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math/0102087 | This holds for MATH by assumption, since acyclic fibrations are weak equivalences. Extend the conclusion from MATH to presheaf topoi MATH. Let MATH be any object of the small category MATH, and consider the adjunction MATH where MATH is the ``evaluation at MATH" functor and MATH its left adjoint (the left NAME extension). Let MATH. MATH is a MATH-diagram in MATH that is, at every object of MATH, a copower of MATH. Recall that MATH is the ``same" as MATH. Since any copower of MATH is a cofibration in MATH, and since coherent axioms are evaluated ``objectwise" in functor categories (compare REF ), MATH will belong to the class MATH. Suppose MATH. A fortiori MATH. By adjunction, MATH, so MATH. But since the class MATH is defined again by coherent axioms, MATH, for every object MATH of MATH, implies MATH. Consider now an arbitrary topos MATH. Choose a site MATH of definition for MATH, and consider the inclusion MATH, where MATH is sheafification. It induces an adjunction MATH. (Out of laziness, retain the same letters to denote these adjoints.) Take any MATH. Since sheafification (being an inverse image part of a topos morphism) preserves the coherently defined class of cofibrations, by adjointness one has MATH whence MATH. Since sheafification must preserve weak equivalences too, MATH. |
math/0102087 | (The reader looking for relevance to abstract homotopy theory may wish to skip to the examples following the proof.) REF is trivial, given that points of a topos preserve and any conservative collection of points reflect the truth of coherent axioms. REF is the mere fact that coherent deductions remain valid in any topos. REF is a ``degenerate case" of the Conceptual Completeness theorem of NAME - NAME - CITE. MATH and MATH, by assumption, have enough models and have the same models in MATH. Hence they have the same class of coherent consequences. This means that their associated syntactic sites MATH, MATH (see NAME - CITE or NAME - CITE) are equivalent. (In fact, MATH and MATH can be constructed to be the same - not only isomorphic, but identical - since MATH and MATH were assumed to share the same language MATH.) But this implies they have the same models in any topos. |
math/0102087 | MATH is trivial: if MATH is locally presentable, then it is (by definition) cocomplete and has a set of dense generators, a fortiori a set of generators. Their coproduct will do as a single generator (in the sense of the proposition), since MATH has a zero object. MATH : by the NAME theorem there exists a ring MATH and an adjoint pair MATH where MATH is full and faithful, and the left adjoint MATH is exact. That is, MATH is equivalent to a localization of MATH. (Here a localization of a category is a full, reflective, isomorphism-closed subcategory with the reflector preserving finite limits.) Such localizations biject with NAME topologies on MATH, that is, collections MATH of right ideals of MATH with certain closure properties, via associating with MATH the full subcategory of MATH of MATH-closed modules. (See for example, CITE.) A module MATH is MATH-closed iff it is orthogonal to the maps MATH, MATH, that is, MATH is an isomorphism for all MATH. MATH is locally presentable; by the theorem of orthogonal reflection in locally presentable categories (see for example, NAME - REF ), so is its full subcategory of MATH-closed modules. |
math/0102087 | Apply REF . If MATH is NAME abelian, so is MATH. The homology functor MATH (here MATH stands for just the discrete set) commutes with filtered colimits and MATH is locally presentable. The class of isomorphisms is accessible in any locally presentable category. REF gives cREF, and REF grants the generating cofibrations. |
math/0102088 | Firstly, for all MATH, MATH . Therefore, MATH and MATH have the same critical points with the same the extremal properties. As a result, MATH and MATH decreases for MATH. Secondly, from the evolution equation, MATH . This shows that MATH satisfies a parabolic equation, MATH. By NAME oscillation theorem, CITE, the number of zeros of MATH is non-increasing in MATH. Thus, for all MATH, MATH has at most REF critical points in MATH. On the other hand, by symmetry of MATH along MATH and MATH, we must have MATH at the symmetry. Hence for each MATH, MATH has exactly one maximum and one minimum for MATH. The desired results for MATH follow easily. The proof for MATH is similar by simply observing that MATH and the equation for MATH is MATH. |
math/0102088 | By definition of MATH and its expansion in MATH, we have MATH . Note that MATH, so MATH . Applying the Mean Value Theorem, there is a MATH such that MATH . According to the preceding proposition, MATH. As a consequence, MATH . The inequalities can be obtained by putting the above into the expansion of MATH. The proof for the inequalities of MATH is similar by observing that MATH and applying the Mean Value Theorem to MATH. |
math/0102088 | We consider the entropy MATH for the normalized flow REF . Then MATH . Suppose there is a MATH such that MATH. Then MATH blows up in finite time and so does MATH. This contradicts the boundedness of MATH. Thus, MATH and MATH is decreasing in MATH as in the embedded case, CITE. Using this property, we conclude that there is a sequence, after passing to a subsequence, MATH such that MATH. Thus, using the uniform bounds on MATH, we see that MATH as MATH which in turns implies MATH and hence MATH converge. |
math/0102088 | We consider the functional MATH and already have MATH . Here, MATH . By NAME Inequality, we have MATH . Combining the above, together with MATH, we have MATH for sufficiently small MATH. |
math/0102088 | Modifying an argument in CITE, we consider the following quantity, MATH. MATH . Therefore, MATH is the width (measured perpendicular to the longest axis) of a leaf of the curve defined by MATH. By NAME 's Inequality, MATH for some constant MATH, because MATH is decreasing (proved in REF ). Thus, MATH is uniformly bounded below. Since MATH it follows that MATH has a uniform upper bound. |
math/0102088 | Suppose on the contrary that MATH for all MATH. Locally express the tip of a leaf as a concave graph. Specifically, we write MATH such that MATH. By assumption and the preceding lemma, there is MATH such that MATH for all MATH. Note that the equation on MATH corresponding to REF is MATH . Under this local parametrization, we have MATH . Therefore, we have the parabolic equation MATH . Fixed a small MATH, we consider the sub-interval MATH. By concavity of the graph, we have MATH . Hence, REF is uniformly parabolic. By parabolic regularity theory, we know that all higher derivatives of MATH are uniformly bounded in MATH. In particular, the curvature at the tip, when MATH, is uniformly bounded. By REF , we know that MATH converges to the MATH-circle, contradicting REF and the maximum principle. |
math/0102088 | Suppose otherwise, then there is MATH such that actual area of a leaf MATH and MATH. We represent the curve at time MATH, MATH as a concave graph as before. There is MATH such that for MATH and MATH, we have the evolution equation MATH which is equivalent to REF ; see, for example, CITE. Again, by similar argument as above, MATH is uniformly bounded for MATH and MATH. Thus, we obtain a uniform bound for MATH. By continuity, as MATH, MATH a finite value, which is a contradiction. |
math/0102088 | Since MATH for all MATH, we have MATH. By continuity, there is MATH such that MATH for MATH. Therefore, the area of a leaf of the curve at time MATH is still positve. According to the above proposition, solution to REF still exists for MATH. On the other hand, it is easy to see that the rate of change of the area of a leaf is less than MATH, CITE. Thus, there is a time MATH that all the MATH leaves disappear. |
math/0102090 | Write MATH. Then MATH and MATH. If MATH, then there are only two possibilities: MATH and MATH. Otherwise MATH and easy computations give us MATH, so MATH. |
math/0102090 | Put MATH. Assume that MATH. By CITE there is an exceptional divisor MATH such that MATH and MATH. Put MATH. Regard MATH as a germ near MATH. Let MATH be an inductive blowup of MATH. Write MATH where MATH is the exceptional divisor, MATH and MATH are proper transforms of MATH and MATH, respectively. By REF , MATH. On the other hand, MATH. By REF , MATH. Clearly, MATH. Applying REF to our situation we obtain MATH for some MATH. This yields MATH, a contradiction. |
math/0102090 | By REF , MATH is lc at MATH. In this situation there is an analytic isomorphism (compare REF ) MATH where MATH and MATH. Take MATH so that MATH and consider the weighted blow up with weights MATH. We get the exceptional divisor MATH with discrepancy MATH . Thus MATH . If MATH, this gives as MATH and equalities MATH. In the case MATH, MATH we have MATH, that is, MATH. |
math/0102090 | We prove our lemma by induction on MATH. Put MATH. Clearly, we may assume that MATH. Consider minimal dlt MATH-factorial modification MATH (see CITE). By definition, this is a birational morphism MATH such that MATH is MATH-factorial and MATH is dlt, where MATH is the proper transform of MATH and the MATH are prime exceptional divisors (if MATH is dlt, one can take MATH). Since MATH and because MATH, MATH cannot contain MATH. Therefore the proper transform of MATH coincides with its pull-back MATH. Replace MATH with MATH. From now on we may assume that MATH is dlt and MATH is MATH-factorial. There is an exceptional divisor MATH such that MATH and MATH. Regard MATH as a germ near a point MATH. Assume that MATH. Let MATH be a component of MATH (passing through MATH). Then MATH is lc CITE and MATH (see REF ). Then it is easy to see that MATH. Taking into account MATH (see CITE), we get our assertion. Now consider the case MATH. Then MATH is klt. Since MATH is a germ near MATH, MATH for some MATH. Take MATH to be minimal with this property. Then the isomorphism MATH defines a MATH-algebra structure on MATH this gives us a cyclic MATH-cover MATH . The ramification divisor of MATH is MATH. Hence MATH and MATH has only log terminal singularities CITE. Put MATH. Then MATH (see CITE). Replacing MATH with its MATH-factorialization we get the desired log pair. |
math/0102090 | Otherwise MATH is a curve and the pair MATH is lc but not klt along MATH. Taking a general hyperplane section we derive a contradiction with REF . |
