From REF we may associate to each MATH and MATH a multiplier MATH supported on the interval MATH obeying REF with MATH such that MATH where MATH is the non-negative quantity MATH . From the signed version of REF we have MATH . Summing REF in MATH, we obtain MATH . On the other hand, from the definition of MATH and duality we can find for each MATH a MATH-normalized function MATH such that MATH . We thus have MATH . We can write the left-hand side as MATH . By the NAME inequality we thus have MATH . To prove REF it suffices to show that MATH . It will be necessary to dyadically decompose the operator MATH around the base frequency MATH. For each MATH, decompose MATH where MATH is a bump function adapted to the region MATH and supported in MATH. We can then decompose MATH where MATH . Thus MATH has NAME support in MATH, is bounded in MATH, and is rather weakly localized in physical space near MATH. For MATH the multiplier MATH has good spatial localization properties, but for MATH the multiplier MATH begins to spread MATH along a wider interval than MATH. However in the case MATH we have the easily verified pointwise estimate MATH from kernel bounds on MATH. By the triangle inequality it thus suffices to show that MATH for all MATH. Fix MATH. We square this as MATH . By symmetry it suffices to show that MATH . We first consider the contribution when MATH. It suffices to show that MATH for all integers MATH, since the claim then follows by summing in MATH and applying REF. Fix MATH. By NAME 's test (that is, estimating MATH) and symmetry it suffices to show that MATH for all MATH. Fix MATH. From REF we conclude that MATH implies MATH. For those values one has a bound of MATH as can be easily verified from REF when MATH and REF when MATH. The claim then follows by summing. It remains to control the contribution when MATH. It suffices to show that MATH for all MATH and MATH, since we may then sum in MATH to obtain MATH and REF follows by summing in MATH and applying REF. It remains to show REF. Fix MATH, MATH. Let MATH denote the set of all MATH which make a non-zero contribution to REF. We now make the key observation that by REF we have intervals MATH are disjoint from MATH. Namely, non-vanishing of MATH implies REF . Moreover, by similar reasoning, if MATH and MATH are in MATH and MATH, then MATH and MATH are disjoint, as one can see from REF if MATH and from (a slight variant of) REF if MATH. For future reference we summarize: The intervals MATH are disjoint from each other and from MATH. We first verify REF in the case MATH. In this case we see from REF and a calculation that MATH so by REF reduces to MATH which is easily verified. Now consider the case MATH. By REF we have MATH where MATH . From the decay of the kernel of MATH we can control MATH pointwise by the NAME maximal function: MATH . From the previous REF reduces to MATH so by NAME and the NAME maximal inequality it suffices to show that MATH . Let us first consider the portion of the MATH norm in the region MATH . In this region we have from the MATH boundedness of MATH and the decay of MATH that MATH (in fact one can get much better bounds than this, especially if MATH) so by NAME and the above key observation (pairwise disjointness of MATH and disjointness from MATH) again this contribution to REF is bounded by MATH which is acceptable. It remains to estimate the contribution outside of MATH. Since we have MATH this follows simply from REF . This completes the proof of REF for the trees which satisfy REF. Now we consider the set of trees which satisfy REF. For simplicity of notation we may again assume this set is equal to MATH. The proof is a reprise of the previous case. We may associate to each MATH a multiplier MATH supported on the interval MATH obeying REF such that MATH where MATH is the non-negative quantity MATH . By duality we can find for each MATH a MATH-normalized function MATH such that MATH . We thus have MATH . By the NAME inequality we have MATH . To prove REF it thus suffices to show that MATH . We dyadically decompose the operator MATH around the base frequency MATH: MATH where MATH is a bump function adapted to the region MATH and supported in MATH. Define MATH . By the triangle inequality it thus suffices to show that MATH for all MATH. (While a better power of MATH can be achieved, we shall not be ambitious here to do so.) Fix MATH. By squaring and using symmetry it suffices to show that MATH . Using REF we see that it suffices to prove for each MATH . Fix MATH. Observe MATH and the pointwise bound MATH with the NAME maximal function MATH. The latter followed from the kernel bound MATH for the kernel MATH of the operator MATH. Let MATH be the set of all MATH such that MATH and MATH. By the above and the NAME maximal theorem it suffices to show MATH . This however follows from REF provided we can show MATH can be split into two subsets, each of which has the property that the intervals MATH with MATH in the subset are pairwise disjoint. This however follows from REF . This completes the proof of REF for the trees which satisfy REF. |
math/0107060 | By the homeomorphism MATH from MATH to MATH, one sees that MATH is a closed subset of MATH if and only if MATH is a closed subset of MATH where MATH is the diagonal MATH of MATH. Since MATH is NAME, then its diagonal is closed and MATH as well. By definition of the quotient topology, MATH is closed if and only if MATH is a closed subset of MATH. It suffices then to notice that MATH to complete the proof. |
math/0107060 | Obvious. |
