paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0009124 | Since MATH is integrable, MATH is NAME equivalent to MATH, CITE. |
math/0009124 | Let MATH be the natural projection MATH. It suffices to show that MATH is isomorphic to MATH. In the notation of the proof of REF , the isomorphism is MATH. Here MATH is the MATH-component of MATH, the form MATH is the symplectic form on MATH, and MATH is extended to a genuine differential form on MATH using the direct product structure. |
math/0009124 | Weak NAME equivalence is just morphism in the weak NAME MATH-category. |
math/0009124 | We focus on the properties of MATH. For MATH the reasoning is identical. To show that MATH is a submersion, we need to verify that this map is onto and so is its derivative MATH. The surjectivity is a pure set-theoretic fact. (Let MATH. Take any MATH and any MATH with MATH. Then MATH.) This argument uses only the assumption that all maps involved are onto. Hence, the surjectivity also holds in the category of linear spaces and, in particular, MATH is onto. A similar argument shows that the natural projections MATH are also submersions. Let us check that for MATH, the fiber MATH is MATH-connected. The projection MATH sending MATH to MATH is a submersion whose fiber over MATH is MATH. Hence, all fibers and the range of this submersion are MATH-connected. It suffices now to apply the following proposition, which we prove in the next section. Let MATH be a submersion. Assume that MATH and all fibers of MATH are MATH-connected. Then MATH is also MATH-connected. It remains to define the isomorphism MATH. First note the factorizations MATH . Using functoriality of the pull-back, CITE, we set this isomorphism to be MATH where the isomorphisms are induced by the isomorphisms of the algebroids from the original weak NAME morphisms. This concludes the proof. |
math/0009124 | Assume first that MATH and the homotopy MATH is just the identity map. Let MATH . This is a compact set. It is easy to see that there exists a neighborhood MATH of MATH in MATH, a small MATH, and an embedding MATH such that MATH, where MATH and MATH. Here and in what follows, we identify MATH with a neighborhood of MATH in MATH. For small MATH, the homotopies MATH take values in this neighborhood. It is also easy to see that MATH can be changed for small MATH so as to make MATH independent of MATH in MATH. (In other words, the MATH-component of MATH is constant for all MATH and MATH.) From now on we assume that MATH have this property. (The resulting maps can be chosen of the same smoothness class as the original ones.) Let MATH be a smooth monotone increasing function which is identically equal to MATH for MATH near MATH and to MATH for MATH near MATH. The required lift is then given by the formula MATH . Let us now prove the lemma for a general smooth homotopy MATH. Consider the pull-back submersion MATH with the domain MATH . By transversality, MATH is smooth and MATH is a submersion. A lift of MATH pulls back to a continuous lift of MATH. (In fact, this pull-back is of the same smoothness class as the original lift, as one can check in local coordinates in the domain of MATH). Thus MATH, MATH, and MATH pull back to maps satisfying the hypotheses of the lemma and covering MATH. For these-pull backs the existence of the lift MATH has already been proved. To obtain a lift of MATH, it remains to compose this lift with the natural map MATH. |
math/0009124 | We will prove the lemma for MATH. (This implies the lemma for MATH, for MATH can always be extended from MATH to some MATH-cube containing MATH and MATH can also be extended to MATH.) The proof is by induction in MATH. For MATH, that is, MATH, the assertion follows from the assumptions that the fibers of MATH are MATH-connected and MATH. Assume that the assertion is proved for MATH. Set MATH and MATH for MATH. Let also MATH and MATH. By the induction hypothesis, MATH has a filling MATH and hence, by REF , this filling extends over MATH for some small MATH. Consider the maximal open in MATH interval MATH (or MATH) over which a filling MATH of MATH exists. If this interval is MATH the proof is finished. Hence, let us assume that the interval is MATH with MATH and arrive to a contradiction. We will present the proof for MATH. The argument for MATH requires only minor modifications. By the induction hypothesis, MATH admits a filling, which extends to a filling MATH over some small interval MATH, by REF . Now we have two fillings, MATH and MATH, of MATH. These are maps MATH which coincide on MATH and hence give rise to a map MATH covering MATH. The induction hypothesis applies to this map, for MATH is a cube of dimension MATH and MATH. Thus it has a filling, which can be thought of as a homotopy MATH between MATH and MATH which covers MATH. REF applies to the homotopy MATH for MATH, the homotopy MATH for MATH, and the connecting homotopy MATH. As a result, we obtain a filling of MATH for MATH. This contradicts the choice of MATH. |
math/0009124 | As in the proof for MATH, we argue inductively. For MATH and MATH, the image MATH is comprised of two mapped MATH-spheres contained in the fibers of MATH. Since these fibers are MATH-connected, the filling does exist. Assuming that the lemma is proved for MATH, we establish it for MATH. Let us think of MATH as the union of two hemispheres: MATH. REF has already been proved for the discs and, hence, it applies to MATH. As a result we obtain two fillings MATH over the hemispheres. Along the ``equator" MATH of MATH, MATH are fillings of the same map MATH. Hence, MATH give rise to a continuous map MATH covering MATH. By the induction hypothesis, this map has a filling and this filling can be viewed as a connecting homotopy MATH between MATH and MATH. Applying REF to MATH, MATH, and MATH, we obtain the required filling for MATH. |
math/0009124 | We only outline the argument, for it is very similar to the proof of REF . As in that proof we replace MATH by MATH. Arguing inductively, we assume that the lemma is proved for the cubes of dimension less than MATH. As above, we set MATH and MATH, where MATH, and consider the maximal open interval in MATH over which the lift exists and which contains MATH. Denote this lift by MATH. Let us assume that the interval has the form MATH with MATH and arrive to contradiction. As in the proof of REF , we focus on the case MATH. (The case MATH is handled similarly.) By the induction assumption, we can lift MATH to MATH and then, using REF , extend it to a homotopy MATH, MATH, for a small MATH. Thus we have two lifts of MATH, namely, MATH and MATH. Together these lifts give rise to a map MATH covering MATH. Applying REF , we find a filling for this map. This filling is nothing else but a homotopy between MATH and MATH covering MATH. By REF , the lift of MATH extends to the interval MATH, which contradicts the choice of MATH. |
math/0009124 | Let MATH, where MATH, be a smooth map and let MATH be a smooth homotopy of MATH to a point. We will find a continuous homotopy MATH of MATH to a point. By REF , there exists a lift MATH of MATH. Furthermore, by REF , MATH can be extended to a little closed collar MATH of MATH in MATH. Let MATH be this extension. Denote by MATH the inner boundary of MATH in MATH. Thus MATH is the sphere MATH over which we have two lifts MATH and MATH of MATH. These lifts comprise a map MATH covering MATH, which, by REF , have a filling. This filling is a homotopy MATH between MATH and MATH covering MATH. Applying REF , we obtain the required lift MATH. |
math/0009131 | REF |
math/0009131 | REF . |
math/0009131 | In view of REF and the fact that MATH and MATH are isomorphisms the three assertions are of course equivalent. REF follows from the identities MATH and MATH by an easy induction. |
math/0009131 | This is REF. |
math/0009131 | First note that we have the identities MATH which together give MATH . Then MATH . |
math/0009131 | The first equality can be found in CITE. The second equality is REF. |
math/0009131 | Because of REF we may assume that MATH is of the form MATH or MATH. We will therefore argue by induction on weight and degree and assume that the assertion holds for all MATH of either less degree or less weight than MATH. (The assertion is certainly trivial for the vacuum, the - up to a scalar factor - unique element of weight REF and degree REF.) In case MATH it follows from REF and induction that MATH . In case MATH for some MATH we argue as follows: MATH . |
math/0009131 | Let MATH denote the component of MATH of degree MATH. We must show that for each MATH the elements MATH where MATH runs through all partitions of MATH, form a set of generators of the MATH-module MATH. Claim: The restriction map MATH is a surjective and degree preserving ring homomorphism and maps MATH to MATH. In particular, we may assume that MATH. Indeed, surjectivity and homogeneity are obvious, hence it is enough to check that MATH is multiplicative. We have MATH where MATH means the sum over terms satisfying the degree REF . The first term of the last line equals MATH, and it suffices to show that no summand of the second term occurs. Indeed, we may decompose any MATH as MATH for some transposition MATH with MATH and MATH. The degree matching condition requires MATH . But the right hand side equals MATH so that REF is never fulfilled. Finally, it is clear that MATH, which proves the claim. Note that MATH is not multiplicative with respect to the convolution product. Assume from now on that MATH and consider the MATH-module MATH. It has a MATH-basis consisting of monomials MATH where as before MATH is a partition of MATH and MATH is the associated partition of MATH given by MATH and MATH for MATH. Here the assumption MATH ensures that MATH. On the other hand, consider MATH . By the definition of the cup product, the elements MATH are all contained in MATH and can therefore be expressed as linear combinations of the MATH. We must show that the associated coefficient matrix is invertible over MATH. This will be achieved by comparison with a third, rational basis of the vector space MATH, provided by the elements MATH . Claim: Let MATH be the matrix defined by MATH . Then MATH . Let MATH be the order on the set of partitions MATH of MATH corresponding to the lexicographical order of the sequences MATH, so that for example the partition MATH is the largest and MATH is the smallest. Since NAME classes and the components of the NAME character satisfy the universal identities MATH it follows that MATH . This shows that MATH is a lower triangular matrix with diagonal entries MATH . The claim follows directly from this. Claim: Let MATH be the matrix defined by MATH. Then MATH . Recall that by REF we have MATH for any polynomial MATH, where the degree MATH component MATH of the differential operator MATH is given by MATH . If MATH is applied to a monomial MATH with MATH, then the smallest component with respect to the lexicographical order is that arising from the choice MATH in REF . More precisely, MATH . It follows by induction that MATH . This shows that MATH is a lower triangular matrix - if we reorder the MATH according to MATH - with diagonal entries MATH . The claim follows from this. Claim: MATH . Of these two equalities the first is an immediate consequence of the two previous claims. The second is a well known identity. In fact, it amounts to realizing that each integer MATH appears both in the numerator and denominator with multiplicity MATH where MATH is the number of partitions of MATH. |
