paper stringlengths 9 16 | proof stringlengths 0 131k |
|---|---|
math/0009158 | Transfer MATH to MATH and pull back to MATH. Then MATH becomes a holomorphic section of a trivial vector bundle and so each component of MATH is a holomorphic function that decays at MATH. But by the removable singularity theorem of NAME, each component extends uniquely to a holomorphic function on MATH and by the maxi... |
math/0009158 | Recall first that MATH . Here MATH is the canonical-bundle of MATH, and the trivial bundle MATH is embedded in MATH by multiplication by MATH. Let us for the moment write MATH so that MATH. This decomposition is preserved by the connection since MATH is parallel, so that MATH decomposes as a pair of operators MATH (tak... |
math/0009159 | By construction, the MATH-graded chain complex MATH has a bounded filtration REF with the associated graded complex given by MATH . Then by the standard technique of CITE, we derive the existence of a spectral sequence MATH with MATH, and MATH . The higher differentials are induced by MATH. The expression of MATH, as d... |
math/0009159 | The result is the consequence of the results on the geometric limits in CITE CITE, together with the dimension formulae for various moduli spaces. Suppose given a solution MATH in a one-dimensional component of MATH, with asymptotic values MATH for MATH. The geometric limits as MATH are determined in CITE. Among these ... |
math/0009159 | For any MATH, let MATH be another MATH-dimensional submanifold in MATH obtained from the construction of the holonomy map and let MATH be a strata transversal section of MATH over MATH. We will prove that the difference between MATH and MATH defines an element in MATH which is homologous to zero. Over MATH, we can choo... |
math/0009159 | Let MATH denote the boundary embedding maps of MATH into MATH. Then the set of MATH-structures on MATH which agree with MATH over MATH, denoted by MATH, form an affine space over MATH with MATH and MATH. For any MATH, after a generic perturbation, the moduli space MATH is an oriented, compact, smooth manifold with dime... |
math/0009160 | MATH is MATH-compact. By NAME 's theorem we infer that MATH is compact when endowed with the cartesian topology. We can now identify MATH with a net in MATH to obtain a sub-net MATH of MATH which converges to an element MATH of MATH. Since MATH is MATH-closed in MATH, MATH induces a bounded linear operator from MATH in... |
math/0009160 | Let MATH be a basis for MATH isometrically equivalent to the usual MATH-basis. Since MATH is MATH-continuous and MATH, there exist vectors MATH in MATH such that MATH for all MATH. There also exist scalars MATH, MATH, MATH, MATH, such that MATH for all MATH and MATH. Finally, there exist scalars MATH, MATH, MATH, such ... |
math/0009160 | We first choose MATH so that MATH and MATH for all MATH and MATH. We then choose MATH for all MATH, and a sequence MATH of positive scalars such that MATH. REF yields a MATH-continuous linear operator MATH, MATH, such that MATH for MATH, and MATH, for MATH. We next apply REF for MATH, MATH, MATH, MATH and MATH, to obta... |
math/0009160 | We apply REF for MATH, MATH, MATH and MATH, to obtain a MATH-continuous linear operator MATH, MATH, so that MATH for all MATH. Our assumptions yield that MATH is the desired projection. |
math/0009160 | Assume MATH is infinite and let MATH be an enumeration of MATH. (The argument for finite MATH is implicitly contained in the proof of the infinite case.) It is clear that MATH is isometrically equivalent to the usual MATH-basis. Set MATH, MATH and let MATH be a null sequence of positive scalars. We shall inductively co... |
math/0009160 | Clearly, MATH is linearly independent. When MATH is finite the assertion follows immediately from REF as MATH is isometric to MATH. If MATH is infinite let MATH be an enumeration of MATH and set MATH. Of course MATH is isometrically equivalent to the usual MATH-basis, and applying the NAME representation and the NAME t... |
math/0009160 | We regard MATH in its MATH-topology and set MATH. Since MATH is separable and MATH is non-separable, MATH is an uncountable MATH-subset of MATH. It follows that MATH is an uncountable Polish space in its relative MATH-topology. We will show that MATH contains a MATH-compact subset MATH homeomorphic to the NAME set MATH... |