math/0107060 | Since MATH is a NAME, then it is described by cells and attaching maps. There exists topological spaces MATH with MATH with the weak topology and MATH (for MATH belonging to some set of indexes) continuous maps which describe how to go from MATH to MATH; we have the following co-cartesian diagram of topological spaces: MATH where MATH is the inclusion of MATH into MATH as its boundary MATH. Let us describe inductively MATH as a globular NAME. We begin by setting MATH. Then we apply inductively the functor MATH on the co-cartesian diagram above, to obtain a new co-cartesian diagram by REF : MATH . First of all, it is easy to see that MATH induces a homeomorphism from MATH onto the boundary MATH of MATH, therefore is the inclusion morphism we expect. We now have to check that MATH is a correct attaching map for globular NAME. For MATH (MATH, MATH), we have MATH. We have to see that it is non-decreasing. Let MATH and MATH be two elements of MATH such that MATH. We have the following cases: CASE: MATH then MATH, thus is less or equal than MATH, CASE: MATH then MATH, thus is greater or equal than MATH, CASE: MATH (the case MATH is trivial since it implies MATH, which is the previous case) then MATH and MATH. Thus, MATH. That MATH is non-contracting is due to the fact that MATH. |
math/0107060 | This is due to the fact that MATH is homeomorphic to MATH. |
math/0107060 | It is an immediate consequence of REF . |
math/0107060 | We prove that attaching globular MATH-cells to any locally compact local po-space still defines a local po-space. As points are trivial local po-spaces, the theorem will follow from an easy induction. First we say that a local po-structure is small if for all MATH and MATH in the open covering defining the local po-structure, MATH and MATH coincide on MATH. It is easy to see that all local po-spaces MATH admit (in its equivalence class of coverings) a small local po-structure: if MATH is a po-neighborhood, then any subset of MATH which is a neighborhood of MATH is also a po-neighborhood; hence one can assume that MATH for some MATH and hence that the partial order on MATH is induced from MATH. The po-neighborhoods satisfying this extra condition are called small po-neighborhoods and they give a neighborhood basis for the topology on MATH, since the intersection of two small po-neighborhoods are again a small po-neighborhood. Moreover, the covering by the small po-neighborhoods defines the local partial order. Let MATH be a local po-space: it is defined by a covering MATH of open sub-po-spaces of MATH together with MATH, for all MATH, the local neighborhood and the corresponding partial order. We now only consider any of its small representatives in its equivalence class of local po-structures (we still call MATH). MATH is a local po-space, which is actually a (global) po-space. So its covering is MATH with corresponding MATH. Let MATH be an attaching map of a globular MATH-cell MATH. We construct the topological space MATH as defined by REF . Let MATH be the canonical surjective map. We have a commutative diagram in the category of topological spaces: MATH where MATH is the inclusion map and MATH, MATH are defined by the push-out diagram. Of course, MATH is injective since MATH is injective. We identify MATH with MATH restricted to MATH and also identify MATH with MATH since it is the composition of the inclusion map from-MATH to MATH with MATH. As MATH is compact, by REF , point REF, we know that MATH is a closed map and MATH is NAME (this holds true by induction again). Therefore MATH is closed since it is compact. Thus by point REF , MATH can be identified with the open subset MATH of MATH and MATH with the closed subset MATH of MATH. Similarly, MATH is a closed subset of MATH so by point REF , MATH can be identified with the open subset MATH of MATH and MATH can be identified with the closed subset MATH of MATH. We use these identifications below. Take now MATH; we are going to construct a neighborhood MATH of MATH in MATH together with a local po-structure on MATH, making MATH into a local po-space with the local po-structure MATH: CASE: Suppose MATH (see REF ). We define MATH that we noticed is identified with an open subset of MATH, and a binary relation MATH on MATH such that MATH if MATH. MATH is obviously a partial order. CASE: Suppose MATH (see REF ). We have noticed that MATH can be identified with an open subset of MATH. MATH is an open subset of MATH containing MATH since MATH is in MATH and MATH is injective on this part. Therefore, by REF , there exists MATH open of MATH containing MATH such that MATH. We define the partial ordering MATH to be the same as MATH on MATH. CASE: The only remaining possibility is that MATH (see REF ). Let us first subdivide the segment MATH; take six elements of MATH. We let (see REF ), CASE: MATH, with partial order MATH defined by, for MATH and MATH or MATH and MATH, it is the usual partial order induced by MATH and for MATH and MATH, MATH. CASE: MATH, with the usual partial order. CASE: MATH, with the usual partial order. We notice that if we identify MATH with MATH, MATH, MATH and MATH is a small local po-structure on the circle and the canonical surjection from the po-space MATH to this local po-space is a morphism of local po-spaces. We define MATH (similarly MATH) closed subset of MATH. The partial orders MATH induce partial orders MATH on MATH. As MATH is locally compact, we can find MATH a closed neighborhood of MATH contained in MATH. Consider the composite map MATH: MATH . It is a closed continuous map as a composition of two closed continuous maps. There exists (a non-necessarily unique) MATH such that MATH. Necessarily, MATH belongs to some MATH. We have MATH thus by REF there exists an open neighborhood MATH of MATH such that MATH . Let MATH be the subset MATH of MATH and MATH be the subset MATH of MATH. This is a partition of MATH. Notice that we can identify elements of MATH with elements of MATH and elements of MATH with elements of MATH. By construction, MATH. We now define a binary relation MATH on MATH as follows: CASE: for MATH, MATH if MATH, CASE: for MATH, MATH if MATH, CASE: for MATH and MATH, CASE: if MATH, MATH if MATH and MATH, (MATH is the unique parameter in MATH such that MATH for some MATH), CASE: if MATH, MATH if MATH, CASE: if MATH we can never have MATH. CASE: for MATH and MATH, CASE: if MATH, MATH if MATH and MATH, CASE: if MATH we can never have MATH, CASE: if MATH, MATH if MATH. This defines a partial order indeed. Reflexivity and transitivity are obvious. We now check antisymmetry. Let MATH and MATH be two elements of MATH such that MATH and MATH. If MATH and MATH both belong to MATH or MATH it is obvious that this implies MATH, since the relation MATH coincide with one of the partial orders MATH or MATH in that case. Suppose MATH, MATH with, CASE: MATH, we have by REF and MATH and MATH and MATH, which is of course impossible, CASE: MATH, MATH is impossible by definition, CASE: MATH, MATH is impossible by definition. It follows that MATH defines a small local po-structure since by construction, for MATH, the partial orders MATH and MATH coincide on the intersection MATH (if non-empty). It then suffices to set MATH. |
