paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0003032 | Let us denote the centralizers in MATH of the actions MATH and MATH by MATH and MATH, respectively. The centralizer MATH contains multiplications by all elements of MATH. For, if one takes any basis in MATH, the multiplication by an element MATH takes elements of the basis into elements of MATH, which are linear combin... |
math/0003032 | The action MATH corresponding to the ring MATH is cyclic by definition since the ring coincides with the orbit of MATH. By REF , if MATH were cyclic, it would be algebraically conjugate to MATH, which, by REF , implies that MATH. |
math/0003032 | By REF any irreducible action MATH of MATH by automorphisms of MATH is algebraically conjugate to an action of the form MATH for a lattice MATH. Let MATH be the centralizer of MATH in the semigroup of linear endomorphisms of MATH. We fix an element MATH with MATH and consider conjugation of the action MATH by multiplic... |
math/0003032 | First, let us point out that it is sufficient to prove the proposition for irreducible actions. For, if MATH is not irreducible, it has a nontrivial irreducible algebraic factor of dimension, say, MATH. Since every factor of an ergodic automorphism is ergodic, we thus obtain an action of MATH in MATH by ergodic automor... |
math/0003032 | Assume that MATH is not hyperbolic. As MATH is simultaneously diagonalizable with MATH and has real eigenvalues, it has an eigenvalue MATH or MATH. The corresponding eigenspace is rational and MATH - invariant. Since MATH is irreducible, this eigenspace has to coincide with the whole space and hence MATH. |
math/0003032 | Suppose the matrix MATH corresponds to the ideal class MATH with the MATH - basis MATH. Then MATH . Since MATH, there exists a matrix MATH having the same eigenvalues which corresponds to a different ideal class MATH with the basis MATH, and we have MATH . The eigenvectors MATH and MATH are chosen with all their entrie... |
math/0003032 | We refer to REF ' from (CITE, ``Corrections."). According to this corollary the measure MATH is an extension of a zero entropy measure for an algebraic factor of smaller dimension with NAME conditional measures in the fiber. But since MATH contains a MATH-automorphism it does not have non - trivial zero entropy factors... |
math/0003032 | First of all, REF is invariant under metric isomorphism, hence MATH also satisfies this condition. But ergodicity with respect to NAME measure can also be expressed in terms of the eigenvalues; hence MATH also satisfies (MATH). Now consider the joining measure MATH on MATH. The conditions of REF are satisfied for the i... |
math/0003032 | Consider a hyperbolic matrix MATH with irreducible characteristic polynomial and real eigenvalues such that the origin is the only fixed point of MATH. Consider a subgroup of MATH isomorphic to MATH and containing MATH as one of its generators. This subgroup determines an embedding MATH. Since MATH and by REF , all mat... |
math/0003032 | Since MATH is a measurable factor of MATH, every element which is ergodic for MATH is also ergodic for MATH. Hence MATH also satisfies condition MATH. As before consider the product action MATH which now by the same argument also satisfies MATH. Take the MATH invariant measure MATH on MATH. This measure provides a join... |
math/0003032 | By REF , MATH and MATH are algebraic factors of each other. This implies that MATH acts on the torus of the same dimension MATH and hence both algebraic factor - maps have finite fibres. Now the statement follows from REF . |
math/0003032 | Since action MATH satisfies condition MATH REF we can apply REF and conclude that we only need to consider the case when MATH and MATH are isomorphic over MATH up to a time change. But then, by REF , MATH and MATH are not algebraically isomorphic up to a time change and hence, by REF , they are not measurably isomorphi... |
math/0003032 | First we notice that a matrix in block form MATH with MATH commutes with MATH if an only if MATH commute with MATH and can thus be identified with elements of MATH. In this case MATH can be identified with a matrix in MATH. Since MATH (compare CITE), the norm of the determinant of the MATH matrix corresponding to MATH ... |
