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math/0002203	Let MATH denote the curvature form of the NAME metric MATH on MATH . Then by REF, we have MATH where MATH is the complex conjugation of MATH with respect to MATH . MATH restricts to a metric MATH and induces a MATH-decomposition MATH . One obtains MATH where MATH is the second fundamental form of the sub-bundle MATH and MATH is its complex conjugate with respect to MATH . Since MATH we have MATH hence MATH . MATH is positive semidefinite and MATH is negative semidefinite, so MATH is negative semidefinite.
math/0002203	Let MATH be the dual of MATH. We have the projection MATH . Note that MATH as a NAME bundle of the system of NAME bundles corresponding to the dual variation of NAME structures MATH. The monodromy of MATH around MATH is again unipotent. We have the projection MATH where MATH is the q-th subbundle in the extended NAME filtration of MATH . This presentation of MATH as a quotient of a subbundle of a variation of NAME structures allows to apply CITE So the NAME forms of the induced NAME metric on MATH represent the corresponding NAME classes of MATH . From REF we get in particular MATH, and hence MATH .
math/0002203	If the fibres of MATH are connected and if MATH is semistable, this is nothing but REF. In general, let MATH be a finite extension of the function field MATH, containing the NAME hull of the algebraic closure of MATH in MATH, and let MATH be the normalization of MATH in MATH. Consider the normalization MATH of MATH, a desingularization MATH and the induced morphisms MATH and MATH. We will enlarge MATH such that MATH is étale over MATH, hence for MATH . If one chooses MATH large enough, MATH will be the disjoint union of semistable families over MATH, hence MATH is a reduced normal crossing divisor, and MATH is an isomorphism. By the generalized NAME formula CITE, MATH and by REF , MATH . The exact sequence REF , for MATH instead of MATH, and induction on MATH allow to show the same for the relative differential forms, that is, MATH . Hence the pullback of the exact sequence REF to MATH is isomorphic to MATH . Writing MATH and MATH for the edge-morphism, we find MATH . Moreover, the inclusion MATH and flat base change give an inclusion MATH and the diagram MATH commutes. In particular, if MATH lies in the kernel of MATH, the sheaf MATH lies in the kernel of MATH. Since we already know REF for semistable morphisms with connected fibres, we find MATH.
math/0002203	Let MATH and MATH be as in REF, and let MATH be a component of the normalization of MATH. If MATH are the induced morphisms, then MATH . This, for all MATH, implies that MATH.
math/0002203	Let MATH be a point. Then MATH is ample, hence MATH is semi-ample. By CITE, the sheaf MATH is nef. Since the same holds true over all MATH, finite over MATH and unramified in MATH, one obtains REF from REF .
math/0002203	If MATH, the sheaf MATH is trivial, hence MATH and REF obviously holds true. Hence we will assume MATH. For some MATH there exists an effective divisor MATH, disjoint from MATH with MATH. By REF and by flat base change, we are free to replace MATH by any MATH, finite over MATH and unramified over a neighborhood of MATH. Hence we are allowed to assume that MATH or that MATH . Consider the MATH-fold fibre product MATH . MATH is flat and NAME and smooth over some open subscheme. Let MATH be a desingularization such that the general fibre MATH of MATH is isomorphic to MATH. For MATH using flat base change, and the natural maps MATH one finds MATH and both are isomorphism over some open dense subset of MATH. REF induces a surjection MATH . In particular, since MATH, the sheaf MATH is a subsheaf of MATH. Let MATH denote the divisor with MATH. For some divisor MATH, supported in fibres of MATH one has MATH . Blowing up MATH with centers in fibres of MATH we find a normal crossing divisor MATH with MATH and such that MATH. For MATH one obtains MATH . Since we assumed MATH, the exponent of MATH is non-negative. By REF, the sheaf MATH is nef and REF implies that MATH is nef. MATH is contained in MATH and using REF ones finds MATH . On the other hand, MATH contains MATH . Over some sufficiently small open dense subset MATH . By REF, and from CITE or CITE, one has MATH . The semicontinuity of MATH in CITE, implies that for MATH small enough, MATH . Hence MATH and MATH have the same rank and MATH is nef. We obtain MATH or MATH, as claimed.
math/0002203	It remains to verify REF. Comparing the first NAME classes of the sheaves in REF one finds MATH. Hence for some invertible sheaf MATH of degree MATH one has an inclusion MATH . Recall that MATH, and that MATH where MATH is the relative fix locus of MATH. In particular, if MATH denotes the effective divisor with MATH, the divisor MATH is larger than MATH. In order to prove that MATH is a subsheaf of MATH, and that the latter is invertible, we just have to show that MATH is isomorphic to MATH, or that MATH. This is obvious since MATH and since MATH is the relative fix locus of an invertible sheaf. REF says in particular, that MATH has no invertible subsheaf of positive degree. Using REF one obtains the same for MATH and MATH is nef. NAME duality and the projection formula imply MATH . To prove the ampleness, claimed in REF, choose some MATH such that MATH is ample and consider a finite covering MATH of degree MATH, étale over a neighborhood of the discriminant divisor MATH. Then MATH, for some ample invertible sheaf MATH on MATH, and both, MATH and MATH are compatible with pullbacks. Replacing MATH by MATH, we may assume thereby that MATH, for some MATH, and that MATH is ample. Repeating the argument used above, for MATH instead of MATH, one finds MATH to be nef, hence MATH to be ample.
math/0002203	To handle both cases at once, define MATH, if MATH and MATH, otherwise. Assume that MATH . Then by REF, MATH is an invertible subsheaf of MATH of degree MATH in particular it is ample. For MATH, we will construct by induction an invertible subsheaf MATH of MATH of degree MATH . If MATH, the sheaf MATH is ample, and by REF it can not lie in the kernel of MATH. On the other hand, by REF, MATH . The invertible sheaf MATH thereby is a subsheaf of MATH of degree MATH. For MATH we obtain a subsheaf MATH of degree MATH, contradicting REF.
math/0002203	The sheaf MATH is nef, hence if REF does not hold true, it is ample. Since MATH is semistable, MATH is compatible with further pullbacks. Replacing MATH by a covering, we may assume that MATH. By REF MATH hence MATH is ample, for a point MATH. REF implies that MATH, obviously a contradiction.
math/0002203	Choose an invertible sheaf MATH on MATH of degree MATH. If MATH one finds MATH. By REF MATH hence MATH is ample. REF implies that MATH contradicting the choice of MATH.
math/0002205	An abelian variety is simple if and only if its endomorphism ring contains no zero-divisors. Thus, if MATH is simple and MATH is not, there must exist an element of MATH that does not come from MATH. But it follows from the NAME theorem CITE that MATH if MATH. This proves the first statement of the lemma. If MATH is ordinary and MATH is a proper subfield of MATH, then it follows from the NAME theorem that MATH is a matrix algebra over MATH. In particular, MATH contains a zero-divisor, so that MATH is not simple.
math/0002205	Let MATH be a primitive MATH-th root of unity in an algebraic closure of MATH and let MATH. Note that MATH contains MATH. Since MATH is a NAME extension of MATH it is also a NAME extension of MATH, and it follows that MATH and MATH are linearly disjoint over MATH, so that MATH. Let MATH. Since MATH is a NAME extension of MATH, we see that MATH lies in MATH, and hence also in MATH. Suppose MATH. Then since MATH is a subfield of the proper subfield MATH of MATH, the lemma's hypothesis shows we must have MATH. If we let MATH be the minimal polynomial of MATH over MATH, then MATH. Suppose MATH. Then MATH, so MATH is a subextension of the abelian extension MATH, and is therefore NAME. Let MATH be its NAME group, and suppose MATH is a non-identity element of MATH. Let MATH, so that MATH lies in the multiplicative group generated by MATH. Suppose MATH is a positive integer less than MATH. Then the hypothesis of the lemma shows that MATH, so we must have MATH. Thus MATH must in fact be a primitive MATH-th root of unity, which shows that MATH. It follows that MATH, and this last field is MATH because MATH.
math/0002205	If MATH is an integer in MATH then clearly MATH is a proper subfield of MATH. On the other hand, if there exists some MATH such that MATH then there exists a smallest such MATH, and by REF this MATH lies in MATH. This proves the first statement of the proposition. It is clear that MATH is absolutely simple if and only if MATH is simple for every finite extension MATH of MATH. The second statement of the proposition follows from this fact and from REF .
math/0002205	Let MATH be the NAME endomorphism of MATH and let MATH be the field MATH. Because MATH is ordinary, the field MATH is a NAME of degree MATH over MATH. The ordinariness of MATH also implies that MATH is a NAME for every positive integer MATH, and that MATH splits over the degree-MATH extension of MATH if and only if MATH is a proper subfield of MATH. Suppose MATH is not absolutely simple. Then there is a positive integer MATH such that MATH is a proper subfield of MATH; let us take MATH to be the smallest such integer, and let MATH be the imaginary quadratic field MATH. By REF , either MATH and MATH, or MATH and MATH, or there is a primitive MATH-th root of unity MATH in MATH such that MATH. Let us first show that if the third possibility is the case and if MATH then MATH must equal MATH. Suppose, to obtain a contradiction, that we are in the third case and that MATH is greater than MATH but not equal to MATH. Then the degree of MATH over MATH is greater than MATH, so MATH must be MATH. Let MATH be the nontrivial automorphism of MATH that fixes MATH. The proof of REF shows that we may choose our primitive root of unity MATH so that MATH. Applying MATH to this equality, we find MATH, so that MATH. The only element of the NAME group of the cyclotomic field with this property is complex conjugation. But then the fixed field MATH of MATH must be totally real, and we have reached a contradiction. So we must find, for MATH and MATH, the conditions on the coefficients MATH and MATH in the minimal polynomial of MATH that are equivalent to MATH lying in a quadratic subfield. Note that the characteristic polynomial of MATH is of the form MATH, and that such a quartic polynomial is the square of a quadratic polynomial if and only if MATH. It is not difficult to explicitly calculate the characteristic polynomial of MATH for each MATH we are considering, and we find that MATH . We have assumed that MATH is simple over MATH, so the characteristic polynomial for MATH is irreducible; this means in particular that the quantity MATH is nonzero. Thus, if MATH is not absolutely simple then one of REF , or REF must hold. Note that if two of these cases were to hold simultaneously, then MATH would equal a multiple of MATH, contradicting our assumption that MATH is coprime to MATH. Thus exactly one of REF through REF must hold. Finally, the formulas for MATH given above make it easy to verify the theorem's statement about the degree of the minimal splitting field of MATH.
math/0002205	Let MATH be the polynomial whose existence is guaranteed by REF . Then MATH satisfies the last four of the five hypotheses of REF ; we will show that it satisfies the first hypothesis as well. First we will consider the case in which MATH. Since MATH we find that the polynomial MATH defined in REF may be written in the form MATH . Now, MATH is either MATH or is not a multiple of MATH, so the coefficient of MATH in MATH is nonzero. In particular, MATH is not of the form MATH. For the case in which MATH we use the easily-proven fact that the reduction of MATH modulo MATH is equal to MATH times the reduction of MATH modulo MATH. Since MATH modulo MATH is irreducible, and since MATH is not irreducible over MATH, the polynomial MATH must have an odd coefficient somewhere between MATH and MATH. Again we see that MATH is not of the form MATH. Thus MATH satisfies all the hypotheses of REF , so by REF the roots of MATH are NAME numbers that correspond to an isogeny class of absolutely simple ordinary varieties over MATH.
math/0002205	Suppose, to obtain a contradiction, that MATH corresponds to an isogeny class that is not absolutely simple. Then by REF there is a positive integer MATH such that MATH is a proper subfield of MATH. Let MATH be the smallest positive integer with this property. Since MATH is ordinary, the field MATH is a NAME, and its maximal real subfield MATH is a proper subfield of MATH. REF shows that MATH must be MATH, so MATH is an imaginary quadratic field. REF shows that either the minimal polynomial MATH of MATH lies in MATH or MATH for some primitive MATH-th root of unity. The first possibility cannot occur, because it would imply that MATH, contradicting REF . Therefore the second possibility must be the case. We find that the maximal real subfield of MATH is a subfield of MATH, and since MATH is not itself the maximal real subfield of a cyclotomic field (by assumption), we find that the maximal real subfield of MATH must be MATH, so that MATH is either a quadratic field or MATH itself. But MATH is the compositum of MATH and MATH, so the degree of MATH over MATH is at most MATH. This contradicts our assumption that the degree of MATH over MATH is MATH, where MATH .