math/0009160 | We first choose an infinite MATH-convergent subsequence MATH of MATH and set MATH. Clearly, MATH. If MATH then MATH is MATH-closed in MATH by REF. We deduce from REF that MATH is the range of a MATH-continuous contractive projection in MATH. It is easy to see that if MATH is an enumeration of MATH then the subspace MAT... |
math/0009160 | If MATH is non-separable the assertion follows from REF . If MATH is separable, then MATH is countable and MATH is isometric to MATH. Let MATH be a basis for MATH isometrically equivalent to the usual MATH-basis, and choose a MATH-convergent subsequence MATH of MATH according to REF . Let MATH. Then it is easy to see t... |
math/0009160 | Let MATH be a basis for MATH isometrically equivalent to the usual MATH-basis. Let MATH, MATH, and let MATH be a sequence of positive scalars such that MATH. The argument in the proof of REF now yields a sequence MATH of MATH-continuous contractive projections in MATH with MATH and such that MATH whenever MATH. Given M... |
math/0009161 | REF is just REF in the present notation. CASE: This is immediate from REF since MATH defines MATH outside MATH. Note that the boundary points MATH of MATH correspond to MATH and MATH, respectively. CASE: NAME of MATH near MATH means asymptotic expansions at the interiors of MATH (that is, for MATH with MATH bounded awa... |
math/0009163 | The conditions are obviously necessary. So we only need to prove the sufficiency. Assume that MATH and MATH is not a constant on MATH. Then there exists a curve MATH such that MATH is a smooth point of MATH, and MATH. Let MATH where MATH. Then MATH and MATH that is, MATH . By REF , there exists a MATH such that MATH . ... |
math/0009163 | We will repeatedly use the projective dimension theorem CITE which says that if MATH and MATH are MATH-dimensional and MATH-codimensional projective varieties, respectively, then MATH. In particular, MATH is not empty if MATH. Let MATH . Then MATH must have co-dimension MATH because of the condition MATH. Therefore the... |
math/0009164 | The connected component of a point MATH in MATH is by definition the maximal connected subset of MATH, which contains MATH. The connected set MATH lies in MATH and contains MATH. Therefore, MATH. |
math/0009164 | Let MATH and MATH be any open neighborhood of MATH. Designate the set MATH by MATH. From property MATH follows that there exist MATH, MATH, such that MATH, MATH. But MATH. Therefore, MATH, MATH. |
math/0009164 | Obviously, MATH, MATH. On the other hand, MATH, MATH. So MATH, MATH, since the sets MATH are closed by the definition. |
math/0009164 | This statement is a direct corollary of the previous statement and the following theorem: if the domains MATH and MATH in the plane do not meet, but MATH, then MATH is connected (see REF ). |
math/0009164 | Fix MATH. It is known (see REF ), that for every MATH points MATH and MATH can be connected with the MATH-chain in the set MATH, i. e. final sequence MATH of points could be found in MATH to comply with the condition MATH . For every MATH select MATH-chain in MATH connecting points MATH and MATH. Since MATH, for every ... |
math/0009164 | Fix MATH and open neighborhood MATH of MATH. Let us show that MATH. The set MATH divides the plane into two connected components. Therefore it consists more than of one point. Take MATH such that MATH and MATH. According to REF there exists MATH. From REF it follows that MATH. Besides, MATH. The set MATH is zero-dimens... |
math/0009164 | Let MATH. There exists MATH, such that MATH. According to REF , we can find MATH, MATH, such that MATH, MATH. Designate by MATH the segment with end-points MATH and MATH, MATH. Obviously, MATH, MATH. The set MATH is connected and MATH. We fix a point MATH. Since MATH for every MATH, the set MATH is connected. |
math/0009164 | Fix MATH. For MATH designate by MATH the connected component of MATH in the set MATH. Then in accord with REF the family MATH forms a base of opened neighborhoods of the point MATH in the topology MATH. On the other hand let us denote MATH for every MATH. Then by easy direct verification we receive MATH for every MATH.... |
math/0009169 | We just have to apply the isomorphisms REF : MATH and similarly MATH . |
math/0009169 | For the first line, remember that MATH if MATH, and zero otherwise. But here we also have that MATH. For the second line, using the properties of the isomorphisms MATH and MATH we obtain that MATH . For the last line we have used the properties of the NAME invariants of MATH REF . |