math/0102090 | Assume the converse. Then there is an exceptional divisor MATH such that MATH. Since MATH is plt, MATH. If MATH is a curve, then MATH is lc but not klt along MATH. As in the proof of REF we derive a contradiction. Thus we may assume that MATH is plt in codimension two. By Adjunction CITE this implies that MATH. Hence MATH is a point. Again by Adjunction MATH is lc but not klt near MATH. As above, we have a contradiction with REF . |
math/0102092 | The equivalence relation in question does not alter the class MATH or MATH in the situations stated. |
math/0102092 | The shadow colored diagram represents MATH. It is known CITE that the cocycle MATH which is defined above represents a generator of MATH. All possibilities of MATH are evaluated by MATH to give the generator MATH, and the result follows. The mirror image is checked similarly. |
math/0102092 | We represent MATH as a closed MATH-braid as depicted in REF . If the top two segments receive the same color, the cocycle invariant is trivial. There are MATH such shadow colors. If the two top segments receive distinct colors, then perform the quandle homology moves as indicated in the figure, and produce a copy of a generator in REF for each set of three crossings. By REF , for any choice of MATH, the figure represents the generator of MATH. Hence for any non-trivial coloring MATH, MATH. The result follows. |
math/0102092 | There are three cases to color the top strings. All three colors are distinct, two colors are the same, and all colors are the same. If all colors are the same, such colorings contribute MATH to the invariant, and there are MATH such shadow colorings. If all colors are distinct, then such a coloring is depicted in REF left. Such a color exists if and only if MATH is even. Note that the color of the middle string, MATH, stays in the middle strings. Hence we need to consider two types of shadow colors, REF MATH is not a color of the unbounded region, REF it is. REF are shown in REF respectively. For each case, the figure shows that such a colored diagram is homologous to a copy of a generator for each set of MATH crossings. Hence this contributes MATH to the invariant. By symmetry of the diagram, if two colors are the same, it can be assumed that the top strings receive, say, MATH in this order from left to right, as depicted in the left hand side of REF . In this case MATH must be a multiple of MATH for such a color to exist, as can be seen from the figure. Hence the first case (when MATH is not a multiple of MATH) follows from the above argument, and the cases for MATH remain. A block of one contribution of MATH (that is, MATH) is depicted in the figure. There are MATH such colorings, and there are MATH such shadow colorings. There are three cases for shadow colors, as depicted in REF left, respectively. For each of these cases, the figures show that they are homologous to three copies of a generator, contributing MATH to the invariant (as each contribution is counted modulo MATH). The second case follows. |
math/0102092 | REF shows the first half of the statement. Note that there are MATH trivial colorings on strings and MATH corresponding shadow colorings, and MATH shadow colorings corresponding to non-trivial colorings on strings. In REF , it is shown that for a particular shadow coloring, a set of three crossings in the diagrams of MATH contributes a single copy of a generator of MATH, and changes the crossings from positive to negative. These three negative crossings cancel with the next set of three positives. Hence this particular shadow coloring contributes MATH from MATH crossings. If MATH, then the rest is a single crossing, and MATH is a right-handed trefoil knot, giving another generator. For other choices of colorings on the regions, simply change the starting point of replacement and use the same argument. Apply type II NAME moves if necessary, to reduce the given colored diagram to MATH case. This gives the case MATH, MATH. If MATH, then apply the operation in REF MATH times, to obtain MATH, which is a left-handed trefoil, representing the negative of the generator. Hence the contribution is MATH, giving MATH. |
math/0102092 | The method depicted in REF applies in the same manner as in the proof of the above proposition. |
math/0102092 | The idempotency is obvious. For any MATH, let MATH be a unique element MATH such that MATH. Then let MATH. Then it follows that MATH, and the uniqueness of MATH with this property is obvious. The self-distributivity follows from the MATH-cocycle condition by computation, as follows. MATH and MATH . We remark here that the computation above can be seen in knot diagrams in REF . Go along the string that goes from top left to bottom right in NAME type III move and read off the cocycles assigned at crossings. Then it picks up the cocycles in the above two computations, for before and after the NAME move, respectively. |
math/0102092 | There is a MATH-cochain MATH such that MATH. We show that MATH defined by MATH gives rise to an equivalence. First we compute MATH which are equal since MATH. Hence MATH defines a quandle homomorphism. The map MATH defined by MATH defines the inverse of MATH, hence MATH is an isomorphism. The map MATH satisfies MATH by definition. |
math/0102092 | Let MATH be a quandle isomorphism with MATH. Since MATH, there is an element MATH such that MATH, for any MATH. This defines a function MATH, MATH. The condition that MATH is a quandle homomorphism implies that MATH by the same computation as the preceding lemma. Hence the result follows. |
math/0102092 | First, if MATH, or MATH, then the above defining relation for MATH implies that MATH. For the MATH-cocycle condition, one computes MATH and on the other hand, MATH so that we obtain the result. The underlines in the equalities indicates where the relation REF is going to be applied in the next step of the calculation. |
math/0102092 | Since MATH for any MATH, there is a function MATH such that MATH for any MATH. From Equality REF we obtain MATH . Hence we have MATH. |
math/0102092 | By assumption there is MATH such that MATH. Define a binary operation on MATH by MATH . Then by Equality REF , this defines a desired quandle operation. |
math/0102095 | First consider the restriction of MATH to MATH. Note that the MATH spaces are isomorphic to the MATH spaces, via multiplication by a function that is constant away from the punctures and given by MATH near the punctures. Conjugating our restricted operator with these isomorphisms, we obtain an operator MATH from MATH to MATH. Moreover, near a puncture MATH or MATH, the operator MATH looks like MATH . In particular, if the operator MATH has asymptotics MATH, the new operator MATH has asymptotics MATH, where MATH . As a consequence of this small perturbation, the asymptotics are not degenerate anymore, and we obtain a NAME operator (see CITE). Its index is given by MATH . It is easy to see that MATH and MATH. Let us now consider the summand MATH. The images under MATH of these sections have their support near a puncture and decay exponentially near that puncture, hence MATH maps the new elements into MATH. The space MATH is finite dimensional. Thus the operator MATH is NAME. Its NAME index is the sum of the index of the operator restricted to MATH and the dimension of the vector space MATH, which is MATH. Hence we obtain the desired formula. |
math/0102095 | By construction, MATH away from the neighborhoods of the gluing. On the other hand, for MATH, we have MATH . But MATH because MATH (respectively, MATH) when MATH (respectively, MATH) is not zero. Therefore, MATH . Hence, it follows from our constructions that MATH where MATH when MATH. Therefore, if MATH is sufficiently large, MATH . Hence, the operator MATH is invertible. Let MATH. By construction, MATH is a right inverse for MATH, and it is uniformly bounded in MATH. |
math/0102095 | Let MATH be the uniformly bounded right inverse for the glued operator MATH, as in REF . We define MATH to be the composition of the map MATH and the projection map MATH from the domain of MATH to its kernel, along the image of MATH. In other words, MATH. We claim that the restriction of MATH to MATH is an isomorphism for MATH sufficiently large. By REF , the dimensions of both spaces agree, so it is enough to show that MATH is injective. By contradiction, assume that for any large MATH, we can find MATH and MATH such that MATH and MATH, for some MATH. Since MATH, we see that MATH by the same kind of computation as for MATH at the beginning of the proof of REF But since MATH, it follows that MATH when MATH. Using this in the original equation gives MATH. But this contradicts MATH. |
math/0102095 | Note that the statement is true by definition if MATH, when MATH and MATH or when MATH and MATH, since we simply cap off a puncture. Next, note that the statement also holds if MATH or MATH instead of zero, that is, in the case of gluing a cylinder. Indeed, we can reduce this case to the previous one after capping off the other end of the cylinder, by associativity of the gluing operation MATH for orientations. Consider now the case of NAME surfaces with an arbitrary topology and arbitrary number of punctures, equipped with operators MATH and MATH such that their asymptotic expression near the punctures are complex linear. The operators MATH and MATH are then homotopic to a complex linear operator in the class of operators with fixed asymptotics at the punctures. Hence, their determinants inherit a natural orientation from their complex structure. The same is of course true for the glued operator and it is clear that gluing complex orientations yields the complex orientation since the gluing map is homotopic to a complex linear isomorphism in this case. Finally, we reduce the general case to the above situation by gluing cylinders MATH and MATH to the punctures at which we want to glue MATH and MATH. We choose MATH and MATH so that the glued operators MATH and MATH have complex linear asymptotic expressions at the negative and positive punctures, respectively. Since we can cap off all positive or negative ends of MATH and MATH, we may assume that there are no such punctures. We have a homotopy MATH and hence MATH due to consistency of gluing in the complex linear case. The latter is equal to MATH . |
