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math/0106152 | This is a restatement without reference to basepoints of the definition of weak equivalence of simplicial sets. |
math/0106152 | Without loss of generality, we may assume that MATH is a level map. Given a finite diagram of local systems MATH on MATH, there is a MATH such that each MATH is represented by a local system MATH on MATH. Then for each MATH, MATH is represented by MATH. Now MATH in MATH is represented by MATH in MATH, so MATH is represented by MATH. Also, MATH in MATH is represented by MATH in MATH. But MATH commutes with finite colimits, so MATH and MATH are isomorphic since they are represented by the same local system on MATH. An identical argument shows that MATH commutes with finite limits. |
math/0106152 | Without loss of generality, assume that MATH is a level map. Note that MATH is an abelian category CITE. Since MATH is exact, it suffices to consider a local system MATH on MATH such that MATH and conclude that MATH. A pro-local system MATH is zero if and only if for every MATH, there exists MATH such that the map MATH is trivial. For any MATH, choose MATH so that MATH is trivial. Let MATH and MATH be local systems on MATH representing MATH and MATH respectively. Choose MATH large enough so that MATH is a trivial map of local systems on MATH. Since MATH is an epimorphism, there exists some MATH and a map MATH such that the map MATH factors through MATH. Since MATH is trivial, the map MATH is trivial when restricted to the components of MATH in the image of MATH. Now the image of MATH in MATH contains the image of MATH, so MATH becomes trivial when pulled back to MATH. Hence the map MATH is trivial. This means that the pro-local system MATH is zero. |
math/0106152 | One direction is clear because level isomorphisms are pro-isomorphisms. Assume that MATH is an isomorphism. We may also assume that MATH is a level map. Define MATH and MATH. Here we identify pro-sets with pro-spaces of dimension zero. Then MATH is isomorphic to MATH since MATH is an isomorphism. Let MATH be the map induced by MATH and the projection MATH. Pullbacks can be constructed levelwise in pro-categories, so for all MATH, MATH and MATH is induced by MATH and MATH. Note that MATH induces an isomorphism MATH. Hence MATH is the desired map. |
math/0106152 | We only prove that MATH is essentially surjective in the sense that every object MATH of MATH is isomorphic to MATH for some MATH in MATH. We leave the rest of the proof to the interested reader. We will use only the surjectivity in this work. With no loss of generality, we may assume that MATH is a level map. By REF , we may also assume that MATH is a level isomorphism. Let MATH be a local system on MATH represented by a local system MATH on MATH. There exists MATH and a commutative diagram MATH of local systems on MATH. Choose one point MATH in each component of MATH. Let MATH be the image of MATH in MATH; this is a choice of one point in each component of MATH. Let MATH be the image of MATH in MATH. By evaluating the above diagram at MATH, the map MATH factors as MATH . The maps MATH and the local system MATH determine a local system MATH on MATH by setting MATH with the MATH-action induced by MATH. Let MATH be the local system on MATH represented by MATH. Note that MATH is isomorphic to MATH. Hence MATH is isomorphic to MATH. |
math/0106152 | The axioms for a proper simplicial model structure are verified in REF through REF. Limits and colimits exist by REF . The two-out-of-three axiom is REF . Retracts preserve weak equivalences because weak equivalences are defined in terms of isomorphisms. Retracts preserve fibrations because retracts preserve lifting properties. REF is the retract axiom for cofibrations. REF are the factoring axioms, while REF is the non-trivial lifting axiom. The axioms for a simplicial model structure are demonstrated in REF shows that the model structure is proper, and REF states that it is not cofibrantly generated. |
math/0106152 | The ``if" direction follows from the fact that strong fibrations are fibrations by REF and the fact that retracts preserve fibrations. Conversely, suppose that MATH is a fibration. Use REF to factor MATH as MATH where MATH is a trivial cofibration and MATH is a strong fibration. The retract argument CITE finishes the proof. |
math/0106152 | Suppose that MATH satisfies the conditions of the proposition. Then MATH is a cofibration by assumption, and it is a weak equivalence by REF . Therefore, MATH is a retract of a trivial cofibration, so it is also a trivial cofibration. Now suppose that MATH is a trivial cofibration. Use REF to factor MATH as MATH where MATH satisfies the conditions of the proposition and MATH is a strong fibration. By REF , MATH has the left lifting property with respect to MATH. The retract argument CITE finishes the proof. |