math/0107060 | Let MATH and MATH be two globular NAME and NAME be a morphism of NAME. The globular cellular decomposition of MATH yields a set of characteristic maps MATH satisfying: CASE: The mapping MATH induces an homeomorphism from MATH to its image. CASE: All the previous globular cells are disjoint and their union gives back MATH. CASE: A subset of MATH is closed if and only if it meets the closure of each globular cells of MATH in a closed set. where MATH runs over a well-ordered set of indexes MATH (one can suppose that MATH is a finite or transfinite cardinal). One can suppose that the mapping MATH is non-decreasing. Let MATH. Let MATH be an ordinal with MATH. If MATH is a limit ordinal, let MATH. If MATH for some ordinal MATH, then let MATH. Notice that MATH is closed in MATH. We are going to prove by transfinite induction on MATH the statement MATH: For any globular NAME MATH and for any set of characteristic maps MATH as above, a morphism of globular NAME from MATH to MATH induces a morphism of local po-spaces from MATH to MATH. Necessarily the equality MATH holds therefore MATH is true. Now let us suppose that MATH holds for MATH and some MATH. We want to check that MATH then holds as well. If MATH, then MATH is either the two-point discrete space, or a loop. So MATH holds. So let us suppose MATH. There are two mutually exclusive cases: CASE: The case where MATH is a limit ordinal. Let MATH. Then MATH for some MATH and the induction hypothesis can be applied; the result follows from the fact that the direct limit is endowed with the weak topology. CASE: The case where MATH for some cardinal MATH. Then MATH with the above notations. With the notation and identification as in the proof of REF , one has three mutually exclusive cases: CASE: MATH: in this case, the induction hypothesis can be applied; CASE: MATH: let MATH be a po-neighborhood of MATH in MATH; then MATH is an open of MATH; there exists a basis of MATH by global po-spaces so there exists a po-neighborhood MATH of MATH in MATH such that MATH; CASE: MATH: let MATH be the canonical closed map from MATH to MATH; by induction hypothesis, MATH is a morphism of po-spaces; therefore there exists a po-neighborhood MATH of MATH in MATH and a po-neighborhood MATH of MATH in MATH such that MATH . So MATH and by REF , there exists an open MATH of MATH such that MATH. Then let us considering the MATH of the proof of REF . Since MATH is continuous, MATH is open and MATH . |
math/0107060 | The map MATH is clearly bijective. Let MATH be the canonical map from MATH to MATH and let MATH be an open subset of MATH. Then MATH is an open subset of MATH, therefore MATH is continuous. So MATH is compact and therefore MATH is an homeomorphism. The fact that MATH preserves the structure of local po-spaces is obvious. |
math/0107060 | The proof goes by NAME 's lemma. |
math/0107060 | We mimic the proof of CITE: it suffices to prove that the diagram below (in the category of sets) is cocartesian for all MATH, MATH since colimits (hence push-outs) are taken point-wise in a functor category into MATH. For all MATH, the inclusions are in fact bijections, and the diagram is then obviously cocartesian. For MATH, the complement of MATH in MATH is the set of copies of cubes MATH, one for each cube of MATH. This means that the map MATH induces a bijection from the complement of MATH onto the complement of MATH. This implies that the diagram is cocartesian for MATH as well. |
math/0107060 | By definition, a morphism a precubical set sends a MATH-cube to another MATH-cube. So the realization as globular NAME induce clearly a morphism of MATH. |
math/0107060 | If such a MATH exists, then MATH is a NAME from MATH to MATH. Reciprocally, if MATH is a NAME from MATH to MATH, then the map MATH is constant for any MATH. Therefore MATH factorizes by MATH, giving the required MATH. |
math/0107060 | Let MATH be a continuous path in MATH from MATH to MATH (MATH being a NAME, its path-connected components coincide with its connected component). Then MATH satisfies the condition of REF . |
math/0107060 | We mimic the classical proof as presented for instance in CITE: the main idea consists of using the fact that the canonical projection from MATH to MATH is a NAME equivalence, having as inverse both MATH and MATH. Let MATH be the category having the same object as MATH and such that MATH . Let MATH be a functor such that for any MATH, MATH is invertible in MATH. The factorization MATH is obvious on the objects. To complete the proof, it suffices to verify that for two NAME morphisms MATH and MATH, then MATH. By definition, there exists MATH from MATH to MATH such that MATH and MATH. Let MATH be the canonical projection from MATH to MATH. Then MATH, MATH and MATH. Therefore MATH has as inverse both MATH and MATH. Thus MATH. |
math/0107060 | There exists a MATH-small category MATH satisfying the universal property of the theorem and constructed as follows: the objects of MATH are those of MATH. The elements of the MATH-small set MATH where MATH and MATH are two MATH-dimensional globular NAME are of the form MATH with MATH where MATH are morphisms of MATH and MATH are NAME equivalences and where the notation MATH for MATH NAME equivalence is a formal inverse of MATH (see for example REF for the construction). Let us consider the following commutative diagram MATH with the notation MATH for the codomain of MATH, MATH for the domain of MATH, and for MATH a subset of some globular NAME MATH, MATH where MATH is the smallest globular cell containing MATH. We see immediately that MATH where MATH means the cardinal of MATH and where MATH is the smallest infinite cardinal, that is, that of the set of non-negative integers. Since MATH is an homeomorphism and in particular is bijective, then MATH . This diagram remaining commutative in MATH, it shows that we can suppose MATH and MATH with an expression like MATH. By an immediate induction, we see that with a morphism of the form MATH lying in MATH, we can suppose that all intermediate objects are of cardinal lower than MATH which is a MATH-small cardinal. Therefore MATH is MATH-small as well. |