math/0002205	Since MATH modulo MATH is irreducible, MATH itself is irreducible in MATH, and since all of its complex roots are real, MATH defines a totally real number field MATH. Let MATH be a root of MATH in MATH. The discriminant of the polynomial MATH is totally negative because the roots of MATH all have magnitude less than MATH, so MATH defines a totally imaginary quadratic extension MATH of MATH. If MATH is a root of MATH in MATH, then MATH contains MATH because MATH. Thus MATH is an algebraic number of degree MATH. Furthermore, if MATH is an embedding of MATH into MATH, then MATH is a root of MATH, and the quadratic formula shows that MATH. Thus MATH is in fact a NAME number of degree MATH. Since MATH is a root of MATH, the polynomial MATH must be the minimal polynomial of MATH. This shows that MATH is an irreducible polynomial whose roots are NAME numbers, and to show that MATH is an ordinary NAME polynomial we need merely check that its middle coefficient is coprime to MATH. But this follows from REF , because the middle coefficient of MATH differs from the constant term of MATH by a multiple of MATH. Now we must check that a root MATH of MATH satisfies the hypotheses of REF . The first of these hypotheses is identical to the first hypothesis of the lemma we are proving, and is therefore satisfied. We will show that MATH is not the maximal real subfield of a cyclotomic field. It will suffice to show that MATH is not NAME over MATH. The defining polynomial MATH of MATH reduces modulo MATH as a linear times an irreducible, so the prime MATH splits in MATH into two primes with different residue class degrees, so MATH cannot be NAME. Finally, we prove that MATH has no proper subfields other than MATH. For suppose MATH had a proper subfield MATH other than MATH. Let MATH be the prime of MATH over MATH whose residue class degree is MATH. Let MATH be the prime of MATH lying under MATH. Let MATH be the residue class degree of MATH over MATH and let MATH be the residue class degree of MATH over MATH. Then we have the three statements: CASE: MATH, CASE: MATH, and CASE: MATH . REF shows that strict inequality must hold in one of REF ; but then, since both of the field extensions MATH and MATH are assumed non-trivial, we find that MATH must be less than MATH. This contradiction shows that MATH has no proper subfields other than MATH.
math/0002205	This follows from the techniques of CITE.
math/0002205	If MATH we can simply choose the appropriate value of MATH from REF , so let us assume that MATH. REF (below) shows that there exist monic degree-MATH polynomials MATH in MATH and MATH in MATH such that MATH is irreducible, such that MATH is a linear times an irreducible and has nonzero constant term, and such that the coefficients of MATH in MATH and MATH are equal to the reductions (modulo MATH and MATH) of the corresponding coefficients of the modified NAME polynomial MATH. Once we have fixed MATH and MATH, we can choose values of MATH in the set MATH such that the polynomial MATH reduces to MATH modulo MATH and to MATH modulo MATH. Note that the constant term of MATH is coprime to MATH because MATH and MATH have nonzero constant terms. Thus, if MATH is a power of MATH or MATH then the constant term of MATH is coprime to MATH. If MATH is not a power of MATH or MATH, then the constant term of MATH may have a factor in common with MATH. If this is the case, replace MATH with either MATH or MATH, whichever one lies in the interval MATH; this changes the constant term of MATH by MATH, so that the constant term is now coprime to MATH but so that MATH still reduces to MATH modulo MATH and to MATH modulo MATH. One can calculate that MATH and since MATH differs from MATH by a polynomial of degree at most MATH, we see that MATH may also be written MATH . In particular, MATH satisfies the first condition of REF . Thus MATH satisfies four of the five conditions listed in the statement of REF . We are left to show that all of its roots are real, and that they have absolute value less than MATH. But this follows from REF ; to apply the lemma we must verify that the quantity MATH is less than MATH, and this follows from the fact that MATH is at most MATH for MATH, and that MATH is at most MATH. Thus, the MATH we have written down satisfies all the conditions of the lemma.
math/0002205	For MATH we choose MATH and MATH from REF . For MATH we argue as follows: REF (p. REF) of CITE shows that there exists a monic irreducible polynomial in MATH of degree MATH with zeroes for the first six coefficients after the leading MATH. We take this polynomial for our MATH. The same corollary shows that there is a monic irreducible polynomial MATH in MATH such that the first six coefficients of MATH are equal to those of the reduction of MATH modulo MATH; we take MATH to be MATH.
math/0002205	Let MATH be the arithmetic function defined by MATH, where MATH is NAME 's MATH-function, let MATH be the number of isogeny classes of abelian surfaces over MATH, let MATH be the number of isogeny classes of simple ordinary abelian surfaces, and let MATH be the number of isogeny classes of absolutely simple ordinary abelian surfaces. REF shows that MATH this upper bound is obtained by combining the estimates that REF gives for the number of ordinary and non-ordinary isogeny classes of abelian surfaces. The same theorem shows that the number of isogeny classes of ordinary abelian surfaces over MATH is at least MATH . The isogeny classes of ordinary elliptic curves over MATH correspond to the integers MATH such that MATH and MATH, so there are at most MATH such isogeny classes. A non-simple isogeny class of ordinary abelian surfaces is determined by its two factors, so there are at most MATH such reducible isogeny classes. Thus we have MATH . Now we must estimate the number of simple ordinary isogeny classes that are not absolutely simple. For this we use REF . First note that if MATH is the NAME polynomial for an ordinary abelian surface over MATH then MATH, and if MATH then MATH . Thus, the number of NAME polynomials of ordinary abelian surfaces that satisfy REF is at most MATH. Also, for every nonzero integer MATH in the interval MATH there is at most one NAME polynomial with MATH that satisfies REF; for every nonzero integer MATH in the interval MATH there is at most one NAME polynomial with MATH that satisfies REF; and for every nonzero integer MATH in the interval MATH there is at most one NAME polynomial with MATH that satisfies REF. We find that there are at most MATH simple NAME polynomials MATH with MATH that are not absolutely simple. Combining these estimates with the lower bound for MATH given above, we find that MATH . Now suppose MATH is given. If MATH then the conclusion of the theorem is clearly true for all MATH, so we may assume that MATH and that MATH. With this lower bound for MATH, our bounds for MATH and MATH show that MATH and MATH . Thus MATH . The denominator is less than MATH, so we have MATH as was to be shown.
math/0002205	Combining the estimates for MATH and MATH given in REF, we find that the quantity MATH is less than or equal to MATH so MATH . An easy induction shows that MATH, and certainly MATH, so we have MATH . Since MATH we have MATH, and combining this with the fact that MATH we find that MATH . But MATH is greater than MATH for MATH, so the right-hand side is at most MATH. This proves the inequality of the proposition.
math/0002205	Let MATH and let MATH denote the lattice MATH. If MATH is a point in MATH let MATH denote the ``brick" MATH . Let MATH denote the set of all MATH such that MATH. The proof of REF (see especially p. REF) shows that MATH where MATH is the mesh of MATH (see p. REF), which is MATH. Since the covolume of MATH is MATH, we find that MATH . Thus MATH and using the fact that MATH we find that MATH . Now suppose MATH is a lattice point in MATH, and consider a typical element MATH of MATH. As MATH ranges over all of MATH, the vector MATH ranges over a set of MATH elements of MATH that reduces modulo MATH to all of MATH. REF above shows that of the MATH polynomials MATH we obtain from the vectors MATH arising from elements of MATH, at least MATH satisfy REF . So for each element of MATH we obtain at least MATH polynomials satisfying REF , and REF . Thus the total number of such polynomials is at least MATH, and by the results of the preceding paragraph this number is at least MATH which is greater than MATH . Next we estimate the number of polynomials MATH that satisfy REF but that fail to satisfy REF . There is a bijection between the set of such polynomials and the set MATH, and REF gives upper and lower bounds for the size of the latter set; in particular, we find that the number of such polynomials differs from MATH by at most MATH which is MATH. Since MATH is at least MATH, this last quantity is at most MATH, which is less than MATH, because MATH when MATH. Thus the number of polynomials that satisfy REF but not REF is at most MATH . Finally we estimate the number of polynomials MATH that satisfy REF but that fail to satisfy REF . Now, a polynomial MATH has all of its roots on the circle MATH if and only if MATH, so there are at most MATH polynomials meeting REF but failing REF . It is very easy to show that MATH when MATH. Now, the number of polynomials meeting all five hypotheses of REF is at least as large as the number that satisfy REF , and REF , less the number that satisfy REF but that fail either REF or REF . We find that the number of polynomials meeting all five hypotheses is at least MATH and this is the statement of REF .
math/0002205	The lemma follows easily from the well-known exact formula MATH where MATH is the NAME function.
math/0002216	Suppose MATH and MATH non-contracting. Let MATH and MATH be two morphisms of strictly positive dimension and MATH. Then MATH therefore MATH cannot be MATH-dimensional. If MATH then MATH if MATH and if MATH for two different reasons. Therefore MATH cannot be MATH-dimensional as soon as MATH.
math/0002216	The characterization of REF gives the solution.
math/0002216	Let MATH and let MATH be the property : ``for any MATH-category MATH and any MATH, then MATH." One has MATH, MATH and necessarily MATH by definition of a morphism of cuts. Therefore MATH holds. Now suppose MATH proved for some MATH. One has MATH since MATH is a morphism of cuts and MATH . Therefore the natural transformation MATH from MATH to MATH can be identified with MATH which is precisely MATH. Therefore MATH is proved. At last, if MATH, then MATH .
math/0002216	The natural map MATH for MATH from MATH to MATH is of the form MATH for MATH with MATH for MATH and MATH. Therefore for MATH, MATH by REF . And by construction of MATH, one obtains MATH.
math/0002216	Consider the MATH of CITE. One has MATH where, for the last formula, MATH are other operators which is not important to explicitly define here : the only important thing is that MATH remains MATH-dimensional if the argument is MATH-dimensional. Hence REF . As for REF , it is enough to check it for MATH. And in this case, MATH is a thin MATH-cube satisfying MATH . Once again, we refer to CITE for the precise definition of the operators involved in the above formulas. The only thing that matters here is the dimension of MATH. By CITE, we know that MATH when MATH is a composite of MATH and such that for MATH negative, MATH. Hence REF .
math/0002216	Using the freeness of MATH, the construction in the proof of REF yields the MATH-functor MATH from MATH to MATH. The hypotheses stated in CITE were too strong indeed. If moreover the shell is fillable in the above sense, one concludes still as in the proof of REF .
math/0002216	One proves by induction on MATH the following property MATH : `` For any MATH-simplex MATH of the globular simplicial nerve of any MATH-category MATH, the map MATH from MATH to MATH induces a MATH-functor and moreover an element of MATH." Let MATH be a MATH-simplex of the globular nerve of MATH. Then MATH is a MATH-functor from MATH to MATH, and therefore it can be identified with the MATH-morphism MATH of MATH. Therefore MATH . Therefore MATH is proved. Now suppose that MATH is proved for MATH. Let MATH be a MATH-simplex of the globular simplicial nerve of some MATH-category MATH. If MATH, then MATH . If MATH, that is, if MATH, set MATH with MATH. For a given MATH such that MATH, set MATH as word. Then let MATH with MATH. The relation between the sequence of MATH and the sequence of MATH is as follows : MATH . And we have MATH . Therefore MATH. And by REF , MATH is the constant MATH-functor from MATH to MATH which sends any face of MATH on MATH. Therefore MATH is a MATH-shell in the cubical nerve of MATH which is fillable. By REF , the labeling MATH of MATH induces a MATH-functor from MATH to MATH and MATH is proved. By construction, the equality MATH holds for any MATH-simplex MATH of the globular nerve and for MATH. It remains to check that for such a simplex MATH, MATH for MATH. Consider a face MATH of the MATH-cube. If MATH, then MATH . If MATH, that is, if MATH, set MATH with MATH. For a given MATH such that MATH, MATH and set MATH as word. Then let MATH with MATH. One has to calculate MATH for some MATH. The situation can be decomposed in three mutually exclusive cases: CASE: MATH. In this case, there exists a unique MATH such that MATH, MATH and MATH . Then MATH and MATH CASE: MATH. In this case, MATH and MATH . Then for some MATH, MATH CASE: MATH. Now MATH and since MATH, then there exists a unique MATH such that MATH and we have MATH . There are two subcases : MATH and MATH. In the first situation, MATH . In the second situation, MATH .
math/0002216	Let MATH and MATH be two morphisms of cuts from MATH to MATH. One proves by induction on MATH that MATH and MATH from MATH to MATH coincide. For MATH, MATH. The only natural transformation from MATH to itself is MATH, therefore MATH. Suppose MATH proved for some MATH. Then for any MATH, and for any MATH, MATH . Now with MATH, MATH . So the MATH-morphism MATH is the value of the constant map MATH of REF (denoted by MATH in REF). Let MATH be the unique MATH-category such that MATH and with MATH, MATH, MATH and MATH. And consider MATH. Suppose that MATH and suppose that at least two MATH are equal to MATH. Then there exists a MATH-morphism MATH of MATH such that MATH with exactly one MATH equal to MATH and such that MATH. Then MATH by the previous calculation. Since MATH is the unique morphism of MATH with MATH-source MATH, then MATH and therefore MATH . Suppose now that MATH with perhaps some MATH in the set. Then MATH and therefore MATH . The MATH-functor MATH from MATH to MATH induces a non-contracting MATH-functor MATH from MATH to MATH with MATH (MATH being the value of the constant map MATH by REF ) and MATH which sends MATH on MATH. So by naturality, MATH . Therefore for any MATH, MATH. By hypothesis, MATH. So MATH and MATH induce the same labeling of the faces of MATH and MATH is proved.
math/0002216	The natural transformation MATH from MATH to MATH can be lifted to a natural transformation MATH from MATH to MATH since the cut MATH is regular. Since MATH and since MATH is one-to-one in positive degree, there is at most one solution for this lifting problem. MATH . Let MATH. For MATH, the natural transformation MATH is of the form MATH for some MATH and some MATH. Therefore MATH . So MATH for any MATH and by REF , MATH. Let MATH be the unique element of MATH such that MATH . Then MATH is a solution.
math/0002216	It is a consequence of the naturality of MATH and MATH and of REF .
math/0002216	The equality MATH implies MATH, implies by REF that for MATH, MATH therefore MATH. Conversely, if for MATH, MATH, then MATH and therefore MATH.