math/0009169 | Let us write MATH. By applying REF we obtain MATH which proves the lemma. |
math/0009169 | Similar to the proof of REF . |
math/0009170 | We define MATH recursively. Suppose MATH are such that MATH satisfies MATH. Note that since MATH is Hermitian, so is MATH. We need to find MATH so that MATH satisfies MATH up to order MATH. But this happens if and only if MATH. Then MATH is a solution. |
math/0009170 | The MATH-linearity and the injectivity of MATH are obvious since MATH and MATH are deformations of MATH and MATH. To prove surjectivity, let MATH be given. Then MATH whence MATH. Thus defining MATH, we have MATH up to order MATH. Since MATH starts with order MATH, we can repeat the argument to find a MATH such that MAT... |
math/0009170 | For the existence, note that since MATH is f.g.p.m., it follows that we can identify MATH, for some MATH and MATH idempotent. Let MATH be an idempotent deforming MATH and consider the (right) f.g.p.m. over MATH given by MATH. By REF (choosing MATH to be MATH in the upper right corner and zero elsewhere), we can use the... |
math/0009170 | As in REF , we choose MATH, a projection deforming MATH, and consider the MATH-module MATH, which we know to define a deformation MATH of MATH. Let MATH be the MATH-valued inner product on MATH obtained from MATH restricted to MATH. A simple computation shows that MATH is a deformation of MATH and hence MATH is a Hermi... |
math/0009170 | Since all deformations of MATH are equivalent we choose a deformation MATH of the projection MATH. Moreover, we choose a unitary MATH with MATH and MATH which is possible due to REF . Now MATH is again a projection with classical limit MATH. Thus it defines a deformation MATH which is equivalent to the first one by an ... |
math/0009170 | Existence and uniqueness of (Hermitian) deformations follow from REF and the observation before the theorem. Suppose now MATH is local/differential/of NAME type. Choose a deformation MATH of MATH and let us consider the MATH-action on MATH induced by MATH as in REF . Note that if we write MATH, it follows from REF that... |
math/0009170 | As we have remarked, the algebras MATH and MATH are NAME equivalent, and MATH is a MATH-bimodule defining this equivalence. By symmetry of NAME equivalence, it follows that there exists an idempotent MATH, for some MATH, so that MATH (MATH as row vectors) as a left MATH-module and MATH as a unital algebra. In the Hermi... |
math/0009170 | From REF it easily follows that MATH satisfies the natural NAME rules (see the last two equations in CITE) and this implies the first statement. A simple computation shows that MATH. Note that MATH is equivalent to the first equation in CITE and the last statement follows directly from CITE . |
math/0009170 | Note that if MATH and MATH, we have MATH, where MATH is the Hamiltonian vector field corresponding to MATH. Observe that the curvature tensor corresponding to MATH, MATH, satisfies MATH, for all MATH and MATH. This implies the result. |
math/0009170 | For MATH, write MATH. It is not hard to check that MATH . It is then clear that MATH. The bracket MATH defines an action of the NAME algebra MATH on MATH by derivations and this implies that MATH. Hence the NAME rule for MATH yields MATH, MATH, and this immediately shows that the bracket MATH on MATH coincides with MAT... |
math/0009170 | Let MATH. Then a straightforward computation shows the relation MATH . It follows immediately that MATH is a positive algebra element and that MATH is full. The fullness of MATH follows from MATH . We observe that REF can be easily shown as in CITE. In order to prove REF , we recall that this just means REF for the com... |
math/0009173 | The proof is easy. |
math/0009173 | This is a straightforward computation. |
math/0009173 | This is proved in REF and also follows from the proof of the main theorem, REF (see REF for details). |
math/0009173 | Clear. |
math/0009173 | See REF. |
math/0009173 | This is obvious. |
math/0009173 | This follows from nilpotency. It is easy to see that we need only show that MATH for any MATH and MATH. Equivalently, we have to show that the componentwise addition of all MATH is not MATH. First, we generalize MATH and MATH. Clearly, every MATH can be written as MATH for some MATH. We say that MATH for MATH if MATH a... |