math/0107060 | Let us consider the MATH-small diagram of categories MATH . Then the direct limit of this diagram exists in the large category of MATH-small categories: see REF . By reading the construction in the proof of this latter proposition, one sees that the direct limit is actually a category with MATH-small objects and MATH-small homsets. |
math/0107060 | Obvious. |
math/0107060 | Let MATH be a family of topological spaces. Then MATH . Note that for all MATH, there exists a unique MATH such that MATH. Let MATH be a bipointed topological space and for all MATH, let MATH be a morphism in MATH. Let MATH be the set map from MATH to MATH defined by MATH, MATH, and MATH (for MATH and MATH). Take MATH such that MATH . We have three possibilities: CASE: MATH and MATH, CASE: MATH and MATH, CASE: MATH. In the latter case, MATH and therefore there exists MATH such that MATH and MATH belong to MATH. Then MATH. The set map MATH is well-defined and continuous because it is the quotient in MATH of the direct sum MATH by the identifications MATH and MATH. Therefore MATH is the direct sum of the MATH for MATH running over MATH in MATH. So the functor MATH preserves the direct sums. Let MATH and MATH be two continuous maps from MATH to MATH. Let MATH be the coequalizer of MATH in MATH. Then there exists a surjection MATH which is clearly an homeomorphism. Let MATH be a bipointed local po-space and let MATH be a morphism in MATH from MATH to MATH such that MATH. Then MATH factorizes through MATH because this latter is the coequalizer of MATH in MATH. It is easily checked that MATH is a non-decreasing map and therefore a morphism. So MATH preserves the coequalizers. This entails the result by REF. |
math/0107060 | The category MATH has a generator: the one-point space; it is cocomplete and well-copowered. The result follows from the Special Adjoint Functor theorem CITE. |
math/0107060 | Since MATH is compact, MATH . This isomorphism specializes to MATH where MATH is the set of non-decreasing continuous maps MATH from MATH to MATH such that MATH and MATH. The first member is in natural bijection with the morphisms of bipointed po-spaces from MATH to MATH, hence the result. |
math/0107060 | Let MATH. By definition, MATH is a non-decreasing continuous path from MATH to MATH. Let MATH be the canonical projection of MATH onto MATH. Since MATH is open and connected, and MATH and MATH are continuous, MATH is open and connected. Thus we can set MATH. Due to the peculiar ordering we have on MATH, MATH being non-decreasing implies that there exists a unique MATH such that for MATH, MATH (that is, its first component is constant on MATH). Let MATH and let MATH be an open of X containing MATH. Let MATH be a compact subset of MATH. Then MATH and for every MATH, MATH. Therefore the map MATH from MATH to MATH is continuous. Therefore the map still denoted by MATH from MATH to MATH is continuous. One has MATH and for all MATH, MATH is the dimap MATH. Let MATH . Then MATH yields a set map from MATH to MATH with MATH and MATH. So it suffices to check the continuity of MATH to complete the proof. Consider the set map MATH from MATH to MATH defined by MATH . Let MATH be a compact subset of MATH and MATH be an open subset of MATH such that MATH for some MATH. Then for any MATH, MATH. Therefore the set map MATH defined by MATH is continuous, and the set map MATH is continuous as well. Since the NAME functor is a right adjoint, it then commutes with products. So MATH is a continuous map from MATH to MATH. Since MATH is the image of MATH by the canonical isomorphism MATH is continuous as well. |
math/0107060 | Let MATH and MATH be two continuous maps from MATH to MATH such that MATH and MATH are NAME to MATH. Let MATH from MATH to MATH be a NAME from MATH to MATH with MATH, MATH and MATH. Consider the set map MATH from MATH to MATH defined by MATH with the notations of REF . Then MATH and in the same manner one gets MATH. So it suffices to prove the continuity of MATH to prove the uniqueness of MATH up to homotopy. We have already proved in REF the continuity of MATH and MATH. Therefore it suffices to prove the continuity of the set map MATH from MATH to MATH. This latter map is the composite of MATH . The last map MATH is the image of the identity map of MATH by MATH and therefore is continuous. At last the set map MATH is the image of the identity map of MATH by MATH and therefore is also continuous. So MATH is an homotopy between MATH and MATH. Now set MATH from MATH to MATH. With the proof of REF , we see immediately that MATH is continuous. It remains to prove that MATH is NAME to MATH. We have already seen in the proof of REF that for MATH, MATH for MATH. For MATH (respectively, MATH), one has by REF (respectively, MATH) and therefore Equality REF is still true for any MATH. So consider the path MATH of MATH. Then MATH is an element of MATH and we have MATH. But MATH. Therefore MATH. So MATH is NAME to MATH with the NAME MATH from MATH to MATH defined by MATH. |
math/0107060 | This is a consequence of REF . |
math/0107060 | Let MATH be a NAME from MATH to MATH such that MATH and MATH. Then MATH and MATH (respectively, MATH and MATH) coincide on MATH (respectively, MATH). Therefore MATH induces a bijection of sets from the MATH-skeleton MATH to the MATH-skeleton MATH with inverse the restriction of MATH to MATH. Let MATH and MATH be two elements of MATH. Then MATH (respectively, MATH) induces a continuous map MATH from MATH (respectively, MATH from MATH) to MATH (respectively, MATH). Let MATH be a continuous map from MATH to MATH which is a NAME from MATH to MATH. Let MATH. By hypothesis, this is a morphism of globular NAME from MATH to itself which induces the identity map on MATH. Let MATH. Then MATH and MATH. Moreover MATH is non-decreasing and continuous because it is the composite of two functions which are non-decreasing and continuous as well. Therefore MATH is a set map from MATH to MATH. We have already proved the continuity of similar maps (as in REF ). Therefore MATH. Similarly, we can prove that MATH. Therefore MATH is a weak NAME equivalence. |