math/0002216	Consider the MATH operators of REF . If MATH, then MATH is negative. So MATH is also negative by REF and determines a unique element MATH such that MATH. It is clear that these operators MATH induces the identity map on the reduced globular complex by REF . Since MATH is also negative, then by REF , MATH for some sequence MATH. Therefore by the injectivity of MATH, MATH .
math/0002216	One has MATH with MATH a thin MATH-cube and MATH a thin MATH-cube. The proof made in CITE shows that MATH and MATH are in the image of MATH. Indeed, the existence of MATH and MATH comes from the vanishing of some globular nerve. Therefore MATH and MATH where MATH is a thin MATH-simplex and MATH a thin MATH-simplex. This completes the proof.
math/0002216	For MATH, it is obvious. For MATH, one has MATH where MATH means the simplicial homology of the simplicial nerve of the MATH-category MATH.
math/0002216	The only non-trivial part is the first assertion. Let MATH be the property : ``for any non-contracting MATH-category MATH and any MATH-functor MATH from MATH to MATH, the set map MATH from the set of faces of MATH to MATH is constant." There is nothing to check for MATH. For MATH, if MATH is a MATH-functor from MATH to MATH, then MATH and MATH in MATH. Therefore MATH and MATH . Therefore MATH is true. Suppose MATH proved for some MATH and let us prove MATH. For any MATH, the MATH-functor MATH induces a MATH-functor on the MATH-category MATH generated by the face MATH and its subfaces. One has an isomorphism of MATH-categories MATH. Therefore the restriction of MATH to the faces of MATH is constant by induction hypothesis. Now it is clear that MATH since MATH. Therefore the set map MATH restricted to MATH is constant. Therefore the restriction of the set map MATH to the faces of dimension at most MATH of MATH is constant. We know that MATH where MATH are faces of MATH of dimension at most MATH. So MATH where MATH is a function using only the compositions of MATH. Then MATH where MATH is obtained from MATH by replacing MATH by MATH since MATH is a MATH-functor from MATH to MATH. So MATH with the axioms of MATH-categories. Therefore MATH is proved.
math/0002216	One has MATH by REF . So it suffices to prove the vanishing of MATH as soon as MATH contains morphisms in strictly positive dimension to prove the theorem. Let MATH and MATH be two MATH-morphisms of MATH such that MATH contains other morphisms than MATH and MATH. Then in particular it contains some MATH-morphisms from MATH to MATH which is a composite of MATH-dimensional faces of MATH. Suppose that MATH. Then MATH is obtained from MATH by replacing some MATH equal to MATH by MATH. Let MATH be these MATH. Then MATH as MATH-category. Therefore it suffices to prove that MATH vanishes. The vanishing of MATH is obvious. One has MATH for MATH by REF and MATH because the simplicial nerve of a composable pasting scheme is contractible CITE.
math/0002216	By proceeding as in REF , we see that it suffices to prove that MATH for any pair MATH of MATH-morphisms of MATH and for MATH. However, MATH is non-empty if and only if MATH with our conventions and in this case, MATH . Therefore MATH by REF .
math/0002216	This is due to the fact that for MATH, the natural map MATH is induced by the set map MATH from MATH to MATH REF .
math/0002216	We have to prove that REF implies REF . Let us consider MATH : ``for any non-contracting MATH-category MATH and any MATH-functor MATH and MATH from MATH to MATH such that MATH and such that the image of MATH is a subset of MATH, then MATH yields a MATH-functor from MATH to MATH and MATH for any MATH." Property MATH is obvious. Suppose MATH proved for MATH. For any MATH, MATH is a set map from MATH to MATH satisfying the hypothesis of the proposition, so by induction hypothesis, MATH yields a MATH-functor from MATH to MATH. Let MATH. For MATH, MATH . Therefore MATH is a MATH-shell. So it provides a unique MATH-functor MATH by REF . It remains to check that MATH and MATH to complete the proof. Let us check the first equality. One has MATH where MATH uses only composition laws and where MATH are faces of MATH of dimension at most MATH. Denote by MATH the same function as MATH with MATH replaced by MATH. Then MATH . Now let us suppose that MATH is MATH-dimensional in MATH for some MATH. Then MATH is MATH-dimensional. Either MATH (the initial state of MATH) or there exists a MATH-morphism MATH of MATH such that MATH and MATH. In the first case, MATH is MATH-dimensional. In the second case, MATH is MATH-dimensional. Then MATH is MATH-dimensional as well as MATH . For any face MATH of MATH, there exists a MATH-morphism MATH from MATH to MATH or MATH : let us say MATH. Since MATH is MATH-dimensional, then MATH is MATH-dimensional, as well as MATH . Therefore MATH is MATH-dimensional.
math/0002216	Equalities MATH, MATH, MATH, MATH are consequences of REF . Equality MATH is proved right above. The verification of MATH is straightforward.
math/0002219	Let MATH be a countable dense set in MATH. Using REF we note that for any MATH and any MATH it follows that MATH . Assuming now that for all MATH the condition MATH is satisfied we can choose a countable subset of MATH, MATH and observe that MATH where MATH is defined just like MATH with the difference that the family MATH is indexed over MATH . Using standard approximation arguments and the fact that MATH is dense in MATH we observe for any MATH and any MATH . Finally assume that MATH is a refinement of MATH. Then by REF it follows for MATH that MATH . Let MATH be any countable collection of subsets of MATH. For MATH, if for all MATH REF is true let MATH be as in REF, and, otherwise, we set MATH. Since MATH is countable we can choose a decreasing sequence MATH which is a refinement of MATH. From REF - REF we deduce that for all MATH . Now MATH says that Player I has in the MATH-game a winning strategy, even if he has to choose his finite codimensional subspaces among MATH, and even if Player II ``can cheat a little bit" by choosing his vectors in MATH. From REF we deduce that this is equivalent to the condition that Player II does not have a winning strategy which means that every MATH approximation to a MATH-block-tree has a branch in MATH. We therefore have proven the equivalence of REF . Note also that REF is trivial and since REF means that Player II has no winning strategy even if Player I has to choose form the set MATH it follows that REF implies REF . In order to prove the second part of the Proposition we note that in the case that MATH has a separable dual we can find a universal countable refinement, that is, a countable refinement of the whole set MATH. Indeed, choose a dense sequence MATH in MATH and let MATH . Secondly note that in this case every MATH-block-tree is weakly null, and, conversely, that for MATH, every weakly null tree MATH has a subtree MATH which is a MATH-approximation of a MATH-block-tree.
math/0002219	We consider the following three cases. If MATH is a reflexive space we can choose according to REF a reflexive space MATH with an FDD MATH which contains MATH. If the dual MATH is separable we can use again a result in REF and choose a space MATH having a shrinking FDD MATH. In the general case we choose MATH to be a MATH-space containing MATH, MATH compact and metric (for example MATH endowed with the MATH-topology) and choose an FDD MATH for MATH. We first write MATH as the null space MATH of a finite dimensional space MATH . We choose a finite set in MATH, which norms all elements of MATH up to a factor MATH and choose for each element of this set a NAME extension to an element in MATH. We denote the set of all extensions by MATH and let MATH be the finite dimensional subspace of MATH generated by MATH. We will produce an FDD MATH for MATH so that MATH. Hence REF will hold. Now MATH . Secondly we choose a subspace MATH, MATH, so that MATH is a complemented sum of MATH and MATH, MATH. Note that in general we do not have control over the norm of the projection onto MATH. Given a dense countable subset MATH in MATH, we inflate MATH to MATH. Thus the closure of MATH is MATH. Then we choose as follows a separable subspace MATH of MATH which is REF-complemented in MATH, MATH-norming, and contains all the spaces MATH, MATH. In the case that MATH has a separable dual (thus also MATH is separable) we simply take MATH. In the general case we let MATH be a separable MATH-space containing a MATH-norming set, all the spaces MATH, and all the spaces MATH (considered as subspaces of MATH). For MATH let MATH be the projection from MATH onto MATH, and let MATH be the adjoint MATH if MATH is separable. In the general case we choose MATH to be a sequence of projections of norm REF from MATH onto a finite dimensional subspace of MATH with the property MATH so that MATH is dense in MATH (as a separable MATH-space MATH is complemented in MATH and has an FDD). We are now in the situation of REF of CITE, that is, the following statements hold: MATH . We conclude from REF in CITE that: CASE: Let MATH and MATH be finite dimensional subspaces of MATH and MATH respectively. Then there is a projection MATH on MATH with finite dimensional range so that the following three REF hold MATH . Using MATH we can proceed as in the proof of REF in CITE to inductively define for each MATH a finite dimensional projection MATH on MATH so that for all MATH . Indeed, for MATH we apply MATH to MATH and MATH. If MATH are chosen we apply MATH to MATH and MATH. We deduce REF, and we observe that for MATH, MATH and MATH. Since for MATH and MATH the second equality implies that MATH we also deduce that MATH. Now we let MATH REF and deduce from REF, that MATH is an FDD of a subspace of MATH which, by REF still contains MATH. REF also implies that MATH is dense in MATH. Putting MATH, we note that for MATH and MATH it follows from REF that MATH, and thus, that MATH. We also deduce that for MATH, MATH is dense in MATH using the following Lemma which seems to be folklore.
math/0002219	Let MATH be a subspace of dimension MATH, admitting a continuous projection MATH, so that MATH. Let MATH. By assumption we find a sequence MATH converging to MATH. Let MATH be the (finite dimensional) vector space generated by MATH and choose a basis of MATH of the form MATH. We represent each vector MATH as MATH and put MATH. Note that MATH and that MATH, for all MATH. Furthermore it follows that since MATH and since MATH is basis of MATH, that MATH for all MATH. Therefore it follows that MATH.
math/0002219	We first choose a decreasing sequence of finite codimensional spaces MATH in MATH so that for each MATH the equivalences MATH , and, if MATH is separable, MATH , of REF hold. Then we choose the space with an FDD MATH as in REF . We note that trivially REF of REF implies REF . Since the conclusion of REF implies that every MATH-block tree (recall, MATH) has for given sequence MATH a subtree which is a MATH-approximation of a MATH-block tree for which some branch is REF with respect to MATH, REF implies REF (Player II cannot have a winning strategy). If MATH is separable the statement MATH is exactly the statement of the second part of REF . Thus, we are left with the verification of the implication MATH . Let MATH and MATH. We put MATH, where the constant MATH comes from the conclusion of REF . Every tree MATH in MATH having the property that MATH is a MATH-approximation to a MATH-block tree, and therefore must have a branch in MATH (REF MATH ).
math/0002219	We put MATH . It is easy to see that MATH is closed in the pointwise topology on MATH, since MATH is closed with respect to the product of the discrete topology on MATH. By the infinite version of NAME 's theorem (compare REF ) we deduce that one of the following two cases occurs. MATH . If the first alternative occurs we are finished by the above remark. Assuming the second alternative, we will show that there is a tree in MATH satisfying REF without any branch in MATH. This would be a contradiction and imply that the second alternative cannot occur. If we assume the second alternative we can pick for each MATH a sequence MATH which is not in MATH. Let MATH. Note that for any MATH, MATH . Here MATH is the infinite sequence starting with MATH and MATH and then consisting of the elements of MATH. Using the finite version of NAME 's theorem and the compactness of MATH we can find a vector MATH and a MATH such that MATH . Doing the same procedure again, we can find a MATH and a MATH so that MATH where MATH and MATH are the first two elements of the sequence MATH. Proceeding this way we construct a sequence MATH and a decreasing sequence MATH of infinite subsequences of MATH so that MATH . This sequence will be the first level of a tree and the beginning of the level by level recursive construction of this tree as follows. Assume that for some MATH and every MATH we have chosen a MATH, a pair of natural number MATH, and MATH, and a sequence MATH so that the following REF are satisfied. MATH . Then we can choose for MATH the elements MATH, MATH etc., and the numbers MATH, MATH, MATH, MATH etc. and the sets MATH, MATH, etc. exactly in the same way we chosed MATH, MATH etc. and the numbers MATH etc. for the first level. REF implies that for every branch MATH of the constructed tree there is a MATH so that MATH, for all MATH. Since MATH it follows that (recall that MATH) MATH, which is a contradiction and finishes the proof.