math/0003034 | Take MATH and MATH in the definition of double concordance. |
math/0003034 | There is a copy of MATH in MATH intersecting REF-twist-spin of MATH in MATH. Since MATH splits MATH, there is an invertible concordance from MATH to MATH. Hence MATH is split by MATH and the result follows. |
math/0003034 | Let MATH be a copy of the non-straight arc of MATH in REF-ball MATH and let MATH be a copy of the non-straight arc of MATH in MATH as shown in REF . The closed curve MATH bounds an obvious punctured torus MATH that is the shaded region in REF . Consider MATH as the plumbing of two MATH. Let MATH be the cores of the two... |
math/0003034 | Consider MATH as the complement of the unknot MATH in MATH. The knot MATH in REF is isotopic to MATH in REF . It is obvious from REF that MATH is the link in MATH formed by replacing a trivial REF-tangle in NAME link with MATH (dotted circle in REF ). The proposition now follows from REF . |
math/0003034 | Let MATH be a knot in MATH. If MATH is trivial it is prime itself. Suppose now that MATH is nontrivial. Let MATH be MATH satellite of MATH where MATH is the knot in MATH in REF . By REF , MATH splits MATH. We now only need to show MATH is prime. Since MATH is the unknot in MATH, MATH is prime by REF and to complete pro... |
math/0003042 | CASE: First of all, note that MATH; thus the group MATH also acts on the spaces MATH and MATH (and hence on the set MATH). If REF-tuple MATH is admissible, then MATH is connected, and hence the quotient MATH must be connected too. This implies that MATH has a unique connected component. Since MATH has exactly MATH conn... |
math/0003042 | From MATH we obtain MATH. Thus, MATH. |
math/0003042 | If MATH is admissible, then it is straightforward that MATH has a unique cycle. Vice versa, if MATH has a unique cycle, then REF holds. Since MATH implies REF , the result is a direct consequence of the above lemma. |
math/0003042 | Since MATH is admissible, the arcs of MATH are precisely the arcs of MATH. Let MATH be the sequence of these arcs, following the canonical orientation on MATH, and let MATH, with MATH. We have: MATH,where MATH. Since MATH if MATH is of type I, MATH if MATH is of type II and MATH if MATH is of type III, the result immed... |
math/0003042 | Since the two systems of curves MATH and MATH on MATH define a NAME diagram of MATH, there exist two handlebodies MATH and MATH of genus MATH, with MATH, such that MATH. Let now MATH be the cyclic group of order MATH generated by the homeomorphism MATH on MATH. The action of MATH on MATH extends to both the handlebodie... |
math/0003042 | Obviously MATH. Since MATH is admissible, it satisfies REF . This proves that MATH satisfies REF , for each MATH. Since MATH and MATH, we obtain MATH, for each MATH, which implies REF, or equivalently REF . Moreover, MATH is odd, since MATH. Thus, REF proves that MATH is admissible. The final result is then a direct co... |
math/0003042 | From MATH it immediately follows that MATH has a unique cycle in MATH. Since MATH is odd, REF proves that MATH is admissible. Since MATH, all assumptions of REF hold; hence MATH is admissible for each MATH and MATH is a MATH-fold cyclic covering of MATH, branched over a knot MATH which is independent on MATH. In order ... |
math/0003052 | The only property we have to check to make sure that the collection MATH defines a MATH-algebra structure, is the identity looking like MATH where MATH is a quadratic (non-commutative) polynomial. Since MATH has zero differential (this means MATH in our notation) the left-hand side vanishes. The right hand side vanishe... |
math/0003061 | By CITE we can find simple rank one NAME algebras MATH, MATH such that MATH and MATH. The NAME Theorem for tensor products CITE shows that MATH. Since the algebras involved are all p.i.s.u.n. and satisfy the U.C.T., the result follows from the Classification Theorem CITE. |
math/0003063 | As in the case of REF, we need to add a clause to the theorem for the purpose of induction, that is, we will prove the following statement by induction on MATH. For a sufficiently general curve MATH of degree MATH in MATH, MATH for any curve MATH that meets MATH properly. In addition, there exists a set MATH of countab... |