math/0107060 | The composite MATH is a homotopy equivalence of NAME because MATH is an homotopy equivalence by hypothesis and because of REF . Therefore MATH has an inverse MATH up to homotopy from MATH to MATH. By REF , MATH and MATH are NAME to the identity (respectively, of MATH and MATH). Again by REF , MATH and MATH are NAME. Therefore MATH and MATH . |
math/0107060 | Let us make the proof for MATH. The canonical map MATH sends a MATH on the corresponding constant dipaths of MATH. The map MATH is necessarily continuous since MATH is discrete. In the other direction, let us consider the set map MATH defined by MATH: such an evaluation map is necessarily continuous as soon as MATH is endowed with the compact-open topology. Then MATH and MATH is homotopic to MATH by the homotopy MATH defined by MATH. The map MATH is the image of the identity of MATH by MATH where MATH is induced by the mapping MATH from MATH to MATH and therefore MATH is continuous. |
math/0107064 | For a NAME extension MATH, we have MATH by sending MATH with inverse MATH in the notation above. We denote MATH, and note that the multiplication on MATH induced by composition of endomorphisms is given by the MATH-multiplication: MATH . The unity element is MATH in the notation above. It is easy to see that MATH, where MATH is the multiplication mapping MATH, is a normalized NAME homomorphism, and MATH, MATH are dual bases satisfying REF . |
math/0107064 | The proof can be found in CITE. |
math/0107064 | We show that REF implies REF , the other implication is trivial. Denote by MATH and MATH orthogonal dual bases in MATH for MATH, where MATH. We compute that MATH: MATH for every MATH. The second statement in the proposition is proven similarly with dual bases MATH in MATH and therefore MATH in MATH. |
math/0107064 | For all MATH, we have MATH where MATH and MATH lie in MATH. MATH is linearly independent over MATH, whence over MATH, so MATH, similarly MATH, is finite dimensional. It follows that MATH restricted to MATH is a NAME homomorphism. Since MATH, MATH are dual bases and MATH, it follows that MATH is a separability element. Similarly, MATH is a NAME algebra with NAME homomorphism MATH, and a separable algebra with separability element MATH. |
math/0107064 | We map MATH into MATH, which has inverse mapping MATH into MATH. The proof of the second statement is completely similar. |
math/0107064 | If MATH, then MATH, which provides an inverse to the first map above. The second part is established similarly. |
math/0107064 | For each MATH we have MATH such that MATH since MATH. Then MATH . The second equality is proven similarly. The second statement is proven in the same way by making use of MATH. |
math/0107064 | Clearly MATH. Conversely, if MATH, then MATH. But MATH by the endomorphism ring theorem and the fact that both are dual bases for MATH. Then MATH as desired. Since MATH for every MATH, we obtain the MATH-multiplication on MATH. Then MATH since MATH. For the second statement, we observe: MATH while the opposite inclusion is immediate. The ring isomorphism follows from the identity: MATH for all MATH, since MATH. For there are MATH such that MATH , and MATH such that, for all MATH, MATH while MATH by irreducibility. Then we easily compute that MATH. Then: MATH . |
math/0107064 | Since MATH is a NAME algebra with NAME homomorphism MATH, it follows from the isomorphism, MATH that MATH . We have MATH since the index MATH. |
math/0107064 | We see that MATH, and we identify MATH with MATH. If MATH, we see that MATH implies that MATH by REF , since MATH is a NAME homomorphism on MATH and therefore faithful. If MATH, then by REF MATH implies first MATH, hence MATH. If MATH, then there are MATH REF such that MATH. Then MATH implies that each MATH, since if MATH, then MATH, a contradiction. Hence, MATH is faithful on MATH. |
math/0107064 | We have for each MATH, MATH: MATH whence by faithfulness MATH. Then MATH sends MATH onto itself, so MATH for each MATH, whence MATH. As for MATH, we note that MATH for every MATH, whence MATH on MATH. Commutativity of REF follows from the computation applying REF : MATH for all MATH. |
math/0107064 | We first note that MATH is the identity on MATH, since MATH, whence MATH . Since MATH for all MATH, we have for each MATH: MATH . It follows from REF that MATH is a MATH-bimodule homomorphism (it corresponds to MATH under the isomorphism MATH of MATH with MATH). That MATH is a NAME homomorphism follows from CITE, if we show it is one-sided faithful, for example, MATH implies MATH. But this follows from MATH being faithful and orthogonality of the dual bases MATH and MATH. |
math/0107064 | According to the definitions of MATH and MATH, we have MATH and also MATH, MATH, whence the result. |
math/0107064 | From REF MATH for all MATH. It is clear from REF that a NAME automorphism fixes elements in the center of a NAME algebra. The assertions about MATH are shown similarly. |
math/0107064 | Since MATH is finite dimensional (co)semisimple, MATH is (co)unimodular and there are integrals MATH and MATH such that MATH, MATH and MATH. Moreover, MATH gives a NAME isomorphism MATH, where MATH, since MATH integral in MATH means MATH for every MATH. If MATH is the NAME isomorphism, given by MATH, then MATH is the isomorphism MATH given by MATH . Now define MATH by MATH, where MATH since MATH. Note that MATH is a MATH-bimodule map and MATH. Denote MATH. Since MATH, which is sent by MATH to MATH, it follows that MATH (compare REF ). The homomorphism MATH (given by MATH) is now readily checked to have inverse mapping given by MATH CITE. By counitarity of the MATH-comodule MATH, then MATH factors through MATH and the map MATH given by MATH. Then MATH, whence the MATH-index MATH is MATH. It is not hard to compute that MATH which is MATH since MATH is irreducible. Since MATH is free over MATH with basis in MATH, we see that the first half of the depth REF condition is satisfied. The second half of depth REF follows from noting that MATH is a right MATH-Galois extension of MATH. For the coaction MATH is given by MATH . One may compute the inverse of the NAME map to be given by MATH. Then MATH and the rest of the proof proceeds as in the previous paragraph. |