math/0106152 | We may assume that MATH is a level map. Suppose that the diagrams of REF exist. Using REF , it follows that MATH is an isomorphism of pro-groups. Now suppose that MATH is an isomorphism of pro-groups. Then the diagrams of REF exist for every MATH when MATH is the chosen basepoint of MATH. Let MATH be any point of MATH. Choose a path from MATH to MATH in MATH. This path induces an isomorphism between the diagram of REF based at MATH and the diagram based at MATH. |
math/0106152 | Let MATH be a fibrant approximation of MATH. Then MATH is a fibrant approximation of MATH in the category of pro-spaces. Hence MATH equals MATH and MATH equals MATH. Therefore, it suffices to assume that MATH is fibrant. According to REF , MATH. Since MATH commutes with filtered colimits, MATH . Since MATH is cofibrant and MATH is fibrant, MATH is equal to MATH, and MATH is equal to MATH for every MATH. Therefore, MATH . |
math/0106152 | Let MATH be a space and let MATH be a pro-space. Then MATH . Therefore MATH and MATH are an adjoint pair. Since MATH preserves cofibrations and trivial cofibrations, the functors are a NAME pair. |
math/0106152 | From REF , we know that MATH is equal to MATH, where MATH is a fibrant approximation of MATH. It suffices to show that MATH is weakly equivalent to MATH. For each MATH, let MATH be a functorial fibrant approximation of MATH. Then MATH is a strict weak equivalence. Now let MATH be a strictly fibrant approximation of MATH. Note that MATH is a fibrant approximation of MATH. Recall that MATH is isomorphic to MATH for each MATH. Also, MATH is weakly equivalent to MATH since each map MATH is a fibration. Now MATH is weakly equivalent to MATH since MATH and MATH are levelwise weakly equivalent. Furthermore, MATH is weakly equivalent to MATH. In other words, MATH is weakly equivalent to MATH. Since MATH is a strictly fibrant approximation of MATH, MATH is weakly equivalent to MATH CITE. |
math/0106152 | Let MATH be a pro-space, and let MATH belong to MATH. Then MATH . The second equality comes from the usual adjointness of pro-finite completion, which says that MATH for any simplicial set MATH and any simplicial finite set MATH. |
math/0106152 | We must show that the left adjoint MATH preserves cofibrations and trivial cofibrations. First consider an inclusion MATH of simplicial sets. The map MATH is an inclusion. This follows from the fact that if MATH and MATH are two distinct MATH-simplices of MATH, then there is a finite quotient MATH of MATH such that MATH and MATH are not equal. Now consider an arbitrary cofibration of pro-spaces MATH. Without loss of generality, we may assume that MATH is a levelwise cofibration. The map MATH is an inclusion for each MATH, so the map MATH is also an inclusion. In other words, MATH is a cofibration in MATH. Next consider a trivial cofibration of pro-spaces MATH. We may assume that MATH is a levelwise cofibration. Note that MATH is an isomorphism by REF . CITE observed that MATH since MATH is a finite abelian group. Therefore MATH . Similarly, MATH. Hence MATH is isomorphic to the map MATH, so it is also an isomorphism. |
math/0106152 | Assume there are levelwise weak equivalences MATH and MATH with an isomorphism MATH. We must construct a levelwise weak equivalence isomorphic to the composition MATH. By adding isomorphisms to the systems for MATH or MATH, make the cardinalities of the index sets equal. Choose an arbitrary isomorphism MATH from the index set of MATH to the index set of MATH. Define a function MATH inductively on height satisfying several conditions. Choose MATH large enough so that MATH represents MATH, MATH, and MATH for all MATH. The function MATH defines cofinal subsystems of MATH and MATH. Hence we may assume that MATH is a level diagram indexed by a cofinite directed set MATH where MATH and MATH are levelwise weak equivalences. However, the composition is not necessarily a levelwise weak equivalence because the map MATH is not a levelwise weak equivalence. Since MATH is an isomorphism of pro-spaces, for every MATH, there exists MATH and a commutative diagram MATH . By restricting to cofinal subsystems, we may assume that such a diagram exists for every MATH. Let MATH be the directed set of indecomposable arrows of MATH. The domain and range functors MATH are both cofinal since MATH is cofinite. For each MATH in MATH, factor the map MATH as MATH where MATH is a cofibration and MATH is a fibration. Let MATH be the pullback MATH, and let MATH be the pushout MATH. These objects fit into a commutative diagram MATH . Note that MATH is a weak equivalence because it is a pullback of a weak equivalence along a fibration. Also, MATH is a weak equivalence because it is a pushout of a weak equivalence along a cofibration. Hence the composition MATH is a levelwise weak equivalence. This composition is isomorphic to MATH since MATH, MATH, and MATH. |