math/0009173 | If MATH, then it is clear from MATH that MATH, negating in each component. But then MATH, which is impossible. |
math/0009173 | In the first case, if MATH reversed orientation, then MATH, thus MATH by nilpotency - this is a contradiction. On the other hand, when MATH reverses orientation, MATH must reverse orientation, and thus by nilpotency cannot be defined on all of MATH. Thus MATH, hence the desired result. The second case follows easily fr... |
math/0009173 | The fact that the maps are well-defined, inverse to each other, and preserve MATH is easy to see from construction when orientations are preserved (it helps to draw a picture). Also, when orientations are preserved, MATH is trivially preserved. So it remains to consider orientation-reversing cases. Given MATH, if orien... |
math/0009173 | This follows directly from the argument above. |
math/0009173 | Fix a choice of sign MATH for this proof. Clearly, whenever MATH, MATH, and MATH, then MATH. In this case, MATH, as MATH also has reversed orientation and the same order as MATH and MATH. It remains only to see that, for any MATH with MATH, there are MATH ways of writing MATH for MATH, and they all are of this form. Th... |
math/0009173 | Take some MATH such that MATH. We show in the following paragraph that either MATH divides MATH, or MATH and MATH. This proves the lemma - all that remains is to see that, in the latter case, MATH. If, instead, MATH, then applying the above result also to MATH, we find that both MATH and MATH are perpendicular to MATH.... |
math/0009173 | First, it is clear that MATH divides MATH iff MATH for MATH an integer. In this case, the theorem is satisfied; so suppose not. By applying MATH some number of times to MATH or MATH, it suffices to assume MATH. Now, if MATH is defined on MATH, then MATH together with the Lemma gives the desired result. So assume MATH i... |
math/0009173 | Let MATH and MATH for some MATH (which exists by construction). Suppose that MATH. We will analyze all possible cases by considering the value of MATH. Write MATH where MATH. First, I claim that MATH. Suppose instead that MATH. In this case, MATH, which immediately implies from the proof of REF that MATH preserves orie... |
math/0009173 | Clearly MATH iff MATH. Suppose MATH. Hence MATH. Suppose for a contradiction that MATH. Then, MATH. However, MATH and REF imply that MATH, a contradiction. So, the identity MATH easily follows (since MATH). The rest of the proof is almost exactly the same as the proof of the first part of REF , getting rid of MATH. Aga... |
math/0009173 | The proof is the same for both REF , does not mention MATH, and is given in the following paragraphs. First, we note that MATH iff MATH and MATH reverse orientation, which is true iff MATH and MATH reverse orientation, which is true iff MATH. Now, we show that MATH implies MATH. To reach a contradiction, suppose that M... |
math/0009173 | Again, REF have nearly the same proof, which follows. Suppose, on the contrary, that MATH. Clearly MATH, else MATH would have reversed orientation, which is not possible by REF . So MATH. Now, MATH is defined on MATH since, in REF is defined on MATH and MATH, which follows from MATH, and in REF is defined on MATH. In p... |
math/0009173 | First we note that, if MATH, then MATH, for the following reason. Suppose MATH and write MATH or MATH, depending on whether MATH is bad or good. Then MATH, so REF applies and shows that MATH, as desired. By symmetry, MATH implies that MATH. So, it suffices to show that MATH if MATH and MATH if MATH. By the symmetry of ... |
math/0009173 | Clearly MATH would imply that MATH and MATH reverse orientation, which is not possible since MATH. Suppose instead that MATH. We have that MATH, and hence MATH, is defined on MATH, and hence MATH. Since MATH, we find that MATH, contradicting REF . So MATH. |
math/0009173 | First, note that MATH is defined on MATH since it is defined on MATH, and the former is a subset of the latter on the diagram. If MATH, then set MATH. If MATH and MATH, set MATH. In either case, MATH is defined on MATH. Indeed, MATH is defined on MATH, and MATH. Note that, by REF , MATH and MATH must preserve orientati... |
math/0009173 | This follows easily from REF . |
math/0009173 | This easily follows from REF . |
math/0009173 | Clear. |