math/0107064 | If MATH for some MATH, then we have MATH for all MATH, since MATH by REF . Taking MATH and using the braid-like relations between NAME idempotents and NAME property REF of MATH we have MATH for all MATH, therefore MATH (by REF ). Similarly, if MATH for some MATH, then MATH for all MATH, which for MATH gives MATH for all MATH, therefore MATH. |
math/0107064 | We use the NAME identities together with REF to compute MATH for all MATH (note that the restriction of MATH and MATH, identifying MATH and MATH). |
math/0107064 | First, let us show that the above equality holds true in the special case MATH. Let MATH be the conditional expectation from MATH to MATH given by MATH as in REF . We claim that for any MATH we have MATH if MATH for all MATH. For since MATH, let MATH with MATH and MATH, then MATH and the latter expression is equal to MATH for all MATH only if for each MATH either MATH or MATH. Observe that MATH restricted to MATH coincides with the NAME automorphism MATH of the NAME extension MATH since MATH therefore, using the NAME identity for MATH, we establish the proposition for MATH: MATH since MATH. Next, arguing as in REF we write MATH with MATH, whence MATH . |
math/0107064 | The result follows from multiplying the identity from REF by MATH on the left and taking MATH from both sides. |
math/0107064 | For all MATH and MATH we have by REF : MATH where we use the definition of MATH, the NAME identity, and REF . Thus, MATH. |
math/0107064 | We obtain this formula by replacing MATH with MATH in REF and using REF as well as REF . |
math/0107064 | The statement follows from the direct computation : MATH for all MATH and MATH, using REF . |
math/0107064 | Note that MATH is a coalgebra automorphism by REF . By REF we have, for all MATH and MATH : MATH whence MATH. |
math/0107064 | Using REF and the definition of the antipode we have MATH. The second identity follows similarly from REF and the corollary MATH from REF : MATH that is, MATH satisfies the antipode properties. |
math/0107064 | Follows from REF , and REF. Note that semisimplicity and separability are notions that coincide for finite dimensional NAME algebras CITE. The non-degenerate duality form of REF makes MATH the NAME algebra dual to MATH. |
math/0107064 | NAME and NAME proved that a semisimple and cosemisimple NAME algebra is involutive CITE. It follows from REF that MATH. But we compute: MATH for all MATH, MATH, from which it follows that MATH. Since MATH, we have MATH. Whence MATH, MATH and MATH are traces on MATH, MATH and MATH, respectively. It follows from REF that MATH is a separable basis for MATH; similarly, MATH is a separable basis for MATH, whence MATH and MATH are strongly separable algebras. |
math/0107064 | The above map defines a left MATH-module structure on MATH, since MATH and MATH . Next, REF implies that MATH. Finally, MATH. |
math/0107064 | If MATH is such that MATH for all MATH, then MATH. Letting MATH we obtain MATH, therefore MATH. Conversely, if MATH, then MATH commutes with MATH and MATH therefore MATH. |
math/0107064 | The bijectivity of MATH follows from REF . To see that MATH is a homomorphism it suffices to note that MATH for all MATH and MATH. Indeed, using REF , MATH . |
math/0107064 | Dual to the left MATH-module algebra MATH defined above is a right MATH-comodule algebra MATH with the same subalgebra of coinvariants MATH, since MATH. By REF and the endomorphism ring theorem, MATH is given by the natural map MATH since if MATH for MATH, then for all MATH, MATH . By REF then, MATH is a right MATH-Galois extension of MATH. It follows from the endomorphism ring theorem for NAME extensions (compare end of REF) that MATH is MATH-Galois. |
math/0107064 | Since MATH, we show that MATH is a total integral by showing it is a right MATH-comodule morphism CITE. Denoting the coaction MATH (which is the dual of Action REF) by MATH, we have MATH for every MATH. Since each MATH by REF , it suffices to check that MATH: MATH . Finally, we note that MATH has convolution inverse in MATH given by MATH where MATH denotes the antipode on MATH. |
math/0107064 | The cocycle MATH associated to MATH is trivial, since MATH . It follows from REF and CITE that MATH is a MATH-module algebra with action MATH given by MATH . It follows from REF and triviality of the crossed product that MATH is a smash product of MATH and MATH as claimed. |
math/0107064 | That MATH follows from the definition of MATH and its NAME algebra structure. Conversely, suppose that MATH is such that MATH for all MATH. In a computation similar to that of CITE, we note that MATH in MATH for any MATH: MATH . Letting MATH, we see that MATH commutes with MATH, so that MATH. Applying MATH to this, we arrive at MATH. |
math/0107064 | This follows from REF , if we prove that MATH given by MATH is an isomorphism. Towards this end, we claim that MATH for every MATH. Let MATH. A few short calculations using REF show that MATH such that MATH, since MATH and MATH . Since MATH freely generates MATH (as a NAME homomorphism), there is MATH such that MATH, whence MATH as claimed. Then MATH for all MATH is surjective. An inverse mapping may be defined by MATH for each MATH, where MATH, MATH are dual bases for MATH as in REF. |
math/0107064 | The forward implication follows from REF . The reverse implication follows from REF . |
math/0107069 | Let MATH be the closed proper subvariety whose points correspond to curves which are either not smooth or whose normal bundle is not of the form REF . Since MATH was generically chosen, MATH is not contained in the proper subvariety MATH. Consequence: if MATH is a general fiber of the morphism MATH, then MATH is a smooth curve with normal bundle MATH, and the tangent map MATH has rank two along MATH. In particular, MATH has maximal rank at MATH, and the claim is shown. |