math/0009134 | compare CITE. |
math/0009134 | We refer to REF for a complete proof. In op.cit. , it is shown under which condition the NAME vector MATH induces permutation on the field MATH, for MATH. |
math/0009134 | It is rather easy to check that the affine REF - folds MATH and MATH have the same number of points over fields containing MATH. Hence, any bijective linear transformation between MATH and MATH will be defined over such fields. Among the critical points of MATH, one finds the points MATH with MATH satisfying the equation MATH. The field generated by these points is MATH. Similarly, among the critical points of MATH one finds the points MATH with MATH satisfying the equation MATH. The field generated by these points is MATH. Of course, both the polynomials have further critical points, but we simply look for a linear map that relates the above critical points. This is because the image of points like MATH (compare above) is defined over a subfield of MATH and this field contains MATH. A similar argument shows that the image of points like MATH is defined over a subfield of MATH and this field contains MATH. One can easily express the parameter MATH in terms of MATH and MATH: MATH. The definition of the linear map follows. |
math/0009134 | If MATH is a point in the singular locus of the variety MATH (compare REF), then the coordinates of MATH must satisfy REF . This implies that for MATH, MATH for MATH and MATH. Therefore, the singular locus of MATH is described by the set of points MATH in the singular locus of the variety MATH . The MATH-tuples MATH are chosen among the set of solutions of MATH under the condition MATH. By a direct computation one obtains the following locus MATH . Hence, to each of REF points in MATH correspond MATH points solutions of REF. In this way, one counts globally MATH singular points for MATH. One can direct verify that they are all rational double points. |
math/0009134 | It follows from the above isomorphism and from the fact that MATH (independent of the number of the singular points). For smooth, quintic hypersurfaces in MATH one knows that MATH. |
math/0009134 | The vanishing cycle exact sequence applied to a NAME pencil of quintic hypersurfaces in MATH with degenerating fiber MATH and generic fiber MATH, reads as MATH . Here, MATH is the MATH-vector space of vanishing cycles. Because MATH has only double points, the rank of MATH equals the number of double points of MATH. Hence one gets MATH . The NAME charactersitic of MATH (a non-singular hypersurface of degree MATH in MATH) is given by integrating its third NAME class against MATH and then by interpreting the integral using NAME theorem MATH . Here, MATH. It is well known that for degree MATH smooth hypersurfaces in MATH the r-th NAME class is given by MATH . For degree MATH hypersurfaces in MATH one gets MATH. Using NAME hyperplane theorem, one may deduce that MATH and MATH. Hence, MATH as MATH. In turn, this result implies MATH. Hence MATH. The NAME spectral sequence for blow-up of double points MATH gives the following exact sequences MATH . Here MATH is the exceptional fiber, that is, a union of MATH. From this sequence it follows that MATH. Using REF one concludes that MATH . |
math/0009134 | The equality MATH follows from REF and the fact that the rank-one vector space MATH is generated by the class of a hyperplane section. The equality MATH is a direct consequence of the above lemma and REF (that is, MATH). The equality MATH follows from REF. Finally, MATH is deduced from the difference MATH (compare proof of the last lemma). |
math/0009134 | In order to prove this, we study the behaviour of the critical points of MATH mod MATH. In paragraph REF, we observed that MATH has MATH critical points with value MATH, MATH with value MATH and MATH with value MATH. We list these points in the following table. MATH . One can easily check that these MATH critical points reduce to different points modprimes of MATH above rational primes bigger than MATH. One way to verify this is to compute the discriminant with respect to MATH of the resultant of MATH and MATH with respect to MATH. This number is only divisible by the primes: MATH, MATH, MATH, MATH, MATH and MATH, so these are the only primes mod which the critical points can coincide. A finite calculation shows that the critical points reduce to different points mod primes above MATH, MATH and MATH. For MATH, the resultant of MATH and MATH (say with respect to MATH) is not identically zero. This implies that these polynomials do not have a common factor modulo MATH. By NAME 's theorem one concludes that MATH mod MATH has at most MATH critical points. Hence, for MATH there are precisely MATH critical points: the reductions of the MATH critical points in characteristic zero. Note that the values MATH, MATH and MATH that MATH assumes at the critical points remain different when reduced modulo primes bigger than MATH. It follows that the number of singular points of the affine singular threefold defined by MATH does not increase upon reducing modprimes bigger than MATH. At each of the critical points of MATH one can check that the local expansion of MATH has a non-degenerate degree-MATH-part which remains non-degenerate when one reduces it modprimes of MATH, above rational primes bigger than MATH. It follows that the part of MATH above the affine part MATH remains non-singular after the reduction mod MATH. It remains to verify that MATH mod MATH has no singularities at infinity. In fact, the homogenenized MATH has the following form: MATH. If its partial derivatives were all zero at a point where MATH then, in characteristic MATH, one would have MATH. |
math/0009134 | Suppose MATH. In paragraph REF we showed that the polynomial MATH, with MATH and MATH . NAME polynomials. It follows from REF that the map MATH permutes MATH. Hence MATH assumes each value exactly MATH times and therefore there are MATH solutions to the equation MATH in MATH. From REF it follows that there are no critical points defined over MATH, hence there is no extra contribution coming from the blow-up. The map MATH permutes the elements of MATH, hence MATH assumes each value MATH times. So MATH has MATH solutions in MATH, which correspond to MATH points. In total, one counts globally MATH points. |
math/0009134 | It follows from REF that MATH. The inequality in MATH yields: MATH . Hence, MATH for MATH big enough. Using the trace formula MATH it follows that MATH. For small MATH the MATH are listed in the table MATH. |
math/0009134 | Let MATH denote the vectorspace MATH, and let MATH be the extension of scalars of MATH. Let MATH be the NAME character of MATH with kernel MATH. By REF we have that MATH for all MATH not in MATH. So MATH for all MATH. It follows that MATH and MATH are isomorphic. Let MATH be an isomorphism that intertwines MATH and MATH. So MATH for all MATH and MATH. Let MATH be an eigenvalue of MATH, and MATH its eigenspace. Clearly, MATH can not be a scalar map, so MATH is a proper subspace of MATH. Since MATH and MATH actually have coefficients in MATH, the eigenvalue MATH is contained in an extension of MATH of degree at most REF. The image MATH commutes with MATH, so its action on MATH is a subrepresentation of MATH. From the remarks made above about the splitting behaviour of the characteristic polynomials it follows that the only possible dimension of MATH is REF, and that MATH is contained in MATH. It also follows MATH has only one other eigenspace MATH, with eigenvalue MATH. Denote the representation on MATH by MATH and the representation on MATH by MATH. Let MATH be an element that represents the non-identity element of MATH. Suppose MATH. Then MATH . So MATH is an eigenvector with eigenvalue MATH, and MATH sends MATH to MATH. It follows that MATH. |
math/0009134 | We only show that MATH ramifies at MATH and at both the infinite places. For more details on the properties of MATH, as well as for the proof of its uniqueness up to isomorphism, we refer to CITE. Notice that MATH is totally definite (that is, MATH is a division quaternion algebra isomorphic to two copies of the NAME 's quaternions). In fact, MATH and MATH are totally negative numbers. It remains to show that MATH are the two finite primes where MATH ramifies. In the local field MATH, the NAME symbol (= NAME invariant) is given by MATH where the subscript MATH denotes the NAME symbol. In fact, MATH is not a square mod-MATH. Hence MATH is a field. In other words, at the inert prime ideal MATH, the algebra MATH ramifies. At the ramified prime MATH (ramified for the extension MATH), the NAME symbol is MATH . Hence, the place MATH does not ramify in MATH. It follows from the definition of the algebra that the only possible place where MATH could possibly ramify is MATH. This is in fact the case as a quaternion algebra is supposed to ramify at an even number of primes in a field. |
math/0009134 | We refer to REF , for the proof. The computations written down in op.cit. generalize easily to any quaternion algebra over a number field. |
math/0009134 | Suppose MATH is an element of MATH. Then MATH, MATH, MATH and MATH. Recall the definition of MATH from MATH in terms of harmonic polynomials MATH. From the definition of the MATH it immediately follows that MATH if MATH, MATH, MATH, MATH or MATH, and MATH otherwise. From the definition of MATH it follows that MATH if MATH is congruent to MATH, MATH, MATH, MATH or MATH mod REF, and MATH otherwise. Note that the eigenvector MATH satisfies MATH if MATH, and MATH otherwise. So MATH also has this property, hence the eigenvalue of MATH is in MATH. It is known that eigenvalues of the NAME action on NAME modular forms are contained in a totally real field. This implies that the eigenvalue of MATH is in MATH. |
math/0009135 | This is immediate from the identities MATH for MATH and MATH. |