math/0002219	Fix MATH. We define MATH where each MATH is given as follows. MATH will be the completion of MATH under the norm MATH . By a segment we mean a sequence MATH with MATH, MATH, for some MATH. Thus a segement can be seen as an interval of a branch (with respect to the usual partial order in MATH), while a branch is a maximal segment. Clearly the node basis MATH given by MATH is a REF-unconditional basis for MATH. Furthermore the unit vector basis of MATH is REF-equivalent to MATH, if MATH is any branch of MATH. Thus no extension of the tree MATH to a weakly null tree of infinite length in MATH has a branch whose basis distance to the MATH-unit vector basis is closer than MATH for MATH. Since it is clear that in every subspace MATH of a MATH sum of finite dimensional spaces every weakly null tree in MATH must have a branch equivalent (for a fixed constant) to the unit vector basis of MATH it follows that MATH cannot be embedded into a subspace of a MATH-sum of finite dimensional spaces. Also each MATH is isomorphic to MATH and thus MATH is reflexive. It remains to show that if MATH is a normalized weakly null sequence in MATH and MATH then a subsequence is MATH-equivalent to the unit vector basis of MATH. By a gliding hump argument it suffices to prove this in a fixed MATH. We proceed by induction on MATH. For MATH the result is clear since MATH is isometric to MATH. Assume the result has been proved for MATH. By passing to a subsequence and perturbing we may assume that MATH is a normalized block basis of the node basis for MATH. Let MATH rapidly. For MATH let MATH be the basis projection of MATH onto MATH, MATH. Passing to a subsequence we may assume that MATH and from the definition of MATH we have MATH. Choose MATH so that MATH. Passing to a subsequence of MATH we may assume that there exist integers MATH so that CASE: MATH CASE: MATH for MATH CASE: MATH is within MATH of MATH. CASE: If MATH, MATH, then if MATH, MATH is MATH-equivalent to the unit vector basis of MATH. CASE: If MATH and MATH then MATH is MATH-equivalent to the unit vector basis of MATH. CASE: MATH CASE: If MATH and MATH then MATH for all MATH. CASE: If MATH and MATH then MATH for MATH. REF use the induction hypothesis and the fact that for all MATH, span-MATH, MATH is isometric to MATH. Our conditions are sufficient to yield (for suitably small MATH's) that MATH is MATH-equivalent to the unit vector basis of MATH. We omit the standard yet tedious calculations.
math/0002219	We choose an appropriate sequence MATH depending upon MATH and the basis constant MATH of MATH. MATH is chosen by the lemma for MATH and MATH. We choose MATH by the lemma for MATH and MATH and so on. If MATH the lemma yields for MATH, MATH with MATH and MATH with MATH. We then let MATH and for MATH, MATH. Thus MATH and so REF holds. To see REF we note the following MATH . Thus MATH which can be made less than MATH. This yields REF.
math/0002219	We first show that MATH embeds into MATH for some sequence MATH of finite dimensional spaces. Then to obtain the MATH estimate we adapt an averaging argument similar to the one of CITE. Applying REF to the set MATH we find a reflexive space MATH with an FDD MATH with basis constant MATH which isometrically contains MATH and MATH so that whenever MATH satisfies MATH for some sequence MATH in MATH it follows that MATH is MATH-equivalent to the unit vector basis of MATH. Let MATH be the blocking given by REF . Let MATH, MATH with MATH for all MATH. Choose MATH and MATH as in REF . It follows from REF that (for MATH's sufficiently small) that MATH and MATH . Let MATH. Since MATH it follows that MATH embeds isomorphically into MATH. We now renorm MATH so as to contain MATH isometrically. Thus MATH has MATH as an FDD and there exists MATH so that if MATH is any block basis of a permutation of MATH then MATH . We repeat the first part of the proof. Let MATH. From REF we may assume that there exist MATH so that if MATH satisfies MATH for some MATH then MATH is MATH-equivalent to the unit vector basis of MATH. Moreover we may assume that this is valid for any further blocking of MATH. From now on we will replace MATH by the finite codimensional subspace MATH and MATH by MATH and replace MATH by MATH. We will show that this new MATH can be MATH-embedded into a MATH sum of finite dimensional spaces. Let MATH be the blocking given by REF . Thus (for appropriately small MATH's) from REF we have that if MATH there exist MATH so that MATH where MATH is the expansion of MATH with respect to the FDD MATH for MATH. Chose MATH so that MATH . For MATH and MATH set MATH (using MATH if MATH) and let MATH. Let MATH. We shall prove that MATH-embeds into MATH where MATH as MATH which will complete the proof. To do this we first define maps MATH for MATH. If MATH is the expansion of MATH with respect to MATH we let MATH . Let MATH and MATH, MATH as above. Write MATH as the expansion of MATH with respect to MATH. Let MATH be given by REF (with respect to MATH). From several applications of the triangle inequality and REF we have MATH . Similarly one has MATH . Finally we define MATH, by MATH. Note that MATH using REF. Similarly one deduces that for MATH it follows that MATH.
math/0002219	The implications REF follow from REF and its proof. If we prove REF then REF will follow. Assume that MATH, MATH, MATH and MATH are given as in the statement of REF . By passing to the renorming MATH, MATH, for MATH we can assume without loss of generality that MATH is isometric to a subspace of MATH. We will show that MATH satisfies REF . Since MATH is isomorphic to a subspace of MATH of finite codimension the claim will follow. Thus let MATH be an isometric embedding and let MATH be a normalized weakly null tree in MATH. We will need the following observation. If MATH is a normalized and weakly null sequence in MATH, then there are normalized weakly null sequences MATH and MATH in MATH and MATH respectively so that, MATH and MATH for MATH. To see this use the NAME theorem to choose a normalized sequence MATH in MATH so that MATH. The sequence MATH is weakly null. Indeed, otherwise we could choose a MATH, MATH, a subsequence MATH and a weakly null sequence MATH in MATH so that MATH for all MATH. Thus, MATH, which implies that MATH and therefore that MATH. Since MATH, we get a contradiction. Then we choose MATH so that MATH. By a similar argument we have that MATH is also weakly null. Using the claim we can find a normalized weakly null tree MATH in MATH and a normalized weakly null tree MATH in MATH, so that MATH and MATH for MATH. Given a MATH we can choose a branch MATH so that MATH is MATH equivalent to the unit vector basis of MATH, and MATH is MATH equivalent to the unit vector basis of MATH. This easily implies that MATH is MATH equivalent to the unit vector basis of MATH.
math/0002219	Let MATH denote the set of all closed subsets of MATH for which MATH holds. For MATH we denote the MATH-neighborhood by MATH and observe the following equivalences MATH . Thus MATH. If MATH then MATH is compact and is contained in the open covering MATH. Thus there exists a finite MATH so that MATH and thus MATH which implies by REF that Player I has a winning strategy for MATH. By the uniform continuity of MATH, MATH can be chosen small enough so that MATH contained in a given neighborhood of MATH which finishes the proof of the first part. The remainder of the proposition follows easily from REF .
math/0002219	Since there exists MATH so that the unit vector basis of MATH is in MATH (see CITE) we have that MATH must be this unit vector basis. In turn this condition (see CITE or CITE) implies that MATH contains an isomorph of MATH (MATH if MATH) and so MATH. Let MATH, a reflexive space with an FDD MATH. The condition on MATH yields that for all MATH there exists MATH so that if MATH then there exists MATH so that if MATH then MATH is MATH-equivalent to the unit vector basis of MATH. From this it follows that MATH satisfies the hypothesis of REF with MATH and thus the theorem follows.
math/0002221	Let MATH be the least integer such that MATH. For MATH we have MATH. Then, MATH .
math/0002221	We will show that MATH is of weak type MATH. By similar arguments, one gets that MATH is bounded from MATH into MATH. In this case, one has to use a version of the NAME decomposition in the lemma above suitable for complex measures. For simplicity we assume MATH. Let MATH and MATH. Let MATH be the almost disjoint family of cubes of REF . Let MATH be the smallest MATH-doubling cube of the form MATH, MATH. Then we can write MATH, with MATH and MATH where the functions MATH satisfy REF and MATH. By REF we have MATH . So we have to show that MATH . Since MATH, MATH and MATH, using some standard estimates we get MATH . Let us see that MATH too. On the one hand, by REF and using the MATH boundedness of MATH and that MATH is MATH-doubling we get MATH . On the other hand, since MATH, if MATH, then MATH, and so MATH . By REF , the first integral on the right hand side is bounded by some constant independent of MATH and MATH, since there are no MATH-doubling cubes of the form MATH between MATH and MATH. Therefore, REF holds. Then we have MATH . Therefore, MATH . The corresponding integral for the function MATH is easier to estimate. Taking into account that MATH, we get MATH . Also, we have MATH . Now, by REF we get REF .
math/0002224	As MATH preserves MATH and MATH, it (and all its multiples by a constant) preserves the Riemannian metric defined by MATH (we replace, if necessary, MATH with MATH, MATH with MATH, such that MATH is positive definite) with respect to which MATH is Killing and MATH. MATH is then identified to a MATH-invariant, anti-symmetric endomorphism of MATH, thus equal to MATH, for a function MATH. But MATH as MATH, thus MATH .
math/0002224	We denote by MATH, respectively, MATH, the NAME connection, respectively, the NAME REF connection associated to the positive MATH . NAME vector field MATH. We have MATH . The claimed result now follows from a straightforward computation.
math/0002224	We have to prove that, if MATH is the NAME curvature of the NAME surface MATH, then it satisfies MATH iff MATH is constant (we omit the indices referring to MATH, as we only use the metric, the NAME connection and the operator MATH on MATH in this proof). First we prove the following fact: if MATH, then MATH . We need to prove that MATH is anti-symmetric, thus it is enough to check that MATH, but this is equal to MATH . If the genus of MATH is greater that REF, it admits no non-zero Killing vector field. If MATH is a torus, non-zero vector fields vanish nowhere, but MATH should vanish in the critical points of MATH (for example, the maximum points).
math/0002224	The almost-complex structure MATH on MATH is defined as follows: MATH, and MATH, MATH being the unitary, oriented, generator of MATH. The product metric is, then, given by its NAME form MATH . It is easy to prove that MATH is integrable and that MATH such that the NAME form of the Hermitian metric MATH, is, by definition, MATH, thus parallel CITE, CITE, CITE. The NAME vector field MATH (the metric dual of the NAME form) is a holomorphic vector field CITE, and it is automatically given (up to a positive constant) by the complex geometry of the surface MATH CITE, which can only be of the three kinds enumerated in the Proposition. In the first two cases, it follows from the descriptions of REF, see also REF of the same paper, that MATH always has closed orbits. In the same cases, MATH and MATH generate the tangent space of the fibers of the elliptic fibration, and the Riemannian product situation can only occur if generic fibers are biholomorphically MATH (the first factor corresponds to the orbits of MATH, and the second (of unknown length MATH) to the orbits of MATH). If MATH is primary, then MATH contains no non-trivial finite subgroup, hence MATH is a principal elliptic bundle over a complex curve of positive genus (CITE, see also CITE). As any NAME metric is regular on these surfaces, CITE, REF , it immediately follows that the NAME metric on MATH is regular (in the primary case). If MATH is a NAME surface, then it is necessarily of class REF (see CITE), that is, it is the quotient of MATH by a group MATH, generated by a normal finite subgroup MATH and by a holomorphic contraction MATH of MATH as in the Proposition, in general with MATH, MATH. We first need the orbits of MATH to be closed: as MATH we obtain that MATH and MATH, where MATH are both primary MATH-roots of unity for MATH, that is, MATH, for MATH. Then it is clear that MATH is a finite quotient of MATH. In the primary case, MATH is the primary NAME surface MATH, where the action of MATH is generated by MATH. As the generic orbits of MATH cross the orthogonal hypersurfaces (tangent to MATH) MATH times, it follows that MATH and MATH. As a NAME surface is a finite quotient of a primary one, we get MATH in general, as claimed.