math/0003063 | Let MATH be the canonical divisor of MATH. We argue by induction on MATH. For MATH, we need to verify that MATH, which is more or less obvious. Suppose that MATH. Let MATH be a pencil of curves in MATH whose central fiber MATH is a union of MATH and MATH and let MATH. Let MATH and MATH be defined as before. Again, the ... |
math/0003064 | We show first REF MATH and then REF MATH. The isomorphism MATH can be proved analogously to REF and so is omitted. CASE: Observe that in the case where MATH holds with some nonsingular matrix MATH over MATH, the statement of the lemma obviously holds. Hence by considering MATH as MATH, we can assume without loss of gen... |
math/0003064 | Suppose that MATH is a local ring and MATH is a unimodular row. Thus there exist MATH such that MATH . Since MATH is local, the set of all nonunits is an ideal. From REF , there exists a MATH with MATH such that MATH is a unit. We assume without loss of generality that MATH is a unit. If MATH, then MATH is a unit, whic... |
math/0003064 | This proof mainly follows that of REF. If MATH is in MATH, then we can select the zero matrix as MATH. Thus we assume in the following that MATH is in MATH. Since the determinant of REF is in MATH, there exists a full-size minor of MATH in MATH by NAME 's expansion of REF and by REF . Let MATH be such a full-size minor... |
math/0003064 | Since MATH is local, REF are equivalent by REF . Thus we only prove REF and vice versa. CASE: Suppose that MATH is MATH-stabilizable. Then the MATH-module MATH is projective by REF. Further it is free by REF. Let MATH and MATH be matrices over MATH with MATH. Then the MATH-module MATH is free of rank MATH since MATH is... |
math/0003064 | Since the following implications are obvious: by virtue of REF , we only show that REF implies REF . Suppose that REF holds. Let MATH, MATH, MATH, and MATH be matrices over MATH with MATH such that MATH and MATH are MATH-nonsingular (recall that MATH is causal). By REF , MATH has both right-/left-coprime factorizations... |
math/0003064 | We show first the ``Only NAME part and then the ``NAME part. (Only If). Suppose that MATH is stabilizable. Then by REF , for every prime ideal MATH in MATH, MATH is MATH-stabilizable. By REF , MATH has both its right-/left-coprime factorizations over MATH. Suppose that MATH holds over MATH with MATH, where the matrices... |
math/0003064 | It is obvious that REF implies REF implies REF . Hence we only show that REF implies REF . CASE: Suppose that REF holds. Let MATH be a maximal ideal of MATH. Since MATH is local, the set of all nonunits in MATH is an ideal. Hence there exists a MATH with MATH such that MATH. Thus there exists MATH in MATH such that MAT... |
math/0003064 | Suppose that MATH is a unique factorization domain. Since the ``NAME part is obvious, we prove only the ``Only NAME part. (Only If). Let MATH be in MATH. Suppose that MATH is projective. If all MATH are zero, the proof is obvious. Thus in the following we suppose that at least one of MATH is nonzero. Since MATH is a un... |
math/0003064 | By virtue of REF , we suppose without loss of generality that MATH and MATH are matrices over MATH and MATH and MATH in MATH with MATH and MATH. Let MATH and MATH be the following matrices: MATH . Then we can see that there exists a unimodular matrix MATH with MATH and that MATH. Let MATH be the ideal generated by the ... |
math/0003064 | As in the proof of REF , if all MATH are zero, the proof is obvious. Thus in the following we assume that at least one of MATH is nonzero. Then there exist a nonzero MATH in MATH and MATH in MATH for MATH to MATH such that MATH and MATH. Thus there exist MATH for MATH to MATH with MATH. If MATH was a zerodivisor, the p... |
math/0003064 | By the same reason as in the proofs of REF , we assume that at least one of MATH is nonzero. (If). Suppose that REF holds. Then there exist MATH for MATH to MATH such that MATH. By appropriate changes of MATH, we assume without loss of generality that all MATH are nonzero with MATH and all MATH are zero subject to MATH... |