math/0009135 | Let MATH and MATH be pointed matroids on MATH and MATH respectively. A pointed strong map MATH arises from a set function MATH mapping MATH to MATH. This function yields a homomorphism of exterior algebras MATH determined by MATH. According to CITE, the image of each circuit of MATH is dependent in MATH. Using REF this implies that MATH sends MATH into MATH, inducing a homomorphism MATH. Since MATH, MATH restricts to a homomorphism MATH. |
math/0009135 | Let us write MATH for MATH. Using the fact that tensor product is a sum in the category of connected graded algebras, together with REF , we obtain a surjective homomorphism MATH. Using REF and CITE, one can show that the domain and target have the same dimension in each degree. Thus the two algebras are isomorphic. |
math/0009135 | Consider the pointed parallel connection MATH. The underlying matroid MATH is MATH, which is precisely MATH. Then, by REF , the MATH. On the other hand, by REF , MATH is also isomorphic to MATH, which again by REF , is isomorphic to MATH. Now, according to REF , we may change the base points of MATH and MATH without affecting the affine MATH algebras. The pointed parallel connection MATH of these new pointed matroids will have underlying matroid MATH isomorphic to the the sum of two isthmuses (neither marked) with an ordinary parallel connection MATH of MATH and MATH along the new marked points of each. Again, we have MATH. Now we change the base point of MATH to one of the isthmuses, and recognize the resulting pointed matroid as MATH. We apply REF once more to obtain the result. |
math/0009135 | Suppose MATH is an isomorphism of MATH to MATH. To begin with, we can then assume without loss that MATH and MATH have the same ground set. The isomorphism MATH determines an isomorphism MATH, and MATH . We need only show that MATH . Let MATH. Then, for MATH, MATH . Since the truncations have rank MATH, we also have, for MATH, MATH. Since MATH is an algebra homomorphism, it commutes with MATH, and thus MATH for MATH. This completes the proof. |
math/0009136 | As noted in REF the equivalence of REF follows form the computation REF - REF there. Namely, since MATH defines MATH, it is nonsingular, so MATH for some real valued function MATH (possibly after replacing MATH by MATH). Then MATH . Therefore, REF holds. In the last equality in REF , we have again used that MATH (see REF). The equivalence of REF is contained in REF , where various characterizations are given for a flow to be geodesible. But the arguments show that REF is equivalent to the flow of MATH being geodesible by a metric as in REF . This completes the proof of REF . NAME REF characterizes geodesible flows by a homological condition. His arguments are easily modified to obtain a characterization of the conditions in REF (actually, the versions on the compact set MATH, rather than on an open hypersurface, as in REF ). We return to the analogue of the setup in REF for a domain in MATH. Denote by MATH the one dimensional foliation whose leaves are the orbits of MATH. (This is actually a foliation of all of MATH). We briefly recall the notions we need from the theory of foliation currents; for details, see REF or REF. Denote by MATH the space of one forms that are MATH on MATH, with the usual NAME space topology. Its strong dual MATH comprises the one currents on MATH. MATH and MATH are mutually dual (that is, MATH is reflexive). The closed convex cone MATH of foliation currents for MATH is the closure, in MATH, of the set of all finite linear combinations with positive coefficients of currents given by pairing at MATH with MATH, for MATH (so-called NAME currents). The crux of the matter is that MATH has a compact base: MATH and MATH is compact (where MATH is viewed as an element of MATH); this is analogous to REF . We say that a MATH-chain is tangential to MATH if it is a sum of MATH-chains, each of which is a MATH-chain in a leaf of MATH. |
math/0009141 | As MATH is finite dimensional, the functor MATH where the right MATH-action on MATH is given by MATH is an equivalence of categories CITE. As the antipode of MATH is bijective CITE we have the following equivalences of categories MATH that is, MATH is NAME equivalent to MATH. It follows from the NAME theory that there exists an algebra isomorphism MATH. Taking into account that MATH, we obtain that MATH. |
math/0009141 | It is obvious that MATH, MATH are algebra maps. For MATH we have MATH and MATH so MATH is coassociative if and only if MATH . Using the pentagon REF , we find that this is equivalent to MATH and this equality holds for any MATH. In a similar way we can prove that MATH is also coassociative. |
math/0009141 | Taking into account the multiplication rule of MATH we have MATH and MATH so we have to prove the equality MATH in MATH. Let's fix the indices MATH, MATH, MATH and MATH and evaluate REF at MATH. REF is then equivalent to MATH which can be obtained by applying the definition of the convolution product and the dual basis formula in the right hand side. We shall prove now that MATH is the inverse of MATH, where MATH is the antipode of MATH. Since MATH is isomorphic to MATH, it is enough to prove that MATH. As MATH, we have to prove the formula MATH which holds, as for indices MATH, MATH we have MATH . |
math/0009141 | CASE: We shall use the notations introduced above. First we shall prove that MATH (respectively, MATH) are unitary subalgebras in MATH and subcoalgebras of MATH (respectively, MATH). This will follow from the formulas: MATH and MATH for all MATH, MATH. We prove REF, (being similar, REF is left to the reader). MATH that is, MATH is a subalgebra of MATH. On the other hand MATH that is, MATH is a subcoalgebra in MATH. A similar computation yields REF, proving that MATH is a subalgebra of MATH and a subcoalgebra in MATH. As MATH is unitary, MATH, that is, MATH and MATH are unitary subalgebras of MATH. Moreover, from the construction we have that MATH and hence we can view MATH. We shall define now the counit and the antipode of the NAME algebras MATH and MATH. They are given by the formulas: MATH and MATH for all MATH. We shall prove that MATH is a NAME algebra; the fact that MATH is a NAME algebra is proved in a similar way. First, we remark that, as MATH is a subalgebra of MATH, the comultiplication from REF can be rewritten as MATH . Now, for MATH we have MATH that is, MATH. A similar computation shows that MATH, and MATH is a counit of MATH. MATH is a right convolution inverse of MATH since MATH . From the fact that MATH is finite dimensional, it follows that MATH is an antipode of MATH. We shall prove now that the subalgebra MATH of MATH generated by MATH and MATH is finite dimensional. We shall use the pentagon equation. We have: MATH . Applying MATH to the pentagon equation we obtain MATH for all MATH, MATH. This formula gives that dim-MATH. CASE: Let us prove now that MATH is an isomorphism of NAME algebras. MATH is an isomorphism of vector spaces since MATH. REF can be rewritten as MATH which means that MATH for all MATH, MATH, that is, MATH is an algebra isomorphism. Let us prove now that MATH is also a coalgebra map. We recall the definition of the comultiplication MATH: MATH for all MATH, MATH. It follows that MATH that is, MATH is also a coalgebra map. Hence, we have proved that MATH is an isomorphism of bialgebras and, as MATH and MATH are NAME algebras, it is also a isomorphism of NAME algebras CITE. CASE: We remark that MATH for all MATH, MATH. The fact that MATH is an algebra map can be proved directly by using this formula; another way to proceed is to use the universal property of the NAME double MATH for REF , with MATH, MATH is the usual inclusion and MATH is the composition MATH, where MATH is the isomorphism from REF and MATH is the usual inclusion. We only have to prove that the compatibility REF holds, that is, MATH for any MATH and MATH, which turns out to be MATH or, equivalently MATH . This equation holds, as MATH, for any MATH. Now let MATH be the canonical element of MATH. Then MATH . CASE: Consider the map MATH for all MATH. We will show that MATH is a right-left MATH-Hopf module, where the structure of right MATH-module is simply the multiplication MATH of MATH. Indeed, for MATH we have MATH and MATH so MATH is a left MATH-comodule. The compatibility relation MATH holds for all MATH and MATH. Hence, MATH and the coinvariants MATH are the right MATH-coinvariants of MATH. From the right-left version of REF modules it follows that the multiplication of MATH, MATH defines an isomorphism of MATH-Hopf modules and, in particular, of right MATH-modules. We recall that MATH is a right MATH-module via MATH, for all MATH, MATH, MATH. It follows that MATH is free as a right MATH-module and, if MATH is finite dimensional, MATH . In a similar way we can show that MATH, where MATH is the multiplication of MATH and MATH for all MATH. Moreover, MATH. If we apply once again REF modules (this time the left-right version) we obtain the other part of the statement. CASE: MATH, MATH and MATH. It follows that MATH is a basis of MATH and hence the restriction of MATH to MATH gives an algebra isomorphism between MATH and MATH that is also a coalgebra map since MATH for all MATH. The last statement is obtain for MATH and MATH, MATH for all MATH. |
math/0009141 | CASE: We have proved in REF that if the spaces MATH and MATH are nonzero, then MATH is unitary. Conversely, if MATH is unitary, the isomorphisms of vector spaces from REF of REF show us that MATH and MATH are nonzero. CASE: Assume now that MATH is finite dimensional. From the construction of REF we have that MATH. In particular, for any positive integer MATH there exist scalars MATH such that MATH . We shall prove now that MATH is unitary, that is, MATH. As MATH is finite dimensional, MATH can be embedded into a matrix algebra MATH, where MATH-dim-MATH. We consider MATH . As MATH is invertible, using the NAME theorem in MATH, the identity matrix MATH can be represented as a linear combination of powers of MATH. Hence, using REF, we obtain in MATH a linear combination MATH, for some scalars MATH. Therefore MATH that is, MATH is unitary. |
math/0009141 | CASE: MATH is a right MATH-invariant; hence, MATH. If we apply MATH to this equality we get that MATH, for any MATH. As MATH is a basis of MATH, we obtain that MATH, for all MATH. If we apply the right MATH-module map MATH to this, we obtain that MATH is a right integral of MATH. CASE: Left to the reader. |
