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math/0012231 | We shall reduce REF-Partition (see CITE) to this problem. Let MATH be an instance of REF, with the MATH positive integers summing to MATH. Then MATH can be partitioned into MATH sets each consisting of REF elements with sum MATH if and only if the root system MATH embeds into the root system MATH. |
math/0012233 | Since MATH when MATH, we have: MATH . It follows that MATH is bounded. To prove that MATH is bounded, we may assume without loss of generality that MATH. For any MATH, let MATH be the unique MATH such that: MATH . Since MATH, there exists a constant MATH such that MATH for any MATH. Set MATH. From the inequality MATH we deduce by letting MATH: MATH . We claim that this implies the existence of a constant MATH such that: MATH . Since REF is obvious when MATH is bounded, we may assume again without loss of generality that it is not the case, and hence MATH since MATH is non-decreasing. Assume that REF is false. Then there exists for any integer MATH a real number MATH such that: MATH . It follows that MATH when MATH and hence MATH. Write MATH where MATH. We have: MATH a contradiction. So, REF is true and we have for MATH: MATH a fact which implies that MATH is bounded. Let us set, for any MATH: MATH . We thus define positive linear forms on MATH and we claim that MATH . For simplicity, set MATH and MATH for MATH. Since we have MATH, there exists for any MATH a constant MATH such that MATH for any MATH. Hence: MATH . For any MATH with MATH, we deduce that: MATH and hence, since MATH when MATH: MATH where MATH. But we have: MATH so that we get by letting MATH: MATH . On the other hand, since we have MATH (compare CITE[REF , p. REF]), we get: MATH . The last equality is deduced from the fact that CITE[REF , p. REF]: MATH . We thus have: MATH and hence: MATH . This proves REF. To show that MATH and achieve the proof of the theorem, it suffices to prove that MATH. But we have by CITE[p. REF]: MATH so that we get by taking MATH and MATH: MATH . It follows that: MATH . |
math/0012233 | This theorem easily follows from the equivalence, for any positive and non increasing function MATH on MATH such that MATH for any MATH, of the two following assertions: CASE: MATH when MATH; CASE: MATH when MATH. The proof of this equivalence uses classical abelian and tauberian theorems. For completeness, we give a proof based on REF (see also CITE). CASE: Assume that MATH, and make the change of variable MATH in the integral defining MATH. We get MATH where MATH . The function MATH is entire, while MATH only converges for MATH and satisfies MATH . Setting MATH we then obtain MATH . Then MATH is a convergent integral for MATH such that MATH. By the NAME tauberian theorem, we get: MATH and hence: MATH . The result follows from REF . CASE: Let us first assume that MATH. Fix MATH with MATH and choose MATH such that: MATH . We have for any MATH, MATH . If these inequalities were true for any MATH we would have: MATH where MATH are non increasing positive functions. Using NAME 's inequality we would get: MATH and hence by letting MATH: MATH . Setting MATH with MATH and multiplying the above inequalities by MATH, we get the result when MATH. Now since the inequaliy REF is only true for MATH, we take MATH and MATH equal to MATH and MATH for MATH, and for MATH: MATH . Since MATH and MATH are nonincreasing positive functions such that: MATH we now use NAME 's inequality to get the conclusion. Indeed, the additional constants that arise with these modifications do not change the final computation of the residue of MATH at MATH. If MATH the same proof still works, replacing MATH by the zero function. |
math/0012233 | The algebra MATH lies in MATH, where MATH is as before the unbounded derivation MATH. We have for any MATH: MATH . Assume first that MATH is an integer. Then MATH and MATH. Therefore we get: MATH . On the other hand, for any MATH, the operator MATH is trace-class since MATH and MATH are bounded operators, and MATH is trace class. Therefore MATH . The proof is thus complete when MATH is an integer. Now if MATH, we choose an integer MATH and a real number MATH such that MATH. Then an easy computation shows that we have: MATH . Therefore, it suffices to show that MATH, where MATH, MATH and MATH. But the operator MATH is bounded. Indeed, one can for instance use the integral expression of MATH given by: MATH to deduce that for MATH, the operator MATH is bounded with norm MATH, while for MATH, one can use the relation MATH to conclude that MATH is bounded as allowed. Now the end of the proof goes as follows. By CITE[REF , p. REF], we have: MATH and hence: MATH . The last equality is a consequence of the summability of the operator MATH. |
math/0012233 | Let us first prove that the function MATH is well defined and bounded. Since we have for any MATH, MATH we deduce that: MATH . There exists a constant MATH such that MATH, therefore we have for any MATH: MATH . We thus deduce from REF that MATH and hence: MATH . Thus, MATH makes sense for any MATH, and defines a linear form MATH on MATH. To prove the corollary, we may assume without loss of generality that MATH. In this case, MATH is a positive NAME measure with support in MATH. Moreover, since MATH for any MATH, we have by REF : MATH and the scale invariance of MATH implies that MATH. By REF again, we get: MATH and the proof is complete. |
math/0012233 | Let MATH be the functions defined on MATH by: MATH . These functions are then continuous on MATH and vanish at infinity. Moreover, MATH so that MATH is a MATH function satisfying MATH . Let us first prove that the function MATH is well defined and bounded for MATH. We set for any MATH, MATH. We thus define a non decreasing continuous function such that MATH, and hence: MATH . But there exists a constant MATH such that MATH, and since MATH is non decreasing, we get: MATH where the last equality follows from the change of variable MATH. We deduce that MATH is bounded, and the inequality MATH shows that the function MATH is well defined and bounded. On the other hand we have for any MATH: MATH and hence MATH . By using the equality REF, it is easy to check that the right hand side is a NAME integral of continuous functions with values in MATH. On the other hand, MATH is a continuous form on MATH and so: MATH with MATH. Since MATH is bounded, we get: MATH . Now by REF , the left hand side is equal to MATH . We thus get: MATH and the proof is complete. |
math/0012233 | Take MATH in the previous theorem. |
math/0012233 | Since MATH, we have MATH. But MATH as MATH by hypothesis, so that there exists MATH such that MATH and hence MATH. |
math/0012233 | Let MATH be as in REF . The projections MATH and MATH are MATH-compact, and REF gives the result. |
math/0012233 | If MATH and MATH are parametrices for MATH and MATH respectively, then MATH is a parametrix for MATH. So the composite of two MATH-Fredholm operators is a MATH-Fredholm operator. In addition: MATH . |
math/0012233 | For any MATH, let MATH be the spectral projection of MATH corresponding to the interval MATH. Since we have MATH, for any MATH, the map MATH extends to a contraction MATH. Set MATH for MATH. We thus define an operator MATH which satisfies by construction: MATH . From this relation, we deduce that MATH and hence MATH on MATH. It follows that MATH and, since MATH is a contraction, it is an orthogonal projection in MATH such that MATH. Moreover, we have: MATH . From the inequality: MATH for any MATH, we get the existence of an inverse MATH for MATH, such that MATH. We have MATH and hence, since MATH: MATH . Since MATH is injective, MATH strongly when MATH, and it follows from the relation MATH that MATH strongly when MATH. By the NAME dominated convergence theorem in MATH, we deduce from REF that: MATH and finally MATH. The proof is complete. |
math/0012233 | The operator MATH is MATH-Fredholm because MATH. To prove the proposition, we may assume that MATH. Indeed, let MATH be such that: MATH are in MATH (and hence in MATH for any MATH) and set: MATH . We have: MATH where the first relation is immediate and the second one uses the equality MATH. Replacing MATH by MATH, MATH by MATH and MATH by MATH, we are thus reduced to the case where MATH. When MATH and MATH are in MATH, we get from the relations MATH and MATH the equality: MATH . To prove that MATH, it thus suffices to show that: MATH where MATH and MATH. To this end, set MATH. We clearly have: MATH and hence: MATH . If the intertwining operator MATH from MATH to MATH were invertible (the inverse would then be automatically in MATH by the bicommutant NAME theorem), we would get by cyclicity of the trace: MATH and Calderon's formula would be proved. Although MATH is not necessarily invertible here, it is injective with dense range from MATH to MATH. It turns out that this is enough to prove REF, by REF , and therefore REF is proved. |
math/0012233 | CASE: Trivial. CASE: Note first that MATH (and hence MATH) belongs to MATH, for any MATH. Indeed, we have: MATH . On the other hand we have MATH for any MATH. Indeed we get by easy computations: MATH where MATH by the previous observation. Moreover, we have: MATH and hence MATH. It follows that MATH and finally that: MATH as allowed. But we have: MATH . Since MATH and MATH are in MATH, we finally get: MATH and REF is proved. CASE: Obvious. CASE: We have MATH and hence MATH . It follows that: MATH and the result follows since we know that MATH. |
math/0012233 | The proof follows the lines of CITE. CASE: We first point out that MATH is in evidence a cyclic cochain. Let MATH. Then we have: MATH . Therefore the cochain MATH is a NAME cocycle on MATH. From REF , we deduce that the operator MATH is MATH-Fredholm in MATH with parametrix given by MATH. Moreover, MATH as well as MATH are in MATH. Therefore REF gives: MATH . Computing MATH and using the relation MATH, we obtain: MATH and hence the conclusion. CASE: That MATH is cyclic is again obvious. Let MATH. Then we have: MATH . Hence MATH is a cylic cocycle on MATH. To compute the MATH-index of MATH, we again apply the Calderon formula. From the relations REF, we deduce that MATH is MATH-Fredholm in MATH with parametrix given by MATH. Moreover, MATH and MATH are in MATH. Therefore REF gives: MATH . The computation of MATH in REF gives: MATH . But, MATH . On the other hand we have: MATH . Therefore, MATH and a similar result holds for MATH and we get: MATH . Hence we get using the trace property of MATH: MATH and thus we finally obtain: MATH which completes the proof. |