math/0002224	Consider a NAME structure on MATH, with the usual notations. see REF We suppose, with no loss of generality, that MATH is primary, thus MATH is a MATH-bundle over a complex curve MATH, such that the NAME vector field MATH is tangent to the fibers, which all are of equal length MATH (and we can suppose MATH). We want to prove that any function MATH satisfying MATH is constant. The function MATH, defined by MATH where MATH is a fiber of the projection MATH, satisfies MATH, thus, from REF , it is constant (recall that MATH is of positive genus). If MATH (which we can assume by adding a constant to the bounded function MATH), then MATH (see REF ) is another MATH . NAME vector field on MATH, that has thus circular orbits see REF of an equal length MATH as MATH is a primary NAME manifold, it is regular, see REF. Suppose MATH. We can also compute MATH where MATH is the orbit associated to MATH (a point in the orbit space MATH of MATH - for topological reasons, the projection MATH is still a principal MATH-fibration). MATH is also a constant, and we would like to relate it to MATH, and to the length MATH of the orbits of MATH; the reason for that is the following Lemma: CASE: If two MATH . NAME vector fields coincide on a common orbit, they coincide everywhere. CASE: Suppose that MATH and MATH are two arbitrary MATH . NAME vector fields, and that the length of the orbits of each of them (measured in the corresponding metrics) is equal to REF. If MATH and MATH, then MATH everywhere. Remark. The point REF. is a particular case of REF.; it is explicitly formulated as it will often be used throughout the proof of REF . The volume form of the original NAME metric is MATH, for MATH, and the volume form of the deformed metric is MATH, where MATH (see REF ). We denote by MATH the volume of MATH, equal as the fibers of the MATH-bundle MATH are of length REF to the volume of MATH. The volume MATH of MATH is then equal to MATH . On the other hand, from NAME 's inequality, we have MATH . But the integral of MATH on MATH is equal to the integral of MATH on MATH, thus to MATH. We get then MATH . We have thus: MATH . In each of REF . or REF., both MATH and MATH have orbits of length REF. On the other hand, we also have MATH, thus we can apply the implication REF for MATH, too (starting from MATH): MATH . We have thus equality in REF , which implies that MATH is constant. It is known that the group MATH of MATH automorphisms of MATH is a NAME group CITE, CITE. We consider the connected subgroup MATH associated to the NAME subalgebra generated by MATH and MATH. This group acts on MATH by MATH automorphisms, and its orbits are connected. We consider the decomposition of MATH in orbits of MATH; they are of three kinds, according to their dimension: CASE: circles MATH; CASE: immersed surfaces MATH; CASE: open orbits. Suppose MATH is a MATH-orbit of dimension REF in MATH; it contains all the circles MATH that intersect it, and it is immersed, hence it projects, via the bundle projection MATH, onto an immersed connected curve MATH. There are two cases: CASE: MATH is an open segment; CASE: MATH is a circle. In both cases, we define MATH to be a unitary continuous vector field in MATH, and MATH to be its horizontal lift to MATH; we have then MATH, and, as MATH from REF , we get MATH thus MATH is constant on the orbits of MATH. If MATH is a circle, then it has a finite length MATH. Assume that MATH is not constant on MATH; then MATH has a regular value MATH. Fix a point MATH, such that MATH, and fix some other points MATH close to it, such that MATH are still regular values of MATH on MATH. All these numbers MATH close to MATH are images of compact immersed curves, all diffeomorphic and horizontal (that is, tangent to MATH). Consider MATH to be their connected component containing the points MATH; then MATH are precisely the orbits of MATH starting from MATH; they all have thus equal length (which is an integer multiple of MATH, as they all cover MATH). Note that this length is measured by the Riemannian metric MATH, and that for MATH the lengths need not be equal any longer: Indeed, the curves MATH are still the same, but the unitary vector fields on them, for MATH, are MATH see REF, and MATH is precisely non-constant in a neighbourhood of MATH. On the other hand, the group MATH, the orbit MATH and the circles MATH can also be considered starting from the NAME metric MATH, in which case an analog reasoning yields that they have equal length, in the metric MATH (the regular points of MATH are also regular points for MATH), which leads to a contradiction. We obtain thus MATH on MATH. We can suppose that MATH is sufficiently close to MATH (by replacing MATH, if necessary, with MATH, for MATH small), such that the orbits of MATH and the ones of MATH are homotopic circles in the torus MATH. We can even suppose that one particular orbit MATH of MATH lies in a small tubular neighbourhood of an orbit MATH of MATH. The projection of this neighbourhood onto MATH induces then a diffeomorphism MATH, such that MATH (because MATH on MATH). Thus MATH thus we can apply the point REF . This implies that MATH (respectively, MATH, which is the same thing). The same reasoning can be applied if the projection MATH of MATH in MATH is a segment of finite length for the metric induced by MATH (and also for the metric induced by MATH, if we suppose MATH and MATH to be sufficiently close to each other). Suppose now MATH is an immersed open curve in MATH of infinite length. We have MATH where MATH is a function, equal to MATH REF. In the following lines, MATH stands for MATH; it is unitary and contained in MATH: MATH thus, by summation, we get MATH, but MATH and it still lies in MATH, hence it is collinear with MATH; but we know from REF that MATH. We have thus MATH . We recall that MATH, thus MATH, too, are constant on the orbits of MATH, which are curves of infinite length. If MATH on such an orbit, the function MATH satisfying the equation above is not bounded, but this is impossible as MATH is compact. Thus MATH is constant on MATH. The rest of the argument used for the case when MATH had finite length can also be applied here. We have thus proven that MATH cannot admit any REF-dimensional orbits. We are going to prove now that the number of REF orbits (which are vertical circles) is finite. Suppose we have an infinite number of such orbits; then we can extract a sequence of points MATH, such that MATH, are orbits of MATH. Then so is MATH, as the union of all REF-dimensional orbits in MATH is closed. We want to prove that MATH is constant on MATH. The normal bundle of MATH in MATH is also the restriction to MATH of the plane bundle MATH, and will still be denoted by MATH. It is a complex line bundle, and its metric depends on the MATH . NAME vector field in MATH that induces a NAME metric on MATH. Consider two such vector fields MATH and MATH (where MATH on MATH), and fix two arbitrary different points MATH. The MATH (integral) actions corresponding to MATH, respectively, MATH on MATH induce two different diffeomorphisms MATH, respectively, MATH, where MATH are contractible surfaces in MATH, locally defined around MATH, tangent to MATH at MATH and transverse to the orbits of MATH. The choice of such surfaces is not essential, as we are interested in the differential at MATH of MATH, respectively, MATH. Consider MATH, which is a diffeomorphism of a neighbourhood MATH of MATH in MATH into MATH. If we denote by MATH the intersections of MATH with MATH, we obviously have MATH . If MATH is a diffeomorphism from a neighbourhood of MATH in MATH into MATH that has a sequence, converging to MATH, of fixed points, then the differential MATH has at least one eigenvector corresponding to the eigenvalue MATH. By subtracting the inclusion MATH of MATH in MATH, we get a function MATH which has a sequence, converging to MATH, of zeros. Then the kernel of its differential at MATH is non-trivial. From REF and the previous Lemma, we conclude that it exists a non-zero MATH such that MATH . But these differentials are equal to the differentials of the MATH (integral) actions induced by MATH, respectively, MATH, on MATH, and these actions preserve the complex structure of MATH. Then MATH satisfies REF as well, so the two differentials MATH and MATH coincide. On the other hand, the MATH action induced by MATH preserves the corresponding metric MATH on MATH, and hence so does MATH. The same holds for MATH and MATH, so we get MATH. As MATH were arbitrarily chosen, MATH is constant on MATH, thus everywhere (REF.). So the only remaining situation is when MATH is a union of open orbits and a finite number of circular orbits of MATH; as MATH is connected, there needs to be only one open orbit MATH (dense in MATH). We study now the structure of the NAME algebra MATH. We will suppose that MATH acts effectively on MATH. Every element MATH can be written as MATH where MATH is a function, and MATH. Because MATH is a constant, we get a linear homomorphism MATH induced by the integral of the functions MATH along the fibers of MATH. The kernel of this homomorphism is a hyperplane MATH, and it contains all the brackets MATH, for any MATH; indeed, the function MATH is precisely the derivative along MATH of MATH, denoted by MATH, hence its integral on the orbits of MATH vanishes. CASE: The bracket with MATH, MATH, induces an automorphism of MATH (still denoted by MATH), which is MATH-diagonalizable, and whose eigenvalues are pure imaginary (hence non-zero); CASE: If MATH is the real part of an eigenvector of MATH, then the bracket MATH is a non-zero multiple of MATH. Because all orbits of MATH have length MATH, it means that the exponential of MATH, MATH, is contained in the isotropy subgroup of any point in the open orbit MATH, but the intersection of all these isotropy groups is trivial, as MATH acts effectively on MATH. In particular, MATH acts trivially on MATH, so the exponential of the endomorphism MATH is the identity. It follows that its eigenvalues are imaginary (integer multiples of MATH), and that its NAME decomposition reduces to the diagonal part. On the other hand, we know from REF that the only MATH . NAME vector fields commuting with MATH are multiples of MATH; therefore, if MATH for MATH, then MATH and MATH is non-singular, hence all its eigenvalues are non-zero. Remark. It follows that MATH is always even, and that MATH is the (commutative) product of a complex structure MATH on MATH with a diagonal matrix with real eigenvalues. CASE: If MATH is the real part of an eigenvector of MATH, then MATH is a multiple of MATH. On the other hand, MATH so MATH commutes with MATH, hence it is collinear to it (see REF ). Consider the case when the function MATH corresponds to the real part MATH of an eigenvector of MATH. Then MATH has dimension REF, as its NAME algebra is generated by MATH, and MATH. We will obtain a contradiction, hence the Theorem will follow. Denote by MATH the function associated to MATH; we have MATH . We also have MATH (where MATH are eigenvalues of MATH) hence MATH, restricted to any orbit of MATH, is a solution of the differential equation MATH in particular it is a sinusoid function: MATH for MATH an arc length parameter (for the NAME metric induced by MATH) on the fiber MATH, and its only critical points are the maximum and the minimum. Let us compute, from REF , the bracket MATH: MATH . As this has to be a constant multiple of MATH, MATH, it follows that, on a circular orbit of MATH (where MATH), we have MATH independently on the circular orbit. But, from REF , MATH, where MATH is a global constant, and MATH depends only on the orbit. It follows then that to all circular orbits MATH of MATH corresponds the same value of MATH, the amplitude of the sinusoid MATH. The only critical points of MATH are then its maximums and its minimums, obtained only on the circular orbits, with the values MATH; indeed, on the open orbit MATH, MATH has to be linearly independent of MATH, thus MATH. The function MATH has the following properties: CASE: it has only a finite set of critical points; CASE: any of these (isolated) points is either a maximum or a minimum. Then, after deforming MATH if necessary (in order to get a function with non-singular Hessian at these critical points), we obtain a NAME function MATH, with a finite set of MATH critical points (where MATH is the number of circular orbits of MATH), which are local maximums or local minimums; The topology of MATH is thus obtained by glueing MATH points to MATH REF-cells, which implies, as MATH is connected, that MATH and MATH is homeomorphic to MATH, which contradicts our hypothesis.
math/0002225	The claimed identities are conformally invariant : for REF it is obvious, and the conformal invariance of REF follows from REF ; to see that for REF , let MATH be a MATH-orthonormal oriented basis of MATH, such that MATH is a MATH-orthonormal basis on MATH giving the orientation as above. Then MATH, and, if we take MATH, the identity REF becomes MATH where angle brackets denote the scalar product induced by MATH. The tensors MATH, in the above form, are independent of the chosen metric MATH CITE, which depends on the normal vector MATH, supposed to be MATH-unitary. If MATH, for MATH, then the corresponding metric MATH, and also MATH, thus the identity REF for MATH is equivalent to the one for MATH. Remark. As MATH is the trace-free component of the Riemannian curvature contained in MATH, and is symmetric, it is enough to evaluate it on pairs MATH which are unitary and orthogonal for the metric MATH, therefore the check of REF will prove the Theorem. As MATH are MATH-eigenvectors in MATH (the space of trace-free endomorphisms of MATH), they are determined by the following formulas, where MATH is any oriented orthonormal basis of MATH: MATH where MATH is supposed to be a local extension, around a region of MATH, of the MATH-orthonormal frame used in REF . As MATH is self-dual, MATH is identically zero, thus, in the points MATH, we have MATH . It is a standard fact that, if MATH is umbilic, there is a local metric MATH in the conformal class MATH of MATH, such that, for MATH, MATH is totally geodesic. Without any loss of generality, because of the conformal invariance of the claimed identities (see above), we fix such a metric. Then we have MATH which, together with REF , implies that MATH, and thus proves the point REF in the Theorem. On the other hand, REF , together with REF , yield MATH . Let us compute now the normal derivative of MATH in a point MATH; we suppose that MATH are locally extended by an orthonormal frame, and that they are parallel at MATH we omit, for simplicity of notation, the point MATH in the following lines: MATH from REF . This is then equal to: MATH from the second NAME identity. Then we have MATH from analogs of REF . Then MATH from REF MATH from REF . Finally, from REF , we get MATH . This proves REF and the point REF in the Theorem. For the point REF , we use REF ; from the codifferential of MATH, only the derivative along the normal vector, MATH, can be non-zero, as MATH vanishes along MATH. This proves the Theorem.