math/0107073 | Given any adapted coframe MATH we need only show that there is a system of adapted coordinates MATH in which MATH at the origin of coordinates. Let us start by picking any adapted coordinates MATH. Certainly we have MATH, for some MATH, no matter what adapted coordinates we picked. Now change coordinates by MATH . Then in the new coordinates, MATH . We still have the last row to deal with. Try the change of coordinates MATH and you find that this accomplishes the task at hand (only at the origin of coordinates). Therefore every adapted coframe from MATH arises from adapted coordinates. Given any two adapted coframes, take such coordinates near each of them, and as diffeomorphism use these coordinate functions. |
math/0107073 | To see that the map MATH is well defined, note that MATH is a MATH plane because MATH is a MATH plane, and the kernel of MATH is entirely contained inside MATH, and has dimension MATH. We now have established the commutative diagram MATH . Pick MATH any adapted coframing on an open subset of MATH. The map MATH determines the plane field MATH by MATH . The same equation holds on MATH: MATH where MATH. This gives MATH . Therefore there is an invertible matrix MATH so that MATH . A similar argument establishes that there is an invertible matrix MATH so that MATH . Taking the exterior derivatives of these equations, and plugging in the structure equations, we find by NAME 's lemma that there are functions MATH so that MATH or in other words that MATH that is, a change of adapted coframing. Therefore MATH has full rank, and the result is clear. |
math/0107073 | This is an immediate consequence of the NAME - NAME theorem; see CITE. |
math/0107073 | The structure equations show that each coframing in MATH is determined up to complex linear multiples. Moreover, the elements MATH of an adapted coframe from MATH are basic for the projection to MATH, so form a coframe on MATH at MATH. REF-forms MATH vanish on MATH. Therefore the MATH identify MATH with MATH, and identify MATH with MATH, a complex subspace. These MATH are determined up to complex multiples, as are the MATH together. Therefore all choices of coframes from the fibers of MATH determine the same almost complex structure MATH on MATH. |
math/0107073 | Given any immersion MATH of an oriented manifold MATH of dimension MATH, we can let MATH be the map associating to each point of MATH the tangent space MATH . To have MATH solve MATH means precisely that MATH for all MATH. Clearly we get the diagram MATH which differentiates to show that MATH so that MATH . It also shows that MATH intersects the tangent space to the fiber MATH transversely. Take any local adapted coframing from the bundle MATH, say MATH, and pull it back via MATH. We find that MATH, and that MATH is a coframing on MATH (actually, only on an open subset of MATH, since it is only a local coframing), because of the transversality of MATH with the fibers. But then on MATH we have MATH so that MATH for some complex valued functions MATH on MATH. By the structure equations, MATH . We see that there are no MATH terms appearing, and therefore (by the NAME - NAME theorem) these MATH define a complex coframing giving a complex structure on MATH. If we change the choice of coframing, then we obtain the same complex structure, because the structure group acts via a complex representation. |
math/0107073 | The tangent spaces to the stalks are described in terms of any adapted coframing MATH by the equations MATH which are complex linear, so the stalks are almost complex submanifolds. The MATH planes are described (invariantly) by the equation MATH for any adapted coframing MATH on MATH. These are complex linear equations, so the MATH planes are complex planes. To see that the stalks are actually complex manifolds, plug in the structure equations to see that MATH on these stalks. Therefore by the NAME - NAME theorem, the stalks are complex submanifolds. |
math/0107073 | The invariants preventing a NAME - NAME tableau are first order, so equations with vanishing first derivatives in the functions MATH must have NAME - NAME tableau. Let us prove the other direction. First, we consider taking our equation and carrying out a simple change of coordinates. By translating the coordinates, we can arrange that the point MATH we are interested in is the origin of coordinates. By rotation in these variables, we can arrange that any chosen tangent plane which satisfies the equation at first order is taken to the plane MATH. Now we can take any function MATH and change coordinates to MATH. As long as MATH at the origin, this change of coordinates preserves our conditions, and we can easily see how it changes our system of equations. MATH . The quadratic MATH has no impact on the lowest order terms inside MATH, so that its effect is felt only in the MATH term. We can use this term to wipe out the MATH and MATH linear terms in MATH, and to wipe out MATH linear terms. But as for MATH linear terms, we can only wipe out those which are symmetric: the MATH terms in MATH . Now we will begin a more abstract approach. Suppose that we have two equations of the form of REF, say MATH and MATH. Each is equipped with a foliation MATH with complex structures on the fibers. By the NAME - NAME theorem ``with parameters" (that is, making obvious modifications to the NAME proof which is clearly presented in CITE, to allow for families of complex structures, using smooth dependence of solutions of elliptic equations on parameters), locally we can find a diffeomorphism MATH which is holomorphic on the fibers of the MATH. It is easy to see (in local coordinates) that the pseudogroup of diffeomorphisms of MATH which are holomorphic on the fibers acts transitively on coframes of MATH of the form MATH with the MATH being MATH-forms on the fibers, and the MATH vanishing on the fibers. Therefore we can arrange that some local adapted coframes MATH of MATH and MATH of MATH agree at some point MATH. By perhaps altering the choices of the coframes, they must therefore satisfy equations of the form MATH . (The last two rows follow from differentiating the first four.) All of the functions MATH vanish at the chosen point MATH. Differentiating, we find MATH with symmetries in the lower indices from NAME 's lemma. If we take the equation MATH to be a complex structure, that is, the NAME - NAME equations, then we can easily see from these structure equations that near the point MATH, the numbers MATH correspond to the numbers MATH . But by explicit coordinate manipulations above, we managed to kill these terms. |