math/0012233 | CASE: By CITE[page REF], we have MATH where MATH is an infinitely smoothing MATH-operator and each MATH is given by a continuous familly with compact support in MATH with MATH a distinguished foliation chart trivializing MATH. Since MATH is trace-class with respect to MATH, CITE[REF , page REF] and MATH we may assume that MATH where MATH is a distinguished foliation chart trivializing MATH. We thus may work locally assuming that MATH is foliated by MATH, for MATH and MATH is a continuous familly of scalar pseudodifferential operators of order MATH on MATH (the proof for matrices is the same). Here MATH is the standard MATH-torus and MATH is the unit disk in MATH. For any MATH, let MATH be the usual Laplacian on the flat torus MATH. Since MATH is an ideal in MATH, we only have to show that the constant familly MATH defines an element in MATH. Indeed we have MATH where MATH is a continuous family of REF pseudodifferential operators on MATH, and hence defines an element of MATH by CITE[ page REF ]. Since we trivially have for any MATH: MATH we get: MATH . The right hand side of this equality converges to MATH . Henceforth, MATH belongs to MATH and we get MATH . CASE: We may work locally and assume again that MATH. For any smooth function MATH, set MATH where MATH is any classical tangential pseudodifferential operator of order MATH with principal symbol equal to MATH. Since two classical pseudodifferential operators of order MATH with the same principal symbol coincide modulo MATH and since MATH we deduce that MATH is well defined. It is clear that MATH is a positive linear form on MATH and is in fact a positive measure on MATH. Let MATH be the disintegration of MATH with respect to the projection MATH CITE[page REF]. For any isometry MATH of MATH, the measure MATH is invariant under the action of MATH on the fibers of MATH because MATH is a trace. By uniqueness of the disintegration of MATH we get MATH so that MATH is proportional to the volume form on MATH for almost every MATH. We thus have MATH where MATH is a bounded MATH-measurable positive function on MATH. Let us prove now that the measure MATH is proportional to MATH. For any continous function MATH on MATH, we know by REF that the NAME trace of the continuous family MATH where MATH is now the Laplacian on the standard sphere, is given by MATH the constant MATH being independent of MATH. On the other hand, we have MATH where the constant MATH does no more depend on MATH. We thus get the existence of a constant MATH such that MATH . It follows that MATH and the computation of REF shows that MATH. |
math/0012233 | We only have to prove REF , the rest of the proposition being a rephrasing of REF in the present situation. We first point out that MATH is affiliated with MATH and that MATH is in MATH because it is affiliated and bounded. On the other hand the principal symbol of MATH commutes with those of all order REF pseudodifferential MATH-operators, so that REF are satisfied. Let now MATH be a parametrix for the elliptic MATH-operator MATH so that MATH are regularizing operators, say live in MATH. The existence of MATH is proved in CITE. Then we have MATH so that MATH and MATH are in the same NAME ideal. But MATH and the conclusion follows. |
math/0012233 | Forget MATH but keep it in mind! Set MATH. The function MATH converges simply to REF when MATH goes to MATH, and by the NAME theorem, we deduce that MATH converges weakly to the identity operator. Since MATH is a normal trace and MATH is trace-class, we have: MATH . But, MATH and since MATH is an even function, we also have MATH. Hence we deduce: MATH . Now the operator MATH belongs to MATH for any MATH and the operator MATH commutes with the grading involution MATH, thus: MATH . Set MATH for the derivation induced by MATH on MATH and let us apply the operator MATH to the equality MATH. We get: MATH which finishes the proof. |
math/0012233 | Since MATH is even, this lemma only involves MATH and we can assume MATH. Let us first formally replace MATH by MATH. We obtain: MATH . But setting MATH if MATH and extending MATH to an even function, we obtain a well defined function MATH satisfying the assumptions of REF . So: MATH and the proof is complete. It remains thus to show that: MATH belongs to MATH and that MATH is a MATH. We now use the NAME transform: MATH . But recall that MATH, and so: MATH . Thus MATH . Note that MATH is bounded since we could assumed MATH in this proof. Thus MATH. On the other hand, we have: MATH where MATH is some constant. Let us see that there exists a constant MATH such that MATH . To this end, let MATH be such that MATH and denote by MATH the spectral projection of MATH corresponding to the interval MATH. We have MATH . Since there exists a constant MATH such that MATH, we get MATH and hence MATH. But, MATH . The first and the last inequalities follow from CITE[REF , page REF]. Since MATH, there exists a constant MATH such that MATH and hence MATH . It follows that MATH and hence MATH when MATH. |
math/0012233 | Let MATH be an even function in MATH which equals REF in a neighborhood of REF as above. Then REF shows that MATH . REF then gives with the bounded operator MATH: MATH . This completes the proof. |
math/0012233 | CASE: We have by straightforward computation: MATH . But MATH belongs to the ideal MATH, since we have: MATH and MATH . In the same way, we have: MATH . But again, MATH belongs to MATH and thus the operator MATH is trace class. Therefore MATH. Thus, MATH. The proof for the cochains MATH is similar. Now we obviously have: MATH . CASE: We first point out that MATH . Therefore, in the expression of MATH we can move MATH to the left and in particular: MATH . On the other hand, MATH . Now, since MATH, we deduce MATH . Therefore and since MATH is bounded, the operator MATH belongs to MATH. Therefore, MATH . Hence we can replace MATH by MATH when necessary in the expression of MATH in REF . On the other hand, we have: MATH . Thus and since MATH is cohomologous to MATH, we deduce that: MATH where MATH corresponds to the terms where the factor MATH appears at least twice. The first remark is that MATH . Therefore, we have: MATH . This latter is nothing but a representative for MATH for any MATH. Therefore: MATH for some cochain MATH. To finish the proof, we thus need to show that MATH is a coboundary. But this is a consequence of the fact that MATH in NAME cohomology. More precisely, consider for instance the NAME cocycle MATH . Then MATH can be represented for MATH by the NAME cocycle MATH . But again, this is precisely, MATH and hence corresponds to the MATH-term in the expression of MATH. Thus all the terms where MATH appears twice are coboundaries. The same argument using the NAME cocycles MATH shows that all the other terms in MATH are coboundaries. The proof is thus complete. |
math/0012233 | If we compute the NAME character of MATH and integrate it against the NAME current then we obtain the pairing of our NAME cocycle (which is cyclic here) with the MATH-theory of MATH by REF . Now REF enables to conclude. |
math/0012233 | The first interpolation formula is proved in CITE, page REF and the third equality also goes back to CITE. |
math/0012233 | By imbeding MATH in MATH and using the simple fact that MATH we can assume that MATH has no minimal projections. If MATH are two projections in MATH such that MATH and MATH then the projection MATH belongs to MATH and satisfies MATH. We thus have MATH thus REF gives the first inequality. The second one follows similarily from REF , see for instance CITECITE. |
math/0012234 | We simply apply the proof in CITE[page REF] and see that the cochain constructed in this way is MATH-equivariant whenever MATH is. More precisely, let MATH be a MATH-invariant element, say MATH . Set for MATH, MATH . MATH is then a NAME cochain which is MATH-equivariant by direct inspectation. If MATH and MATH then MATH is a MATH-equivariant (cyclic) cochain, since MATH. Following CITE[page REF], we verify using MATH that: MATH . Thus we obtain MATH with MATH a MATH-equivariant cyclic cochain on MATH. Let now MATH be an invertible MATH-invariant element of MATH. Then again following CITE[page REF] and by classical equivariant MATH-theory arguments, we can assume that MATH where MATH is MATH-invariant. In fact there exists MATH such that MATH and hence MATH is well defined using the exponential in MATH and we obtain by straightforward computation: MATH and if MATH then MATH . The proof is then complete since MATH is MATH-equivariant. |
math/0012234 | Using classical MATH-theory techniques, we can assume that MATH is unital (with a trivial action of MATH on MATH). Also we only need to show that if MATH are MATH-conjugated, then MATH . By REF , there exists an odd MATH-equivariant cyclic cochain MATH such that MATH . A direct computation then shows that all but one of the terms in MATH disappear and we get: MATH . The MATH-equivariance of MATH then yields, MATH and since MATH is MATH-invariant, we obtain the result. The above computation also shows that the pairing only depends on the MATH-equivariant cyclic cohomology class of MATH. Furthermore, the pairing respects the ideal MATH, for: If MATH is an other irreducible representation of MATH then we obviously have MATH . |
math/0012234 | Let MATH be any MATH-projection with MATH a finite dimensional representation of MATH. Denote by MATH the MATH-invariant NAME operator MATH. Then MATH is a parametrix for MATH modulo MATH which is in addition MATH-invariant. By the equivariant Calderon formula (See REF below), we obtain: MATH . The computation of MATH in the non equivariant case CITE[page REF] enables then to conclude. |
math/0012234 | The proof in the non equivariant case generalizes straightforward (See for instance CITE [page REF]), so we will be sketchy. We set MATH and MATH. Let MATH and MATH be the projections on the eigenspace associated with REF for MATH and MATH respectively. These are defined for instance using NAME integrals around REF. Then MATH and MATH are MATH-invariant projections and MATH, MATH preserves their domains. We also have MATH . This enables to restrict the NAME operator MATH to MATH so that it realizes a MATH-equivariant isomorphism between MATH and MATH. Hence: MATH and thus, MATH . On the other hand, in restriction to MATH and with values in MATH, MATH intertwines MATH and MATH, for any MATH and so the conclusion. |
math/0012234 | CASE: Straightforward computation. In particular one easily sees that MATH does not depend on MATH for MATH. CASE: Consider MATH then REF in the definition of a precycle are satisfied and this is already proved in [REF, page REF]. Recall that the gobal NAME differential decomposes with respect to the non canonical (since MATH is not a subbundle of MATH) isomorphism MATH induced by MATH in the following way CITE MATH where MATH is the longitudinal differential and MATH is a component of degree MATH that brings the non integrability of the bundle MATH. It is a well known fact that if MATH is the smooth section of MATH defined by MATH where MATH is the projection onto MATH along MATH and MATH is the identification MATH, then MATH is exactly the interior product by MATH or rather its MATH-component. Note that although MATH, MATH takes into account the component MATH. Computing MATH we obtain the following equality MATH . We can therefore apply [REF, page REF ] and the method that enables to construct a cycle out of a precycle so that we only have to prove that MATH given by MATH satisfies REF . Take MATH and MATH then using a partition of unity on MATH we can assume that MATH where MATH and MATH are distinguished open sets and MATH as in CITE. Then MATH where MATH corresponds to MATH under the holonomy transformation induced by MATH. Now the action of MATH consists in integrating first in the leaf direction (this corresponds for fibrations to a graded trace CITE ) and then applying MATH. Thus we get MATH . Now a variant of the NAME theorem and because MATH is holonomy invariant (MATH), we obtain MATH . Note that the MATH obtained here is not an element of MATH but of MATH, the curvature method works as well CITE . Let now MATH and let us prove that MATH. Because MATH is holonomy invariant we see that MATH with obvious notations. Then once again we localize to a distinguished chart in MATH and notice that integration over the plaques commutes with the transverse differential up to longitudinal differentials so that the two conditions MATH and MATH enable to conclude. |