math/0102095 | This property follows from the consistency of the orientations under disjoint union applied to the MATH - punctured spheres used to cap off operators over arbitrary surfaces with punctures. Comparing the orientations via MATH is the same as comparing the orientations on the disjoint unions MATH and MATH of MATH and MATH. This gives an orientation of the (virtual) vector space MATH which is given by orientations of the two vector spaces of the difference. Thus we can compare the orientations induced by either of the two possible orders of the components with that given one. It is not hard to check that this differs exactly if both indices MATH and MATH are odd. But MATH . The proof for the case of changing the ordering at the positive ends is almost the same; in that case, we have to use spheres with one negative puncture and the index is MATH . |
math/0102095 | By definition, MATH coincides with the complex orientation. On the other hand, MATH . But the latter is a complex orientation multiplied by MATH, by definition of MATH and MATH. Comparison with the definition of MATH gives the desired result. |
math/0102095 | By definition, MATH coincides with the complex orientation. On the other hand, MATH coincides with MATH because of associativity. In order to apply REF to the gluing MATH, we have to permute several punctures in the right term : MATH must be exchanged with MATH. This gives precisely the sign MATH. After the gluing MATH, we obtain a complex orientation, and comparison with the definition of MATH gives the desired result. |
math/0102095 | First, if MATH is a closed NAME surface, we can deform MATH into a MATH-linear operator MATH by a path of operators. The isomorphism induced by MATH from the kernel and cokernel of MATH to the kernel and cokernel of MATH is MATH-linear, so it sends MATH to MATH and hence MATH to MATH. In the general case, consider the action of MATH on MATH. On one hand, MATH preserves the complex orientation as above. On the other hand, the action of MATH clearly commutes with the gluing maps, and by REF extends as the identity on each capping REF-punctured sphere. Therefore the coherent orientations must be preserved as well. |
math/0102095 | First, we have to explain how to orient the determinant bundles of the moduli spaces MATH using the construction of the previous section. Let MATH. Consider the linearized operator MATH splitting as MATH. The family of operators MATH for MATH defines a continuous map MATH . The operator MATH is homotopic to MATH, a stabilization of the operator MATH with the complex vector space MATH. Hence, their determinant spaces are canonically isomorphic and the determinant bundle over MATH corresponding to the NAME operator MATH is isomorphic to MATH. In particular, the orientations from REF induce orientations on the determinant bundle of MATH. The mapping class group of MATH acts naturally on MATH. Let MATH be the moduli space of this action. By REF , the coherent orientations are preserved by this action and descend to the moduli spaces MATH. Next, we have to show that, given compact subsets MATH and MATH the gluing maps MATH are orientation preserving up to a sign MATH where MATH is determined by REF . In order to prove this, we have to relate the differential of MATH to the linear gluing map of REF . Note that the construction of the gluing map requires the NAME section MATH to be transverse to the zero section. In other words, the linearization MATH of this section at MATH must be a surjective operator. This transversality can be achieved in different ways, such as perturbing the almost complex structure MATH or perturbing the right hand side of the NAME equation with an element of MATH having compact support in the complementary of the punctures. In all cases, the linearization of the perturbed section MATH is still an element of MATH modulo some zero order terms not affecting our constructions. Therefore, the definition of coherent orientations is independent of the precise way we achieve transversality for the moduli spaces. Let MATH for MATH. Using cutoff functions near the closed orbits MATH, we construct a pre-glued map MATH into the glued cobordism MATH. Gluing the two conformal structures we also construct conformal data MATH on the glued surface MATH. Then the glued data MATH satisfy MATH where MATH as MATH. Under the above transversality assumption, the linearized operator MATH is surjective and has a uniformly bounded right inverse, as in REF . An actual holomorphic curve MATH is obtained using NAME iterations (see CITE), in a neighborhood of MATH of size controlled by MATH. In particular, the difference in norm of the linearizations MATH and MATH, and the glued operator MATH obtained from MATH, for MATH as in REF, are arbitrarily small for MATH sufficiently large. The differential of the gluing map MATH is the composition of the linearizations of the pre-gluing map and of the NAME iteration map. The linearization of the pre-gluing map involves gluing sections of MATH for MATH using some cut-off functions. It is therefore of the same form as the linear gluing map MATH from REF. On the other hand, the differential of the NAME iteration map approaches the projection MATH to the kernel of MATH along the image of its right inverse MATH as MATH becomes large. The above discussion shows that this projection is very close to the projection MATH of REF for MATH large. This shows that the differential of MATH and the linear gluing map are very close for MATH large, so that they induce the same map on orientations. |
math/0102095 | First note that, in view of REF for the construction of coherent orientations, it is enough to prove the theorem for the moduli space MATH, where MATH is the MATH-fold covering of the simple orbit MATH. The index of the operator MATH is given by MATH. Then MATH is the pullback of MATH under the branched covering MATH. In particular, this operator has a MATH - symmetry of rotations. Hence, MATH acts on the kernel and cokernel of MATH, and these vector spaces split into irreducible representations over MATH. If MATH is odd, there are no elements of even order, so the action of MATH preserves orientations. Hence, coherent orientations are invariant under a change of asymptotic direction. If MATH is even, the possible irreducible representations include REF representations of dimension REF (trivial and sign change) and MATH representations of dimension REF, generated by rotations of angle MATH, MATH, MATH. The trivial representations correspond to the kernel and cokernel of MATH. Therefore, the index difference MATH has the same parity as the multiplicity of the orientation reversing representation over MATH in the kernel and cokernel of MATH. Hence, rotating the asymptotic direction of MATH reverses the coherent orientation if and only if MATH is odd. This is exactly the case described in the theorem (see for example, REF). |
math/0102095 | MATH . Let MATH be the odd multiplicity of MATH. Recall that MATH acts by rotation on the kernel and cokernel of MATH and that these vector spaces split into the invariant part and some irreducible representations over MATH generated by a rotation of angle MATH, MATH. The invariant part consists of the pullback of the kernel and cokernel of the operator MATH and hence it is oriented. We choose the orientation on each irreducible representation MATH in such a way that MATH. MATH . If the multiplicity MATH of MATH is even, the kernel and cokernel of MATH split into the invariant part, some irreducible representations over MATH generated by a rotation of angle MATH, MATH, MATH, and the part MATH on which the generator of MATH acts by MATH. The invariant part and the MATH summands are oriented as above. On the other hand, the kernel and cokernel of the operator corresponding to the double orbit split precisely into the invariant part and MATH. Since the choice of an orientation for the simple orbit is equivalent to an orientation of the invariant part, the choice of an orientation for the double orbit is just what we need to orient all the other multiples. MATH . In the case MATH we have to find a way to orient the part MATH on which the rotation acts by MATH. Choose the operator MATH so that it splits into the standard NAME operator MATH and an operator MATH of the form MATH near the positive puncture. Since MATH has nondegenerate asymptotics, its NAME index is given by MATH. This kind of operators were well studied in CITE. We now summarize some facts about MATH from that paper, assuming MATH. Otherwise, we consider its formal adjoint, exchanging kernel and cokernel. CASE: Let MATH be an eigenvector with eigenvalue MATH of the operator MATH. It has a well-defined winding number MATH in the fixed trivialization. Each winding number appears exactly twice and increases with the eigenvalue. Two eigenvectors with the same winding number are pointwise linearly independent in MATH or a multiple of each other. CASE: The operator MATH is surjective. Elements in MATH have the form MATH, where MATH, MATH and MATH, where MATH . CASE: NAME, we may choose a basis of MATH such that the leading terms are in MATH correspondence to the set of eigenvectors MATH satisfying the above conditions. Therefore, the NAME index of MATH can be computed in terms of these data. We have MATH where MATH if the winding of the smallest positive eigenvalue is the same as that of the biggest negative, and MATH otherwise. Let MATH be the double cover of the orbit MATH. Let MATH be the pullback of the operator MATH under the branched covering MATH. The kernel of MATH splits into the pullback of MATH and the part MATH that is not invariant under the MATH action. By assumption, MATH is good, so that MATH. Therefore, MATH splits into summands MATH generated by REF eigenvectors of equal winding number. Since these REF eigenvectors are pointwise linearly independent, the corresponding summand MATH inherits a natural orientation from the complex orientation of MATH. Hence, we obtain a natural orientation on MATH. |
math/0102096 | We have to show that MATH contracts a finite number of curves. The verification is tedious, but similar to CITE, pp. REF. |
math/0102096 | For if MATH passed through a terminal quotient point MATH, REF would imply that MATH, a contradiction. |
math/0102096 | If MATH is not terminal, by REF, there is a divisorial contraction MATH, extracting a divisor MATH for which MATH. By REF, MATH is the weighted blow up we are speaking of, so we get that MATH. NAME if MATH is terminal, it is part of the definition that MATH. |
math/0102096 | It is awkward to try to prove this by the same method as REF , because of possible curve maximal centers. Fortunately, you can check that the proof of REF pg. REF, proves the statement. |