math/0009141 | Let MATH be a basis of MATH and MATH the canonical element. We have to prove that the NAME algebra MATH extracted from REF of REF is isomorphic to MATH, with the initial NAME algebra structure. MATH, MATH is an injective algebra map. We identify MATH . From the construction, MATH is the subalgebra of MATH having MATH as a basis; that is, there exists an algebra isomorphism MATH. It remains to be proven that the coalgebra structure (respectively, the antipode) of MATH extracted from REF is exactly the original coalgebra structure (respectively, the antipode) of MATH. As the counit and the antipode of a NAME algebra are uniquely determined by the multiplication and the comultiplication, the only thing left to be shown is the fact that, via the above identification, MATH. This means that MATH or equivalently, MATH . Now we compute MATH . On the other hand MATH . Hence, we have to show the formula MATH . Let us set the indices MATH, MATH, MATH, and evaluate REF at MATH. REF is then equivalent to MATH and this is easily verified using the dual basis formula. It follows that MATH and MATH as NAME algebras. |
math/0009141 | In REF of REF we proved that there exists a NAME algebra isomorphism MATH for any MATH and MATH, which means that all the NAME algebras associated to the elements of an orbit of the action REF are isomorphic. We shall now prove the converse. First we show that two finite dimensional NAME algebras MATH and MATH are isomorphic if and only if MATH and MATH are isomorphic as objects in MATH. The "only if" part follows from REF of REF. It remains to prove the "if" part. Let MATH be a NAME algebra isomorphism. Then, MATH, MATH is an isomorphism of NAME algebras and MATH for all MATH, MATH is an algebra isomorphism between the two NAME. On the other hand, if MATH is a dual basis of MATH, then MATH is a dual basis of MATH and hence MATH, and this proves that MATH is an isomorphism in MATH. Let MATH, MATH. Using REF we obtain that MATH if and only if MATH in MATH, where MATH is the image of MATH under the algebra isomorphism MATH. Now, the two matrix algebras MATH and MATH are isomorphic if and only if MATH and the NAME theorem tells us that any automorphism MATH of the matrix algebra MATH is an inner one: there exists MATH such that MATH. Hence we obtain that MATH in MATH if and only if MATH and there exists MATH such that MATH, that is, MATH is equivalent to MATH, as needed. |
math/0009143 | It clearly suffices to deal with the case where MATH and both have mean zero. Let MATH . Take a function in MATH, of mean zero, and consider its NAME expansion MATH where MATH is the partial sum over all frequencies of norm less than MATH. Then the correlation function MATH is a sum of terms MATH . By NAME, the second and third terms above are bounded respectively by MATH and by MATH where MATH stands for the MATH-norm. The first term equals MATH . We claim that this is an empty sum, hence vanishes, for our choice of MATH: Indeed, by our condition if MATH we have MATH since MATH. Thus we find that for MATH as above that MATH . Since we can choose MATH as MATH (subject to REF) we get MATH and hence we have mixing. To prove exponential decay of correlations for NAME observables MATH, recall that if MATH satisfies the NAME REF of order MATH then for some constant MATH (see for example, CITE for the proof in the case of functions of one variable ) and hence taking MATH to be the smallest integer less than MATH gives MATH as required. |
math/0009143 | This is the well known Euclidean algorithm: Assume that MATH with MATH. Primitivity is equivalent to MATH being co-prime. Then find an integer MATH so that MATH with MATH. We then arrive at a new vector MATH. Now find an integer MATH so that MATH with MATH to get another vector MATH and so on. This proceeds until we get to the point that we have computed the greatest common divisor of MATH and MATH (which in our case equals MATH) with at most MATH steps. This gives us either the vector MATH, in which case we are done, or the vector MATH. In the latter case just note that MATH. |
math/0009143 | One can easily check that homogeneous quasi-morphisms on MATH which vanish on parabolics are NAME functions with respect to metric MATH. Let MATH be such a quasi-morphism with MATH. Take MATH sufficiently small and consider any sequence MATH with MATH. It follows that MATH for some MATH and all MATH. Applying REF , we get that MATH is mixing with exponential decay of correlations on NAME observables. |
math/0009143 | The proof is divided into REF steps. CASE: We claim that every hyperbolic matrix MATH is conjugate in MATH to a matrix MATH with MATH. Indeed, assume that MATH, where MATH . One calculates that MATH, where MATH. The discriminant MATH of the quadratic form MATH equals MATH. Since MATH is hyperbolic, MATH is irrational. In this case (see CITE) there exists a primitive vector MATH with MATH . Hence we obtain MATH as required. CASE: As a consequence we get that every hyperbolic matrix MATH decomposes as MATH, where MATH and MATH is parabolic. Indeed, in view of REF we can assume without loss of generality that MATH with MATH. Write MATH so MATH. One can always choose MATH so that MATH. CASE: For an integer MATH put MATH, where the maximum is taken over all hyperbolic matrices MATH whose trace lies in MATH. Since a quasi-morphism is a class function on the group, and the number of hyperbolic conjugacy classes with given trace is finite, the function MATH is well defined. REF yields inequality MATH for all MATH, where we set MATH for MATH, and brackets stand for the integral part of a real number. Arguing by induction we get that MATH for all MATH. This completes the proof. |
math/0009144 | Since MATH, we can find a unitary automorphism MATH of MATH over MATH which is equal to MATH on MATH and MATH, and equal to MATH on MATH. Pulling MATH back by MATH, we reduce to the case MATH, so that the caloron configuration is now a loop in MATH. But this space is contractible, so the result follows. |
math/0009144 | The space of calorons on a given framed bundle, with given boundary data MATH is contractible. It is easy to see that any continuous path joining MATH to MATH gives rise to a norm-continuous path of NAME operators between the NAME spaces in MATH. Since each of these is NAME by REF , it follows that the index is constant on this path. |
math/0009146 | By the proof of REF we have that for any MATH, MATH, where, as before, MATH is the image of MATH as a subspace of MATH, and in fact, by REF and since MATH is stable, it results MATH. Thus we have, up to a change of basis, MATH where MATH and MATH are subspaces of MATH of dimension MATH. It is easily checked that MATH and since MATH, it must be MATH: this implies that MATH and therefore we can assume MATH for any MATH. Moreover MATH otherwise, up to the action of MATH, it would be MATH and by REF , MATH would not be stable, because there would exist a vector MATH such that MATH. Therefore MATH are linearly independent and we can suppose MATH for some basis MATH of MATH. |
math/0009146 | The exactness of REF is proven in (CITE, pag. REF). In order to prove the exactness of REF , we proceed by mimicking the proof of the existence of the NAME complex given in CITE. Let MATH and let MATH be the projection onto the first space. The morphism MATH defines a section MATH, given by MATH where the MATH's are the entries of MATH and MATH are the coordinates of MATH. The zero locus of MATH is MATH and the NAME complex associated is given by: MATH . We have that MATH and, since MATH is stable, then MATH REF . By REF , the sequence: MATH is exact for any MATH. Since each fiber of MATH is isomorphic to MATH, it results: MATH where MATH is the higher direct image functor associated to MATH (see REF). Moreover MATH, that yields the exact sequence: MATH . The exactness of the sequence REF follows by gluing REF tensored by MATH with REF : in fact, in both the sequences the morphisms MATH are canonically defined. |
math/0009146 | Let MATH and let us consider a resolution of MATH: MATH where each MATH is a direct sum of line bundles. Let us split the sequence into short exact sequences: MATH then, applying the functor MATH, we get: MATH for any MATH such that MATH. Thus MATH for any MATH. Moreover MATH if MATH and thus by REF , MATH is reflexive. |
math/0009146 | The injective map MATH induces the exact sequence: MATH where MATH is a MATH-rank sheaf such that MATH: thus, dualizing this sequence and observing that, by REF , MATH, it results MATH, in fact by REF all these sheaves are torsion-free. Moreover MATH and therefore REF follows. |
math/0009146 | By REF , we can suppose MATH . Moreover the same technique used above can be applied to prove the thesis for all MATH. Thus it suffices to show that MATH. It is easily checked that by dualizing the sequence REF we get the sequence MATH . Thus we just need to prove that if MATH is such that MATH then MATH: this is a direct computation. |
math/0009146 | The first statement of REF easily implies that the map MATH is injective. Since MATH is torsion-free, so is MATH. Let now MATH be a torsion-free sub-sheaf of rank MATH. then MATH: since MATH REF , it results MATH, that is, MATH. Thus MATH is MATH-stable. NAME, let MATH be a non-stable matrix. Then, by REF , we can write MATH where MATH is a vector of length MATH. Thus MATH defines a sub-sheaf MATH that is contained in the exact sequence: MATH . It is easily checked that MATH. |
math/0009146 | Let us prove first that any MATH is uniquely determined by a morphism of sequences: MATH . This is a direct consequence of the fact that, by the vanishing of MATH and MATH we get the exact sequence: MATH . Thus if MATH is simple, then the only automorphisms of MATH are the homotheties, that implies REF . |
math/0009147 | Let MATH . We want to show that MATH. It is easy to see that MATH is a closed subset and that it is closed under addition and conjugation. Let MATH and MATH. Then MATH so MATH. Hence MATH is also closed under multiplication. So it is a MATH-subalgebra of MATH. Since MATH for each MATH. So MATH. |
math/0009147 | First notice that for each MATH . From this we get that for each MATH . By REF we see that there is only a finite number of different sets MATH. So we can for each MATH choose finite sets MATH and MATH such that MATH . From this we get MATH . So MATH for each MATH. We can therefore define MATH by MATH where MATH is as in REF . Since the MATH's are mutually disjoint the MATH's are mutually orthogonal projections. By REF we have that MATH . So MATH is generated by MATH. Since for each MATH we have by REF that MATH . |
math/0009147 | Let MATH. By REF MATH for each MATH, and since MATH, we have MATH. So MATH . Hence MATH is a partial isometry. Since the left NAME cover graph is left-resolving, we have that if MATH either MATH or MATH. If MATH by REF , and if MATH . So MATH for MATH. By REF MATH . So the partial isometries MATH satisfy the NAME relations and therefore there exists a MATH-homomorphism from MATH to MATH sending MATH to MATH. |