math/0012234 | For holonomy invariant distributions REF is clear and the proof for measures easily generalizes to deal with distributions. Let us then restrict to the case where dim-MATH as a current. Let MATH be any MATH-equivariant transverse distribution constructed for instance using any MATH-invariant metric on MATH. Then the differential MATH is the MATH component of the NAME differential in the decomposition MATH induced by MATH. Let MATH and let MATH, denote as before respectively by MATH and MATH, the smooth kernels associated with MATH and MATH for any MATH. Assume for simplicity that MATH is orientable with a chosen orientation. Fix MATH and following the notations of CITE let MATH be a local trivializing chart of MATH around MATH. If MATH then its differential in the transverse direction can be described as follows: We use the trivialization MATH to deduce a trivialization of MATH that identifies it with a copie of MATH in MATH. MATH is then a smooth function with compact support on the trivial groupoid MATH, where the last copie of MATH is the one corresponding to MATH in MATH. The differential MATH of MATH in the transverse direction MATH yields an element of MATH. Finally, MATH is the differential MATH in the direction MATH. Since MATH is defined globally as the projection of of the DeRham differential onto MATH, the local expression is independent of the chart used to define it. Now, for a trivialization of MATH around MATH, we chose as a local neighborhood MATH which is diffeomorphic to MATH with again MATH a q-plane in MATH corresponding to MATH in MATH. If MATH with MATH then MATH and MATH induces a germ of holonomy that is exactly the action of MATH on MATH thanks to REF . This shows that the transverse copie of MATH in the trivialisation MATH of MATH coincides with that of MATH and hence MATH . So MATH . Now recall that MATH . Let MATH and assume that MATH where MATH and MATH where MATH and MATH correspond to the decomposition induced by MATH. The isometry MATH transforms MATH onto MATH with MATH, in the same way we have MATH with MATH. Let us compute MATH using the local trivializations, we obtain MATH . Thus we obtain the following equality MATH . In the same way we show that MATH . Now if MATH is any twisted basic current then it is invariant under holonomy and so MATH . It remains to treat the terms involving the curvature MATH in MATH . These terms are of the form MATH where MATH contains factors with MATH. Thus REF and hence the above proof works again. |
math/0012234 | Using REF it is sufficient to prove that all the basic cocycles are MATH-equivariant. But this is the content of REF . |
math/0012234 | This is an immediate consequence of the NAME theorem together with the additivity of MATH. |
math/0012234 | Let MATH be an open tubular neighborhood of MATH in MATH so that the fibres are included in the leaves of MATH and let us identify as usually MATH with the normal vector bundle to MATH in MATH. Recall that the NAME extension is the composition of a NAME map MATH with the extension MATH . Thus thanks to the excision property, we only need to prove that the basic cocycles do commute with any NAME map. In CITE it is proved that for vector fibrations the curvature MATH of the vector bundle MATH is killed in the computations. Now if MATH satisfies MATH (for a fiberwise measure MATH) then it realizes a NAME map by setting CITE : MATH . In this formula, MATH and MATH is its projection in MATH. Let MATH be a closed twisted basic current and let MATH be the associated MATH-cyclic cocycle. Let MATH, then the transverse differential satisfies MATH . Now MATH, MATH . Thus if MATH, then we obtain MATH . All the terms arising from the transverse differential of MATH are annihilated by REF above and we obtain the result. |
math/0012234 | We will use REF , which can be summarized by the commutativity of the following square MATH and the fact that MATH is a MATH-isomorphism with inverse MATH. Recall that MATH is the NAME extension as described in CITE . We deduce: MATH . On the other hand, we have defined MATH as the image of MATH by the additive map MATH . So, we get: MATH . Let us prove now that MATH commutes with the NAME extension MATH, say that: MATH . This is a consequence of the MATH-equivariance of the NAME projection. More precisely: Let MATH be a MATH-stable tubular neighborhood of MATH in MATH which is diffeomorphic to MATH. We denote as before on MATH and MATH the foliations inherited by MATH and MATH from the original foliation MATH. Recall that the NAME system MATH on MATH is given by the pullback MATH of a NAME measure MATH on the leaf manifold MATH (say: MATH with a discrete transverse topology). Let the restriction of MATH to MATH be the product of a leaf-measure MATH on MATH by a fiberwise NAME measure MATH, and choose MATH, MATH such that: MATH . Then with MATH given by: MATH we have MATH . The NAME isomorphism MATH is then induced by the tensor product with any NAME rank REF projection, and is easily described by: MATH where MATH is the projection of MATH under MATH. MATH is well defined and compactly supported, and it brings MATH-half densities into MATH-half densities so that using REF we only need to prove that: MATH . But a straighforward computation of the kernels yields to the same result. Now, we obtain MATH and so, MATH where MATH is the evaluation at MATH, given by MATH. We can also define the evaluation MATH that we denote by MATH. Hence if MATH is the extension of MATH to the tensor product by MATH, then we get: MATH . |
math/0012234 | Let MATH be the map from MATH to MATH given by: MATH . Then, the NAME 's index theorem for the cyclic cocycle associated with MATH is (up to a sign that we include in the definition of the NAME character) exactly the equality: MATH . The chern character can be extended to MATH, with values in MATH by the map MATH given by: MATH . Hence, if we trivially extend MATH to MATH, then: MATH . On the other hand, MATH and MATH . So we obtain: MATH and finally, MATH . |
math/0012234 | Here MATH, so: MATH and hence, the corollary is proved. |
math/0012234 | See CITE . |
math/0012235 | Since MATH is characteristic, we have MATH and the assertion follows. |
math/0012235 | Let MATH and denote by MATH the universal covering. This manifold is obtained from the universal covering MATH of MATH by adding copies of MATH, one at each preimage of the point at which we performed the connected sum of MATH and MATH. It follows easily from the obvious NAME - NAME sequence that we have a splitting MATH which is orthogonal with respect to the intersection form. The action of the deck transformation group on the second summand (from the left) is given by MATH, and the restriction of the intersection form MATH with values in the group ring to the second summand is completely described by MATH, where the dot denotes the intersection product on MATH. Now choose some MATH such that MATH. Consider the element MATH. An easy computation shows that MATH and MATH. If MATH denotes the projection, we have MATH. The element MATH is contained in the first summand of the above decomposition. Choose a map MATH in the homotopy class MATH. Then MATH, and, since MATH, we have MATH. |
math/0012235 | Let MATH and denote by MATH the universal covering. Again we will make use of the decomposition MATH . As in the proof of REF , we only have to find an element MATH in the second summand of this decomposition such that MATH and MATH. For this purpose, consider the elements MATH in the group MATH, here we use, as usual, the basis of this group given by the homology classes of the factors. Then MATH, MATH and MATH. Now let MATH . Then MATH. Furthermore, a short calculation shows that MATH, and the proof is complete. |
math/0012235 | MATH : Assume that we are given an immersion MATH as in the statement of the proposition. Pick a representative of MATH in the group ring MATH and let MATH denote its mod - REF reduction. By assumption, the image of MATH under the canonical map MATH is zero. In the non - homogeneous description of the standard chain complex of MATH with MATH - coefficients, the group MATH of one - cycles consists of all finite linear combinations MATH, MATH (see for instance REF). The boundary of an element MATH is given by MATH . If we use the identification of MATH and MATH as MATH - modules, induced by MATH, the kernel of the map MATH is identified with the kernel of the projection MATH which is by definition the subgroup generated by all boundaries MATH. Consequently, the element MATH is a sum MATH for some elements MATH. By REF , we can therefore construct an element MATH such that MATH and MATH. This implies that there is a representative MATH of MATH in MATH such that all the coefficients MATH are even numbers. By REF , we can now find an immersion MATH for some MATH which represents MATH and has reduced self intersection number zero. The result then follows from REF . MATH : clear, since MATH is spin, so the class MATH is characteristic. MATH : Assume that we are given a MATH - null embedding MATH representing a class MATH, where MATH is characteristic. Let MATH and denote by MATH the universal covering. By REF , there is an immersion MATH representing MATH such that MATH. Let MATH denote the corresponding homology class. Then MATH. We have a decomposition MATH . Let MATH be the corresponding decomposition of the element MATH. Using that MATH, it is easy to see that MATH, and clearly MATH, MATH. If we can prove that MATH, we obtain MATH and any immersion corresponding to MATH will fulfill our requirements. In order to show that MATH, write MATH, where MATH and MATH. An easy calculation shows that MATH. Hence we have MATH . Now we know that MATH is characteristic, that is, MATH is a characteristic class for MATH. Using REF , we therefore obtain (for fixed MATH) MATH since MATH is characteristic. As this is true for every MATH, all the coefficients in the above expression are even and therefore MATH as desired. |