math/0106152 | Every levelwise weak equivalence is a weak equivalence in the sense of REF . |
math/0106152 | The class of trivial cofibrations contains the class of strictly trivial cofibrations by REF . |
math/0106152 | NAME are the same as strict cofibrations. |
math/0106152 | The condition on MATH ensures that its strictly fibrant replacement MATH is also a fibrant replacement. To calculate morphisms from MATH to MATH in either homotopy category, consider morphisms from MATH to MATH modulo the simplicial homotopy relation. Hence the morphisms are the same. |
math/0106152 | CITE showed that MATH has all equalizers (respectively, coequalizers) provided that MATH does. It suffices to construct arbitrary products (respectively, coproducts) when these exist in MATH. Let MATH be a set and let MATH be a pro-object for each MATH in MATH. Let MATH be the cofiltering index category for MATH. Define MATH to be the cofiltering system with objects MATH and index category MATH consisting of pairs MATH where MATH is a finite subset of MATH and MATH is an element of MATH. A morphism of MATH from MATH to MATH corresponds to an inclusion MATH and morphisms MATH in MATH for all MATH in MATH. Use of finite subsets MATH of MATH is essential because we use the fact that finite products commute with filtered colimits. Direct calculation shows that for any MATH in MATH, MATH . Thus arbitrary products exist. To construct the coproduct, define MATH to be the cofiltering system with objects MATH and index category MATH. Note that MATH is cofiltering since each MATH is. In order to show that MATH has the correct universal mapping property, it suffices to see that MATH for MATH any object of MATH (that is, MATH is a constant system in MATH). This can be checked directly, using the fact that colimits indexed on product categories commute with the relevant products. Thus arbitrary coproducts exist. |
math/0106152 | Suppose that MATH is a retract of MATH, where MATH is a level map that belongs to MATH levelwise. Hence there is a commutative diagram MATH where the horizontal compositions are identity maps. We must show that MATH also belongs to MATH. Choose a level representative for MATH. By adding isomorphisms to the systems for MATH or MATH, make the cardinalities of the index sets equal. Choose an arbitrary isomorphism MATH from the index set of MATH to the index set of MATH. Define a function MATH inductively on height satisfying several conditions. First, choose MATH large enough so that MATH and MATH represent respectively the maps MATH and MATH. Also, choose MATH large enough so that MATH. Finally, choose MATH large enough so that for all MATH, MATH with a commuting diagram MATH . Now the function MATH defines cofinal subsystems MATH and MATH of MATH and MATH where MATH and MATH have the same index sets as MATH and MATH. Repeat this process on MATH to obtain another function MATH inducing cofinal subsystems MATH and MATH of MATH and MATH. The result is a level diagram MATH representing (up to isomorphism) MATH as a retract of MATH. Since MATH and MATH are identity maps, MATH can be chosen so that the composites MATH and MATH are structure maps of MATH and MATH for all MATH. Since MATH belongs to MATH levelwise, the same is true for MATH. Define a pro-space MATH by starting with the system MATH and replacing the single map MATH with the pair of maps MATH. Define MATH similarly. Note that MATH and MATH are cofinal subsystems of MATH and MATH respectively, so MATH is isomorphic to MATH and MATH is isomorphic to MATH. Thus it suffices to show that the level map MATH belongs to MATH. The subsystem of MATH on objects MATH is also a cofinal subsystem, and the same is true for the subsystem on objects MATH in MATH. Beware that the subsystem of MATH on MATH is not isomorphic to MATH because the structure maps are different. The same warning applies to MATH and MATH. Restrict MATH to the subsystem MATH. This last map belongs to MATH levelwise, so it belongs to MATH. Hence MATH also belongs to MATH because MATH is closed under isomorphisms. |