math/0009147 | We will prove the statement by induction over the length of MATH. First assume that MATH. Then the statement follows directly from the definition of the left NAME cover graph. Assume next that we have proved the statement for MATH, and that MATH. Let MATH. If MATH, then MATH. So there exists a unique path MATH such that MATH and MATH and furthermore MATH. Since MATH and MATH, MATH. Thus there exists an unique edge MATH, such that MATH and MATH and furthermore MATH. Since MATH, MATH is a path on the left NAME cover graph and MATH, MATH and MATH. If MATH is another path such that MATH and MATH, then MATH, MATH, MATH and MATH. So MATH and MATH. Hence MATH. If there exists a path MATH such that MATH and MATH , then MATH is a path such that MATH and MATH, and MATH is an edge such that MATH and MATH. So MATH and MATH. Hence MATH. |
math/0009147 | Observe that MATH for all MATH if MATH, and that MATH for all MATH if MATH. So MATH if MATH and MATH if MATH. Since MATH is generated by MATH and MATH for MATH there exists a MATH-homomorphism MATH from MATH to MATH sending MATH to MATH. For each MATH define MATH by MATH . Since MATH we have that MATH for each MATH. Since the left NAME cover graph is left-resolving we have MATH so MATH is a partial isometry. We see that MATH . If MATH, MATH and if MATH, MATH so MATH for all MATH. By REF we have that MATH for all MATH. So according to REF MATH extends to a MATH-homomorphism from MATH to MATH sending MATH to MATH and MATH to MATH. |
math/0009147 | According to REF there exists a MATH-homomorphism MATH such that MATH, and according to REF there exists a MATH-homomorphism MATH such that MATH and MATH. We have that MATH where we for the last equality use that MATH, and that MATH if there does not exists an edge with range MATH and label MATH. So MATH, and since MATH. Thus MATH and MATH are each other's inverse and MATH. |
math/0009148 | The first assertion is proved using CITE and the fact that the full dimensional cones of the NAME fan of MATH are exactly the cones corresponding to monomial initial ideals of MATH. The second assertion is easily proved by noticing that MATH is a monomial ideal none of whose generators contain the variable MATH. |
math/0009148 | First suppose that MATH does not have minimum negative support. Then there is MATH such that MATH is strictly contained in MATH. This means that MATH for MATH, and MATH. Say MATH. If MATH, then MATH does not hold. If MATH, then MATH does not hold. If MATH, then MATH does not hold, and if MATH then MATH does not hold. All of this means that such a MATH cannot exist, and thus MATH has minimum negative support. We show that MATH has minimum negative support by contradiction. Assume it does not have minimum negative support. Then there is MATH such that MATH. But then the negative support of MATH is strictly contained in the negative support of MATH, a contradiction. The proofs for the other two vectors are similar. |
math/0009148 | This follows from the same arguments that proved REF . |
math/0009148 | To see that our candidate for standard pair satisfies the criterion of REF, we have to show that the only integer point in a certain polytope is the origin. This follows exactly from the same arguments of REF if MATH is a monomial ideal. Otherwise, we shrink MATH so that the same arguments will work when we use the weight vector MATH. We also need to find elements in MATH that belong to that polytope when one of the defining inequalities is removed. Those elements will be MATH, MATH and MATH. |
math/0009148 | Fix MATH. Let MATH such that MATH if MATH; MATH otherwise. Observe that MATH for MATH. For MATH we define, following CITE: MATH . Following the proof of REF, we see that, for MATH there exists a positive integer MATH such that MATH contains a nonzero integer vector MATH. It must satisfy MATH. The reason for this is that, since MATH is a monomial ideal, there exists a unique solution of the integer program MATH where MATH (see CITE), and we can choose MATH as that solution. The vectors MATH are almost what we want, except that we cannot a priori guarantee that MATH, even if we look at all the polytopes MATH for MATH, that is, even if we look at the (possibly unbounded) polyhedron: MATH . However, we may assume MATH since we can always choose MATH small enough so that a feasible point that satisfies MATH is better than one that satisfies MATH. The following notation is very convenient: MATH . Notice that MATH. Let us first deal with the case when the hyperplane MATH is parallel to MATH, that is, when there exists MATH such that MATH. We know that MATH for some MATH. Since MATH, we have MATH which implies MATH, so that MATH. But MATH is an integer. By construction of MATH, this implies MATH and MATH (remember that the only integer coordinates of MATH are the first four). Now MATH and MATH imply that MATH. Now fix MATH such that the MATH-th row of MATH is not a multiple of the third one, and suppose that the integer program MATH is unbounded, and every bounded subprogram has its solution on the hyperplane MATH. Then MATH is an infinite set. Notice that MATH is not contained in the half-space MATH, since the defining inequalities of MATH given by rows that are multiples of the first row of MATH are of the form MATH. This follows from similar arguments as those in the preceeding paragraph. But now the set MATH contains infinitely many lattice points on the hyperplane MATH, is not itself contained in this hyperplane, but is a subset of MATH. This is impossible. Thus, if MATH satisfies MATH, the integer program: MATH must be bounded. Let MATH be the set of all such indices MATH, with MATH the (unique) solution to the corresponding integer program. We can now follow the proof of REF to show that MATH is an associated prime of MATH. Now let MATH be such that MATH is a minimal prime of MATH containing MATH. Then the vectors MATH for MATH satisfy all the desired properties, and the cardinality of MATH is MATH. The only thing we still have to show is that the rows of MATH indexed by MATH are linearly independent. To see this, let MATH be a standard pair of MATH, and look at the set: MATH which, by REF, is a polytope. If the rows of MATH indexed by MATH are not linearly independent, then MATH matrix whose rows are those rows of MATH has a nontrivial kernel. Hence there exists MATH such that MATH for all MATH. Since all the MATH are nonnegative, this means that MATH contains at least half of the line MATH, contradicting the fact that MATH is a bounded set. This concludes the proof. |
math/0009148 | We compute canonical series with respect to the weight vector MATH, as in CITE, assuming that MATH is a monomial ideal. We will deal with the case when MATH is not monomial later. The logarithm-free canonical solutions of MATH are of the form MATH where MATH is a fake exponent of minimum negative support. The fact that MATH is a fake exponent means that there exists a standard pair MATH of MATH, with MATH, such that MATH is the unique vector satisfying MATH and MATH. The only fake exponent with minimum negative support in MATH is MATH, whose canonical solution is MATH, and this function satisfies MATH. Let MATH be a fake exponent with minimum negative support that does not differ with MATH by an integer vector. Call MATH its canonical solution. If MATH belongs to MATH, it is clear that we must have MATH, that is, MATH must be a constant function with respect to MATH. In particular, we need MATH. If MATH is an element of MATH, then it must satisfy the inequalities MATH . But the set MATH intersects the lattice MATH only at MATH (see CITE). Switching the inequality signs, we conclude MATH, so that MATH. Now, if MATH is not a monomial ideal, take MATH and MATH as in REF . We can choose MATH so that the polytopes MATH and MATH have the same integer points. Now the previous reasoning applies when we compute canonical series with respect to MATH instead of MATH. |
math/0009148 | It is clear that the functions described above belong to the kernel of MATH. Suppose first that MATH is a logarithm-free solution of MATH that is constant with respect to MATH. We compute canonical series with respect to the weight vector MATH. If this cannot be done (that is, if MATH is not a monomial ideal) we replace this weight by MATH from REF with MATH small enough so that the ideas still work. Now MATH is a linear combination of logarithm-free canonical series (with respect to the weight MATH), each corresponding to a fake exponent with minimum negative support. Say MATH, where MATH and MATH are the exponents with minimum negative support. By taking initials, we see that at least one of those exponents must have its first coordinate equal to zero. Call that exponent MATH. But then, by the proof of REF , the canonical series corresponding to MATH is MATH, and this function belongs to our candidate spanning set. Subtracting MATH to MATH and repeating the process, we conclude that MATH is a linear combination of the functions in our candidate spanning set. Our task now is to show that no logarithmic solution of MATH can be constant with respect to MATH. Let MATH be a (possibly logarithmic) solution of MATH and suppose that MATH. The function MATH is a linear combination of canonical series. We write MATH where in each MATH we collect all canonical series appearing as summands in MATH whose corresponding exponents differ by integer vectors. Then there exist MATH exponents with minimum negative support and first coordinate equal to zero, such that MATH, where MATH. Also notice that each MATH must be constant with respect to MATH. We must show that each function MATH must be logarithm-free. Pick one of those functions MATH and the exponent MATH. We will now drop the index MATH for convenience in the notation. Write MATH in the form of REF . In this case MATH for any MATH by construction of MATH. Now we apply REF to the exponent MATH. Let MATH, write MATH for the vector MATH and let MATH be maximal with respect to the MATH-th coordinate. Remember MATH, where MATH contains only terms in MATH that are either less than MATH or incomparable to MATH. We know that MATH, since MATH. Then MATH . All the terms that come from MATH by applying the product rule are either zero or must be cancelled by something from MATH. As a matter of fact, MATH has a nonzero term which is a multiple of MATH if MATH, or of MATH otherwise. The numerators of these fractions are nonzero by construction of MATH. Then we have a sub-series MATH of MATH such that MATH . This means that MATH is a polynomial in the variable MATH, which contradicts the fact that MATH contains no term in MATH. This implies that MATH, so that MATH contains no MATH, and this is true for all MATH. Now pick any MATH, and MATH maximal with respect to the MATH-th coordinate. As before, MATH. Of course, since MATH is itself a hypergeometric function constant with respect to MATH, we may assume that MATH has no term in MATH. This and the homogeneity REF imply that there is a sub-sum of MATH of the form MATH, where MATH belongs to the kernel of MATH, and there are no other terms in MATH in MATH. From our previous reasoning, we know that MATH for all MATH. Since the rows of MATH indexed by MATH are linearly independent (and the columns of MATH span the kernel of MATH), we conclude that MATH. In particular, MATH. This completes the proof that MATH is logarithm-free. |