math/0012235 | First, we show this in the special case that MATH is a connected orientable surface. Then MATH, so MATH is the multiplication by some number MATH. We have to prove that MATH. For this purpose, suppose that MATH is some element in MATH. Since MATH, we have MATH. Now it is an immediate consequence of NAME 's Theorem that the second NAME - NAME class of a REF - manifold and its orientation class are related by MATH where MATH denotes the homology operation dual to the second NAME square and MATH is the mod - REF orientation. By naturality, we obtain that MATH since MATH. This implies that every element in the image of MATH is divisible by MATH, hence MATH must be even. So we are done if we can show that every multiple of MATH is in the image of MATH. Suppose that MATH. Let MATH and let MATH denote the projection to one factor. Then, by the NAME Theorem, MATH is onto. Pick a preimage MATH of MATH and let MATH. Since MATH is spin (and hence the stable normal bundle is spin), there exists a normal MATH - structure MATH on MATH having first NAME class MATH, hence MATH. Moreover MATH=REF, so MATH, and MATH defines an element MATH. By definition, MATH. So we have proved that MATH is in the image of MATH, and this completes the proof of the lemma in the special case that MATH is a surface. As to the general case, note that for every MATH, there is a connected oriented surface MATH and a map MATH mapping the orientation of MATH to MATH. Since MATH is the multiplication by MATH on MATH, we have MATH and this proves the lemma in the general case. |
math/0012235 | Suppose we are given some MATH and some MATH mapping to MATH under MATH. By definition, the subgroup MATH is just the first stage MATH of the filtration of MATH used in the AHSS converging to MATH, and MATH. Now suppose that MATH. Recall that MATH by REF . We can find a map MATH such that MATH, and MATH can be chosen to be cellular. Let MATH denote the terms of the AHSS for MATH. Again we have MATH, so MATH maps MATH to the non - zero element of MATH. Now MATH, hence we obtain that MATH. On the other hand, we have a commuting diagram MATH and as MATH, we have MATH, in particular MATH. This shows that MATH, a contradiction. |
math/0012235 | By REF , we only have to show that the image of MATH under the map MATH is zero. Let MATH denote the terms of the AHSS for MATH and denote by MATH the terms of the spectral sequence for MATH. The CW - decomposition of MATH yields filtrations MATH of MATH and MATH of MATH, and we have a filtration preserving homomorphism MATH. As the image of MATH in MATH is zero, MATH (recall that MATH). Let MATH denote the image of MATH. We claim that MATH, that is, that the induced map MATH maps the equivalence class MATH of MATH to zero. In fact, MATH is a quotient of MATH and, by REF , MATH, so we have a commuting diagram: MATH . Pick a lift MATH of MATH. According to the definition of MATH and by REF , we have MATH, so MATH has order at most MATH. Now REF implies that the upper horizontal arrow is the multiplication by MATH. Hence it maps MATH to zero and we obtain that MATH, as claimed. Now, by definition, MATH is simply the image of MATH, where MATH denotes REF - skeleton of MATH. Hence, by REF , we can conclude that actually MATH, and the lemma is proved. |
math/0012235 | It is immediate from the definition of MATH that we have the inclusion image MATH kernel. So assume that we have a class MATH such that MATH. Let MATH be a triple representing MATH. As in the proof of REF , we then have MATH . But by REF , the image of MATH in MATH is precisely the kernel of MATH, and therefore we can find an element MATH having the same image in MATH as MATH. Let MATH denote the image of MATH in MATH. Then MATH is in the kernel of MATH and MATH. By REF , we can conclude that MATH is in the image of MATH, and since MATH by construction, we are done. |
math/0012235 | Since a MATH - null embedding can be lifted to the universal covering, it is clear that every class which can be stably represented by a MATH - null embedding is spherical. To prove the converse, assume that we are given a REF - manifold MATH and a spherical and characteristic class MATH. First let us assume that there is a spherical class MATH such that MATH. We will use the abbreviation MATH. Pick a map MATH inducing an isomorphism MATH and choose a normal MATH - structure MATH on MATH having first NAME class MATH. By definition, the homomorphism MATH maps the bordism class MATH to MATH. As MATH is spherical, this is zero. Hence REF yields a MATH REF - manifold MATH and a spin REF - manifold MATH together with a map MATH such that MATH . Here MATH denotes the MATH - structure induced by the spin structure on MATH. Note that MATH. We can also assume that MATH is simply connected and that MATH is an isomorphism. Furthermore we can arrange for MATH to be a primitive class by adding a copy of MATH with the unique MATH - structure having NAME class MATH. Let MATH and consider the obvious fibration MATH. The normal MATH - structure MATH and the map MATH define a MATH - structure MATH on MATH. It is not difficult to see that the existence of the class MATH implies that MATH is a REF - equivalence. Similarly the MATH - structure MATH on MATH obtained by gluing MATH and MATH and the map MATH which is MATH on MATH and trivial on MATH define a MATH - structure MATH. As the class MATH is primitive and MATH is an isomorphism, this MATH - structure is also a REF - equivalence. Of course the bordism group of MATH - structures is simply MATH, and therefore the two MATH - structures MATH and MATH are cobordant. REF now implies the existence of numbers MATH and of a diffeomorphism MATH compatible with the normal MATH - structures obtained by MATH respectively MATH and the canonical MATH - structure on MATH. In particular, MATH maps MATH to MATH. By construction, MATH has support in the simply connected part MATH and can therefore be represented by a MATH - null embedding. Pulling back this surface via MATH we obtain a MATH - null embedding MATH representing MATH as desired. As to the general case, assume that MATH is a spherical and characteristic homology class. Denote the usual generator of MATH by MATH and consider REF - manifold MATH. Let MATH. The homology class of a generically embedded MATH in MATH is spherical and has intersection number one with MATH. By what we just proved, this implies that MATH can be stably represented by a MATH - null embedding. As MATH is characteristic, REF can be applied and we obtain that also MATH can be stably represented by a MATH - null embedding. |
math/0012235 | By REF , a characteristic class MATH which can be stably represented by an embedded sphere fulfills MATH. Let us now assume that conversely, the congruence MATH holds. By REF we can assume that the class MATH can be represented by a smoothly embedded surface MATH such that MATH is trivial. By REF, the NAME invariant of MATH vanishes. Now we could proceed as in the proof of REF to obtain an embedded sphere in MATH for some MATH representing MATH. However, there is a slightly different argument using transversal spheres instead of framed surgery along circles. Since the NAME invariant of MATH is zero, we can find a homologically non - trivial circle MATH with MATH, here our notation is the same as in CITE. Pick an embedded disk MATH with boundary MATH whose interior intersects the surface MATH transversely with algebraic intersection number MATH. After performing boundary twists, we can assume that MATH. Fix a framing of MATH in MATH and let MATH denote the integer valued obstruction to extending this framing to a section of the normal bundle of MATH. According to the definition of MATH, we have MATH, hence MATH is even. Now we will use transversal spheres to remove the intersection points of MATH and MATH, as follows. After passing from MATH to MATH and attaching one of the factors to MATH (note that this does note change MATH, since the attached sphere has trivial normal bundle), we can assume that MATH has a transversal sphere which does not meet MATH, namely the second factor of MATH. We can use parallel copies of this transversal sphere to remove all the intersection points between the interior of MATH and MATH. As MATH, the homology classes of all these copies add up to zero. Hence we obtain a new surface MATH which still has homology class MATH and genus MATH such that MATH intersects MATH only in MATH. Since the surface MATH is obtained from MATH by cutting out disks and gluing in other disks instead, the circle MATH is still non - trivial. Furthermore, we did not change the disk MATH, and hence the obstruction to extending a framing of MATH over MATH is still MATH. The arguments given so far show that we can arrange for the interior of the disk MATH to be disjoint from MATH. After adding another copy of MATH, we can find an embedded sphere MATH with self intersection number MATH disjoint from MATH and MATH, note that MATH is even. Tubing this sphere into the disk MATH gives a disk MATH with boundary MATH, still disjoint from MATH, such that the framing obstruction vanishes. By doing surgery along MATH, we can now obtain a new surface representing MATH with genus MATH. Repeating the argument, we can therefore construct an embedded sphere as desired. |
math/0012235 | Again it is clear that a class which can be stably represented by a MATH - null embedding is spherical. Now let MATH be a REF - manifold with a characteristic spherical class MATH. First let us suppose that the NAME - NAME invariant MATH is zero. Then there is some MATH such that the manifold MATH is smoothable. Applying REF to this manifold shows that the class MATH can be stably represented by a MATH - null embedding. By the very definition of ``stably representable", the same is true for MATH. In the case that MATH, consider REF - manifold MATH and the characteristic class MATH. Then MATH, and by the first part of the proof, we can represent the class MATH stably by a locally flat MATH - null embedding. By REF , the same is true for MATH. |
math/0012237 | Let MATH be the first time node MATH switches from `off' to `on', and let MATH be the first time it receives a signal. Further, let MATH be the MATH-th time node MATH receives a signal, and let MATH, MATH. The random variable MATH is exponentially distributed with parameter MATH. Furthermore, the random variables MATH, MATH are i.i.d. and also independent of MATH. So we have: MATH . |
math/0012237 | We use stochastic domination in proving both bounds. The lower bound is easy: Note that before the first receival time at node REF all nodes MATH must recover at least once. So MATH stochastically dominates MATH, where MATH are i.i.d. exponentially distributed random variables with mean REF. It follows that MATH which proves the lower bound. The upper bound uses a little trick: Suppose we add an extra node REF at the left of node REF, with recovery rate MATH. Denote this new system by II and the old system by I. Let MATH be the first time in system II that node REF receives a signal. It is clear that system II is an extension of the old one, in the sense that the nodes MATH `do not feel the change', so that obviously MATH. Finally consider the system, denoted III, obtained from system I by putting an extra node MATH at the right of MATH, with recovery rate MATH. (So, in system III the input signals are sent to MATH which, if it is `on', sends them to MATH, etc). Let MATH denote the first time node REF receives a signal in system III. By REF , MATH has the same distribution as MATH. So we have MATH . The following computation is for system III. Let MATH be a non-negative integer. Let MATH be the event that an input signal is sent in the time interval MATH, MATH the event that node MATH has no recovery in the interval MATH, but does have a recovery in MATH, MATH the event that each of the nodes MATH which is off at time MATH has a recovery before time MATH, and MATH the event that an input signal is sent to MATH in the interval MATH. It is easy to see that the conditional probability of MATH given all information up to time MATH is at least MATH which is larger than MATH, uniformly in MATH, for sufficiently large MATH. Moreover, if all the events MATH happen, node MATH will receive a signal in the interval MATH (and hence in MATH). So, for each integer MATH, we have MATH, from which the required result follows. |