math/0106152 | Apply REF to the class of all cofibrations in MATH. |
math/0106152 | For MATH, the map MATH factors as MATH . Also, the map MATH factors through MATH. This immediately proves two of the three cases. For the third case, suppose that MATH and MATH are weak equivalences. Then MATH is an isomorphism for all MATH. By REF , MATH is also an isomorphism for all MATH. |
math/0106152 | For simplicity write the group operations additively, even though the groups are not necessarily abelian. We may assume that MATH is a level map. If MATH is an isomorphism, then MATH is also an isomorphism by REF . Now suppose that MATH is an isomorphism. By REF applied twice, for every MATH, there exist MATH and a commutative diagram MATH where the diagonal maps are not necessarily group homomorphisms. However, the composite map MATH is in fact a group homomorphism. For every MATH and MATH in MATH, MATH because of commutativity in the top square. Also, MATH because of commutativity in the bottom square. Finally, MATH. Therefore, there is a commutative diagram of groups MATH . By REF , MATH is an isomorphism. |
math/0106152 | First suppose that MATH is a weak equivalence. Since MATH is an isomorphism, REF gives the conclusion for MATH. For MATH, REF implies that, for every MATH, there exists a MATH and a commutative diagram MATH of local systems on MATH. In particular, for every basepoint MATH in MATH, there is a commutative diagram MATH . This proves the ``only if" part of the claim. Now suppose that the diagrams in the statement of the proposition exist. By REF , MATH is an isomorphism. For MATH, we use the fact that a local system MATH on a space MATH is determined up to isomorphism by its value MATH as a MATH-module for one point MATH in each component of MATH. For every MATH, there exist MATH such that for every basepoint MATH in MATH there are commutative diagrams MATH . A diagram chase like that in the proof of REF shows that the map MATH is actually a map of MATH-modules, even though the diagonal maps in the left diagram above are not maps of MATH-modules. This defines a commutative diagram MATH of local systems on MATH. Hence MATH is an isomorphism by REF . |
math/0106152 | For any MATH in MATH and any MATH, choose MATH such that MATH. For every point MATH in MATH, there is a commutative diagram of solid arrows MATH . Since the top horizontal map is an isomorphism, this diagram can be extended to include the dashed arrow. Thus MATH satisfies the condition of REF , so it is a weak equivalence. |
math/0106152 | Let MATH be the map MATH, so MATH is a MATH-fibration by the definition of strong fibrations. We use the lifting property characterization of MATH-fibrations. For the purposes of induction, assume that the maps MATH and MATH are MATH-fibrations for all MATH. In particular, they are all MATH-fibrations since MATH. Let MATH be a MATH-cofibration. A lift in the diagram MATH is the same as a lift for MATH . A lift exists in the last diagram because MATH is a MATH-fibration. Hence lifts can be extended inductively to obtain lifts for MATH. Thus MATH is a MATH-fibration. The projection MATH is the pullback of MATH along the base MATH. Therefore, it is also a MATH-fibration since right lifting properties are preserved by pullbacks. Now MATH, so MATH is a MATH-fibration since it is a composition of such maps. |
math/0106152 | One direction was proved in REF . Use the notation of REF . Suppose that MATH is a MATH-equivalence and MATH is a fibration for all MATH. It suffices to show that MATH is also a MATH-equivalence. For the purposes of induction, assume that MATH is a MATH-equivalence for MATH. As in the proof of REF , MATH is a MATH-fibration by the inductive assumption. Hence MATH also is a MATH-fibration. The maps MATH and MATH are MATH-equivalences. Since MATH, it follows that MATH is also a MATH-equivalence. |
math/0106152 | There exists a lift in the diagram MATH because the image of MATH in MATH contains the image of MATH. Now consider a diagram MATH . We have already shown that this diagram exists for MATH. We show by induction that this diagram exists for all MATH. The induction step amounts to finding a lift in the above diagram. We construct a lift one simplex at a time. Therefore, it suffices to assume that MATH and MATH. The obstruction to finding a lift is an element of MATH, and this obstruction lies in the image of MATH. When MATH the map MATH is trivial, so there is no obstruction to lifting. By induction on MATH, there is a diagram MATH . A lift exists in this diagram since the left vertical arrow is a MATH-cofibration and the right vertical arrow is a MATH-fibration. The map MATH is the desired lifting. |