math/0003064 | We first show REF MATH and then REF the opposite inclusion. CASE: For every MATH, MATH holds, which implies that MATH. Hence MATH. CASE: Suppose that MATH is an element of the quotient ideal MATH. Then for every MATH, there exists MATH such that MATH holds and so MATH. Since this equality holds for every MATH, MATH has... |
math/0003064 | Let MATH be fixed. We first show REF MATH and then REF MATH. They are sufficient to prove this theorem. CASE: Let MATH be an arbitrary but fixed element of MATH. Then there exists a matrix MATH over MATH with MATH. Then for every MATH, we have MATH, so that MATH. This implies MATH. Hence we have MATH. CASE: Let MATH be... |
math/0003064 | Let MATH denote the elementary factor of the matrix MATH with respect to MATH. Also let MATH denote the reduced minor of MATH with respect to MATH. In the case where MATH is a unique factorization domain, the generalized elementary factor of the plant MATH with respect to MATH is equal to the principal ideal MATH. Thus... |
math/0003065 | See CITE or CITE. |
math/0003065 | We give the proof of the cofibration case, as the fibration case is strictly dual. Let MATH and MATH. Then a map MATH and its adjoint MATH are related by MATH, where MATH is the pushout of MATH along MATH. If REF holds and if MATH is a cofibration, we can construct MATH so that it is a weak equivalence to a fibrant obj... |
math/0003065 | See CITE. |
math/0003065 | Let MATH. REF for MATH given above shows that MATH is computed from MATH and MATH using only colimits and finite products, and both of these commute with filtered colimits and reflexive coequalizers. Hence MATH factors through a functor MATH. To show that MATH and MATH are equivalences, it suffices to show that if MATH... |
math/0003065 | That limits, filtered colimits, and reflexive coequalizers exist and are created in MATH is immediate from REF. That the free algebra functor is left adjoint is a standard property of monads CITE. Existence of colimits follows from CITE; or note that colimits of a diagram MATH can be constructed explicitly as the refle... |
math/0003065 | It is clear using REF that MATH factors through the subcategory. It remains to show that MATH is an equivalence. Let MATH denote the full subcategory of finitely generated free MATH-algebras; every object in this subcategory is isomorphic to MATH for some MATH. Consider the sequence of functors MATH the right-hand arro... |
math/0003065 | Let MATH. Then MATH is the equalizer of MATH, or equivalently of MATH, where the two arrows send MATH to MATH and MATH respectively, where MATH and MATH denote the algebra structure maps. |
math/0003065 | Define MATH to be the evident map MATH, and MATH to be the evident map MATH. Then MATH is easily seen to be a theory, and the evident functor MATH an equivalence. |
math/0003065 | This is immediate from the existence of the endomorphism theories. |
math/0003065 | Apply REF in each simplicial degree. |
math/0003065 | For convenience, we write the proof only in the case MATH and MATH are singleton sets; the general case is only notationally more difficult. We first show that MATH is in fact a natural transformation between functors defined on the full subcategory of free objects in MATH. Consider a map MATH between free objects. Thi... |
math/0003065 | Let MATH be as in REF . For this proof, we will write simplicial operators as acting on the right. Say that a MATH in non-degenerate if it is not of the form MATH for some non-identity MATH and some MATH. We claim: if MATH, MATH are non-degenerate elements such that MATH for some MATH, then CASE: MATH and MATH, and CAS... |
math/0003065 | We have that MATH; thus, we must show that MATH is a free degeneracy diagram of MATH-graded sets. First, suppose that MATH and MATH are a discrete theory and algebra. The degeneracy diagram MATH extends to a functor MATH, using the fact that MATH is a monad and and MATH and algebra: the ``face" maps are given by MATH a... |
math/0003065 | The right adjoint MATH is the identity on the underlying simplicial sets, and hence preserves weak equivalences and fibrations, and thus the left adjoint preserves cofibrations. |