math/0009173 | CASE: This follows immediately from REF in the cases that MATH and MATH. (Note that it is impossible to have MATH by REF .) In the case that MATH and MATH or in the case that MATH and MATH, it is clear that MATH. CASE: It follows that only one of the MATH can be in MATH from REF . If MATH, then no MATH is outer by REF ... |
math/0009173 | That MATH, and MATH are disjoint from MATH follows from REF , and the definition, respectively. The other facts regarding disjointedness are obvious. To check that the union is all of MATH, we apply REF .ii, which shows that MATH. It is clear, though, that MATH. This proves the desired result. |
math/0009173 | CASE: This follows immediately from REF . CASE: This follows immediately from REF .ii, REF, and REF. CASE: This follows immediately from REF . CASE: This follows immediately from the definition of MATH. CASE: This follows from REF , since we know MATH is disjoint from MATH. |
math/0009173 | For every MATH, we see from the Lemma, REF , that MATH. Furthermore, MATH and MATH where these are defined (Lemma, REF). Now take any MATH. First note that either MATH or MATH because MATH. Note that MATH because MATH and MATH are disjoint from MATH REF . Now, if MATH, then using REF , MATH (of type a or b, depending o... |
math/0009173 | The terms in the expansion of MATH are MATH for each MATH, as well as the terms MATH for MATH and MATH for each MATH. |
math/0009173 | CASE: Let MATH. The negative case is similar (see comments at the end of the proof of this part). Let MATH and let MATH be the part of the formula for MATH which will not change upon applying MATH, MATH, and MATH (where applicable). Let MATH if MATH and MATH otherwise. Set MATH. First we show that MATH . First, by REF ... |
math/0009175 | The first assertion can be easily checked. The second part follows from the computation in CITE. For completeness sake we give the argument here: Let MATH be the NAME of MATH along MATH. Then MATH has the presentation MATH . Obviously, we have a epimorphism MATH mapping MATH to MATH, MATH to MATH, and MATH to MATH. It ... |
math/0009175 | Observe that if MATH is induced from MATH, that is, MATH (so we can view MATH also as an operator on MATH), then essentially MATH and we deduce that MATH (compare CITE). Therefore, it will be sufficient to find MATH such that MATH. Take MATH of REF . Choosing MATH and MATH, we see that REF is in the spectrum of MATH, a... |
math/0009175 | We perform a standard construction where one attaching map will be given by the MATH of REF , compare for example, CITE. Let MATH be a finite MATH-dimensional NAME with MATH, for example, the MATH-complex of the finite presentation given above. Let MATH be the wedge product of MATH and MATH. The corresponding map MATH ... |
math/0009175 | Choose a finite MATH-dimensional simplicial complex MATH homotopy equivalent to the NAME MATH of REF . Then embed MATH into MATH CITE and thicken MATH to a homotopy equivalent MATH-dimensional compact smooth manifold MATH with boundary MATH, such that moreover the inclusion of MATH into MATH is a homotopy equivalence C... |
math/0009179 | If the lemmas does not holds then there exists critical points in MATH. Let MATH be the closest one of MATH. MATH, otherwise there will be another critical point between MATH and MATH. But this implies MATH, where MATH, which is absurd. |
math/0009179 | If the intersection number is at least REF, the interior of MATH will contain at least two intervals MATH and MATH in the family MATH, for some MATH. Since MATH, for all MATH, the intervals MATH belong to MATH, which is absurd, since MATH satisfies MATH. |
math/0009179 | Let MATH be the maximal interval such that MATH, MATH and MATH is monotone in MATH. We claim that MATH. Otherwise there exists a critical point MATH to MATH in the boundary of MATH such that MATH is in the interior of MATH. Then, by REF , the interior of MATH contains MATH, MATH, in the left or right side of MATH. This... |
math/0009179 | Similar to previous lemma. |
math/0009179 | Immediate consequence of the previous lemmas. |
math/0009179 | This follows of the non existence of wandering intervals. |
math/0009179 | Use that MATH and MATH are bi-Lipchitz in a neighborhood of zero. |
math/0009179 | We will use the smallest interval trick. Let MATH be the smallest interval in MATH. Observe that MATH, where MATH depends on MATH. Taking MATH, the interval MATH satisfies MATH. Consider the maximal interval MATH which contains MATH such that MATH is monotone in MATH and MATH. We claim that MATH. Otherwise, there exist... |