math/0106152 | NAME the commutative diagram with exact rows MATH . |
math/0106152 | Consider a commutative diagram of pro-spaces MATH such that MATH is a trivial cofibration and MATH is a strong fibration. We may assume that MATH is a level map that is a levelwise cofibration. By choosing an appropriate cofinal subsystem for MATH as in the proof of REF , we may additionally assume that the square diagram is a level diagram. This choice preserves MATH as a levelwise cofibration. We construct a lifting by induction. Suppose that for all MATH, there exists MATH and a map MATH such that for all MATH, MATH and there is a commutative diagram MATH . To extend the map MATH to level MATH, we must find a lift in the diagram MATH for some MATH such that MATH for all MATH. Write the map MATH as MATH. Because MATH is a strong fibration, MATH is a MATH-fibration for some MATH. Choose MATH so that MATH for all MATH. Now select MATH so that MATH and there exist commutative diagrams MATH for all MATH and all basepoints MATH in MATH. This is possible since MATH is a weak equivalence and there are only finitely many conditions on the choice of each MATH. Finally, choose MATH so that MATH and there exists a commutative diagram MATH . NAME factor each map MATH as MATH, where MATH is a trivial cofibration and MATH is a fibration. Choose a basepoint MATH in MATH. Since MATH is a weak equivalence, there exists a basepoint MATH in MATH such that there is a path in MATH from MATH to the image of MATH. For MATH, there is a diagram of solid arrows MATH where the vertical arrows are induced by the choice of path. Since the vertical arrows are all isomorphisms, the dashed arrow also exists. For similar reasons, there is also a commutative diagram MATH . Now we have MATH where the dashed arrow exists because MATH is a trivial cofibration and MATH is a fibration. So we need only find a lift for the diagram MATH . REF tells us that this diagram satisfies the hypotheses of REF . Hence the desired lift exists. |
math/0106152 | We may assume that MATH is a level map. We construct the factorization inductively. Assume for sake of induction that the factorization is already constructed on all indices less than MATH. Recall the height function MATH from REF . Factor MATH as MATH where MATH is a MATH-cofibration and MATH is a MATH-fibration. Let MATH be the projection map induced by MATH. This extends the factorization to level MATH. Now MATH satisfies the definition of strong fibration by construction. Also, MATH is a levelwise cofibration by construction. Finally, MATH is a weak equivalence by REF . |
math/0106152 | Use the strict model structure to factor MATH as a cofibration MATH followed by a strictly trivial fibration MATH. Then MATH is also a trivial fibration by REF . |
math/0106152 | Consider a commutative diagram of pro-spaces MATH such that MATH is a cofibration and MATH is a trivial fibration. We may assume that MATH is a level map that is a levelwise cofibration. By choosing an appropriate cofinal subsystem for MATH, we may additionally assume that the square diagram is a level diagram. This choice preserves MATH as a levelwise cofibration. Use the strict model structure to factor MATH as MATH, where MATH is a levelwise cofibration and MATH is a strictly trivial fibration. Note that MATH is a weak equivalence by the two-out-of-three axiom as proved in REF . In the diagram MATH the lift in the lower left square exists because of the strict model structure, and the lift in the upper right square exists by the definition of fibrations. The composition MATH is the desired lift. |
math/0106152 | For any simplicial set MATH, let MATH be the filtering system of finite subspaces of MATH. For an object MATH of MATH, define MATH to be MATH and MATH to be MATH. Using the fact that the system MATH is cofinal in the system MATH, the required isomorphisms MATH and MATH can be verified directly. |