math/0012237 | Suppose such a sequence does exist. There are two possibilities: either there exist MATH and MATH with MATH for all MATH or there exist no such MATH and MATH. In the latter case we have (by the rules above) that MATH for all MATH and all MATH. Since all MATH are smaller than some number MATH, every MATH equals REF at time MATH, a contradiction. As to the former case, let MATH and MATH be as stated there. Let MATH be the smallest number larger than MATH with MATH for all MATH. From the rules given above (and the assumption for this case) it follows that MATH for all MATH and so MATH at every time, in particular at time MATH: again a contradiction. Since both cases lead to a contradiction, the proposition has been proved. |
math/0012237 | The most natural (and rather standard) way to see this is by use of a space-time diagram. This enables us to couple two on-off systems with the same recovery rates but different input interval distributions, say MATH and MATH. We give a short outline of the argument: Let MATH denote the points of a renewal process with interval distribution MATH. (That is, MATH are i.i.d. random variables with distribution function MATH). Now assign to each node MATH, independently of the other nodes and of the above renewal process, a NAME point process with intensity MATH. These NAME points are interpreted as potential recovery points. This means that if MATH is such a point for node MATH, and node MATH is in state MATH just before time MATH, it switches to state REF at time MATH (otherwise the point is ignored). The MATH and MATH can be defined in a natural way in terms of the above NAME processes and the renewal process. If we now replace MATH by MATH, we can compare the new situation with the old one with the help of a suitable natural coupling: use the same realization of the above mentioned NAME point processes and take an obvious coupling of MATH and MATH. Details are left to the reader. |
math/0012237 | Using REF we have MATH, which obviously stochastically dominates MATH. |
math/0012237 | CASE: The c.a.d.l.a.g. property follows immediatley from the definition of the functions MATH. CASE: Suppose that for some MATH and MATH for all MATH. Then, because of REF , there is a MATH such that MATH for all MATH. Hence, by definition of MATH, MATH. However, because signals are dense, there is a MATH with MATH. Let MATH be the smallest of such MATH. So we have MATH, MATH and MATH, which contradicts property (iiib) of a signal/recovery sequence. CASE: Suppose that for some MATH and some MATH for all MATH, and MATH. By REF there is a MATH with MATH. Let MATH be the smallest. So we have: MATH and MATH. This clearly implies that MATH but at the same time MATH is in the interior of the set MATH. This contradicts property (iiib) of signal/recovery systems. CASE: Suppose MATH and MATH for some MATH and MATH. So MATH. But then (by REF of a signal/recovery system) MATH, which is in conflict with the above mentioned fact that MATH. |
math/0012237 | As before, let MATH denote the exponential distribution with mean MATH. For each MATH and MATH we have (using REF again) MATH . Note that this last expression is the probability that in a finite on-off system with MATH nodes with recovery rates MATH, and where the input signals are generated according to a NAME process with intensity MATH, the last node receives a signal before time MATH. This probability is clearly larger than or equal to the probability that each of REF below happens: CASE: No input signal is sent in the interval MATH. CASE: Every node is in state REF at time MATH. CASE: An input signal is sent in the interval MATH. This probability is MATH . For every MATH this is a lower bound for MATH. Now use REF and take MATH to complete the proof of REF . |
math/0012237 | We have, for any MATH, MATH where the first two expressions in the right-hand side refer to a system with leftmost and rightmost nodes MATH and MATH, and MATH and MATH, respectively. The inequality is obvious from the definition, the first equality follows from REF . Remind that MATH denotes the exponential distribution function with mean MATH. Note that the last expression in the right-hand side of REF is the probability that in a (size MATH) on-off system to which input signals are sent according to a NAME process with intensity MATH, and with recovery rates MATH, the last node receives a signal in the time interval MATH, and the computations below refer to that system. We will choose MATH appropriately, depending on MATH. First of all, it follows from REF that there exists a sequence MATH with the properties that MATH, MATH for all MATH, and MATH. Now take MATH. Let MATH and MATH be the infimum and supremum of the interval MATH. It is clear that the last expression in REF is larger than or equal to the probability that each of the following events REF occur: CASE: No input signal is sent in MATH. CASE: Each node in the system which had value MATH at time MATH, has recovered before time MATH. CASE: An input signal is sent in the interval MATH. This probability is MATH . The right hand side in the last inequality does not depend on MATH and goes to MATH as MATH. This completes the proof of REF and of REF . |
math/0012238 | Given MATH, or MATH, one first shows that REF is tensorial in MATH, or MATH. The product rule then uniquely defines the map MATH, or MATH. It is easy to check their required properties. Finally, if MATH is a section of MATH then MATH is a section of MATH since MATH . |
math/0012238 | Integrating REF over the compact NAME surface shows that MATH is the adjoint elliptic operator to MATH. Using the invariance of the index of an elliptic operator under continuous deformations and the NAME Theorem in the complex holomorphic setting, we obtain MATH . |
math/0012238 | REF shows that local parallel sections MATH of MATH induce local holomorphic sections MATH of MATH. For the uniqueness part take any other holomorphic structure MATH on MATH which thus differs from MATH by a section MATH. Since MATH and MATH both annihilate parallel local sections of MATH, also MATH annihilates such sections. But at each point MATH we may always choose a basis of MATH consisting of elements in MATH, which then can be locally extended to parallel sections and thus MATH. |
math/0012238 | By definition, MATH is a holomorphic curve if there exists a complex structure MATH on MATH so that MATH . Dualizing this relation and using REF we obtain the lemma with MATH. |
math/0012238 | Recall REF that MATH is a holomorphic curve if and only if the mixed structure MATH leaves MATH invariant. But this is equivalent REF to MATH stabilizing MATH, that is, to MATH being a holomorphic subbundle of MATH. REF then shows that the holomorphic structure MATH on MATH from REF is given by MATH on MATH. If we denote by MATH the NAME field of MATH then REF gives MATH . We used that MATH which holds since MATH stabilizes MATH. The second equality follows from the fact that MATH stabilizes MATH and MATH. |
math/0012238 | Let MATH and take MATH, that is, MATH, then we get MATH . Therefore MATH takes values in MATH if and only if MATH leaves MATH invariant which, by REF, is equivalent to MATH being a holomorphic curve. But MATH is open inside MATH so that the tangent space to the fiber MATH at MATH is given by MATH. Since MATH the quotient projection MATH is projection onto the MATH-eigenspace of MATH and thus MATH . Note that MATH is a section of MATH which, by REF, is the same as MATH. Let MATH be MATH-stable for some complex structure on MATH and assume MATH is holomorphic curve with respect to the induced complex structure MATH on MATH. Then, by REF , the holomorphic structure on MATH is given MATH and thus MATH . |
math/0012238 | Choose any lift MATH of MATH, that is, MATH. Then MATH so that MATH is a MATH valued MATH-form. Since MATH and MATH restricts to MATH on MATH's, we obtain MATH which implies that MATH in case MATH. For MATH we use REF , instead of MATH, to get MATH . If MATH, then MATH is also a lift of MATH. Moreover, the equation MATH has the unique solution MATH. |
math/0012238 | We proceed by induction. Assume that we have constructed extensions MATH for MATH with the desired properties. Let MATH be an extension of MATH such that MATH. Any other extension is of the form MATH, where MATH. The condition MATH then becomes MATH . Applying MATH we get MATH by the induction assumption. Thus the right hand side of REF is a MATH valued MATH-form. Moreover, this MATH-form has no MATH-part since, for MATH, MATH . On the lowest level, MATH, we obtain the same conclusion MATH due to REF and the assumptions on MATH. Since MATH is invertible, we now can solve REF uniquely for MATH. Let MATH and MATH be two extensions of MATH to jet complex homomorphisms. Then their difference is MATH valued and in the kernel of MATH, and thus it is zero. This shows the uniqueness of extensions. |
math/0012238 | Given MATH we clearly have MATH by the properties of MATH and MATH. This immediately implies MATH on MATH-forms and thus MATH . For the converse we note that MATH is linear over real valued functions and thus defines a MATH-form with values in MATH. But MATH so that MATH is MATH valued. Moreover, MATH which shows that MATH is a section of MATH. Notice also that MATH when restricted to MATH. Thus we have the splitting MATH defining MATH with MATH, which gives MATH . Assume now that MATH is the difference of two adapted connections. Then MATH so that MATH. Using again that MATH for an adapted connection, we get MATH which shows that MATH has no MATH-part. |
math/0012238 | REF gives us MATH for some bundle map MATH with MATH. Using REF this immediately yields MATH and hence also the corresponding statements for the MATH-commuting and anticommuting parts. |
math/0012238 | Again we proceed by induction. We already have a complex structure on MATH. Assume now that we have MATH up to MATH with the desired properties. To construct MATH, first choose any complex structure MATH on MATH extending MATH, that is, MATH, and inducing MATH on MATH. That MATH, with MATH, is another such extension is equivalent to MATH and the fact that MATH and MATH anticommute. Now put MATH and determine MATH so that MATH holds. In this relation MATH is determined by MATH, which we already know. Furthermore, MATH which implies that MATH is a section of MATH. But MATH is invertible, so we can solve REF uniquely for MATH. The necessary properties for MATH are now easily checked: restricted to MATH since MATH is complex linear. Thus MATH. Finally, MATH and MATH anticommute because MATH . It remains to show that MATH is equivalent to the vanishing of MATH on MATH. Since MATH we conclude inductively that MATH . Thus, MATH takes values in MATH and, to show its vanishing, it suffices to calculate MATH restricted to MATH: MATH . We used MATH and the fact that MATH, which implies MATH on MATH by induction. The converse is proven similarly. |