math/0002225	The proof is different in the cases MATH and MATH; one of the reasons is that conformal flatness reduces, in higher dimensions, to the vanishing of the NAME tensor, while in dimension REF it is a higher-order condition more difficult to handle. The first step (common to all cases) is to prove that a small deformation (seen just as a compact submanifold of M, CITE) of such a compact null-geodesic MATH is still a compact null-geodesic, and to characterize the global sections of the normal bundle of MATH as locally determined by NAME fields. Let MATH be a (immersed) null-geodesic in MATH. Let MATH be a vector field along MATH. Then the condition MATH (where MATH) is independent of the metric MATH with respect to which we take the derivative MATH. The relation between two NAME connections (or, more generally, NAME structures) of metrics in the same conformal class, is given by CITE: MATH where MATH is a REF-form, and the rising of indices in MATH is made using the same (arbitrary) metric MATH as in the scalar product MATH. The Lemma immediately follows. Denote by MATH the normal bundle of MATH in M, and by MATH its subbundle represented by vectors orthogonal to MATH. Fix a metric MATH in the conformal class MATH. Let MATH be a NAME field along MATH, satisfying to the NAME equation MATH . It represents an infinitesimal deformation of MATH through null-geodesics if and only if MATH. MATH induces a section in the normal bundle MATH, or MATH if MATH in a point, hence everywhere. We want to prove that this section is independent of the connection MATH: The NAME equations on a null-geodesic MATH-M induce a second order linear differential operator MATH on MATH, which, restricted to the sections MATH such that MATH, depends only on the conformal structure MATH of M. In particular, MATH restricted to MATH is conformally invariant. For a NAME connection MATH of a local metric on M, we locally define the following differential operator on the sections of MATH: MATH by MATH. Because MATH is a null-geodesic, MATH induces a (local) differential operator on MATH, and we need to relate MATH to the corresponding operator MATH induced by another connection MATH. First we write MATH then we recall that another NAME connection MATH is related to MATH by REF , such that we get MATH which is identically zero modulo MATH, provided that MATH (the latter condition being independent of the NAME connection, according to the previous Lemma). Using the fact that MATH is the union of two contractible sets MATH (on each of which the NAME equation, with any initial condition - the same for MATH and for MATH - in MATH, has a unique solution - and these solutions necessarily coincide on the connected intersection MATH), we immediately get: Let MATH be an immersed null-geodesic, diffeomorphic to a projective line MATH. Then any local NAME field MATH with MATH induces a global normal field MATH on MATH. This has important consequences about the normal bundle of MATH in M, as NAME fields provide it with global sections; in particular, MATH is a line bundle admitting nowhere-vanishing sections, hence it is trivial; on the other hand, MATH is a MATH - rank bundle over MATH, admitting sections with any prescribed REF-jet (induced, again, by some appropriate NAME fields), hence MATH . For MATH, this is the general form of a vector bundle over MATH, according to REF; the condition MATH arises from the existence of sections of MATH with prescribed REF-jet. We are going to show later that all MATH are equal to REF. First we prove: NAME close to a compact, simply-connected one are also compact and simply-connected, and they are generically embedded. We consider the projectivized bundle MATH of the isotropic cone MATH. It is a standard fact CITE that any null-geodesic MATH has a canonical horizontal lift MATH (depending only on the conformal structure), such that MATH, where MATH is the projection, and MATH. Note that MATH is always embedded, even if MATH may have self-intersections (it is always immersed). The lifts of the null-geodesics of M consist in a foliation of MATH, which has a compact, simply-connected leaf, namely the lift MATH of our compact, simply-connected null-geodesic MATH. By NAME 's stability Theorem CITE, then there is a saturated tubular neighbourhood of MATH, diffeomorphic to MATH (where MATH is a polydisc), such that the leaves close to MATH are identified, via the above diffeomorphism, to the (compact and simply-connected) curves MATH. So all null-geodesics close to MATH are compact and simply connected. If MATH has self-intersections at the points MATH, we blow-up M at those points, and the lift of MATH is now embedded. So must be then the lifts of the null-geodesics close to MATH, as they are now deformations of the lifted (hence, embedded) curve. But, generically, such curves avoid the finite set of points MATH; the corresponding null-geodesics must have been embedded from the beginning. From now on, according to the previous Proposition, we may suppose that MATH is a compact, simply-connected, embedded null-geodesic. We compute the normal bundle of MATH in M, using the relation REF and the projection MATH, as follows: We have the following exact sequence of bundles: MATH where MATH is the normal subbundle of MATH represented by vectors tangent to the fibers of MATH and MATH is the normal bundle of MATH in MATH. In a point MATH, the fiber of MATH is equal to MATH, so the tangent space to it is isomorphic to MATH, for the projective variety MATH. Thus MATH as MATH. The central bundle in the exact sequence REF is trivial, because MATH is a leaf of a foliation. REF imply that the NAME numbers MATH are subject to the following constraint: MATH thus, as MATH, we have MATH. We have then: The normal bundle of a compact, simply-connected, null-geodesic MATH in M is isomorphic to MATH and all its global sections are induced by NAME fields MATH such that MATH. Moreover, the deformations of MATH as a compact curve coincide with the null-geodesics close to MATH. The last statement follows from the expression of the normal bundle, and REF: the normal bundle satisfies MATH, thus the space of deformations of MATH as a compact curve has dimension equal to MATH, which is precisely the dimension of the space of null-geodesics, defined locally, over a geodesically convex open set, as the space of the leaves of the horizontal foliation of MATH CITE. We conclude using the fact that all null-geodesics close to MATH are deformations of this one (as a compact, and simply-connected, curve). We return to the proof of REF . Consider first the case when the dimension of M, MATH. We are going to show that the NAME tensor of M, MATH, is identically zero (a special sub-case is MATH, when MATH). For simplicity, suppose first that MATH, and consider the fiber of MATH at an arbitrary point MATH: it has a non-degenerate conformal structure, induced from M, and the isotropy cone spans the whole fiber MATH (this still holds for MATH, but not for MATH). Let MATH be an isotropic line. We have: Let MATH be an open set of the null-geodesic MATH, on which local metrics MATH are well defined. If a (locally defined, over MATH) line subbundle MATH is parallel (or stable) for MATH, then it is parallel for MATH as well. The proof is a straightforward application of REF . Let MATH be linearly independent isotropic lines in MATH. According to the previous Lemma, and to the fact that MATH is simply-connected, their parallel transport over MATH does not depend on any NAME connection of a metric in the conformal class. We get, thus, a global splitting MATH where the line bundles MATH are all isotropic and parallel. All these bundles are isomorphic to MATH. As their sections are also sections of MATH, they cannot vanish at more that REF point, for each MATH, thus MATH. On the other hand, the sum of all MATH's has to be MATH, thus MATH. Let MATH be a section of MATH; from REF , it is locally represented by a NAME field MATH, for the metric MATH. From the NAME equation, by taking the scalar product with MATH, we get: MATH and it is easy to see that, because of the fact that all scalar products involving MATH and MATH are REF, the term MATH of the curvature REF satisfies the above relation identically. This equation holds, at MATH, for any isotropic vector MATH, but we may consider also other compact, simply-connected, null-geodesics MATH, containing MATH, and close to MATH (namely, small deformations of the compact curve MATH). For any REF-plane MATH-M, we denote by MATH the sectional curvature of MATH: MATH and we have seen that, if MATH is totally isotropic, MATH depends only on MATH (and on the metric MATH only via the scalar product MATH . ). If MATH, the NAME tensor at MATH, MATH, is determined by the sectional curvatures MATH, for MATH a small neighbourhood of the totally isotropic arbitrary REF-plane MATH in the NAME of totally isotropic REF-planes at MATH. Remark. A similar statement holds in dimension REF, but in that case, the NAME of totally isotropic REF-planes has REF connected components; as a consequence, MATH is determined by the sectional curvatures of MATH-planes, and MATH by the sectional curvatures of the MATH-planes REF . This reduces to the claim that MATH if and only if MATH which is a problem of linear algebra. If we consider the space MATH of all curvature tensors MATH satisfying MATH, then this is a vector space, which is invariant to the action of MATH (which is the NAME algebra infinitesimal action corresponding to the action of MATH - note that the NAME of totally isotropic planes is preserved by this action). But there are only REF MATH-irreducible components of the space of curvature tensors, and we have seen that for the NAME tensors MATH, the totally isotropic planes always have zero sectional curvature. Then either any NAME tensor satisfying REF is zero at MATH, or MATH contains the whole space of curvature tensors. The latter possibility can easily be excluded by an example of a curvature tensor MATH satisfying: MATH . From REF and the previous Lemma we conclude that MATH for any MATH contained in a compact, simply-connected, null-geodesic; but we know from REF that the set of such points contains a neighbourhood of MATH, thus, by analyticity of MATH, it vanishes identically. The proof is similar in dimension REF (note that, in the self-dual case, we can retrieve the result by applying REF ; this is how we shall proceed for the case of dimension REF, using the NAME correspondence); the difference with the higher-dimensional case is that the splitting REF is canonical, MATH corresponding, say, to the MATH-plane MATH containing MATH, and MATH to the MATH-plane MATH containing MATH. It is important now that each of MATH is isomorphic to MATH, because the vanishing of MATH implies MATH, and the vanishing of MATH implies MATH REF . Thus the manifold MATH is conformally flat. Consider now the particular case where MATH. We are going to use the NAME correspondence, then REF , to prove that M is then conformally flat. Note that we cannot use directly REF and the above proven result for self-dual manifolds, as the ambient self-dual manifold MATH can only be defined for a civilized (for example, geodesically convex) CASE: We cover MATH with geodesically convex open sets MATH, such that: MATH where MATH is still geodesically convex (with respect to some particular NAME connection). This is possible by choosing MATH, small enough CITE. Then we choose a relatively compact tubular neighbourhood MATH of MATH, such that its closure is covered by the MATH's. We may choose this tubular neighbourhood small enough to be contained in the projection MATH, from MATH, of a saturated neighbourhood (see REF ) of the lift MATH. We consider then the twistor spaces MATH, the spaces of null-geodesics of MATH. The compact, simply-connected, null-geodesics close to MATH identify (diffeomorphically) the neighbourhoods of MATH with the space MATH of the deformations (contained in MATH) of MATH as a compact curve. We can see then (a small open set of) MATH as an open set common to all the MATH's: Following NAME, we define the self-dual manifolds MATH as the spaces of projective lines in MATH, with normal bundle MATH. Then MATH is an umbilic hypersurface in MATH. The local twistor spaces MATH admit contact structures, which coincide on MATH, and contain projective lines MATH corresponding to points MATH. If we denote by MATH the twistor space of MATH, then MATH is identified to an open set in MATH and, at the same time, to an open set in MATH, in particular the twistor lines MATH and MATH (corresponding to the same point MATH) are identified. Thus MATH and MATH coincide and we denote by MATH this (non-compact) curve in MATH, and by MATH the canonical contact structure of MATH (restricted from the ones of MATH). Remark. We already have obtained that the integral MATH-cone ( that is, the union of twistor lines passing through MATH and tangent to MATH, see the comment after REF ) corresponding to MATH is a part of a smooth surface: the union of the lines MATH, MATH. Thus, from REF , the NAME tensor MATH of the self-dual manifold MATH vanishes on the MATH-planes generated by MATH. But this is nothing new: we know, from REF , that MATH vanishes on MATH. We intend to apply REF to prove that MATH vanishes on points close to MATH, but generically in MATH. We do that by showing that the integral MATH-cones corresponding to planes MATH close to MATH are parts of smooth surfaces, then we conclude using REF . First we choose Hermitian metrics MATH on MATH, such that they coincide (with MATH) on MATH. We have a diffeomorphism between MATH and MATH, so we choose relatively compact open sets in MATH, covering MATH, with the following properties: As the metrics MATH induce metrics on MATH, we first choose a small enough distance MATH such that CASE: MATH, there is a sub-covering MATH of MATH such that the ``tubular neighbourhoods" MATH are compact (MATH is the distance between MATH and MATH, and MATH is the ``orthogonal projection" - for the Hermitian metric - from MATH to MATH; it is well defined because of the condition below); CASE: MATH is less than the bijectivity radius of the (Hermitian) exponentials for the points of MATH in MATH, and for the points of MATH in MATH (if MATH . ). We have then For any MATH, MATH such that the curves MATH, MATH are tangent to the same direction in MATH, the respective curves MATH coincide. We first note that the projection MATH from MATH to MATH is equivalent to the MATH - orthogonal projection of the direction of MATH to a direction in MATH, so MATH; thus MATH belongs to both MATH and MATH, and we use again the twistor space MATH to conclude that MATH and MATH are ``restrictions" to MATH of the same projective line (as they both have the same tangent space at MATH) MATH, for a point MATH. Now we have a tubular neighbourhood MATH of MATH, of radius MATH, such that, for any REF-plane MATH, the conditions in REF are satisfied (considering any of the local twistor spaces MATH). We recall that, via the NAME correspondence, a point MATH in the twistor space of MATH, MATH, is identified to the point MATH in the twistor space of MATH, still denoted by MATH. They correspond to the null-geodesic MATH, respectively, to the MATH-surface MATH, such that MATH CITE, CITE. The planes MATH above are included in MATH, and they correspond to null-geodesics in MATH contained in MATH CITE, CITE. By REF , we conclude that the NAME tensor MATH of MATH vanishes along all null-geodesics of MATH, close (in MATH) to MATH and included in the MATH-surface MATH, determined by MATH. This means that MATH vanishes everywhere on MATH. By deforming MATH, we obtain that MATH vanishes on a neighbourhood of MATH in MATH, hence MATH is conformally flat. It follows from REF that MATH, hence M, is conformally flat.
math/0002226	The proof is trivial.
math/0002226	It is follows from the obvious relations: MATH .
math/0002226	On account of REF we rewrite the equation MATH, as an operator equality in the space MATH: MATH . Substituting REF we get MATH . It is not hard to prove that MATH . Then from REF we have MATH . Substituting REF we get REF .
math/0002226	Using REF we rewrite the differential REF in the form of the operator equality in the space MATH: MATH . In view of REF is equivalent to the following operator equality in the space MATH: MATH . It is obvious from REF that it is sufficient to prove that the right-hand side of REF vanishes. From REF it follows that the latter is true.
math/0002234	Let MATH be an isometric dilation of MATH, and REF hold. Suppose that MATH is a subspace of MATH, and MATH, is an isometric dilation of MATH. Since for any MATH and MATH we have MATH, and MATH is invariant under MATH. Therefore, MATH and MATH. Thus, MATH is a minimal isometric dilation of MATH. For the rest of this Proposition it is sufficient to prove that if MATH is a minimal uniform isometric dilation of MATH then REF is true. The right-hand side of the equality in REF (denote it by MATH) is an invariant subspace in MATH under operators MATH for all MATH. If MATH is a uniform isometric dilation of MATH then MATH is also a uniform isometric dilation of MATH. Indeed, for any MATH and MATH we have MATH . Besides, REF implies that MATH is a uniform isometric dilation of MATH. If MATH is a minimal uniform isometric dilation of MATH then MATH for all MATH, and the proof is complete.
math/0002234	Let MATH be a unitary dilation of a linear pencil MATH, and REF hold. Suppose that MATH is a subspace of MATH, and MATH, is a unitary dilation of a linear pencil MATH. In the same way as in REF we show that MATH is invariant under MATH and MATH for all MATH. Therefore, MATH and MATH. Thus, MATH is a minimal unitary dilation of MATH. For the rest of this Proposition it is sufficient to prove that if MATH is a minimal uniform unitary dilation of MATH then REF is true. The right-hand side in REF (denote it by MATH) is a reducing subspace in MATH for MATH. In the same way as in REF we can show that MATH, is a uniform unitary dilation of MATH. If MATH is a minimal uniform unitary dilation of MATH then MATH for all MATH, and the proof is complete.