math/0102096 | We begin by describing the general setup for the proof. CASE: General setup Let MATH be a sufficiently large and divisible positive integer; fix a finite dimensional very ample linear system MATH . Note that we write MATH to signify the canonical class of MATH, as opposed to the usual notation MATH. We use the notation MATH throughout this proof. Denote MATH the image of MATH in MATH, so that MATH . By construction MATH while MATH for all valuations MATH. CASE: We summarize in the following diagram, and explain below, our notation for the various spaces and morphisms which we use in the course of the proof MATH . We now introduce in detail the various spaces and morphisms. Please draw your own picture, and do your own calculations to justify the description we give; otherwise you will not follow the proof. CASE: Denote MATH the weighted blow up with weights MATH and exceptional divisor MATH. Similarly denote MATH the weighted blow up with weights MATH. It is easy to check, for instance by performing the blowing up explicitly, that MATH has a singular point MATH of type MATH and is elsewhere nonsingular. CASE: Denote MATH the blow up of the maximal ideal at MATH with exceptional divisors MATH, MATH. The abuse of notation means to suggest, for instance, that the rational map MATH (not a morphism!) is an isomorphism at the generic point of MATH, and contracts MATH. Thus the notation MATH denotes the ``same" divisor in two different varieties MATH and MATH. We let the context decide which is meant; however, when we wish to be precise about the ambient variety, we write for instance MATH meaning the divisor MATH on the variety MATH. We do this for other varieties and divisors, as well. This is justified since many of the quantities we are interested in, such as discrepancies and multiplicities, depend only on the divisor, not on the ambient variety. CASE: It is easy to check that MATH, and MATH, MATH intersect in a line MATH. Also, MATH itself is nonsingular apart from three distinct ordinary nodes MATH, each looking like MATH CASE: Denote MATH the blow up of the three MATH-s; it has three exceptional divisors MATH, all isomorphic to MATH with normal bundles MATH. It is easy to check that MATH is nonsingular. We also denote MATH the obvious morphisms and MATH CASE: It is important to understand that the maps MATH are not morphisms. Indeed for instance we can resolve the map MATH by a diagram MATH where MATH is a small resolution of all the MATH; MATH blows up all the three points MATH, introducing exceptional curves MATH, while MATH is an isomorphism. The morphism MATH contracts MATH, with normal bundle MATH, to a singular point MATH. The images of the MATH are three lines MATH passing through MATH. We use these lines later in the proof. Similar remarks and notation apply to the map MATH. CASE: Now we start with our given divisorial contraction MATH and we want to show that MATH or MATH; assuming the contrary we will derive a contradiction. The proof divides out in cases, depending on the position of the center MATH of MATH on MATH: MATH . In all cases, we define MATH, MATH by the formula MATH . Another way to say this is that MATH and MATH are the multiplicities of MATH along the MATH-s and MATH-s. The assumption that MATH means that MATH. We treat each case separately. CASE: Let MATH be a generic surface through MATH. It is easy to compute MATH hence MATH which is another way to say, for instance, that MATH, and similarly for the discrepancies of the MATH. Suppose now that MATH, say. We apply NAME connectedness theorem, REF and especially REF , to the morphism MATH and the divisor MATH (where the formula defines MATH) and we conclude that there is a ``line" MATH (by this we mean that MATH maps to a line under the morphism MATH) such that MATH . One of the three MATH-s, say MATH, does not lie on the line MATH. It is easy to construct a contraction MATH having MATH as its unique exceptional divisor (MATH has canonical but not terminal singularities so this is not a divisorial contraction in the NAME category). In fact there is a morphism MATH, resulting from a free linear system MATH on MATH, which in turn is the proper transform of a linear system MATH on MATH such that MATH . Define now MATH . Then MATH where MATH. We now apply NAME connectedness to the morphism MATH and the divisor MATH (where the formula defines MATH). It is important here to understand that the divisor MATH is not contracted by MATH; the reason we can apply NAME connectedness is that MATH. This is why we introduced MATH in the first place; it would have been tempting to run the argument with MATH. Now MATH is contracted by MATH; the fibers of the rational map MATH are the lines trough MATH. Furthermore MATH . Since MATH is a curve, MATH, and a simple calculation then shows that MATH, therefore MATH . This implies that MATH is not connected in the neighborhood of a general fiber of MATH. The contradiction finishes the proof in REF . CASE: Let us define MATH, MATH by the formulas MATH where ``(other)" means a combination of the MATH-s. Another way to define these numbers would have been to set MATH . It is convenient (and, ultimately, straightforward) for us to calculate MATH and MATH in terms of MATH, MATH, in fact we only need the CASE: Claim MATH . To prove the claim, note MATH therefore MATH from which we conclude MATH. Similarly, MATH and the claim follows. CASE:REF Assume that MATH but is not one of the MATH-s. This implies that MATH is the unique singular point. The divisor MATH is strictly canonical, that is, canonical but not terminal, in a neighborhood of MATH (the valuation MATH corresponding to the exceptional divisor of the divisorial contraction which we have been studying all this time, has discrepancy MATH). Because MATH, the divisor MATH is not canonical in a neighborhood of MATH. By REF , both MATH, MATH are MATH hence MATH and either MATH or MATH, a contradiction which concludes this case. CASE:REF Assume now that MATH is one of the MATH-s, say MATH. The proof just given breaks down because typically in this case the center MATH of MATH on MATH is not the singular point MATH. Therefore we argue directly on MATH. It is important to be aware that the divisors MATH and MATH are not MATH-Cartier at MATH, but the sum MATH is. Consider the divisor MATH on MATH. Note MATH hence MATH is MATH-Cartier but not canonical at MATH, hence assuming as we may that MATH so that MATH, and using REF , we have MATH . We now calculate MATH in two different ways. On one hand MATH while, on the other hand, using MATH: MATH where MATH; hence MATH gives MATH . Similarly MATH gives MATH . Now recall the curves MATH; by REF, and similarly for MATH, while MATH. It follows from REF that MATH, hence combining inequalities we get MATH and from this we conclude MATH, a contradiction. |
math/0102096 | REF . |
math/0102096 | This is standard, and is the same as the proof in CITE. If MATH is NAME and NAME a NAME fibre space, a birational map MATH is defined by a mobile linear system MATH. By the NAME inequalities REF , if MATH is not an isomorphism then MATH has a maximal center MATH or MATH, hence a maximal singularity MATH or MATH by REF . By REF , there is a birational map MATH, where either MATH or MATH, that is a NAME link, and by REF untwists the maximal center MATH or MATH, so that MATH has smaller degree. Thus after a number of steps, either MATH or MATH. |
math/0102096 | In REF . |
math/0102096 | Let MATH be a curve and assume that MATH is a maximal center for MATH. This implies that MATH. In the proof, we reach a contradiction in several steps: CASE: A raw argument shows that MATH. CASE: MATH can not be a space curve. CASE: MATH can not be a plane curve. CASE: Choosing general members MATH, MATH of MATH and intersecting with a general hyperplane section MATH we obtain MATH . This implies that MATH. CASE: REF : space curves If MATH is a space curve, then by REF it must be a rational normal curve of degree REF, contained in a hyperplane MATH. Let MATH be a general quadric vanishing on MATH,MATH the mobile part of MATH; write MATH where MATH is nef. Note that, because MATH is cut out by quadrics, MATH . We reach a contradiction by showing that MATH. For simplicity we treat two separate cases, namely: CASE: MATH, CASE: MATH. CASE:REF Here we assume that MATH. It follows that MATH is nonsingular and MATH (all calculations of intersection numbers are performed on MATH). Indeed it is easy to see that MATH is a nonsingular complete intersection of a quadric and a quartic, therefore MATH. Then: MATH shows that MATH. A simple calculation then gives: MATH . This shows that MATH and finishes the proof in this case. Note that we only need MATH; the additional room in the argument, is what ultimately makes it possible to treat the next REF essentially by the same method. CASE:REF Now we assume that MATH. Write as in REF MATH the resolution of singularities of MATH, constructed in the proof of REF . Using that MATH is generated by quadrics, and that a general quadric MATH is nonsingular, it is easy to see that the proper transform MATH must itself be nonsingular. Denote MATH and write MATH where MATH and MATH are MATH-curves (here, following the notation of the proof of REF , MATH, MATH and MATH, MATH, MATH are the exceptional divisors of MATH). There are now two subcases (up to relabelling the exceptional divisors), depending on how the curve MATH ``sits" in the singularity MATH. The crucial observation is that, because MATH is a nonsingular curve, MATH intersect transversally a unique exceptional divisor. The cases are as follows: CASE: The proper transform MATH intersects MATH. In this case, MATH meets MATH, MATH and is disjoint from all the MATH-s. Here MATH is a MATH-singularity, and MATH. CASE: The proper transform MATH intersects MATH. In this case, MATH meets MATH, MATH, MATH and is disjoint from MATH and MATH. Here MATH is a MATH-singularity, and MATH. We claim that in both subcases MATH . Indeed it is easy to see, as in REF (MATH), that MATH, and by the projection formula MATH in the two NAME REFEF and REFEF, respectively. Finally MATH implies MATH, a contradiction which concludes REF . CASE: REF : plane curves Here we assume that MATH is a plane curve of degree MATH (by REF , MATH), other than a line passing through MATH. Here too, as in REF , it is helpful and convenient to treat two cases, namely: CASE: MATH and MATH. CASE: MATH and MATH. CASE:REF We first deal with the easy case MATH (following well known ``ancient" methods of NAME and NAME). Consider as usual a general element MATH, denote MATH the mobile part of MATH; write MATH where MATH is nef. We aim to show that MATH. It is easy to see that MATH is a nonsingular complete