math/0003065 | Let MATH be a natural transformation of functors MATH such that MATH is product preserving, MATH is a fibrant simplicial set and MATH is a weak equivalence for all MATH; we can use REF or REF . This functor MATH extends to MATH by REF. Now consider MATH where the fiber product is defined using MATH and MATH is defined ... |
math/0003065 | Consider the diagram MATH . The left-hand side is obtained by applying MATH to the square used in the proof of REF. By REF the functor MATH must be representable by some right MATH-module, and therefore the horizontal maps on the right-hand side are obtained using REF, and the right-hand square commutes. The top and bo... |
math/0003065 | If MATH is a trivial cofibration, this is REF. The theorem follows using REF. |
math/0003065 | We can first reduce to the case when MATH is a discrete graded simplicial set, using the diagonal principle REF and the fact that MATH (and similarly MATH) can be obtained as the diagonal of the simplicial object in MATH given by MATH, where MATH is the MATH-th simplicial degree of MATH. Next note that it is enough to ... |
math/0003065 | The equivalence of REF is immediate, since the MATH-th graded piece of MATH is MATH. Since for any MATH, MATH is a cofibrant MATH-algebra, REF implies REF . To show that REF implies REF , let MATH be a simplicial object in MATH defined by MATH; then MATH is MATH-free by REF, and hence is cofibrant, and MATH is a weak e... |
math/0003065 | The functors MATH are represented by an appropriate bimodules MATH and MATH, as described in REF. We claim that the map MATH induced by MATH is a weak equivalence, which means that we can derive the corollary as a special case of REF. To see that MATH is a weak equivalence, it suffices to show that it induces a weak eq... |
math/0003065 | First, note that the pair is a NAME equivalence if and only if the adjunction map MATH is a weak equivalence for every cofibrant MATH-algebra MATH. This is because, given MATH, the adjoint map factors MATH and MATH is a weak equivalence if and only if MATH is. The result now follows from REF, since the adjunction map i... |
math/0003065 | We have already seen that MATH is always right proper REF, so we need only consider left properness. That REF implies REF follows by observing that if MATH, then the square MATH is a pushout square in MATH-algebras in which the top arrow is a cofibration; properness implies that MATH is a weak equivalence if MATH is. T... |
math/0003065 | Recall from REF that MATH is equivalent to a full subcategory of the category of endofunctors on MATH. There is an evident explicit isomorphism MATH natural in MATH, as can be seen by applying REF. More explicit computations show that the monoidal structure on MATH restricts to MATH along MATH. |
math/0003065 | It is enough to show that MATH is the free monoid with respect to the MATH-product on MATH; that is, maps MATH are in bijective correspondence with maps MATH of monoids. Then from REF it follows formally that MATH is the free monoid with respect to the MATH-product, that is, it is a free theory. To make MATH into a mon... |
math/0003065 | Using REF, this amounts to showing that MATH, which is a direct translation of REF. |
math/0003065 | If MATH has the same image under the two maps, then it can have no vertices labelled by MATH, and hence must be a trivial tree. There is exactly one trivial tree for each element of MATH, and MATH contains only these. |
math/0003065 | We first show that it suffices to assume that MATH is a MATH-free theory and that MATH is a MATH-free map of theories. In fact, using the model category structure we see that MATH is a retract of a map MATH, where MATH and MATH are MATH-free. Then there are maps MATH of right MATH-modules, and the composite of these ma... |
math/0003065 | Using the endomorphism theory technology of REF, it is easy to see that MATH. By REF we see that algebras over MATH are the same as MATH-algebras MATH equipped with a map MATH of graded sets, or equivalently, the same as MATH-algebras MATH equipped with a map MATH of MATH-algebras. |
math/0003065 | Suppose that MATH. By REF, MATH is a cofibration between cofibrant theories, and thus MATH carries weak equivalences to weak equivalences by REF. Since there is an isomorphism MATH of underlying MATH-algebras, it follows that MATH is proper by REF. |