math/0002234	By REF the equality in REF is true. Since MATH is a minimal uniform unitary dilation of MATH (see our remark in the beginning of this Section), by REF we have REF. Since REF holds, we get MATH and by REF MATH is a minimal unitary dilation of a pencil MATH (of course, a dilation of a dilation of MATH is again a dilation of MATH). Since for any MATH is a uniform dilation of MATH if and only if MATH is a uniform dilation of MATH, and the proof is complete.
math/0002235	By NAME 's approximation theorem CITE, one can find a convergent power series mapping MATH such that MATH and for all MATH, MATH in MATH. Let MATH so that if MATH then MATH. Since the radius of convergence of the MATH, MATH, is at least MATH, the radius of convergence of any MATH, MATH, is at least MATH. Thus, the family MATH has a radius of convergence at least equal to MATH and for all MATH with MATH, one has MATH, MATH.
math/0002235	We reproduce here the arguments of CITE. Write MATH where MATH is a MATH complex-analytic matrix such that MATH; that is, MATH. By assumption, we know that we have MATH. This implies that one can find an integer MATH such that if MATH is any formal power series which agrees up to order MATH with MATH then MATH. For this integer MATH, according to NAME 's approximation theorem, one can find a convergent power series MATH satisfying MATH and agreeing with MATH up to order MATH. By REF , we get MATH in MATH. Since MATH, we obtain MATH and thus MATH is convergent.
math/0002235	We write the expansion MATH . For the sake of clarity, we shall first give the proof of the Proposition in the case MATH. The case MATH. We restrict all the identities given by REF to the MATH-dimensional subspace MATH . This gives, for any multiindex MATH, MATH where MATH is given by REF . Observe that for each multiindex MATH, MATH is convergent. To show that MATH is convergent, we claim that it suffices to show that there exists MATH and MATH such that the radius of convergence of each MATH is at least MATH, and such that the following NAME estimates hold: MATH . Indeed, if REF holds, then the formal holomorphic power series MATH defines a convergent power series in MATH. (Here and in what follows, for any MATH and for any MATH, MATH denotes the euclidean ball centered at REF in MATH of radius MATH.) Moreover, by REF , we have for any multiindex MATH, MATH in MATH and hence MATH . This proves that under REF , MATH is a convergent power series in MATH. It remains to prove REF . Since MATH is holomorphic in a neighborhood of MATH, in view of REF , one can find MATH and a constant MATH such that for any multiindices MATH, MATH, MATH . In view of REF (in the case MATH), we can apply REF to conclude that there exists MATH such that the family MATH is convergent in MATH and such that REF holds with MATH. This finishes the proof of REF in the case MATH. The case MATH. We proceed now to the proof of REF for general MATH. For this, we need the following lemma. Under the assumptions of REF, for any multiindices MATH, MATH, the formal holomorphic power series MATH is convergent. We prove the Lemma by induction on MATH (for any multiindex MATH). For MATH, the statement follows from REF , as we previously noticed. Let MATH. For MATH, using REF , we obtain for MATH, MATH . If we apply MATH to this equation, we obtain MATH . One can easily check that this implies that there exist a polynomial MATH such that the left-hand side of REF is equal to MATH where MATH, MATH. Furthermore, we observe that the right-hand side of REF can be written in the form MATH where MATH is a convergent power series. Restricting REF to MATH, one obtains MATH . The induction hypothesis tells us that for any multiindex MATH and for any multiindex MATH such that MATH, the formal holomorphic power series MATH is convergent. Thus, we obtain the desired similar statement for MATH, for any multiindex MATH. We come back to the proof of REF . For all multiindices MATH, we put MATH . By REF , the MATH are convergent. We now fix MATH, MATH. We want to prove that MATH is convergent in some neighborhood MATH of MATH. For this, as in the case MATH, it suffices to prove that one can find MATH and MATH such that the radius of convergence of the family MATH is at least MATH and such that the following estimates hold: MATH . We first observe that for any multiindex MATH with MATH, there exists a universal polynomial MATH, such that MATH . This means in particular that the polynomials MATH, MATH, are independent of MATH. Putting MATH and MATH in REF , we obtain MATH . Recall that MATH is fixed. For MATH, consider the convergent power series of the variables MATH defined by MATH . Let MATH. In view of REF , each MATH, MATH, is convergent in MATH. Moreover, for MATH, we have the following estimates MATH . Put MATH . This implies that in MATH, the following estimates hold for some suitable constant MATH: MATH . From this, we see that there exists MATH such that for MATH, MATH . In view of REF , we have for any multiindex MATH, MATH as formal power series in MATH. Thus, in view of REF , we are in a position to apply REF to conclude that there exists MATH such that the family MATH is convergent in MATH, in which, moreover, the desired estimates REF hold. This implies that MATH. The proof of REF is thus complete.
math/0002235	First observe that MATH is a formal power series solution of the analytic system in the unknown MATH, MATH . Thus, an application of NAME 's approximation theorem CITE gives, for any positive integer MATH, a MATH-tuple of convergent power series MATH solution in MATH of REF , and which agrees up to order MATH with MATH. The Lemma follows by noticing that MATH in MATH if and only if MATH is solution of REF . The proof of REF is complete.
math/0002235	Restricting the identity REF to the set MATH where MATH, MATH, are the NAME sets mappings defined by REF , we obtain MATH . Here, MATH is the formal power series defined by REF . We would like to apply REF to the formal REF with MATH, MATH, MATH, MATH, MATH, MATH, MATH and MATH . For this, one has to check that any derivative of the formal holomorphic power series MATH with respect to MATH evaluated at MATH is convergent with respect to the variables MATH, MATH and MATH. All these derivatives involve derivatives of MATH at MATH (which are convergent) and derivatives of the form MATH, for MATH. Because of the reality REF and the definition of MATH and MATH given by REF , we have MATH . This implies that for each MATH, we have MATH with the right-hand side being convergent in MATH by REF . Thus, by REF , we have, for any positive integer MATH, a convergent power series mapping, denoted MATH, which agree up to order MATH with MATH and such that MATH . Since MATH is a convergent power series mapping, in order to show that MATH, it suffices to show by REF that for MATH large enough the generic rank of the holomorphic map MATH is MATH. For this, note that since MATH is minimal at REF, the map MATH is of generic rank MATH, and thus, by the form of MATH given in REF , the holomorphic map MATH is of generic rank MATH (see CITE). Moreover, since MATH is invertible, we have MATH, which implies that the rank of the formal map MATH is MATH (at the origin). From this, we see that the formal map MATH has rank MATH. (The rank of such a formal map is its rank in the quotient field of MATH.) Since MATH agrees up to order MATH with MATH, we obtain that for MATH large enough, the mapping REF is of generic rank MATH. This completes the proof of REF .
math/0002235	By the NAME expansion REF and by REF , we obtain that all the MATH are convergent in a common neighborhood MATH of MATH. Since MATH is holomorphically nondegenerate, by CITE, there exists MATH, MATH, MATH, such that MATH . Since MATH is a formal biholomorphism, this implies that MATH as a formal power series in MATH. Put MATH and MATH, MATH. Observe that since MATH is convergent, MATH for MATH. Moreover, since MATH, MATH, in MATH, by REF , we may apply REF to conclude that MATH is convergent.
math/0002235	We again use the notations of REF. As in the proof of REF , we have, using the expansion REF , MATH . Thus, we know, by REF , that for any multi-index MATH, MATH is convergent in some neighborhood MATH of MATH in MATH. We choose MATH, MATH, of generic maximal rank equal to MATH in a neighborhood MATH of MATH in MATH (see CITE). Then, if we define MATH, MATH, we obtain the desired statement of the Theorem.
math/0002235	The implication MATH is clear. The other implication is equivalent to the following proposition.
math/0002235	Let MATH be as in the proposition. We can assume that, near MATH, MATH where each MATH. To this complex analytic set MATH, as is customary, we associate the following ideal of MATH defined by MATH . By the NAME property, we can assume that MATH is generated by a family MATH. Furthermore, by the CITE, for MATH, there exists an integer MATH such that MATH is in the ideal generated by the MATH, MATH. This implies that for MATH, MATH in MATH. The following result, a consequence of the NAME approximation theorem, is contained in REF . Assume that the height of the ideal MATH (generated by the family MATH in MATH) is equal to MATH. Then, any formal solution MATH, MATH, of the system MATH (in the unknown MATH) is convergent. Thus, to obtain the convergence of our original formal power series MATH, it suffices to check that the height of MATH is equal to MATH. Since MATH is a local regular ring of NAME dimension MATH, by REF, we have the formula MATH where MATH is the NAME dimension of the ring MATH. Since the NAME dimension of such a ring coincides with the dimension of the complex analytic set MATH (compare CITE, p. REF ), which is, here, equal to MATH, we obtain that the height of MATH is MATH. This completes the proof of REF , and hence, the proof of REF .
math/0002235	By REF , there exists a convergent power series mapping MATH such that for each MATH, MATH is convergent, MATH. Then, the graph of MATH is formally contained in the complex analytic set MATH . Let MATH be the decomposition of MATH into irreducible components. For any positive integer MATH, one can find, according to the NAME approximation theorem CITE, a convergent power series mapping MATH defined in some small neighborhood MATH of MATH in MATH, which agrees with MATH up to order MATH (at REF) and such that the graph of MATH, denoted MATH, is contained in MATH. Since MATH is contained in MATH, it must be contained in an irreducible component of MATH. Thus, by the pigeonhole principle, at least one subsequence of MATH is contained in one of such irreducible components, say MATH. There is no loss of generality in assuming that such a subsequence is MATH itself. We first observe that this implies that the graph of MATH is formally contained in MATH. Moreover, since MATH is a formal biholomorphism, the family MATH is also a family of local biholomorphisms. In particular, this implies that the generic rank of the family of holomorphic functions MATH is MATH. As a consequence, if MATH is close enough to REF in MATH and is chosen so that the rank of the preceding family at MATH equals MATH, the implicit function theorem shows that MATH is a MATH-dimensional complex submanifold near MATH. Since MATH is irreducible, it is pure-dimensional; thus MATH is a MATH pure-dimensional complex analytic set formally containing the graph of MATH. By definition of the transcendence degree, this implies that MATH.
math/0002237	It is clear that MATH is a lattice and that MATH is convex in MATH. We now check that MATH is complete. Let MATH. We will show that MATH exists. NAME we may assume that MATH, MATH, MATH. Let MATH. We may also assume that MATH is not the least upper bound of MATH, so let MATH be some upper bound. Fix MATH such that MATH. We will write MATH for the supremum operation in MATH. Let MATH. Then MATH, so by REF we have MATH. Hence the sequence MATH, defined by MATH is weakly increasing. For all MATH we have MATH so since MATH is convex we get: MATH. Let MATH. Clearly, MATH is the least upper bound for MATH. (Let MATH be any upper bound for MATH, say MATH, MATH, then MATH for all MATH.)
math/0002237	See CITE and CITE. (The ``moreover" part is not stated there, but following the proof it is easy to see that the lattice MATH constructed in CITE satisfies MATH and will be complete.)
math/0002237	We let MATH, MATH, and MATH as above. Let MATH for MATH and MATH for MATH. It remains to show that the map MATH is an orthocomplement for MATH. Clearly this map is well defined and an involution, and it agrees with the map MATH on MATH. Also, we have MATH for all MATH. Now let MATH, without loss of generality MATH. We will only check MATH, leaving the dual to the reader. So let MATH, MATH. Then MATH must be in MATH, and there is a MATH such that MATH. Now MATH implies MATH, hence MATH.
math/0002237	Since every ortholattice can be embedded into a complete ortholattice (the NAME completion; see, for example, CITE, CITE) we may assume that MATH is complete. Let MATH. Let MATH be horizontal sum of MATH and MATH, that is, assume that MATH and MATH have the same least and greatest elements (but are otherwise disjoint), and make MATH into a lattice by taking MATH . Note that MATH is a complete MATH-lattice and MATH. Now consider the partial functions MATH, MATH and MATH, defined by CASE: MATH for all MATH. CASE: MATH, MATH, for all MATH. Notice the elements of the set MATH are pairwise incomparable, so the function MATH is trivially monotone. By REF we can find a lattice MATH, MATH in which the functions MATH, MATH and MATH are restrictions of polynomials MATH, MATH and MATH, respectively. Now let MATH, so MATH is an orthoextension of MATH. Now MATH is an orthopolynomial with coefficients in MATH, and clearly MATH for all MATH.
math/0002237	Choose a cardinal MATH of cofinality MATH such that there is a transfinite enumeration (not necesasrily REF) MATH of all functions MATH. Define an increasing transfinite sequence MATH of MATH-lattices satisfying CASE: MATH. CASE: If MATH, then MATH CASE: For every MATH there is a lattice polynomial MATH with coefficients in MATH such that for all MATH: MATH. CASE: For every MATH, MATH contains an isomorphic copy of MATH. CASE: If MATH is a limit stage, then MATH is the direct limit of MATH, (that is, MATH). Finally, let MATH. Note that MATH will contain an isomorphic copy of MATH for every MATH. This finishes the proof of the first statement. To prove the second claim, apply the conclusion from the first claim MATH many times to get the conclusion for all unary functions, and then use REF to take care of all MATH-ary functions.