intersection of a quartic with a hypersurface of degree MATH, therefore MATH. If MATH, then MATH and: MATH shows that MATH. A simple calculation gives: MATH which implies that MATH. The proof is similar when MATH: MATH, and then MATH and again MATH. These calculations finish REF MATH. CASE:REF From now on we assume that MATH, MATH not a line. In this case, restriction to a general element of the test linear system MATH does not lead to a contradiction; it is necessary to use a different test system. Denote MATH the plane spanned by MATH, let MATH, MATH be general hyperplane sections of MATH containing MATH. We work with the ``test system" MATH, even though MATH is usually only a component of its base locus MATH. We are assuming that MATH is general, hence in particular it is terminal and MATH-factorial. This implies that MATH can not be contained in MATH, and MATH is a curve. Unfortunately, we have to divide the proof in several cases according to what MATH is. In the end each case is not very different or harder than any of the other cases, but we could not find a unified presentation. The cases are as follows: CASE: MATH, CASE: MATH, CASE: MATH, CASE: MATH, CASE: MATH. We now treat REF - REF ; at the end we will show that REF do not happen (at least assuming, as we do, that MATH is general), in other words, we will show that MATH is always reduced when MATH is general. Before treating each case individually, we make some general comments and fix the notation for the whole argument. Assuming for now that MATH is reduced, we restrict to MATH and write MATH . Our technique consists in selecting a ``most favorable" component of MATH, calculating an intersection number on MATH using that MATH is nef, and finally get that MATH. When MATH is reduced, it is clear that if MATH is a component of MATH, MATH; in particular MATH, and also MATH. Because MATH is a maximal singularity, MATH, hence possibly after relabelling components of MATH, we can assume that: MATH . (ignore the term MATH if no curve MATH is present). Consider now the effective MATH-divisor MATH . In REF is a line and MATH . We now show that MATH (on MATH); together with the last displayed equation this implies that MATH and finishes the proof in REF . Note that this is intuitively almost obvious: for example in REF MATH is the plane union of a cubic and a line, and we expect these to intersect in REF points (when we only need one!). The problem with saying this is, of course, that the set theoretic intersection MATH can be all concentrated on the singular point MATH. We now study this situation more carefully. Note first that MATH is nonsingular outside MATH. This follows easily from the fact that the base locus MATH of MATH is a reduced curve with only planar singularities, and MATH itself is nonsingular outside of MATH (this is all familiar and easy: if MATH is the blow up of MATH along MATH, then MATH has isolated singularities outside MATH). By our generality assumption MATH, and using the notation of the proof of REF , we have that MATH. Also, MATH is nonsingular away from MATH. Therefore either the set theoretic intersection MATH contains a nonsingular point of MATH, or MATH contains a nonsingular point of MATH. In both cases this point contributes with an integer value MATH to the intersection number MATH, hence our claim that MATH. This finishes the proof in REF . In REF are both conics and MATH . It is easy to see that MATH and MATH intersect in at least REF nonsingular points on MATH, and from this conclude that, in this case also, MATH (the details are very similar to REF and left to the reader). It remains to show that REF can not occur, that is, MATH is always reduced when MATH is general. CASE: MATH is general, every plane section of MATH is reduced. This is a fairly easy exercise. In coordinates MATH is given by MATH . Where MATH and MATH are a homogeneous cubic and quartic not involving MATH. The singular point MATH is of course the coordinate point MATH. Consider the projection MATH on MATH with homogeneous coordinates MATH; it is a generically REF-to-REF map, which is to say that the equation of MATH is quadratic in the variable MATH. Now MATH for a unique line MATH and it is almost immediate that the hyperplane section MATH is nonreduced if, and only if, either one of the following happens: CASE: The line MATH is contained in the discriminant surface MATH. It is very easy to see that, for a general choice of MATH, MATH, this surface contains no lines. CASE: The line MATH is contained in the plane MATH and MATH, MATH, when restricted to MATH, both have a double root at MATH. In any event, this is ruled out by REF. |
math/0102096 | We can choose weighted projective coordinates MATH such that the equations of MATH are as follows: MATH . To understand the proof, it helps to know some explicit features of the geometry of MATH. To begin with, MATH is nonsingular apart from two MATH-points MATH, MATH at MATH and MATH. Denote MATH the projection from MATH; it can be useful to know that the image of MATH, for example, is the hypersurface MATH, as can be readily calculated eliminating the variable MATH from the equations of MATH. Also, denote MATH the projection on the coordinates of degree MATH. The curves of MATH, contracted by MATH, are REF curves MATH with MATH given by MATH. Similarly, the curves of MATH, contracted by MATH, are REF curves MATH with MATH given by MATH. Finally, the curves contracted by MATH are REF just mentioned, plus REF curves MATH with MATH given by MATH; under the generality assumption MATH these are irreducible. Assume that the curve MATH is the center of a maximal singularity MATH of a mobile linear system MATH. By REF , MATH is contained in the nonsingular locus of MATH. Denote MATH, MATH. Choosing general members MATH, MATH of MATH and intersecting with a general hyperplane section MATH we obtain MATH . This implies that MATH. We treat the two cases MATH, MATH separately. CASE: Case MATH . Here MATH is either a line or a conic in MATH. In either case MATH is a nonsingular rational curve and MATH is defined scheme theoretically by the quartics (with the natural embedding of MATH in MATH the curve MATH is a normal quartic). Let MATH be a general member; write as usual MATH (with MATH nef.). We easily calculate on MATH that MATH, and MATH . This implies that MATH and finishes this case. CASE: Case MATH . Here MATH is a line, MATH is a nonsingular rational curve. Denote MATH, MATH two general members of the pencil MATH, MATH the base locus. Denoting MATH, we can also say that MATH. In the end we will show that MATH is reduced; for now let us assume it. We restrict to MATH and write MATH . When MATH is reduced, it is clear that if MATH is a component of MATH, MATH; in particular MATH, and also MATH. Because MATH is a maximal singularity, MATH for all MATH, hence possibly after relabelling components of MATH, we can assume that MATH. Consider the effective MATH-divisor MATH . We calculate the intersection product, on MATH, with MATH: MATH . Note that here MATH is a half-integer. It is completely elementary to check that MATH (in doing this, it helps to note that MATH is nonsingular outside MATH, MATH). Together with the last displayed equation this implies that MATH and finishes the proof. It remains to show that MATH is reduced. CASE: MATH is general MATH is reduced. This is a fairly easy exercise; the situation corresponds exactly to the quartic MATH as treated in the proof of REF . In short, it is easy to see that MATH is nonreduced if, and only if, either one of the following happens: CASE: The line MATH is contained in the discriminant surface MATH. It is clear that, for a general choice of MATH, MATH, this surface contains no lines. CASE: The line MATH is contained in the plane MATH and MATH, MATH, when restricted to MATH, both have a double root at MATH. In any even this possibility is certainly ruled out by REF. |
math/0102096 | Let MATH be a nonsingular point, MATH. If MATH consider a general element MATH. Then MATH is mobile and MATH . This is enough, by REF , to conclude that MATH is not a center. Let us now worry about the case MATH; this can only happen when MATH is a curve contracted by MATH, and as we have already noted at the start of the proof of REF , it is a consequence of the generality assumption MATH that MATH is irreducible. If MATH then write MATH where MATH is the mobile part; dividing by MATH be obtain MATH where MATH, and MATH. Note that MATH is a rational curve on MATH passing trough REF simple double points. Therefore MATH and, computing the self intersection of MATH, we get MATH . Again by REF we exclude MATH as a center. If MATH then the above arguments on the surface MATH are not enough to exclude it as a maximal center. This time we need to consider the linear system MATH. It is easy to check that: CASE: MATH is general, then CASE: MATH has a simple double point at MATH, CASE: MATH is nonsingular along MATH, CASE: MATH has a singularity of type MATH at MATH. Let MATH the blow up of MATH with exceptional divisor MATH. Write MATH and MATH. MATH has a double point at MATH thus MATH is a conic. By NAME connectedness there is a line MATH such that MATH . Therefore, for the generic MATH, by inversion of adjunction MATH is not LC at two distinct points MATH and MATH. We want to use this fact to derive a numerical constraint on MATH. To do this let us first compute the intersection matrix on MATH: MATH (The only nontrivial product is MATH, by the adjunction formula with correction coming from the different). Write MATH where MATH is the mobile part. We have MATH . To find a lower bound for MATH recall that MATH is not log canonical at MATH and MATH therefore by REF , we always have MATH . Combining with REF yields MATH and the discriminant of this quadratic equation with respect to MATH is MATH . This inequality shows that MATH cannot be a center of maximal singularities and concludes the proof of the Theorem. |
math/0102097 | The orbit space MATH is naturally isomorphic to the associated bundle MATH, so it is the total space of a natural fiber bundle with fiber MATH. The natural projection MATH which maps any point to its orbit is a principal MATH - bundle. The NAME connection MATH defining the parabolic geometry on MATH is equivariant for the action of the group MATH and reproduces the generators in MATH of fundamental vector fields. Hence it is also a NAME connection on the principal MATH - bundle MATH and we have obtained a natural parabolic geometry of type MATH on MATH. Since the parabolic geometries on MATH and MATH are given by the same NAME connection, MATH and MATH coincide as functions with values in MATH so we clearly get MATH as a function with values in MATH. From this, REF is obvious, while REF follows from REF . |