math/0003065 | Given a simplicial theory MATH, one can construct a weak equivalence MATH from a cofibrant theory MATH, since simplicial theories are a model category REF. Then MATH is a proper simplicial model category by REF, and the induced NAME pair MATH is a NAME equivalence by REF. |
math/0003065 | Recall that MATH being pointed means that the initial object MATH is isomorphic to the terminal object, denoted MATH. Choose MATH as in the proof of Theorem B, so that MATH is proper and is NAME equivalent to MATH via MATH. The initial object in MATH is MATH, which is not in general the terminal object. But since MATH ... |
math/0003065 | We first show that it suffices to assume that MATH is MATH-free. In general, MATH is a retract of some MATH-free MATH. Let MATH be a cofibration of MATH-algebras. Write MATH and MATH. Then the diagram MATH is a retract of the diagram obtained by applying MATH to it, which is MATH, and the map MATH is a cofibration of M... |
math/0003065 | A model category MATH is cellular in the sense of CITE if it is a cofibrantly generated model category with sets MATH and MATH of generating cofibrations and trivial cofibrations with the property that CASE: the domains and codomains of the elements of MATH are ``compact", CASE: the domains of the elements of MATH are ... |
math/0003066 | Consider an operator of the general form MATH . Then it is easily seen that MATH satisfies the NAME equation if and only if MATH and MATH . These equations are satisfied when MATH and MATH. Moreover these are essentially the only such solutions CITE. |
math/0003066 | The four relations are routine verifications. The fact that MATH then satisfies the NAME equation is a well-known fact about MATH-matrices extended to this slightly more general situation. |
math/0003066 | Recall that MATH . For the second term we note that MATH . Combining this with the binomial expansion of the first term yields the assertion. |
math/0003066 | From REF , the coefficients MATH are non-zero only when MATH and in this case, MATH . Hence MATH . Thus interpreting MATH as an operator on MATH we get MATH . In matrix form, MATH . |
math/0003066 | Define a shift operator MATH by MATH and let MATH act as usual on operators by conjugation. Then, if MATH, MATH . Choose MATH. Then MATH. This shows that MATH is similar to MATH and hence satisfies the MQYBE when MATH. |
math/0003066 | We prove that MATH. In matrix form this is equivalent to MATH . Using the fact that MATH when MATH as required. |
math/0003067 | We first note that we have inclusions MATH which give us an inclusion monomorphism MATH. Hence by NAME duality we have a dual homomorphism MATH, where the identification between MATH and MATH is given by MATH, and we easily check that for MATH, we have MATH as defined in the statement of the proposition. Now the action... |
math/0003067 | We have constructed an equivalence between the wavelet representation MATH of REF and the representation MATH of REF, so it only remains to show that MATH is a direct integral of irreducible monomial representations. Now in general the theory of direct integrals of measurable fields of NAME spaces and direct integrals ... |
math/0003067 | The content of REF shows that under the hypotheses of the Corollary, the proof of REF still works. Keeping the notation of REF , we see that the set MATH will be an orthonormal basis for MATH if and only if the set MATH is an orthonormal basis for MATH. Now MATH where MATH is viewed as a subset of MATH, and since MATH ... |
math/0003067 | Let MATH be a transformation wavelet set for MATH. The proof of REF shows that the wavelet representation MATH is unitarily equivalent to a representation MATH which can be expressed as a direct integral MATH where for each MATH, MATH is an irreducible representation which is unitarily equivalent to MATH defined on the... |
math/0003068 | Any self-dual REF-form MATH on any oriented REF-manifold satisfies the NAME formula CITE MATH where MATH is the self-dual NAME tensor. It follows that MATH so that MATH and hence MATH . On the other hand, the particular self-dual REF-form MATH satisfies MATH . Setting MATH, we thus have MATH . But REF tells us that MAT... |