math/0012238 | From REF we have MATH which implies MATH due to REF . On the other hand REF gives MATH which, together with MATH, implies MATH . |
math/0012238 | We work with the explicit holomorphic jet complex described above. This means that MATH, with MATH a section of MATH. The condition MATH translates into MATH where MATH is any adapted connection REF on the jet complex. Using trivializations near MATH, we may assume that all occurring sections are vector valued functions. Moreover, up to zero order terms, MATH is the directional derivative. The lemma then follows. |
math/0012238 | Let MATH be a local trivializing frame of MATH near MATH. Then MATH with MATH a local MATH valued map. Furthermore, MATH where we used that MATH is complex antilinear. Putting MATH for local sections MATH and MATH of MATH, the equation MATH is locally given by MATH . By standard results in analysis (for example, CITE,CITE), non-trivial solutions to the above equation cannot vanish to infinite order (on a connected set). Let MATH be the order of the smallest non-vanishing derivative of MATH at MATH, that is, MATH with MATH not all zero. Inserting back into REF we obtain MATH which is only possible if MATH for MATH. Thus, MATH and MATH where MATH. Clearly, MATH depends only on MATH and not on the choice of trivialization or coordinates. |
math/0012238 | Using the explicit representation of the jet complex MATH we have MATH where MATH is an adapted connection REF on the jet complex. To calculate MATH we decompose MATH so that MATH for the canonical complex structure MATH of REF . Then MATH and thus MATH . The section MATH is MATH valued and satisfies inductively MATH . Thus we have shown MATH or, by taking out MATH, MATH in case when MATH. Note that the MATH in the last expression generally is not a smooth function. For MATH the last formula reduces to MATH which is trivially true, since the left hand side is smooth. In fact, the MATH then has to be a MATH. |
math/0012238 | If MATH is a NAME point then there is a non-trivial section MATH with MATH. REF then implies that MATH, that is, MATH is not an isomorphism. On the other hand, if MATH is not a NAME point then MATH is clearly injective: MATH implies that MATH vanishes to at least order MATH at MATH, and thus has to be identically zero. From REF we get MATH, so that MATH and thus MATH. Since MATH the claim follows. |
math/0012238 | Let MATH denote the jet complex of the holomorphic bundle MATH. Since MATH has no NAME points the bundle map MATH is an isomorphism by REF and maps the NAME flag MATH to the flag MATH in MATH. It thus suffices to show that MATH is a NAME flag. Put MATH which has the trivial connection MATH coming from the directional derivative in MATH via MATH. Since MATH, but also MATH, we obtain MATH, so that MATH is an adapted connection. But then MATH which implies MATH . The successive quotients MATH get mapped by MATH isomorphically to the kernels MATH and therefore have a complex structure. Now let MATH be the derivatives of the flag MATH and let MATH be the restrictions of MATH to the kernels MATH in the jet complex MATH. Then, on MATH, we have MATH which implies that MATH is an isomorphism of line bundles. Thus MATH is a NAME flag by REF . Finally, MATH is the dual curve to the NAME embedding MATH if and only if MATH. But this follows immediately since the kernel of the evaluation map MATH is MATH and the NAME correspondence is given by MATH. |
math/0012238 | We first verify the axioms of the holomorphic jet complex: dualizing the NAME flag we obtain the sequence of surjective bundle homomorphisms MATH whose kernels are MATH . Thus we have complex structures on each MATH by dualizing the complex structures on MATH. Next we define MATH by the obvious product rule MATH where MATH, MATH and MATH is the flat connection on MATH. Since MATH this is well defined. It is routine to check the required NAME rule for MATH. Furthermore, MATH since MATH is flat. To calculate the restriction MATH of MATH to MATH we take MATH, that is, MATH is a section of MATH vanishing on MATH, and MATH. Then MATH so that MATH . The properties of MATH now imply that MATH are complex linear isomorphisms. To verify the last axiom MATH, where MATH denotes the holomorphic structure on MATH, we take MATH and MATH to calculate MATH . We used REF to see that MATH is again a section of MATH and REF of the holomorphic structure on MATH. But the jet complex up to any level is determined uniquely by the axioms in REF. By REF any NAME curve MATH is full. The linear system MATH induced by the NAME correspondence REF is therefore equal to MATH. If MATH then MATH where the latter follows from REF. Therefore the MATH-th prolongation of MATH is given by MATH, that is, MATH is the identity map. REF then implies that the curve MATH, or equivalently, the linear system MATH, has no NAME points. |
math/0012238 | We have seen in REF that the dualized NAME flag MATH consists of the first MATH jet bundles of the holomorphic jet complex MATH of MATH. REF provides us with a canonical adapted complex structure MATH on MATH. Therefore the dual complex structure MATH on MATH, which we again call MATH, leaves the NAME flag invariant and induces MATH on the quotients MATH. On MATH respectively MATH we can now decompose REF the connections MATH with respect to MATH. Due to REF the dual connection MATH on MATH satisfies MATH and hence is adapted REF then implies MATH which is equivalent to MATH . Since MATH REF implies that MATH and MATH are holomorphic, both as subbundles and curves. To see that MATH is unique we note that the conditions of the theorem imply that the dual complex structures MATH on MATH satisfy the requirements of REF for the first MATH jet bundles. Thus, MATH are the unique adapted complex structures on the holomorphic jet complex of MATH. To verify the unramified NAME relations we integrate the trace of the curvature MATH of the flat connection MATH on MATH. Except MATH, which anticommutes with MATH, all other terms are MATH commuting so that MATH where we used that MATH by type. Multiplying this last relation by MATH we obtain MATH and, taking traces, also MATH . Considering that MATH and MATH, we obtain MATH which, together with REF , implies MATH . Since MATH we obtain from REF MATH which finishes the proof. |
math/0012238 | Let MATH be the NAME flag of MATH and let MATH be the complex structure on MATH given in REF . As usual we consider MATH via right multiplication by the quaternion MATH. Since MATH stabilizes the flag MATH we have MATH with MATH the MATH-eigenspace of MATH on MATH. From REF we know that MATH has zero NAME energy if and only if the twistor lift MATH is a complex holomorphic curve in MATH. The derivatives of the NAME flag MATH commute with MATH and thus induce MATH which are easily seen to be the derivatives of the first MATH osculating curves MATH of the complex holomorphic curve MATH. Since MATH are isomorphisms the curve MATH has the NAME gap sequence MATH with MATH for MATH. |
math/0012238 | From REF we see that equality holds in REF if and only if the dual curve MATH has zero NAME energy. By REF this is equivalent to MATH being the twistor projection of a holomorphic curve in MATH. |
math/0012238 | We apply MATH to the equation MATH and obtain by REF MATH where we also used that MATH for MATH by REF. Inserting REF we thus get MATH and, since MATH for MATH are linearly independent, also MATH which proves the claim regarding MATH. To obtain the statement for MATH we first note that MATH is smooth since MATH when MATH. Thus we can write MATH with a smooth map MATH which implies MATH . Taking into account the special form of MATH and MATH, one gets the stated expression of MATH by a direct calculation. |
math/0012238 | Let MATH be a NAME point and MATH, MATH, a basis of MATH whose vanishing orders at MATH are the NAME gap sequence MATH for MATH. Let MATH be the corresponding local framing of MATH constructed in REF. Then both frames are adapted, meaning that MATH are a basis of MATH and MATH frame MATH in a neighborhood around MATH. In REF we have seen that, away from NAME points, MATH maps the NAME flag MATH isomorphically to MATH. Therefore, expressing MATH as in REF , we see that MATH is an adapted local frame for the flag MATH away from MATH. Our aim is to modify this frame so that it stays adapted to the flag MATH but approaches MATH as MATH approaches the NAME point MATH. Using the same notation as in REF we have MATH with MATH. We decompose MATH into upper, with ones along the diagonal, and lower diagonal matrices. Then the upper diagonal matrix MATH approaches the identity as MATH goes to MATH. With the lower diagonal matrix MATH which fails to be invertible at MATH, we obtain the upper-lower decomposition MATH of MATH. Since MATH is lower diagonal, MATH is also an adapted basis for the flag MATH in a punctured neighborhood of MATH. But then MATH is an adapted frame for the NAME flag MATH near MATH approaching MATH as MATH goes to MATH. Thus the NAME flag MATH approaches MATH. To calculate the limit of MATH on MATH as MATH tends to MATH we again work with the frames MATH in MATH and MATH in MATH. From REF we know that the matrix of MATH on MATH in the frame MATH is given by MATH . Thus the matrix of MATH on MATH in the frame MATH becomes MATH . From MATH and the fact that MATH, MATH and MATH commute with MATH, we obtain MATH which shows that MATH on MATH has a limit as MATH tends to MATH. |
math/0012238 | Let MATH be an endomorphism with MATH where MATH. Then MATH with MATH for non-zero MATH, and hence MATH which means that MATH is skew-adjoint with respect to the given hermitian product. In particular, MATH and MATH, for MATH, are skew-adjoint. |
math/0012238 | It is always possible to choose a quaternionic connection MATH on MATH which makes MATH parallel. Then any other such connection is of the form MATH with MATH a skew adjoint MATH-form on MATH with values in MATH. Using this freedom, we choose MATH such that MATH becomes a complex connection, that is, makes MATH parallel: the equation MATH has the MATH valued MATH-form MATH as a solution. Since any element in MATH squares to a negative multiple of the identity, REF implies that MATH is in fact skew-adjoint. Notice that we have fixed MATH up to the addition of MATH for some real MATH-form MATH. Now NAME multiplication is parallel with respect to MATH if and only if MATH or, equivalently MATH for all vector fields MATH on MATH. To see that this equation has a solution MATH we may, by MATH-linearity, assume that MATH. Let MATH be such that MATH are a local orthonormal frame of MATH. Then MATH for some local real MATH-form MATH. Moreover, MATH anticommutes with MATH so that MATH for local real MATH-forms MATH. Multiplying the latter by MATH, and observing that MATH is skew-adjoint, we conclude MATH. Therefore the right hand side of REF is a multiple of MATH, and we can solve for MATH uniquely. |