math/0102097 | Locally, any vector field on MATH can be lifted to a vector field on MATH. Sections of the subbundle MATH correspond exactly to sections of the subbundle MATH. Since the NAME bracket of two lifts is a lift of the NAME bracket of the original fields, we see that MATH is integrable provided that the space of sections of MATH is closed under the NAME bracket. By definition of the exterior derivative, this is equivalent to MATH for all MATH. By definition of the curvature function, MATH . Since MATH is a NAME subalgebra, we see that MATH for all MATH implies integrability of MATH. Conversely, if we find MATH and a point MATH such that MATH, then choose representatives MATH for MATH and MATH. The tangent vectors MATH and MATH have nonzero projections to MATH. Extending them to projectable sections of the subbundle MATH, we obtain two sections whose NAME bracket does not lie in MATH. Their projections on MATH are sections of MATH, whose NAME bracket cannot be contained in MATH. |
math/0102097 | For MATH consider MATH and the NAME bracket MATH. Then we compute MATH so the assumption on MATH implies that MATH. Hence MATH restricts to a NAME algebra homomorphism MATH, that is, an action of the NAME algebra MATH on MATH. By NAME 's second fundamental theorem, see pp. REF, this NAME algebra action integrates to a local group action. In particular, for any MATH there is an open neighborhood MATH of MATH in MATH and a smooth map MATH such that CASE: If MATH, then MATH and MATH for all MATH. CASE: MATH provided that MATH, MATH and MATH all lie in MATH. Now consider a local leaf space MATH which is so small that there is a smooth section MATH of MATH. Possibly shrinking the leaf space further, we find an open neighborhood MATH of MATH in MATH such that for some set MATH as above we have MATH and MATH for all MATH and all MATH. Then we define MATH by MATH. For MATH the tangent map MATH is evidently given by MATH so it is a linear isomorphism. Possibly shrinking MATH and MATH, we may assume that MATH is a diffeomorphism onto an open subset MATH. We may further assume that MATH where MATH is an open neighborhood of zero in MATH such that MATH is a diffeomorphism from MATH onto an open neighborhood MATH of MATH in MATH and MATH is a ball around zero in MATH. For a fixed point MATH, any vector tangent to MATH can be written as MATH for some MATH and MATH. For sufficiently small MATH, we by construction have MATH, and thus MATH maps this tangent vector to MATH. Thus we see that MATH on MATH. Moreover, MATH always lies in MATH, which implies that MATH is contained in one leaf of the foliation corresponding to the integrable distribution MATH. From REF we conclude that the map MATH is constant on MATH, and since MATH we conclude that MATH. For MATH and MATH we have MATH for all MATH. Since MATH is the fundamental vector field on MATH generated by MATH, the infinitesimal condition MATH immediately implies that MATH, where in the right hand side we use the principal right action on MATH. Since MATH acts freely both on MATH and on MATH we can uniquely extend MATH to a MATH - equivariant diffeomorphism from MATH to the MATH - invariant open subset MATH. Since the family of fundamental vector fields on MATH and the family of the vector fields MATH on MATH have the same equivariancy property, this extension still satisfies MATH for all MATH. |
math/0102097 | CASE: By REF there is a open neighborhood MATH of MATH in MATH and a diffeomorphism MATH from MATH onto an open subset of MATH such that MATH and such that MATH for all MATH. Hence we can form MATH, which restricts to a linear isomorphism on each tangent space. Let us denote by MATH the principal right action of MATH on MATH. We can extend the values of this form in MATH equivariantly to a MATH - valued one - form MATH on MATH by putting MATH . By construction, this form restricts to an isomorphism on each tangent space and satisfies MATH for all MATH. Since the vector fields MATH and MATH are MATH - related, their flows are MATH related. Thus MATH, whenever the left hand side is defined. The curvature condition MATH for all MATH and MATH reads as MATH, where in the last step we have used that MATH is constant. This infinitesimal equivariancy condition easily implies local equivariancy, that is, that MATH, whenever the flow is defined. Hence for MATH such that MATH for all MATH we have MATH for all MATH. This shows that MATH coincides with MATH on an open neighborhood of MATH in MATH, so MATH is smooth on this neighborhood and hence by equivariancy on all of MATH. By construction, MATH reproduces the generators of fundamental vector fields. Hence MATH has the same property in points of the form MATH with MATH, and thus by equivariancy in all points of MATH. We have therefore verified that MATH is a NAME connection, and thus defines a parabolic geometry of type MATH on MATH. Concerning normality, we have observed above that MATH coincides with MATH on an open neighborhood of MATH. On this neighborhood, the curvature function MATH of MATH is given by MATH, where MATH is the inclusion from REF. Now the claim on normality follows from REF , equivariancy of MATH, and the fact that MATH is MATH - equivariant. CASE: Take the map MATH from REF . By REF we may assume MATH to be a MATH - invariant subset of MATH and MATH to be a MATH - equivariant diffeomorphism onto a MATH - invariant open subset MATH. From REF we know that MATH on an open subset of MATH, which has to be MATH - invariant since both MATH and MATH are MATH - equivariant. But this exactly means that MATH defines an isomorphism of parabolic geometries of type MATH from an open subset in MATH to the open subset MATH. Finally, we have seen in REF that isomorphism of the correspondence spaces implies isomorphism of the underlying spaces provided that MATH is connected. |
math/0102097 | By REF we get appropriate parabolic geometries on sufficiently small open subsets of MATH. By REF these locally defined structures fit together to define a principal bundle and a NAME connection on MATH. Also the isomorphisms between open subsets of the correspondence spaces and appropriate subsets of MATH piece together smoothly by REF . |
math/0102097 | We first claim that if MATH, then MATH maps sections of MATH to sections of MATH: Let us compute in jet - modules as it is done in CITE. So consider the first jet prolongation MATH, which as a MATH - module is isomorphic to MATH. By REF MATH is induced by the homomorhism MATH, which is given by MATH for MATH and MATH. Thus, MATH corresponds to the homomorphism MATH. Since MATH the first summand coincides with MATH and by REF the second summand gives MATH, so the claim follows. CASE: Let MATH be the MATH - submodule of MATH generated by MATH, and let MATH be the corresponding bundle, so MATH. By REF we get MATH, so by the above claim MATH maps sections of MATH to sections of MATH. By REF . Since MATH preserves MATH, the corresponding algebraic operator preserves sections of MATH for any choice of NAME structure. Hence the operator MATH preserves sections of the subbundle MATH. Finally, on the image of MATH the operator MATH by definition coincides with MATH, so it preserves sections of MATH. Hence the splitting operator maps sections of MATH to sections of MATH. CASE: Since MATH is stable under MATH insertions and MATH has values in MATH we see that MATH maps sections of MATH to sections of MATH. Since MATH, this together with the claim implies that MATH maps sections of MATH to sections of MATH. Now the result follows exactly as in REF . CASE: Let us denote by MATH the operator corresponding to MATH. Further, put MATH and MATH. Since MATH, we conclude that on the image of MATH, we have MATH . From REF we know that MATH. Since we deal with a regular parabolic geometry, all nonzero homogeneous components of MATH have degree bounded from below by some MATH, and by the NAME identity (see CITE) the lowest nonzero homogeneous component of MATH is harmonic. Hence MATH is congruent to MATH modulo elements of homogeneous degree MATH. Hence MATH is congruent modulo elements homogeneous of degree MATH to MATH, and the latter element lies in MATH, since MATH is stable under MATH - insertions. Hence we conclude that MATH is congruent to a section of MATH modulo elements of homogeneous degree MATH. From the definition of MATH and formula REF above, we then conclude that MATH is congruent to a section of MATH modulo elements of that homogeneity. As above, this implies that MATH is congruent to a section of MATH modulo elements homogeneous of degree MATH. Hence MATH and therefore MATH are congruent to sections of MATH modulo elements of that homogeneity, and iterating this argument, the result follows. |
math/0102097 | REF is only a straightforward reformulation of REF . CASE: Consider the submodule MATH of those maps which vanish if one of their entries lies in MATH. To prove that MATH has values in MATH, by REF we have to show that MATH and MATH is stable under MATH - insertions. By REF is exactly the image of the natural inclusion MATH from REF. By definition of the insertion operator, for MATH, we get MATH. But by REF we get MATH, so MATH is stable under MATH - insertions. To prove that MATH, we use NAME 's algebraic formula for MATH. Let MATH be the inclusion. Recall from REF that we have the decomposition MATH. The Killing form of MATH induces dualities between the first and last, and the second and fourth summands and its restriction to the third summand is non - degenerate. Now we choose a basis MATH of MATH consisting of homogeneous elements, which starts with bases of the first three summands and has the duals of the first two bases in the last two summands. Denoting by MATH the dual basis with respect to the Killing form, we by construction have MATH for all MATH and the first elements of MATH (which lie in MATH) coincide with the last elements of MATH and vice versa. According to REF, one can write MATH as MATH . Here MATH denotes the adjoint action of MATH on MATH, while MATH denotes the natural action of MATH on MATH. The subspace MATH is invariant under any map which acts only on the MATH part, as well as under MATH for each MATH. In particular, any summand in the first sum preserves this subspace, while for the other two sums the same holds for summands in which MATH lies in MATH,since then also MATH lies in this subspace. It remains to show that MATH also preserves this subspace. But by construction, we can rewrite this part as MATH and since MATH, the result follows. |