math/0009179 | Let MATH. Near of MATH, we have MATH. By the real bounds MATH . We learn this argument in CITE. |
math/0009179 | Because MATH, MATH will return to MATH after a bounded number of iterations of monotone and folding parts of the MATH-th renormalization. In particular, by real bounds, the derivative of map MATH is bounded away from infinity. This proves the lemma. |
math/0009179 | Assume by contradiction that MATH is very close to a point MATH with MATH. By the real bounds for the partial monotone parts of the MATH-th renormalization, MATH is very close to MATH with respect to level MATH. But MATH returns to MATH after a bounded number of iterates of MATH. Since MATH is deeply inside MATH and by... |
math/0009179 | Let MATH. Let MATH be the closest point of MATH in MATH such that MATH. Note that if MATH is close to MATH, using REF and that MATH, there will be an attractor in the interval MATH which attracts a critical point. This is an absurd. The point MATH cannot be close to the periodic point in MATH, otherwise, by the real bo... |
math/0009179 | Note that MATH is commensurable with MATH. By the real bounds, the map MATH has bounded derivative. This proves the lemma. |
math/0009179 | Let MATH be as in REF . In the proof of REF we proved that MATH contains a MATH-neighborhood of MATH. Let MATH. Let MATH be the pullback of MATH along the k-cycle. By REF , MATH. Let MATH be a connect component of MATH. There are two cases: if MATH is a connect component of MATH then MATH clearly contains an interval i... |
math/0009179 | Note that MATH is the connect component of MATH which contains MATH. Hence MATH is commensurable with MATH and MATH, for some MATH, by REF . Let MATH be the boundary point of MATH inside MATH. Surely MATH. It is sufficient to prove that the interval MATH is commensurable with MATH. Indeed, if MATH is very small, by the... |
math/0009179 | Observe that MATH is not small with respect to MATH. Otherwise, by the real bounds and bounded combinatorics, there will be an attractor in MATH which attracts the critical point MATH. Suppose that a point in the boundary of MATH is close to the boundary of MATH. By the control on the derivatives in the monotone and fo... |
math/0009179 | If the statement does not holds then there exists MATH, for some MATH, very close to MATH with respect to level MATH, hence MATH. After a bounded number of iterations of MATH, MATH is mapped inside MATH. But this is impossible, because MATH is very close to a point in MATH, which never returns to MATH. |
math/0009179 | Since the maps MATH are compositions of univalents maps which map the real line to the real line and the maps MATH belong to the NAME class, we can use REF (see also REF) in pg. REF. |
math/0009179 | Use the control on the first and second derivatives given by complex NAME lemma. |
math/0009179 | If MATH is close to MATH, use REF ; otherwise, capture MATH in a NAME neighborhood whose diameter is commensurable with MATH and use REF . For details, see CITE. |
math/0009179 | (compare with the proof of REF) The pullback of MATH along the MATH-cycle exists. Hence let MATH, MATH, be as in REF . The complex pullback of MATH, MATH, until MATH by MATH is inside MATH, with MATH, by REF . By REF , MATH cuts MATH in commensurable parts, where MATH is the predecessor of MATH at level MATH. Hence , b... |
math/0009179 | Let MATH be the smallest time satisfying the above properties. We have MATH, for some MATH. Then MATH, by REF (use REF successively, and finally REF). Since MATH is bounded, by REF there exists MATH such that MATH. Hence MATH . Since MATH-jumps and MATH, by real bounds, MATH is contained in MATH. By REF , there exists ... |
math/0009179 | (compare with the proof of REF) Let MATH be the maximal interval of monotonicity of MATH which contains MATH. Since MATH is contained in MATH, the interval MATH is contained in MATH. In particular, MATH is very small. Assume that REF does not hold. By REF , MATH never MATH-jumps for MATH. Hence, by REF , MATH, with MAT... |
math/0009179 | (compare CITE, subsection REF) Let MATH, with MATH, be such that MATH. In other words, MATH is in the 'scale' of level MATH. If MATH, we are done by REF . Otherwise MATH. Note that MATH, where MATH and MATH. If MATH is zero, using arguments as in the proof of REF , MATH is well defined and it is contained in MATH, wher... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.