math/0009148 | That MATH is the fake exponent of MATH corresponding to MATH follows from the fact that MATH (and that we have only modified the third coordinate of MATH). Now we have to show that MATH has minimum negative support. We know that MATH for MATH, so that MATH has at least three integer coordinates. If it has exactly those integer coordinates, or if its integer coordinates are all greater than or equal to zero, then MATH. It follows that it has minimum negative support. To see this, suppose MATH is strictly contained in MATH for some MATH. This means that MATH, for MATH, and MATH. Then MATH, a contradiction. Now assume that MATH has some negative integer coordinates, and write MATH for some MATH. Then MATH has at least four integer coordinates. We claim that in that case, MATH has exactly four integer coordinates, and they are the first four. To show this claim that we will use the numbers MATH from Construction REF. We know MATH, so that MATH. Suppose that MATH for some MATH. Then the MATH-th column of MATH and the third column of MATH are linearly independent, because otherwise, we would have MATH so that MATH. This means that MATH. But now the construction of the numbers MATH implies that the only integer coordinates of MATH must be the third one, the MATH-th one, and maybe the first one (if the first row of MATH is a multiple of the third). We obtain a contradiction. Thus, the only integer coordinates of MATH are the first four. Moreover, MATH has some negative integer coordinates. This can only happen if MATH is a top dimensional standard pair, MATH is strictly contained in MATH, and MATH is a negative integer, where MATH is the only element of MATH. Assume that MATH does not have minimum negative support, and pick MATH such that MATH is strictly contained in MATH. Looking at MATH, we conclude that we cannot have MATH. Then MATH and MATH. It follows that MATH has minimum negative support MATH (and is thus an exponent of MATH). We will show that MATH actually does not have minimum negative support. This contradiction will imply the desired conclusion about MATH. In order to show that MATH does not have minimum negative support, we need to find a vector MATH such that the negative support of MATH is empty. We know that MATH, MATH and MATH. We have the following cases: CASE: MATH , CASE: MATH , CASE: MATH , CASE: MATH , CASE: MATH , CASE: MATH . In case MATH, MATH is contained in MATH. In case MATH, MATH is contained in MATH. In case MATH, MATH is contained in MATH. In case MATH, MATH is contained in MATH. In case MATH is contained in MATH. In case MATH, MATH is contained in MATH. Cases MATH, MATH and MATH follow directly from the construction of MATH. For case MATH, remember that we assumed at the beginning of REF that either the second and fourth rows of MATH are linearly dependent, or the cone MATH is contained in the first quadrant. This means that the only way case MATH could happen is if the second and fourth rows of MATH are linearly dependent, and MATH. Then our assertion about negative supports follows by direct verification. Finally, let us do case MATH. Case MATH will be similar. Since MATH and MATH, we have MATH. Since MATH is a negative integer, we have MATH. The inequalities that MATH satisfies imply that this vector belongs to the second quadrant of MATH. Its second coordinate is strictly less than one. To see this, remember that MATH. The line MATH cuts the vertical axis of MATH above zero and strictly below one (this is because the line MATH cuts the vertical axis at height MATH). It follows that MATH. We have: MATH . We know MATH, since MATH. We also know MATH. The sum of this two numbers is an integer (since MATH is an integer), so it must be greater than or equal to MATH. This implies MATH, and concludes the proof that MATH is contained in MATH. This containment might not be strict, but certainly MATH (or MATH in the other cases). Moreover, we can repeat this process, and keep adding columns of MATH until the third coordinate is a nonnegative integer, while keeping the first, second and fourth coordinates also nonnegative. This shows that MATH does not have minimum negative support, which is the contradiction we wanted. |
math/0009148 | Suppose there is a solution MATH of MATH such that MATH. We will obtain a contradiction. We proceed as in the part of the proof of REF where we show that the functions MATH are logarithm-free. The first step is to use REF to write MATH for every MATH. We apply REF , with the goal of showing that MATH has no terms in MATH for MATH. Let MATH, pick MATH maximal with respect to the MATH-th coordinate, and let MATH from REF . Then MATH . As in the proof of REF , there are nonzero terms when we compute MATH using the product rule. Now, all these terms have either logarithms or denominator a positive integer power of MATH. By construction, MATH, so that MATH. Now, since MATH, MATH is a further derivative of MATH. But MATH has no terms with denominator MATH with MATH. This means that MATH must be cancelled with terms coming from MATH, and this implies (again, as in REF ) that MATH or, equivalently, that MATH has no terms in MATH for MATH. From this we can show that MATH is logarithm-free. Now, MATH has a term MATH, and the only way this term matches with a term of MATH is if MATH has a term MATH. But MATH is logarithm-free, and we obtain a contradiction. |
math/0009148 | Pick any solution MATH of MATH whose derivative with respect to MATH lies in MATH. Write MATH as in REF : MATH for MATH. Here we must have MATH, since MATH lies in MATH. Suppose that MATH. Look at the logarithm-free function MATH . If MATH, the term MATH has logarithms, so it must be cancelled with terms coming from MATH. Thus MATH must have a sub-series MATH such that MATH. Then MATH is constant with respect to MATH, which contradicts the construction of MATH (all its logarithmic terms are either less than or incomparable to MATH). Therefore, MATH or MATH. Now choose MATH with MATH maximal. Suppose MATH. Consider the function: MATH . If MATH the nonzero summand MATH has logarithms, hence it must be cancelled by some other term of the right hand side sum. Since the numerator MATH is constant with respect to MATH, this is impossible. Thus MATH or MATH. Similar arguments using MATH show that MATH maximal with respect to the first coordinate must be either MATH or have MATH, and the same using MATH will give the analogous conclusion when MATH is maximal with respect to the fourth coordinate. Now pick MATH and choose MATH with MATH maximal. Suppose MATH. Then MATH for MATH. We know (looking at the homogeneities REF ) that MATH . Call MATH the coefficient of MATH in MATH. Comparing both sides of the previous equalities, we conclude that MATH . Let MATH be the vector whose MATH-th coordinate is MATH if MATH, and the rest are zeros. If MATH, MATH is a nonzero element of the kernel of MATH, whose first MATH coordinates are MATH. Such an element does not exist. If MATH, MATH, and the first MATH coordinates of MATH are zero. Thus, MATH is a vector in the kernel of MATH and is therefore of the form MATH, for MATH. Since the first MATH rows of MATH lie in different quadrants of MATH and the first four entries of MATH are nonnegative, this is impossible. Hence MATH where MATH is logarithm-free. If we assume that MATH has no term MATH (perfectly legal, since this is a solution of MATH that is constant with respect to MATH), the fact that the vector MATH belongs to the kernel of MATH follows from the homogeneities REF . |
math/0009148 | By REF , an element MATH of the solution space of MATH such that MATH lies in MATH is of the form MATH with MATH a logarithm-free function with integer exponents, no term MATH, and MATH in the kernel of MATH. Notice that once the MATH are fixed, MATH is unique with those MATH, since the difference of two such functions would be a logarithm-free solution of MATH with no term in the kernel of MATH, whose derivative with respect to MATH belongs to MATH. It follows from REF that this difference must be zero. Since the vector of the MATH is in the kernel of MATH, the previous remark implies that the space of solutions of MATH whose derivative with respect to MATH lies in MATH has dimension at most MATH, the dimension of the kernel of MATH. |
math/0009148 | By contradiction, suppose there is a solution MATH of MATH such that MATH where MATH is a linear combination of the functions from REF . We can write MATH where MATH is the sum of the terms in MATH whose exponents and MATH differ by an integer vector, and MATH is the sum of the terms in MATH whose exponents and MATH differ by an integer vector. Clearly, MATH and MATH. But the functions MATH and MATH must be solutions of MATH. To see this, notice that if two monomials MATH and MATH are such that MATH, then the intersection of the MATH-modules obtained by acting with MATH on MATH and MATH is either empty or MATH. Therefore all the MATH must be zero (by REF ) and also MATH must be zero (by REF ). |
math/0009148 | In REF we built one function in MATH for each function in a basis of MATH (which we knew from REF ). Moreover, REF provided at least two linearly independent functions for MATH. REF shows that all of these functions are linearly independent. Therefore MATH and this implies that: MATH where MATH is the image of MATH. The left hand side of REF equals the dimension of the solution space of MATH. The right hand side equals MATH plus the dimension of the solution space of MATH. This concludes the proof. |
math/0009148 | Pick MATH, and MATH as in Construction REF. We can choose MATH small enough so that the numbers MATH satisfy the conditions of Construction REF for all MATH. Call MATH and MATH . Then REF implies that MATH for all MATH. Now the proof of REF implies that MATH. This concludes the proof. |
math/0009155 | The automorphism group of MATH is a semidirect product of MATH and the outer automorphism group of the root system corresponding to MATH. It is easy to verify that, in the cases at hand, a nontrivial outer automorphism acts by sending MATH to MATH. From this, the result is clear. |