math/0102097 | We can split MATH according to MATH - types into MATH. Since the adjoint action of MATH is by complex linear maps, this is a splitting of MATH - modules. Since MATH is a complex NAME algebra, both MATH and MATH preserve complex multilinear maps, so for MATH we get MATH. One easily verifies that MATH is stable under MATH - insertions. Hence by REF , in the regular normal case, the harmonic curvature MATH is of type MATH if and only if the whole curvature MATH is of type MATH. For MATH, we have the vector field MATH on MATH, and in the proof of REF we have seen that MATH. Since the bracket in MATH is complex bilinear this formula immediately implies that the MATH - part of MATH maps MATH to MATH, where MATH denotes the NAME tensor of MATH. Consequently, the integrability of the almost complex structure MATH is equivalent to vanishing of the MATH - component of the curvature function, and this vanishing also implies integrability of the almost complex structure MATH on MATH. If the almost complex structures are integrable then the projection MATH is by construction holomorphic. NAME of MATH immediately implies that for MATH and MATH we get MATH, which implies that MATH is a holomorphic mapping. On the other hand, since MATH reproduces the generators of fundamental vector fields, the map MATH is holomorphic, too, and these two facts imply that the principal right action MATH is holomorphic, so MATH is a holomorphic principal bundle. Given that MATH is a complex manifold, the NAME connection MATH, which by construction is a MATH - form, is holomorphic, if and only if its exterior derivative is a MATH - form, and since the bracket in MATH is complex bilinear, this is equivalent to the curvature being of type MATH, which implies the result. |
math/0102097 | Since the Killing form induces a duality between MATH and MATH, the cotangent bundle MATH is the associated bundle MATH. By definition, the subgroup MATH preserves the line MATH, and one immediately verifies that this property characterizes MATH. Passing to the projectivization, we see that MATH, which implies that MATH is exactly the projectivized cotangent bundle MATH. The tangent bundle to MATH is MATH, while the tangent bundle to MATH is MATH. The tangent map of the projection MATH corresponds in this picture exactly to the projection MATH. The contact distribution MATH is given by MATH. Since MATH is exactly the annihilator of MATH with respect to the Killing form, we conclude that for a point MATH (that is, MATH is a line in MATH), we have MATH, so we exactly get the canonical contact structure on MATH. Since MATH, the subbundle MATH consists of those tangent vectors, which project to zero, so this is exactly the vertical subbundle in MATH. Any connection MATH in the projective class induces a linear connection on MATH, which gives rise to a vertical projection from MATH onto the vertical subbundle. This vertical projection is characterized by the fact that the covariant derivative of a one form MATH is the nontrivial component of the composition of the vertical projection with the tangent map MATH. Factoring to the projectivization we see, that any linear connection on MATH gives rise to a vertical projection on MATH. For projectively equivalent connections MATH and MATH, one easily computes that on the level of one - forms one gets MATH. Passing from MATH to MATH means exactly factoring by the line generated by MATH in the vertical subspace, and the subbundle MATH is characterized by MATH iff MATH, which shows that the restriction of the vertical projection to MATH depends only on the projective class. One verifies directly that this construction really describes the bundle MATH. |
math/0102102 | This is a generalization of the argument used to prove REF. Recall (for example, from CITE) that surgery on a clover MATH with MATH edges corresponds to surgery on a link MATH of MATH components. Given an orientation of the edges of MATH, we can split MATH into the disjoint union of MATH-component sublinks MATH and MATH, where MATH (respectively, MATH) consists of the sublink of MATH assigned to the tails of the edges of MATH (respectively, of the heads of the edges of MATH, together with the leaves of MATH). As long as we avoid assigning all three of the components at a trivalent vertex to MATH or MATH, we will have the desired decomposition of MATH. The corresponding conditions imposed on the orientation of the edges of MATH are: CASE: No trivalent vertex is a source or a sink, CASE: Every edge with a univalent vertex is oriented toward the univalent vertex. These are the same conditions as REF in the proof of REF except that we now require no trivalent sinks also. But this will follow by the same argument as in CITE except that we need to choose the orientations of the cut edges more carefully. In particular we need to avoid choosing the orientation of two cut edges which share a trivalent vertex so that they both point into that vertex. But it is not hard to see that this can be done. |
math/0102102 | For each univalent vertex of MATH, there is a corresponding part of MATH which looks like the left part of REF . Now we can perform a NAME move (see CITE,CITE) so that the four component link MATH in REF is replaced by two component link MATH. If we do this at every univalent vertex of MATH we obtain the link MATH. Now consider the partition MATH given by REF . The corresponding partition of MATH is given by MATH and MATH. It is easy to see that both MATH and MATH are trivial in MATH. This completes the proof. |
math/0102102 | We prove this using a generalization of an argument of CITE. Consider a connected clover MATH of the class MATH. Suppose that MATH is obtained from MATH by surgery on MATH. If MATH has at least one internal trivalent vertex, then MATH and MATH are MATH-equivalent (see the discussion following REF ); in particular MATH. Otherwise, MATH must be a wheel with a certain number MATH of legs and with a total of MATH edges. Thus, the associated link MATH in MATH has MATH components (see Figure below). Using the NAME move in REF at every leaf of MATH we see that MATH is NAME in MATH to a link MATH with MATH components, whose components can be numbered in pairs MATH so that: CASE: MATH (respectively, MATH) bounds a disk MATH (respectively, MATH) in MATH, CASE: MATH, for MATH, each consists of two oppositely oriented clasps, CASE: MATH, for MATH and MATH each consists of a single clasp, and CASE: there are no other intersections among the disks. An example for MATH is shown below: MATH . We can now lift MATH and MATH to disks, MATH and MATH, in the infinite cyclic cover MATH of MATH. The lifts of MATH form a link MATH in MATH which has a linking matrix MATH with entries in MATH. To compute MATH note that we can choose the lifts MATH and MATH so that: CASE: MATH consists of a single clasp, for every MATH, CASE: MATH consists of a single clasp, oriented opposite to that in REF , for every MATH, CASE: MATH, for MATH, consists of a single clasp, and CASE: MATH, for some integer MATH, consists of a single clasp. In REF , MATH (up to sign) is just the linking number of MATH with the imbedded wheel of MATH. Now it follows from this intersection data and the fact that MATH is MATH-framed that we can orient MATH so that the linking matrix MATH is given by MATH . For any matrix MATH over MATH, MATH denotes the conjugate (under the involution MATH) transpose of MATH. The desired result MATH is now a consequence of the following lemma, which is proved by a standard argument going back to NAME, generalized to covering spaces (see for example CITE). |
math/0102102 | Observe that MATH. Consider the following diagram of exact sequences of MATH-modules. MATH-Notice that MATH. Moreover, MATH is freely generated by the meridian disks of MATH, lifted to MATH, and MATH is freely generated by the disks attached by the surgeries. Thus, since the components of MATH are null-homologous in MATH, MATH. Also note that MATH and so we have a mapping MATH induced by MATH, which can be interpreted as expressing the longitudes of MATH as linear combinations of the meridians of MATH in MATH. Therefore this map is given by the linking numbers of MATH and has MATH as a representative matrix. This completes the proof of REF and, as a consequence, REF . |
math/0102102 | If MATH is a root of MATH then MATH. Thus we have MATH, from which the conclusion follows. |
math/0102103 | We need to show that MATH is invariant under the moves MATH and MATH. If MATH, then MATH thus MATH. If MATH, then the following identity MATH implies that MATH. |
math/0102103 | It is well-known (see CITE) that a presentation matrix for MATH is MATH, and similarly for MATH. Since MATH, the lemma follows. |
math/0102103 | REF follows from the NAME formula and the fact that MATH if MATH are matrices over a commutative ring. To prove REF, first note that it is obvious if MATH is triangular. Secondly, it follows from REF that if it is true for MATH and MATH, then it is true for MATH. Thus if will follow from the fact that any such MATH can be written MATH, where MATH is lower triangular (that is, MATH if MATH) and MATH is upper triangular. We prove this by induction on the size of MATH. Write MATH, where MATH is a column vector and MATH is a row vector. By induction we can write MATH, for triangular matrices MATH. Now we define MATH . One checks immediately that MATH. |
math/0102104 | This is immediate from REF. |
math/0102104 | REF follows from REF. For REF , use the following relative version of the NAME - NAME sequence, MATH and the fact that MATH. |
math/0102104 | This follows from the NAME - NAME sequence, REF , after noting that MATH has nonzero NAME number, MATH. The final sentence follows since if MATH and MATH are both infinite, then, by REF, MATH. |
math/0102104 | Since MATH is MATH - dimensional, MATH for MATH. Since MATH, MATH, by REF. By REF, MATH. Hence, by NAME 's REF , MATH. |
math/0102104 | This follows from REF . |