math/0003068 | Let MATH be a fixed smooth back-ground metric, and notice that, for MATH, the function MATH is given by MATH by virtue of the weighted conformal invariance of MATH. Thus MATH behaves under conformal rescaling just like the scalar curvature, despite the fact that it might well only be a NAME function. In order to find a... |
math/0003068 | Let MATH be the conformal class of some MATH-adapted metric MATH. Since the NAME star operator is conformally invariant in the middle dimension, every metric in MATH is also MATH-adapted. Since MATH is a monopole class for MATH, it therefore follows that MATH does not contain any metrics of positive scalar curvature. R... |
math/0003068 | If equality holds in REF , the last inequality in the proof of REF forces MATH to be the harmonic representative of MATH. Moreover, the NAME inequalities used in that proof shows that equality can only hold if MATH and MATH are constant, and moreover equal. Thus the harmonic form MATH has constant norm, so that MATH is... |
math/0003068 | Let us begin by rewriting REF as MATH where MATH denotes the MATH norm with respect to MATH. By the triangle inequality, we therefore have MATH . We now elect to interpret the left-hand side as the dot product MATH in MATH. Applying NAME, we thus have MATH . Thus MATH so that MATH as claimed. If equality holds, moreove... |
math/0003068 | The proof is a direct extension of the computations in CITE, although for MATH we now use the holonomy-constrained NAME invariant of CITE; compare CITE. Now notice that we may assume that MATH, since otherwise there is nothing to prove. This has the pleasant consequence that any NAME invariant of MATH is independent of... |
math/0003068 | We may assume that MATH, since otherwise the result follows from the NAME inequality. Now MATH for any metric on MATH on MATH. If MATH is an NAME metric, the trace-free part MATH of the NAME curvature vanishes, and we then have MATH by REF , with equality only if MATH and MATH both vanish. If MATH carries an NAME metri... |
math/0003068 | For MATH any even integer bigger than MATH million, CITE has constructed a simply connected minimal complex surface MATH of general type (in fact, a hyperelliptic fibration) with MATH and MATH. If we now blow up such a surface at MATH points, where MATH, the resulting simply connected MATH-manifold MATH does not admit ... |
math/0003068 | If MATH, REF becomes MATH and the desired inequality is therefore an immediate consequence. Moreover, if equality holds, REF applies, and MATH is an almost-Kähler manifold with the relevant special properties. |
math/0003068 | As in the proof of REF , there is a monopole class MATH of MATH with MATH and such that MATH . If MATH carried an anti-self-dual NAME metric, REF would then tell us that MATH so that MATH . Moreover, equality could only hold if MATH and MATH both vanished for precisely the same reasons delineated in the proof of REF . ... |
math/0003068 | We again begin with REF , MATH but this time interpret the left-hand side as the dot product MATH in MATH. Applying the NAME inequality, we then obtain MATH . Thus MATH and MATH as claimed. If equality holds, our use of NAME forces MATH which is to say that MATH . In this case, we then have MATH and it then follows CIT... |
math/0003068 | Since REF tells us that MATH it follows that MATH for any metric on MATH on MATH. The result now follows immediately from REF . |
math/0003068 | Since MATH REF tells us that it suffices to prove that MATH . Now observe that the proof of REF tells us that that, for every metric MATH on MATH, there is a spin-MATH structure MATH with non-zero NAME invariant such that MATH . Hence MATH for every metric MATH on MATH. It follows that MATH . To finish the proof, it su... |
math/0003068 | Complex surfaces of NAME dimension MATH, are, by definition, precisely those of general type; for these, the result is just the MATH case of REF . On the other hand, the analogous assertion for NAME dimensions MATH and MATH immediately follows from the fact CITE that any elliptic surface admits sequences of metrics for... |
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