math/0012238 | Let MATH be an orthonormal local frame of MATH. Then MATH . |
math/0012238 | Since the map REF and the hermitian product in its first slot are complex antilinear, MATH is complex linear. Moreover, MATH where we used that MATH, so that MATH is well-defined. Clearly, MATH is non-zero and thus an isomorphism of complex quaternionic line bundles. To show that MATH is holomorphic, we first observe that REF implies that MATH . Then, using the definition of MATH, one calculates MATH for MATH and MATH a local orthonormal basis. Thus, MATH and MATH is a holomorphic bundle isomorphism. |
math/0012238 | The curvature MATH of the connection MATH is given by MATH where we used that the connection MATH is flat. From REF we have MATH and therefore MATH . Inserting this into the expression for MATH we obtain MATH . The first term takes values in MATH, whereas the second term takes values in MATH. Together with REF , this shows that MATH is NAME if and only if MATH is flat for unitary MATH. Finally MATH so that MATH where we used that MATH and that MATH. This shows that also MATH is NAME. |
math/0012238 | We rephrase the basic fact from linear algebra that eigenvectors corresponding to distinct eigenvalues are linearly independent. Denote by MATH the MATH-part of the connection MATH with respect to the complex structure MATH on MATH. From REF we easily calculate that MATH where the antiholomorphic structure MATH collects the MATH-independent terms and MATH is a MATH valued MATH-form. Note that parallel sections of MATH are also in the kernel of MATH. Assume that MATH, MATH, are linearly independent parallel sections of MATH for distinct MATH, so that MATH . Let MATH, MATH, be a parallel section of MATH with MATH distinct from the MATH. We have to show that all MATH. Since MATH and MATH, it suffices to show that MATH are linearly independent. Because parallel sections are holomorphic, this will follow if we can show that MATH is injective. Away from the isolated zeros REF of MATH, the bundle map MATH has kernel equal to the MATH-eigenbundle MATH of MATH. Therefore the kernel of MATH on MATH consists of sections MATH for which MATH . But MATH preserves sections of MATH whereas MATH maps them to sections of MATH, the MATH-eigenbundle of MATH. Since MATH also vanishes at most at isolated points, we conclude that MATH. Note that these last arguments used the fact that we are working on a quaternionic line bundle. |
math/0012238 | We will show that under our assumptions the spectral genus must be zero, which implies that MATH is given in terms of trigonometric functions and thus is a homomorphism CITE. From REF we see that the spectral genus can at most be one. In this case the spectral curve MATH has at most one pair of branch points, not counting MATH and MATH. But then we only need to pass to a double cover REF of MATH to obtain global holomorphic sections from the two branch values. We therefore obtain the improved estimate MATH for harmonic tori of spectral genus one. |
math/0012239 | It is well-known that a torus knot is fibered with fiber being its minimal NAME surface. We will describe how to construct this fiber by plumbing left-handed NAME bands (compare CITE). The monodromy of a left-handed NAME band is a positive NAME twist along its core circle as shown in REF . Note that our convention for monodromy (see REF) differs from NAME 's in CITE. It is proven in CITE (see also CITE) that the monodromy of a surface obtained by plumbing two surfaces is the composition of their monodromies. We can plumb two left-handed NAME bands to get a MATH torus knot with its fibered surface. Simply identify a neigborhood of the arc MATH in one NAME band with a neighborhood of the arc MATH in the other NAME band, transversally as shown in REF . The resulting monodromy will be the product of two positive NAME twists along the curves also drawn in REF . Note that the two curves (one of which is drawn thicker) intersect each other only once and they stay parallel when they go through the left twist on the surface. It is clear that we can iterate this plumbing operation to express the monodromy of a MATH torus knot as a product of MATH positive NAME twists. By attaching more left-handed NAME bands we can construct the fibered surface of a MATH torus knot for arbitrary MATH and MATH. First construct the gate in the back and then plumb a NAME band in the front face of that gate and proceed as above to obtain a second gate. We can iterate this process to get as many gates as we want. This is illustrated for MATH and MATH in REF . Hence the monodromy of the fibration of the complement of a torus knot in MATH is a product of positive NAME twists. These twists are nonseparating by our construction. |
math/0012239 | We describe NAME 's construction given in CITE. We say that a link in MATH is in a square bridge position with respect to the plane MATH if the projection onto the plane is regular and each segment above the plane projects to a horizontal segment and each one below to a vertical segment. Clearly any link can be put in a square bridge position. Suppose that the horizontal and vertical segments of the projection of the link in the MATH-plane are arranged by isotopy so that each horizontal segment is a subset of MATH for some MATH and and each vertical segment is a subset of MATH for some MATH. Now consider REF-disk MATH for each MATH and REF-disk MATH for each MATH, where MATH is a small positive number. Attach these disks by small bands (see REF ) corresponding to each point MATH for MATH and MATH. If MATH and MATH are relatively prime then the result is the minimal NAME surface MATH for a MATH torus knot MATH such that MATH and MATH. Each component of the link MATH is a nonseparating embedded curve on the surface MATH since we can find an arc connecting that component to the boundary MATH from either side of the component. Moreover we can choose MATH and MATH arbitrarily large by adding more disks of either type MATH or type MATH. |
math/0012239 | Consider the handle decomposition of the MATH for a torus knot MATH. REF gives an explicit description of the vanishing cycles. Cancel each REF-handle with a REF-handle so that the result is just REF-handle MATH. |
math/0012239 | Let MATH be a compact NAME surface with boundary. We use NAME 's characterization of compact NAME surfaces. REF no REF-handles and one REF-handle Suppose that the compact NAME surface MATH with boundary is obtained by attaching a REF-handle MATH to MATH along a Legendrian knot MATH, with framing MATH. REF shows the front projection of a Legendrian trefoil knot. First of all, we smooth all the cusps of the diagram and rotate everything counterclockwise to put MATH into a square bridge position as in REF . Now we use NAME 's algorithm (compare REF ) to find a torus knot MATH with its minimal NAME surface MATH such that MATH is an embedded circle on the surface MATH. For example, we can embed the trefoil knot into the NAME surface of a MATH torus knot as shown in REF . Let MATH be a copy of MATH pushed in the positive normal direction to MATH, and let MATH be the linking number of MATH and MATH computed with parallel orientations. We need the following observation to prove our theorem. MATH . When we push MATH in the positive normal direction to MATH, we observe that MATH will be exactly the NAME framing of MATH, by simply counting the linking number of MATH and MATH. Therefore attaching a REF-handle to MATH along a given Legendrian knot MATH in MATH, with framing MATH, is the same as attaching a REF-handle along the same knot MATH (which is isotoped to be embedded in a fiber of the boundary of a MATH ) with framing MATH. But then the framing MATH is the framing MATH relative to the product framing of MATH. In other words, we proved that attaching a Legendrian REF-handle is the same as attaching a NAME REF-handle in our setting. The global monodromy of MATH will be the monodromy of the torus knot MATH composed with a positive NAME twist along MATH. REF no REF-handles Let MATH be a Legendrian link in MATH with components MATH. Suppose that the compact NAME surface MATH with boundary is obtained by attaching a REF-handle MATH to MATH along MATH for each MATH. First smooth all the cusps of the diagram and rotate everything counterclockwise to put MATH into a square bridge position. Then find a torus knot MATH with its minimal NAME surface MATH such that each MATH is an embedded circle on MATH for MATH. Now for each MATH, attach a REF-handle MATH simultaneously to MATH along MATH with framing MATH . The result is going to be a PALF by REF , since the link components are disjointly embedded nonseparating circles in MATH. So we showed the global monodromy of MATH is the monodromy of the torus knot MATH composed with positive NAME twists along MATH's. Note that the NAME twists along MATH's commute since they are pairwise disjoint on the surface MATH. General case: First we represent REF-handles with dotted-circles stacked over the front projection of the Legendrian tangle. Here we assume that the framed link diagram is in standard form (compare CITE). Then we modify the handle decomposition by twisting the strands going through each REF-handle negatively once. In the new diagram the Legendrian framing will be the blackboard framing with one left-twist added for each left cusp. This is illustrated in the second diagram in REF . Next we ignore the dots on the dotted-circles for a moment and consider the whole diagram as a link in MATH. Then we put this link diagram in a square bridge position as in REF (see REF ) and find a torus knot MATH such that all the link components lie on the NAME surface MATH of MATH. Now consider the MATH on MATH with regular fiber MATH as in REF . We would like to extend MATH on MATH to a PALF on MATH union REF-handles. Recall that attaching a REF-handle to MATH (with the dotted-circle notation) is the same as pushing the interior of the obvious disk that is spanned by the dotted circle into the interior of MATH and removing a tubular neighborhood of the image from MATH. Before attaching REF-handles we apply the following procedure (compare CITE): We isotope each dotted-circle in the complement of the rest of the link such that it becomes transversal to the fibers of MATH, meeting each fiber only once. (see REF ). Thus by attaching a REF-handle to MATH we actually remove a small REF-disk MATH from each fiber of MATH, and hence obtaining a new PALF on MATH union a REF-handle. After attaching all REF-handles to MATH, we get a new PALF such that the regular fiber is obtained by removing disjoint small disks from MATH. Moreover the attaching circles of REF-handles are embedded in a fiber in the boundary of the new PALF such that the surface framing of each attaching circle is equal to its Legendrian framing. Then REF implies, that is attaching a Legendrian REF-handle at this the same as attaching a NAME REF-handle. Hence, we can extend our PALF on MATH REF-handles to a PALF on MATH REF-handles MATH . Legendrian REF-handles. The vanishing cycles (hence the monodromy) of the constructed PALF are determined explicitly as follows: We start with the monodromy of the torus knot MATH, extend this over REF-handles by identity and then we add more vanishing cycles corresponding to REF-handles. Finally, we note that the MATH torus knot in REF can be constructed using arbitrarily large MATH and MATH. Therefore our construction yields infinitely many pairwise nonequivalent PALF's, since for chosen MATH and MATH the genus of the regular fiber will be at least MATH. Conversely, let MATH be a PALF, then it is obtained by a sequence of steps of attaching REF-handles MATH, where each MATH is a PALF and MATH is obtained from MATH by attaching a REF-handle to a nonseparating curve MATH lying on a fiber MATH. Furthermore this handle is attached to MATH with the framing MATH, where MATH is the framing induced from the surface MATH. Inductively we assume that MATH has a NAME structure, with a convex fiber MATH. By CITE we can start the induction, and assume that the convex surface MATH is divided by MATH. By the ``Legendrian realization principle" of CITE (pp REF), after an isotopy of MATH, MATH can be taken to be the NAME framing, and then the result follows by NAME 's theorem (NAME has pointed out that, in case of MATH identification of MATH with NAME framing also follows from CITE- CITE). Though not necessary, in this process, by using CITE we can also make the framing of MATH induced from MATH to be the NAME framing if we wish. |