math/0009155 | By NAME duality, MATH. Since MATH, where MATH is nef and big, and MATH, MATH for every root MATH. A similar statement holds for MATH. Let MATH be a root. Then MATH, and so by the NAME theorem, MATH. By the first part of the lemma, MATH and so MATH. Thus MATH if and only if MATH for some effective curve MATH. In this case, since MATH and MATH is nef and big, it follows that MATH, and hence that MATH. The remaining equivalences are clear. |
math/0009155 | In any case MATH has degree zero. Suppose that MATH is not effective. Then MATH for all MATH. Consider the long exact cohomology sequence arising from MATH . Since MATH is NAME dual to MATH, it is zero for all MATH. Thus MATH is an isomorphism for all MATH. It follows that MATH is effective if and only if MATH, if and only if MATH is trivial (since it is a line bundle of degree zero). The remaining statements are clear. |
math/0009155 | Let MATH be a vector bundle on MATH such that MATH for every irreducible MATH-curve MATH. We shall show that MATH is trivial in some analytic neighborhood of the exceptional curve MATH. If MATH are the MATH-curves, and MATH, it follows easily from the fact that the dual graph of MATH is a union of contractible components and that the MATH meet transversally that MATH. Let MATH be a contractible NAME neighborhood of MATH, possibly disconnected, and let MATH. We claim that the natural map MATH is surjective. In fact, the cokernel of this map is contained in MATH. A standard argument using the fact that the singularities are rational double points and the formal functions theorem shows that MATH. Thus MATH is surjective. Lifting a basis of sections of MATH to MATH, we can assume, possibly after shrinking MATH and MATH, that MATH is trivial. Now choose a faithful representation MATH of MATH, which we use to view MATH as a subgroup of MATH for some MATH. Let MATH be the induced vector bundle MATH. By assumption, MATH is the trivial vector bundle MATH for every MATH-curve MATH. Thus, MATH is the trivial vector bundle over MATH. Let MATH be the quotient space MATH, which exists as an affine variety (since MATH is reductive) and hence as a complex manifold. Let MATH be the sheaf of morphisms from MATH to MATH, and similarly for MATH and MATH. From the exact sequence of pointed sets MATH we have a long exact cohomology sequence through the term MATH. If MATH is the class in MATH induced by MATH, then MATH maps to the trivial element in MATH, and hence comes by coboundary from MATH, which we can identify with the set of morphisms from MATH to MATH. If MATH is such a morphism, then since MATH is affine, MATH. Since MATH is normal, it follows that MATH is induced from a morphism MATH. But since there are local cross sections for the map MATH (in the classical topology), again after shrinking MATH we can assume that MATH is in the image of a morphism from MATH to MATH. In particular, this says that MATH is trivial. Thus the triviality of MATH implies that of MATH. Choosing a local trivialization of MATH for which MATH is one of the open sets and the remaining ones do not meet MATH gives a MATH-cocycle defining MATH which also defines a bundle MATH on MATH, and clearly MATH. The uniqueness is a straightforward consequence of the fact that, as MATH is affine, every morphism from MATH into MATH is constant on the exceptional curves and hence descends to a morphism from MATH into MATH. |
math/0009155 | First we claim: With MATH as in the statement of REF , CASE: MATH; CASE: Let MATH be MATH-curves and let MATH. Then MATH . CASE: By NAME, MATH only depends on the topological type of MATH. Thus we may replace MATH by MATH. But MATH and as we have seen, MATH for every root MATH. Hence MATH. CASE: The bundle MATH has a filtration whose successive quotients are either MATH or MATH, MATH. Thus, by REF , MATH, and a similar argument shows that MATH. Returning to the proof of REF , we see that MATH . Thus MATH, with equality if and only if MATH. In particular, MATH . Let MATH. From the exact sequence MATH and the fact that MATH is NAME dual to MATH, it follows that, if MATH, then MATH. Thus MATH REF . Finally, since MATH, the map MATH is injective. As we have seen, MATH. Thus, if MATH, then MATH as well, so that MATH . |
math/0009155 | First, we claim that MATH is topologically trivial. It suffices to check that MATH is topologically trivial. The surjection MATH induces a MATH-bundle MATH, and the corresponding line bundle is easily seen to be MATH, which is trivial. Hence MATH lifts to a MATH-bundle, which is automatically topologically trivial since MATH is simply connected. We have the exact sequence MATH . By REF , MATH. Hence MATH as well, so that MATH is rigid. But, since the trivial bundle is semistable, the only rigid MATH-bundle on MATH is the trivial bundle. Thus MATH is trivial. |
math/0009155 | By REF , it suffices to prove the result for one choice of MATH, which we may assume to be a smooth elliptic curve. By CITE, there exists a MATH-bundle MATH of the form MATH, such that REF MATH, and REF MATH is a lift to MATH of the bundle MATH. Let us show that we can find a MATH-bundle MATH such that MATH. To make the construction, choose any decreasing filtration MATH of MATH by normal MATH-invariant subgroups such that, for every MATH, MATH is contained in the center of MATH. It follows that MATH is a vector group which is a direct sum of root spaces MATH, MATH, say, and MATH acts on MATH in the usual way. By induction on MATH, starting with MATH, it suffices to show the following: Given MATH, suppose that we have found a MATH-bundle MATH such that MATH. Then we can lift MATH to a MATH-bundle MATH such that MATH. By general formalism (we use the notation of CITE), the obstruction to lifting MATH to some MATH-bundle lives in the cohomology group MATH. But MATH and MATH. Thus there is a lift of MATH to some MATH-bundle MATH. Let MATH. Then MATH and MATH are two lifts of MATH to MATH, and as such they differ by the action of MATH. On the other hand, MATH . By the last sentence of REF , since all of the roots in MATH are positive, the restriction map MATH is an isomorphism for every MATH. Thus, we can adjust the lift MATH by the action of MATH so that MATH. This completes the inductive step. |
math/0009155 | It is easy to check via the exact sequence MATH that the obstruction to lifting MATH to a MATH-bundle vanishes if and only if the bundle MATH is trivial. Moreover, using the above sequence and the fact that there is a subgroup of MATH isomorphic to MATH which surjects onto MATH any lift of MATH to MATH is unique up to isomorphism. Also, it is easy to verify that MATH, where MATH is the corresponding MATH-bundle. The exact sequence of tori MATH corresponds on the level of fundamental groups to MATH where the surjection MATH is given by MATH. It follows from REF that MATH corresponds to the line bundle MATH. Thus, MATH corresponds to the line bundle MATH. It follows that, if we set MATH, then MATH is trivial and thus MATH lifts to a MATH-bundle MATH. The regularity of MATH follows from the regularity of MATH. Finally, let us check that the period point of MATH corresponds to the point it defines in the moduli space. Since this point is independent of the NAME class of MATH, we may as well work with MATH. Working instead with the MATH-bundle MATH, note that a MATH-bundle over MATH is the same as an element of MATH. The bundle MATH is characterized by the property that, for MATH, the MATH-bundle arising from the homomorphism MATH is MATH. From this, it is clear that the period point agrees with the moduli point. |
math/0009155 | Every set of MATH disjoint lines can be completed to a set of MATH disjoint lines. On the other hand, such a set defines a blowdown to MATH and hence a diagonal basis MATH with MATH and MATH. The NAME group acts transitively on such bases, by REF , and hence on the set of all sets of MATH disjoint lines. A similar argument handles the case of MATH disjoint lines such that the blowdown is MATH. Now suppose that MATH is a twisted cubic on MATH, that is, a smooth rational curve MATH with MATH. The linear system MATH defines a birational morphism to MATH, and we have seen that all such are conjugate under the NAME group. If MATH is a conic, then it is easy to check that there is a blowdown to MATH such that MATH for MATH the pullback of a hyperplane class and MATH an exceptional curve, and again all such are conjugate under MATH. Finally, if MATH is a smooth rational quartic, then MATH defines a birational morphism from MATH to a quadric in MATH, and since MATH has no MATH-curves, the quadric is smooth. We conclude in this case by the last sentence of the preceding paragraph. |
math/0009155 | For MATH, this is clear by the remarks before the statement. For MATH, the adjoint bundle MATH, where MATH is the set of roots. The result then follows since MATH. |
math/0009158 | Recall first the relation between the cylindrical metric, and the euclidean metrics MATH and MATH (near MATH and MATH, respectively): MATH where MATH . Here we are assuming that MATH is a smooth point, or if not, we pass to a uniformizing chart centred at MATH. It will be clear that MATH-equivariance is preserved throughout, so we can afford to ignore singularities from now on. Denoting the pointwise norms that correspond to the different choices of conformal gauge in REF by MATH, MATH and MATH respectively, we have by REF If MATH, then MATH; CASE: If MATH, then MATH; CASE: If MATH, then MATH. We compare first MATH and MATH. By the weighted NAME alternative, we have MATH and MATH the formal adjoint being taken with respect to the MATH-metric MATH. Therefore by conformal invariance of these differential operators and the rescaling formulae REF above, MATH and MATH . We argue next that the singularity at MATH is removable in all cases except MATH. First note that in all cases, MATH and MATH admit extensions as distributional sections to the whole of MATH CITE. Denoting these extensions by the same symbols, we obtain equations of the form MATH, MATH, where MATH and MATH are linear combinations of derivatives of the NAME distribution MATH. Thus MATH and MATH are sums of terms homogeneous of degree MATH for MATH and this is not compatible with the orders of MATH and MATH and the growth conditions on MATH and MATH that appear in MATH and MATH. So in these three cases, MATH and MATH, and since MATH and MATH are overdetermined elliptic the solutions are actually smooth. Hence we obtain MATH the last following from REF relating the lengths of an element of MATH in the compact and WALE models. We also have MATH . In order to overcome the problem encountered with MATH we note a result of CITE which provides exact comparisons of MATH-Sobolev spaces of MATH with ordinary