math/0002237	Let MATH, MATH small. Let MATH be the sub(ortho)lattice generated by MATH. Since MATH is MATH-power closed, we can find a sublattice MATH and an isomorphism MATH. For MATH let MATH be defined by MATH . Note that MATH . Define a partial function MATH as follows: MATH and MATH undefined if MATH. Note that MATH is monotone if MATH is monotone. Now note that MATH, MATH, , MATH are unary functions from MATH to MATH, so they are all represented by polynomials, and so by MATH and MATH also MATH is represented by a polynomial.
math/0002241	Let MATH denote the set of all countable subsets of the indexing set MATH. Consider the following relation MATH . We need to verify the following three properties of the above defined relation. Existence. If MATH, then there exists MATH such that MATH. Proof. First of all let us make the following observation. Claim. For each MATH there exists a finite subset MATH such that MATH. Proof of Claim. The unit ball MATH (here MATH denotes the norm of the NAME space MATH) being bounded in MATH is, by CITE, bounded in MATH. Continuity of MATH guarantees that MATH is also bounded in MATH. Applying CITE once again, we conclude that there exists a finite subset MATH such that MATH. Finally the linearity of MATH implies that MATH and proves the Claim. Let now MATH. For each MATH, according to Claim, there exists a finite subset MATH such that MATH. Without loss of generality we may assume that MATH for each MATH. Let MATH. Clearly MATH is countable, MATH and MATH for each MATH. This guarantees that MATH and shows that MATH. NAME. If MATH, MATH and MATH, then MATH. Proof. REF implies that MATH. The inclusion MATH implies that MATH. Consequently MATH, which means that MATH. MATH-closeness. Suppose that MATH and MATH for each MATH. Then MATH, where MATH. Proof. Consider the following inductive sequence MATH limit of which is isomorphic to MATH (horizontal arrows represent canonical inclusions). Since MATH for each MATH (assumption MATH), it follows that MATH . This obviously means that MATH as required. According to CITE the set of MATH-reflexive elements of MATH is cofinal in MATH. An element MATH is MATH-reflexive if MATH. In our situation this means that the given countable subset MATH of MATH is contained in a larger countable subset MATH such that MATH. Proof is completed.
math/0002241	Let MATH be a complemented subspace of the sum MATH. Choose a continuous linear map MATH such that MATH for each MATH. Let us agree that a subset MATH is called MATH-admissible if MATH. For a subset MATH, let MATH. CASE: If MATH is a MATH-admissible, then MATH. Proof. Indeed, if MATH, then there exists a point MATH such that MATH. Since MATH is MATH-admissible, it follows that MATH . Clearly, MATH. This shows that MATH. Conversely, if MATH, then MATH and hence, by the property of MATH, MATH. Since MATH, it follows that MATH. CASE: The union of an arbitrary collection of MATH-admissible subsets of MATH is MATH-admissible. Proof. Straightforward verification based of the definition of the MATH-admissibility. CASE: Every countable subset of MATH is contained in a countable MATH-admissible subset of MATH. Proof. This follows from REF applied to the map MATH. CASE: If MATH is a MATH-admissible subset of MATH, then MATH for each point MATH, where MATH. Proof. This follows from the corresponding property of the map MATH. Before we state the next property of MATH-admissible sets note that if MATH, then the map MATH defined by letting MATH is continuous and linear. CASE: Let MATH and MATH are MATH-admissible subsets of MATH and MATH. Then MATH is a complemented subspace in MATH and MATH is a complemented subspace in MATH. Proof. Consider the following commutative diagram MATH in which MATH is the canonical map and MATH is defined on cosets by letting (recall that MATH) MATH . Let us denote by MATH the natural inclusion and consider a map MATH defined by letting (in terms of cosets) MATH . Note that MATH (this follows from the equality MATH). In particular, this shows that MATH is isomorphic to a complemented subspace of MATH. Finally consider the composition MATH and note that MATH . This shows that MATH is a complemented subspace of MATH and completes the proof of REF . Let MATH. Then we can write MATH. Since the collection of countable MATH-admissible subsets of MATH is cofinal in MATH (see REF ), each element MATH is contained in a countable MATH-admissible subset MATH. According to REF , the set MATH is MATH-admissible for each MATH. Consider the inductive system MATH, where MATH (see REF ) and MATH denotes the natural inclusion for each MATH. For a limit ordinal number MATH the space MATH is isomorphic to the limit space of the direct system MATH (verification of this fact is based on REF coupled with the fact that MATH is isomorphic to the limit of the direct system MATH). In particular, MATH is isomorphic to the limit of the inductive system MATH. For each MATH, according to REF , the inclusion MATH is isomorphic to the inclusion MATH. In this situation the straightforward transfinite induction shows that MATH is isomorphic to the locally convex direct sum MATH. By construction, the set MATH is countable and MATH is a complemented subspace of MATH. Note also that for each MATH the set MATH is countable and MATH is a complemented subspace of MATH. This completes the proof of REF .
math/0002241	For countable MATH results follows from CITE and CITE. Let now MATH is uncountable and MATH be a complemented subspace of a locally convex direct sum MATH. By REF , MATH is isomorphic to a locally convex direct sum MATH, where MATH is a complemented subspace of the countable sum MATH where MATH for each MATH. According to CITE, MATH for each MATH. Consequently, MATH is isomorphic to the locally convex direct sum MATH as required.
math/0002242	Let MATH denote the set of all countable subsets of the indexing set MATH. Consider the following relation MATH where MATH and MATH denote canonical projections onto the corresponding subproducts. We need to verify the following three properties of the above defined relation. Existence. If MATH, then there exists MATH such that MATH. Proof. Let MATH. For each MATH consider the composition MATH. Since MATH is a NAME space, it follows that every continuous linear map into MATH, defined on an infinite product of NAME space, can be factored through a finite subproduct (this is a well known fact; see, for instance, CITE). Consequently there exist a finite subset MATH and a continuous linear map MATH such that such that MATH for each MATH. Without loss of generality we may assume that MATH for each MATH (otherwise consider the set MATH). Let MATH and MATH, MATH. Clearly MATH . Next consider the diagonal product MATH and note that MATH is a continuous linear map which satisfies the equality MATH. This shows that MATH. NAME. If MATH, MATH and MATH, then MATH. Proof. This is trivial. Indeed let MATH be a continuous linear map such that MATH. Consider the map MATH defined as the composition MATH. Since MATH it follows that MATH. MATH-closeness. Suppose that MATH and MATH for each MATH. Then MATH. Proof. Consider the following projective sequence MATH limit of which is isomorphic to the product MATH, where MATH. Since MATH, there exists a continuous linear map MATH such that MATH, MATH. Note that MATH for each MATH. Indeed let MATH and consider any point MATH such that MATH. Since MATH we have MATH . In this situation the collection MATH uniquely defines a continuous linear map MATH such that MATH for each MATH (MATH is simply the diagonal product of MATH's). It only remains to note that MATH which completes the proof of the fact that MATH. According to CITE the set of MATH-reflexive elements of MATH is cofinal in MATH. An element MATH is MATH-reflexive if MATH. In our situation this means that the given countable subset MATH of MATH is contained in a larger countable subset MATH for which there exists a continuous linear map MATH satisfying the equality MATH. Proof is completed.
math/0002242	Let us first of all set up a notation. For a subset MATH, where MATH is an indexing set with MATH, let MATH . Let also for MATH and MATH denote canonical projections onto the corresponding subproducts. Let MATH be a complemented subspace of the product MATH. Choose a continuous homomorphism MATH such that MATH for each MATH. Let us agree that a subset MATH is called MATH-admissible if MATH for each point MATH. CASE: The union of an arbitrary family of MATH-admissible sets is MATH-admissible. Let MATH be a collection of MATH-admissible sets and MATH. Let MATH. Clearly MATH for each MATH and consequently MATH for each MATH. Assuming that there is a point MATH such that MATH we conclude that there exists an index MATH such that MATH. Since MATH it follows that there exists an index MATH such that MATH. Then we have MATH. But this is impossible MATH . This contradiction proves the claim. CASE: If MATH is MATH-admissible, then MATH is a complemented subspace of MATH. Indeed, let MATH be the canonical section of MATH (this means that MATH). Consider a continuous linear map MATH. Obviously, MATH for any point MATH. Since MATH is MATH-admissible the latter implies that MATH . This shows that MATH is a complemented subspace of MATH. CASE: Let MATH and MATH be MATH-admissible subsets of MATH and MATH. Then there exists a topological isomorphism MATH which makes the diagram MATH commutative. Obviously MATH. Consider the map MATH. Also let MATH. Observe that MATH. Indeed, if MATH, then MATH. Since MATH is MATH-admissible, we have MATH. Consequently, MATH . Next observe that MATH for any point MATH. Indeed, since MATH is MATH-admissible and since MATH we have MATH . In this situation we can define a map MATH by letting MATH . A straightforward verification shows that MATH is a continuous linear map which satisfies the required equality MATH. Also note that by letting MATH we define a continuous linear map MATH. It is easy to see that MATH . This proves that MATH is a topological isomorphism and finishes the proof of REF . Every countable subset of MATH is contained in a countable MATH-admissible subset of MATH. Let MATH be a countable subset of MATH. Our goal is to find a countable MATH-admissible subset MATH such that MATH. By REF , there exist a countable subset MATH of MATH and a continuous homomorphism MATH such that MATH and MATH. Consider a point MATH. Also pick a point MATH such that MATH. Then MATH . This shows that MATH (this shows, in fact, that MATH is complemented in MATH). In order to show that MATH is MATH-admissible let us consider a point MATH. By the observation made above, MATH. Finally MATH which implies that MATH is MATH-admissible. We now use the above listed properties of MATH-admissible subsets and proceed as follows. By REF , each element MATH is contained in a countable MATH-admissible subset MATH. According to REF , the set MATH is MATH-admissible for each MATH. Consider the projective system MATH where MATH . Since MATH, it follows that MATH. Obvious transfinite induction based on REF shows that MATH . Since, by the construction, MATH is a countable MATH-admissible subset of MATH, it follows from REF that MATH and MATH, MATH, being complemented subspaces of countable products of NAME spaces, are NAME spaces. This finishes the proof of REF .
math/0002242	MATH. By CITE and CITE, MATH is an injective object of the category MATH for any set MATH. Obviously (see, for instance, CITE) product of an arbitrary collection of injective objects of the category MATH is also an injective object of this category. Consequently, the NAME space MATH, MATH, as a complemented subspace of MATH, is injective. Finally, the space MATH, as a product of injectives, is an injective object of the category MATH. MATH. The space MATH can be identified with a closed linear subspace of the product MATH of NAME spaces MATH, MATH. Each of the spaces MATH can in turn be identified with a closed linear subspace of the space MATH for some set MATH, MATH. REF implies in this situation that MATH is a complemented subspace of the product MATH. The required conclusion now follows from REF .
math/0002243	The first thing to notice is that MATH, where MATH is a co-closed MATH-form. Since MATH and MATH are solutions to the NAME we have MATH . Therefore MATH . This last statement and the fact MATH implies that MATH . Since MATH are solutions to the NAME equations we have MATH multiplying by MATH both sides of the equality we get that MATH. Taking the point-wise norm we will have MATH. If we denote by MATH and MATH the set of points where MATH and MATH vanish respectively, and we denote by MATH and MATH their corresponding complements, we will have that MATH, therefore if MATH is not a reducible solution then MATH is a non-empty open set. By a result of NAME (see CITE) we will have that MATH, since it vanishes in an open set.
math/0002243	This is a consequence of REF , the expression for MATH (see REF ) and that MATH, where MATH and MATH are the NAME operators of MATH and MATH, respectively.
math/0002243	The proof of this proposition can be carry out following the steps in the proof of REF pg. REF, replacing the expression for MATH with MATH.
math/0002243	The fact that MATH satisfies the NAME follows from REF . We just need to show that MATH. In order to do this, we will use the metric MATH as the background metric. MATH . To prove that MATH we need to recall that MATH where MATH. The computation is very similar to the one above.
math/0002243	Let MATH. The existence of such metric is equivalent to finding a neighborhood MATH of MATH, and a NAME form MATH in the same NAME class of MATH, such that MATH. It is well known that there exist a MATH-neighborhood MATH of MATH and a function MATH such that MATH, where MATH and MATH denotes the distance (using the NAME metric MATH) on MATH to MATH. Let MATH be the space of smooth functions on MATH that satisfy MATH where MATH, depend on MATH. Observe that if MATH is zero we do not have anything to prove, otherwise MATH, but MATH, where MATH denotes the completion of MATH in the MATH topology. To see this consider the one-parameter family of functions MATH, where MATH is a smooth bump function such that MATH . All these functions are in MATH and satisfy MATH . It is not difficult to see that MATH in the MATH topology. It is important to recall that the set MATH of smooth functions MATH such that MATH, is open in the MATH topology. This two facts allow us to find MATH, MATH close to MATH, such that MATH therefore we have MATH where MATH.
math/0002243	It is easy to see that MATH, as it is to see that MATH is a solution of REF with function MATH replacing MATH. The first equation in REF tell us that MATH is a holomorphic section on MATH. Using NAME 's Theorem we can extend this to a holomorphic section on MATH. All the analysis done in proving REF can be carry out if we replace the strictly positive function MATH in REF by a non-negative function MATH whose zero set has measure zero.
math/0002243	Since all the analysis done in proving REF can be carry out if we replace the strictly positive function MATH in REF by a non-negative function MATH whose zero set has measure zero, existence is a consequence of REF and uniqueness is obtained using REF .

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