math/0107073 | Use the NAME - NAME theorem to produce holomorphic coordinates on MATH. Follow the proof of REF . |
math/0107073 | This is immediate from the NAME - NAME theorem, since the structure REF are involutive. |
math/0107073 | Fattening up the structure group to the complex contact group (see CITE) we obtain the structure equations MATH of a complex contact structure. By the NAME - NAME theorem and the NAME theorem for complex contact structures, such structures are all locally isomorphic. MATH is the contact plane field. The fibers of MATH are given by the complex linear equations MATH, for any section MATH of MATH. Therefore they are complex submanifolds, and clearly tangent to the plane field MATH, therefore NAME. |
math/0107073 | Suppose that we have a holomorphic section MATH which does not vanish at some point MATH. Take affine coordinates near MATH . Then in those coordinates MATH . By taking a linear change of those coordinates, we can arrange MATH . Now by rotating those coordinates, we get other sections MATH not vanishing at MATH . Wedging them together, we get a section MATH not vanishing at MATH . But this line bundle has no global sections, because it has negative first NAME class. |
math/0107073 | We have seen that the points of MATH can be interpreted as NAME submanifolds in MATH: the fibers MATH. The cohomology numbers above prove the existence of a locally complete moduli space MATH of NAME submanifolds, by NAME 's theorem. Using a local section of MATH, we get a map of MATH into the moduli space MATH. The spaces MATH and MATH have the same dimension, and MATH is mapped by a smooth injection. We want to show that this map is an immersion, and therefore a local diffeomorphism. Consider a particular fiber MATH of MATH over a point MATH. Let MATH be any section of MATH, that is, any adapted coframing. We will write it as MATH since there is only one value for MATH. As CITE explains, a tangent vector MATH corresponds to a unique section of MATH which is determined as follows: take the holomorphic section of the normal bundle MATH which projects to MATH. Then project it to MATH . If this vanishes, then the section of the normal bundle must be the image of a section of MATH. As we have seen MATH . Again, this bundle has no global holomorphic sections. Consequently, the vectors MATH are injectively mapped to elements of MATH, so the map MATH is a local diffeomorphism. We can put a complex structure on MATH, pulling back the one from MATH, and the map MATH must be holomorphic for that complex structure, since the map to moduli space MATH is. Choosing any local holomorphic coordinates MATH on MATH, we can pull them back to MATH to find that in terms of any adapted coframing MATH, MATH for some complex valued functions MATH. By a complex linear change of coordinates, we can arrange that at some chosen point of MATH, MATH . Taking exterior derivative, we find MATH so that the MATH structure is torsion-free, and again by the NAME - NAME theorem, and NAME 's theorem for complex contact structures, it is flat. Therefore REF are the NAME - NAME equations. |
math/0107073 | That this map is well defined follows from there being, at each point MATH, some vector MATH. That this map is an immersion follows immediately from our discussion of the map MATH . |
math/0107073 | If the assumed conditions hold, then MATH and the immersion MATH is a local biholomorphism, and under this map MATH pulls back to MATH. Because MATH is compact, it is a covering map. Because MATH is simply connected, this map is a global biholomorphism. |
math/0107073 | Suppose that REF do not hold. Consider a section MATH. Its zero locus is a projective variety, since MATH is. Take MATH a smooth point of MATH. Pick a MATH complex basis MATH of MATH with MATH. Using the holomorphic immersion MATH (which is only defined in a neighborhood of MATH) we find that MATH is mapped to an analytic variety in MATH, with a smooth point at the origin. We can make a MATH complex linear change of basis to arrange that MATH is tangent to MATH. Then in the MATH coordinates, MATH . The variety MATH is a straight line, lying entirely inside MATH . So it is tangent to MATH, and therefore either lies entirely inside MATH or else strikes it at the origin with multiplicity (that is, after small topological perturbation, strikes at least twice). By changing the choice of complex basis MATH we can see that the same is true for any line lying inside MATH and passing through the origin of these coordinates. Returning from the MATH coordinates, either REF MATH contains the component of MATH passing through MATH, or REF else it has multiplicity at least two with MATH for some choice of MATH complex basis MATH. Topologically, this says that MATH . But this contradicts our hypotheses. By positivity of intersections of the MATH, we see then that MATH. Each MATH is a smooth variety, since each vector MATH belongs to a MATH complex basis at each point MATH. If MATH has more than one component, say MATH with MATH, then taking intersection with the other MATH we find that MATH contradicting another of our topological hypotheses. The same for MATH. So MATH. |
math/0107073 | The argument above shows that this map is defined and is a local diffeomorphism. But if MATH is compact, then so is MATH, and so the result is immediate since the sphere MATH is compact and simply connected. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.