math/0102104 | Although REF follow from the proof of REF , we first give simple alternative arguments for them which illustrate the above methods. Since MATH, the complexes MATH and MATH are MATH - equivariantly homotopy equivalent; hence, MATH. Since MATH, we have, by REF, that MATH, proving REF . To prove REF , let MATH and MATH denote the two points of MATH. Then MATH is the union of two copies of the cone on MATH along MATH. By excision, REF , MATH and by the exact sequence of the pair, MATH. Hence, MATH, which proves REF . In the exact sequence which we are considering in REF , MATH is the ambient space and MATH means the reduced MATH - homology of the subcomplex MATH in MATH (MATH). Thus, MATH. Let MATH be the inclusion. A class of the form MATH in the direct sum obviously goes to MATH in MATH, while MATH maps the diagonal subspace of elements of the form MATH isomorphically onto MATH. REF follows (as do REF ). |
math/0102104 | By REF, the sequences MATH and MATH are isomorphic. In other words, under NAME duality, the connecting homomorphism MATH is isomorphic to MATH. Since MATH is the adjoint of MATH, REF implies that MATH . By the exact sequence of the pair, REF , MATH, so MATH. Then, by REF, we also have MATH, which proves the lemma. |
math/0102104 | In this special case, the diagram in REF becomes the following: MATH . |
math/0102104 | Suppose MATH is a MATH or MATH for which MATH fails. Let MATH be a MATH - gon. Since MATH, the NAME REF shows that MATH also fails for MATH. |
math/0102104 | Let MATH be a MATH. By REF , if MATH holds for MATH, then MATH holds for MATH. |
math/0102104 | Suppose MATH and MATH hold, that MATH is a MATH and, as in REF, that MATH. If MATH, then since MATH holds for MATH, MATH for all MATH. By MATH, MATH for MATH. The NAME - NAME sequence then yields that MATH for MATH and that MATH. So, MATH holds for MATH. If MATH, then by MATH, MATH for all MATH. Hence, the NAME - NAME sequence yields, MATH. By MATH, MATH for MATH. It then follows from REF that MATH holds for MATH. |
math/0102104 | Let MATH be a MATH and MATH a vertex of MATH. Write MATH, as in REF. Assume MATH holds. By REF, MATH holds for MATH, that is, MATH for all MATH. From the sequence of the pair MATH we get: MATH. Since MATH holds, the last two terms are nonzero only in the middle dimension. Hence, MATH. |
math/0102104 | This follows from NAME duality. (If MATH, take MATH to get MATH for MATH. If MATH, take MATH, to get MATH for MATH.) |
math/0102104 | We proceed as in REF. Given MATH, set MATH, as before. Assume MATH holds. If MATH, then MATH for MATH (by REF). For MATH, by MATH, we have that MATH . So, for MATH, MATH and MATH (by REF ). By REF, the second equation implies that MATH for all MATH. It then follows from the exact sequence of the pair and REF, that MATH holds for MATH. If MATH, then MATH for all MATH (by REF). The sequence of the pair MATH then gives: MATH . By MATH, MATH for MATH; hence, by REF, it vanishes for MATH. Therefore, MATH for MATH and by REF, MATH for MATH. So, MATH holds for MATH. |
math/0102104 | Assume MATH holds. By REF, MATH for MATH. Hence, in the exact sequence of the pair, MATH the first and third terms vanish for all MATH. |
math/0102104 | Suppose MATH holds. Let MATH be as in MATH and let MATH. Assume, by induction on the number of vertices in MATH, that MATH holds for MATH. (The case MATH being trivial.) We want to prove it also holds for MATH, that is, that MATH for MATH. Consider the exact sequence of the triple MATH: MATH . Suppose MATH. By inductive hypothesis, MATH. By excision, REF , MATH. By REF , MATH. Since MATH holds for MATH and since MATH, MATH. So, MATH. Consequently, MATH. |
math/0102104 | We proceed as in the previous proof. If MATH, then MATH. Since we are assuming MATH holds, REF implies that MATH for MATH. Hence, if we assume by induction that the lemma holds for MATH, then it also holds for MATH. |
math/0102104 | Assume MATH and MATH hold. Let MATH be as in MATH and let MATH. Assume, by induction on the number of vertices in MATH, that MATH holds for MATH. (The case MATH being trivial.) We want to prove that it also holds for MATH, that is, that MATH for MATH. First suppose that MATH. Consider the NAME - NAME sequence for MATH: MATH . By MATH and REF , MATH (since MATH) and hence, MATH (by REF ). By inductive hypothesis, MATH, and consequently, MATH. For MATH, we compare the NAME - NAME sequence of MATH with that of MATH (where MATH): MATH . By MATH, MATH; hence, MATH is injective. By MATH, MATH is injective. Hence, MATH is injective and therefore, MATH. |
math/0102104 | By REF, MATH implies MATH, for all MATH. Suppose, by induction on MATH, that MATH holds for some MATH. If MATH is odd, then by REF , MATH implies MATH. If MATH is even, then by REF , MATH and MATH imply MATH. |
math/0102104 | The proof is by induction on MATH, starting at MATH. If MATH, then MATH is a MATH and the result follows from REF . If MATH, then for any vertex MATH, we have, by inductive hypothesis that MATH for MATH. The proof of REF then shows that MATH and the first part of the proof of REF shows that MATH for MATH. |
math/0102104 | This follows from the previous theorem (taking MATH) and NAME duality. |
math/0102104 | Suppose that MATH and MATH are valence MATH vertices which are connected by an edge. Then the star of that edge is the configuration pictured in the figure below. The indicated vertices MATH and MATH cannot coincide, since if they did MATH would contain an empty MATH - circuit and hence, not be a flag complex. Similarly, the top and bottom vertices cannot be connected by an edge, since MATH would again contain empty MATH - circuits. Let MATH be the boundary of the star in the figure. If MATH is the boundary of two adjacent MATH - simplices, then MATH is the suspension of a MATH - gon. If MATH is the link of a missing vertex, then MATH is the suspension of a MATH - gon. Otherwise, MATH is an empty MATH - circuit, contradicting REF . |
math/0102104 | By REF MATH vanishes for any MATH. Hence, it follows from the NAME - NAME sequence, REF , that we can adjoin MATH to MATH without changing MATH. |
math/0102104 | This follows from the NAME - NAME sequence as before. |
math/0102104 | By REF, MATH is combinatorially equivalent to the boundary complex of a convex polytope with vertices of valence MATH and MATH only. In REF, NAME lists MATH conditions MATH - MATH on assigned angles for such a polytope to be realized in MATH. The conditions MATH and MATH are immediate under our hypothesis, since all angles are MATH. The remaining conditions refer to certain configurations of faces of the polytope, and turn out to be vacuous in our case, since these configurations never appear under our hypothesis. Indeed, since MATH is a flag triangulation, it follows that MATH does not contain empty MATH - circuits, and therefore, MATH does not contain triangular prismatic elements and cannot be a triangular prism. Similarly, since MATH does not contain empty MATH - circuits, every MATH - circuit in MATH is a boundary of either two adjacent triangles or of a square cell, and therefore, MATH does not contain quadrangular prismatic elements. Thus, we have verified conditions MATH, MATH and MATH. To verify condition MATH we note that if two faces of MATH intersect at a vertex, but are not adjacent, then this vertex has to have valence MATH. So this vertex corresponds to a square cell of MATH, and the two faces correspond to opposite corners of the square. The configuration in REF has a third face, adjacent to both previous two, so the corresponding vertex in MATH is connected to these corners. In MATH this square is subdivided by the diagonals and, since MATH does not contain empty MATH - circuits, one of the remaining corners of the square in MATH must be connected to the vertex corresponding to the third face. This means that MATH contains a configuration pictured in REF , which according to that lemma is impossible. |
math/0102104 | If MATH is the suspension of a MATH - or MATH - gon, then the theorem follows from REF . If MATH is not the suspension of a MATH - gon or a MATH - gon and if it has no empty MATH - circuits, then by REF, MATH vanishes for all MATH, where MATH denotes the set of valence MATH vertices. Hence, by REF , MATH also vanishes. In every other case, MATH has an empty MATH - circuit which we can use to decompose MATH as, MATH, as in REF. Since MATH and MATH each have fewer vertices than does MATH, this process must eventually terminate. So, the theorem follows from REF . |
math/0102104 | By REF , we may assume that MATH is connected. Suppose that MATH is piecewise linearly embedded in MATH. By introducing a new vertex in the interior of each complementary region and then coning off the boundary of each region, we obtain a flag triangulation MATH of the MATH - sphere with MATH embedded as a full subcomplex. By MATH, MATH. |
math/0102104 | As in REF , embed MATH in the flag triangulation MATH of MATH obtained by introducing new vertex MATH for each boundary component MATH. Since MATH vanishes, MATH . A simple calculation using REF gives that this is nonzero only for MATH and that MATH . The first formula follows. To prove the second, consider the pair MATH. By excision, its homology is isomorphic to that of MATH. Hence, MATH and by REF , the second term is nonzero only for MATH, in which case, MATH . |
math/0102104 | As in the proof of REF , we can assume that MATH is a full subcomplex of some flag triangulation MATH of the orientable surface of genus MATH. By REF , MATH; hence, the map MATH is injective. Since we are assuming MATH, the result follows. |
math/0102104 | By REF, the double of MATH, MATH, is also a MATH. By REF, MATH is the union of two copies of MATH glued along MATH. So, we have a NAME sequence, MATH . By REF , the kernel of the map MATH into either factor has dimension MATH. Thus, the kernel of the map MATH has dimension MATH. Hence, MATH . Substituting in MATH for MATH (by REF), we get the desired inequality. |
math/0102104 | CASE: Suppose MATH is not a simplex. Then we can find vertices MATH in MATH which are not connected by an edge. Let MATH be the MATH - fold iterated double MATH along MATH, as defined in REF. Then MATH has MATH preimages in MATH and the link of each is isomorphic to MATH. Choose one, say MATH, and set MATH. Since we are assuming MATH, we have, by REF, that MATH. By REF, MATH. Hence, MATH . Since this holds for any MATH, MATH. Case MATH is a simplex MATH. If MATH, then MATH is a suspension and we are done by REF . If MATH, then let MATH (defined in REF). By REF, MATH where MATH. Moreover, there are MATH preimages of MATH in MATH, no two of which are connected by an edge and such that the link of each is isomorphic to MATH. Choose one of these preimages, say MATH, and set MATH. Since MATH contains MATH preimages of MATH, MATH, we can apply REF to MATH to conclude that MATH. |
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