math/0012241 | Let MATH be the open cell containing MATH, that is, MATH, where MATH is the NAME opposite to MATH. Let MATH be the NAME cell containing MATH. The set of elements MATH in relative position MATH is MATH. Let MATH be the set of weights of MATH, that is, roots of the negative unipotent complementary to MATH. Let MATH be the MATH-equivariant isomorphism of the open cell MATH with the product of root spaces MATH for MATH. The image MATH is the product of MATH for MATH. Therefore it suffices to show that each component MATH is regular at MATH and MATH unless MATH. Since MATH is MATH-invariant, MATH . Hence MATH . Therefore MATH . It follows that MATH at MATH if MATH, and is regular in any case. |
math/0012241 | Over the formal disk MATH at MATH, the bundle MATH is trivial and the action of MATH is given by MATH for some MATH. Since MATH we have MATH . In particular, MATH. More generally, if MATH is a MATH-equivariant bundle over MATH, where MATH is any MATH-algebra, then the action is given by an automorphism MATH. Let MATH be a MATH-equivariant bundle over MATH. We wish to show that there exists an automorphism MATH which transforms the MATH-action on MATH to the product action, that is, MATH . Consider the element of MATH defined by MATH. Since MATH is a cocycle in the cohomology of MATH with values in MATH. Similarly MATH which maps to MATH. We claim there exists a MATH-chain MATH such that MATH. We construct MATH order-by-order. Let MATH. Let MATH be the kernel of the truncation map MATH. The exact sequence of groups MATH induces an exact sequence of pointed sets in non-abelian cohomology (see for example, CITE) MATH . Since MATH is a nilpotent, MATH is trivial, by induction on the length of the central series which reduces to the case that MATH is a MATH-module. Therefore, MATH injects into MATH for all MATH. The complexes MATH satisfy the NAME condition: the image of MATH (respectively, MATH) in MATH respectively, MATH stabilizes as MATH. Indeed, let MATH. Extending MATH to a map MATH is equivalent to extending the map MATH; the latter extends because MATH is isomorphic to MATH near the identity. Therefore, MATH is surjective, which implies the same result for the chain complexes. The NAME condition implies that MATH (see CITE for the abelian case) and therefore also injects into MATH. The claim follows since MATH and MATH both map to MATH in MATH. |
math/0012241 | Consider an embedding MATH, and let MATH denote its canonical reduction. Let MATH be another reduction with slope MATH, and MATH be the parabolic reduction of MATH to MATH induced by MATH. Since MATH for any weight MATH, MATH . Since the NAME filtration is the unique filtration of its slope, MATH. This implies that MATH. |
math/0012241 | Suppose MATH is contained in an open face MATH. For each maximal parabolic MATH, there exist a finite set MATH . These inequalities define a rational affine subspace MATH. Let MATH . Since MATH is finite and MATH is compact, MATH is non-zero. For MATH sufficiently close to MATH in MATH we have MATH since the semistability condition for the two sets of markings is the same. To prove the bijection for flat MATH-bundles, consider the manifold MATH with action of MATH given by MATH and group valued moment map (see CITE; one could use loop group actions here as well) MATH . The moduli space of flat bundles is the symplectic quotient MATH . We claim that for MATH and MATH sufficiently small, there exists a homeomorphism MATH . The quotient REF can be taken in stages, first a quotient by MATH and then a quotient by MATH. As in the NAME theorem, it suffices to show that the one-parameter subgroup MATH generated by MATH acts locally freely on MATH. Suppose MATH is fixed by MATH. Then MATH . Since MATH, we have MATH for some MATH and MATH is a representative of MATH, up to multiplication by MATH. We may assume MATH. The fixed point set of MATH is MATH . Its image under the moment map is equal to MATH . The stabilizer MATH has roots MATH with MATH. Therefore, the NAME MATH of the semisimple part of MATH is MATH where MATH are the fundamental weights corresponding to simple roots MATH with MATH. Let us identify the NAME alcove MATH with a subset of MATH, using the exponential map. The torus MATH intersects MATH in the subset defined by the equations MATH . The intersection of REF with MATH is therefore MATH . If MATH belongs to this set then so does MATH, which implies MATH . Hence MATH is a combination of simple roots vanishing on MATH which is a contradiction. |
math/0012241 | The intersection of a dense set with an open dense set is dense, hence MATH is open and dense in MATH. Since the image of a dense set under a surjective map is dense, the image of MATH is dense in MATH. |
math/0012241 | REF follows from the holonomy description REF and surjectivity of the commutator map MATH CITE. CASE: By REF , MATH is non-empty if and only if a semistable bundle MATH with general parabolic structures MATH is parabolic semistable. By the NAME theorem CITE any principle MATH-bundle admits a reduction of the structure group to MATH. Since MATH is simple, MATH. A principal MATH-bundle MATH with MATH is semistable if and only if MATH is isomorphic to the trivial bundle, which completes the proof. |
math/0012243 | We shall define a polynomial map MATH as follows. If MATH, MATH, MATH, are coordinates on the source jet space MATH as in REF , MATH and MATH are coordinates on the target jet space MATH, then MATH is defined by MATH . We claim that the generic rank of MATH, MATH, is equal to MATH, the dimension of MATH over MATH. For this, let MATH . First note that MATH is greater or equal to the generic rank (in MATH) of the MATH matrix MATH. Moreover, it is not difficult to see that the generic rank of the matrix MATH is the same as that of the matrix MATH. Since for MATH, the rank of MATH is clearly MATH, it follows that MATH. This proves the claim. Given a formal power series MATH, since MATH, we can consider the composition MATH as a formal power series in MATH. We write MATH . Observe that if MATH is in the subring MATH then for each MATH, MATH is a polynomial in MATH, that is, MATH. Let MATH be as in REF satisfying REF . For any vector MATH, by REF with MATH, we obtain MATH . As a consequence, we have MATH for any MATH and any vector MATH in MATH. Since MATH is a polynomial, it follows that MATH and hence the formal power series MATH is zero in MATH. To conclude that MATH is identically zero, by for example, REF, it suffices to use the fact that MATH. This completes the proof of REF . |
math/0012243 | The existence of the map MATH and its properties follow easily from the chain rule. The uniqueness of such a map is a consequence of REF . The proof of the last statement of the proposition is straightforward and left to the reader. |
math/0012243 | Recall by REF that an invertible formal map MATH induces a formal invertible map MATH. We leave it to the reader to check that the equality MATH follows from REF , where the pushforward of an ideal is given by REF . If MATH are generators of the manifold ideal MATH, we may choose a formal invertible map MATH such that MATH for MATH and hence the manifold ideal MATH is generated by the coordinate functions MATH. We take MATH for coordinates in the source jet space MATH and MATH for coordinates in the target one, as in REF . It is then easy to check that the ideal MATH is the manifold ideal generated by the coordinate functions MATH for MATH and MATH. It follows from REF that MATH is a manifold ideal and is generated by the MATH for MATH and MATH. Since by construction MATH, MATH, the last part of the proposition follows. |
math/0012243 | CASE: Since MATH, it follows that MATH implies the equality of the ideals MATH and MATH. Conversely, if MATH, then there exists a MATH matrix MATH with entries in MATH such that MATH . Putting MATH in REF and making use of REF , we obtain that MATH and hence MATH. CASE: Since MATH, if the components of MATH are convergent, then MATH is convergent. Conversely, if MATH is convergent, then by REF , there exist MATH, MATH, with linearly independent differentials at REF such that MATH in MATH. As a consequence, there exist a MATH invertible matrix MATH with entries in MATH such that MATH and hence MATH is invertible. By the implicit function theorem, one sees that the equation MATH has a unique convergent solution MATH. It follows from REF that MATH and hence that MATH is a convergent power series mapping. This completes the proof of REF . CASE: The proof of this case is similar to that of REF above by making use of the algebraic version of the implicit function theorem. |
math/0012243 | Since MATH is generic, by making use of the implicit function theorem we can assume that MATH, where MATH is the codimension of MATH and MATH, and that MATH is generated by the components of MATH where MATH is a formal mapping. If MATH is the complexification of a real-analytic (respectively, real-algebraic) generic submanifold MATH, then the formal map MATH is convergent (respectively, algebraic). By the reality of MATH, one has MATH . Corresponding to the splitting MATH, we may write MATH, with MATH and MATH. By the definition of a NAME variety mapping MATH, we necessarily have MATH . Since MATH, the matrix MATH is invertible. As a consequence of the implicit function theorem, there exist a formal mapping MATH such that MATH . It follows from REF that MATH. The lemma follows by taking MATH. If one takes MATH, where MATH is the component of MATH as in REF , then the reader can check that one has MATH. |
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