NAME spaces on MATH for a good choice of weight MATH and exponent MATH. The version we need states that the obvious map on compactly supported functions extends to an isomorphism MATH provided that MATH . Note that the vanishing condition makes sense because MATH for MATH in MATH dimensions and that NAME 's result for functions applies here because MATH is weightless (and so behaves conformally like the trivial bundle). By direct calculation (in which the factor MATH is crucial) we see also that there is an isomorphism MATH if MATH and MATH are related as in REF. Using conformal invariance of MATH we have MATH and in particular MATH . It follows that MATH if MATH . To prove this, it is enough to show that given MATH, there exists MATH such that MATH. Working in normal coordinates MATH near MATH, we have the formula MATH just as in MATH with the Euclidean metric. Then MATH if MATH is constant and MATH in a neighbourhood of MATH. This completes the proofs of all statements pertaining to MATH and MATH in REF . CASE: Fix MATH and MATH as in REF. Then by NAME 's result and conformal invariance, there is a natural map MATH given by mapping MATH to the cohomology class MATH of the image of MATH in MATH. We claim first that this map is injective. Indeed, if MATH for some MATH with MATH, then transferring back to MATH we obtain MATH, such that MATH as MATH and satisfying MATH. But MATH implies that MATH and since MATH is in MATH, we may integrate by parts, getting MATH. This establishes the injectivity. To prove surjectivity, we need to show that if MATH represents a class in MATH, then we can find MATH, such that MATH and MATH . By a previous argument we may assume MATH so that MATH. It is straightforward to show MATH by adapting standard NAME arguments. In particular, we can solve the equation and transfer to MATH, getting a section MATH which vanishes at MATH. This completes the comparison of MATH for conformal blow-ups, and so the proof of REF . The comparison of MATH in REF follows very similar lines, using an analogue of NAME 's theorem to compare weighted NAME spaces on MATH with NAME spaces on MATH. The details are omitted. |
math/0009158 | Use the conformal isometry MATH given by MATH and the conformal invariance of MATH. Because of the conformal weights, we have if MATH is a section of MATH so that a solution MATH in MATH, homogeneous of degree MATH, translates into an exponential solution with factor MATH on the cylinder. In particular the constant solution in MATH gives rise to a solution that goes like MATH along the cylinder. We use a `removable singularities' argument like the one in the proof of REF . If MATH in MATH and MATH has homogeneity MATH in MATH, then MATH is a distribution supported at MATH and homogeneous of degree MATH. If this distribution vanishes at MATH, then by elliptic regularity, MATH is smooth near MATH and hence MATH is a non-negative integer. If the distribution is non-vanishing, then MATH must be a multiple of some derivative of MATH, hence MATH for some integer MATH. Hence MATH and so MATH cannot occur. |
math/0009158 | If MATH, MATH, then from REF , MATH has a phg expansion where the index set is just the positive integers. Translating to MATH and remembering the conformal weight, we get MATH which satisfies MATH and MATH near MATH. Hence by elliptic regularity, the MATH-null space of MATH on MATH agrees with the standard null-space of MATH on MATH. The argument for MATH is the same, for MATH can be identified with the MATH null-space of MATH and this is conformally invariant as an operator between bundles with the same conformal weights as for MATH (compare REF). The argument for comparison with MATH is also closely analogous. |
math/0009158 | CASE: Note first by the discussion in REF that the MATH-norm of MATH is MATH for some MATH. Taking MATH, MATH as MATH. To check that the map is smooth, it evidently suffices to show that the nonlinear terms define a smooth map. Note first by the properties of MATH that MATH where MATH is independent of MATH. Hence from REF we have MATH . Similarly MATH and MATH . (Here MATH is a generic constant bounded independent of MATH but possibly varying from line to line.) This is not quite enough because the MATH also have a real-analytic dependence on the MATH-jet of MATH. However, this is convergent for all MATH with MATH and by multiplication properties of elements of MATH with MATH, these extend to define smooth maps from a fixed ball MATH into MATH. Combining these observations with the previous estimates for the terms in the derivatives of MATH, we obtain REF . CASE: This follows from the main estimate REF and the fact that MATH is obtained by gluing fully elliptic operators. CASE: This is deduced by a simple modification of the arguments used to prove MATH smooth. The details are omitted. |
math/0009158 | Combine the equations in the form MATH where MATH is a polyhomogeneous section which is a linear combination of MATH, MATH and MATH, MATH and MATH . CASE: Interior regularity. That MATH is MATH in MATH now follows from standard regularity results, which are applicable because MATH is already MATH, where MATH CITE. CASE: Boundary regularity. The method we use is closely analogous to that used by NAME in CITE; we are indebted to him for useful discussions on this point. Near the boundary, MATH, where MATH and MATH, say, where MATH has a phg asymptotic expansion. Then REF takes the form MATH and we already know that MATH is uniformly bounded as MATH. Now if we had the indicial operator MATH in place of MATH on the LHS of REF, we could use the fact that MATH has an inverse which behaves well on the MATH-Sobolev spaces to conclude that MATH for every MATH and some fixed MATH. It would follow that MATH is continuous at MATH for all MATH and all multi-indices MATH. Continuing with the argument under the simplifying assumption that MATH not MATH is on the LHS of REF, we can now use the fact that MATH and its inverse also preserve spaces of polyhomogeneous functions. To do so, assume by induction that MATH has a phg expansion up to some order MATH, say. Then because of the factor MATH on the Right-hand side of REF, MATH has an expansion to order MATH, and so MATH also has such an expansion. Hence MATH has a complete phg expansion at the boundary. The result with MATH replacing MATH comes from a suitable approximation argument. The details of this are straightforward but lengthy, and are omitted. |
math/0009158 | Combine REF . The estimate on MATH follows from the fact that the MATH-norm of MATH in REF is of the same order of magnitude as MATH and the estimate REF. The completeness of the family of metrics constructed comes directly from the implicit function theorem. |
math/0009158 | Combine REF with REF . |
math/0009158 | Take MATH to be the conformal blow up at MATH of MATH. Take MATH to be the MATH-manifold obtained by gluing MATH to MATH by identifying MATH with MATH by MATH. Then MATH, viewed as a MATH-manifold, has boundary equal to MATH, while the boundary of MATH is MATH. Applying REF to MATH and MATH now gives the conclusion, in view of the calculations of the deformation cohomology groups for a cylinder given in REF . (We have denoted by MATH the action induced by MATH or MATH on MATH.) |
math/0009158 | It is clear that REF combined with REF gives a NAME conformal structure MATH on MATH. However it is obvious that MATH is NAME, so by CITE, there is a NAME representative of the conformal class MATH and this must necessarily be scalar-flat. |
math/0009158 | Recall first the NAME metric MATH CITE, a weakly asymptotically euclidean scalar-flat NAME metric on the blow-up MATH of MATH at the origin. The complement of the exceptional divisor in MATH is biholomorphic to MATH, so we can construct the multiple blow-up MATH, with its standard complex structure, by gluing a copy of MATH at each of the MATH. More precisely, let MATH denote the conformal blow-down of MATH. Then MATH by REF . So let MATH stand for the MATH-manifold obtained from MATH by conformal blow-up of each of the MATH and conformal blow-down of MATH. By REF , MATH and this is zero, because if MATH in euclidean space then every component of MATH satisfies MATH and so MATH by the growth condition at MATH. We take MATH to be MATH copies of MATH and apply REF , taking all weights equal to MATH since all boundary components are spherical REF . The conclusion is that MATH admits a hermitian-ASD conformal structure with good asymptotic behaviour at the boundary. It remains to verify that there is a weakly asymptotically euclidean NAME representative of this conformal class. The fundamental MATH-form MATH of the metric constructed by REF has the form MATH, where the MATH form a standard basis of left-invariant MATH-forms on MATH. Following the method of CITE, introduce a MATH-form MATH which measures the failure of MATH to be NAME: MATH . From the asymptotic formula for MATH, MATH where MATH is a MATH-form whose length decays exponentially as MATH. It is a local calculation that on a NAME manifold MATH is ASD hence MATH is harmonic. Because MATH and MATH is in MATH, we conclude by the standard integration-by-parts argument that MATH. Hence MATH is closed and decaying and so by CITE represents an element of the relative cohomology group MATH. By other results in CITE, MATH where MATH is also exponentially decaying. It follows that MATH and this is the conformal factor that yields the NAME representative. |
math/0009158 | If MATH is NAME, we can identify MATH with the composite MATH of NAME operators as above. To prove the proposition, it is enough to note the NAME formulae MATH which hold whenever MATH and MATH. (The verification of these is left to the reader.) Suppose MATH; set MATH, so that MATH. If MATH then MATH is invertible, so MATH, that is, MATH. Similarly MATH is invertible, so MATH. This completes the proof of the first part. Suppose now that MATH. With MATH and MATH as before, we deduce first that MATH is a parallel section. In particular, MATH. But MATH also lies in the image of MATH (by definition), so MATH must be zero, by the NAME alternative. Thus MATH and, applying the NAME formula, MATH is parallel. Thus we have identified the kernel of MATH with the space of parallel sections of MATH in the NAME case. |
math/0009158 | Recall the formula MATH from the previous proposition. Let MATH and let MATH. Then MATH as MATH, and so MATH and letting MATH we conclude as before that MATH is parallel. Since MATH at MATH, moreover, MATH. Hence MATH. We make the same argument as above using the NAME formula. This time the boundary term is MATH but we still conclude that MATH is parallel and hence MATH, since MATH at MATH. |
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