math/0106152 | By REF , it suffices to check the axioms only for finite simplicial sets. Most of the axioms are obvious; we verify only the non-trivial ones here. Let MATH and MATH be arbitrary pro-spaces, and let MATH be a finite simplicial set. We use the fact that MATH is equal to MATH for any filtered system MATH of simplicial sets because MATH is finite. It follows by direct calculation that MATH . We now show that the map MATH is a fibration whenever MATH is a cofibration and MATH is a fibration and that this map is a trivial fibration if either MATH or MATH is trivial. We proceed by showing that MATH has the relevant right lifting property. Let MATH be a generating cofibration or a generating trivial cofibration. Note that MATH and MATH are finite simplicial sets. By adjointness, it suffices to show that the map MATH is a cofibration that is trivial if either MATH or MATH is trivial. We may assume that MATH is a levelwise cofibration. For every MATH, MATH is a cofibration. Therefore, the map MATH is also a cofibration. This is a standard fact about simplicial sets. Thus MATH is a levelwise cofibration. In order to show that MATH is trivial whenever MATH or MATH is, it suffices to show that the map MATH is trivial if MATH is trivial and that the map MATH is trivial if MATH is trivial. This reduction follows from the two-out-of-three axiom, the fact that trivial cofibrations are preserved by pushouts, and the commutative diagram MATH . First suppose that MATH is trivial. The map MATH is a levelwise weak equivalence, so it is a weak equivalence of pro-spaces. Now suppose that MATH is trivial. Since MATH is constructed by levelwise product with MATH, REF is easily verified. |
math/0106152 | Left properness follows immediately from the fact that all pro-spaces are cofibrant. We must show that the model structure is right proper. Let MATH be a fibration and let MATH be a weak equivalence. Use REF to suppose that MATH and MATH are level maps with the same cofinite directed index set MATH for which there is a strictly increasing function MATH such that MATH is a MATH-equivalence. Let MATH be the pullback MATH, which is constructed levelwise. We must show that the projection MATH is a weak equivalence. We start with a special case. First suppose that MATH is a levelwise fibration. Let MATH be a basepoint in MATH. This yields a diagram MATH in which the rows are fiber sequences. From the MATH-lemma applied to the long exact sequences of homotopy groups of the fibrations, MATH is also a MATH-equivalence. By REF , MATH is a weak equivalence. Now let MATH be an arbitrary fibration. By REF , there exists a strong fibration MATH such that MATH is a retract of MATH. Note that MATH is a levelwise fibration by REF . Consider the commutative diagram MATH where MATH. This diagram is a retract of squares in the sense that all of the horizontal compositions are identity maps. The map MATH is a weak equivalence by the special case. Since weak equivalences are closed under retracts, the map MATH is also a weak equivalence. |
math/0106152 | REF showed that a map MATH satisfying the conditions of the proposition is a weak equivalence. Now suppose that MATH is a weak equivalence. We may assume that MATH is a level map. Use REF to factor MATH as MATH where MATH is a trivial cofibration and MATH is a trivial fibration. By REF , MATH is also a strictly trivial fibration. In particular, MATH is isomorphic to a levelwise weak equivalence. The proof of REF indicates that MATH satisfies the conditions of the proposition. By an argument similar to the proof of REF , MATH also satisfies the conditions of the proposition. |
math/0106152 | REF is easily verified. |
math/0106152 | Suppose that MATH is a strict weak equivalence. Then it is also a weak equivalence. The maps MATH and MATH are weak equivalences by REF , so MATH is also. Now suppose that MATH is a weak equivalence. By REF , we may assume that MATH is a level map indexed by a cofinite directed set MATH for which there is a strictly increasing function MATH such that MATH is a MATH-equivalence. Consider the subsystem MATH of MATH and the subsystem MATH of MATH. Note that MATH and MATH are cofinal in MATH and MATH. Let MATH be the level map MATH induced by MATH, so MATH is isomorphic to MATH. Since MATH is a MATH-equivalence, the map MATH is a weak equivalence. Hence MATH is a levelwise weak equivalence, so MATH is a strict weak equivalence. |
math/0106152 | Let MATH be a weak equivalence. By REF , we may assume that MATH is a level map indexed by a cofinite directed set MATH for which there is an increasing function MATH such that MATH is a MATH-equivalence. By the NAME theorem, MATH induces a cohomology isomorphism in dimensions less than MATH for any local system on MATH. Hence MATH induces an isomorphism MATH in the colimit for every MATH. Now suppose that MATH satisfies the conditions of the proposition. Factor MATH as MATH where MATH is a cofibration and MATH is a strictly trivial fibration. Since MATH induces cohomology isomorphisms by the first part of the proof, the map MATH still satisfies the hypotheses of the proposition. Therefore, we may assume that MATH is a level map that is a level cofibration. Note that MATH is weakly equivalent to MATH since MATH is constructed levelwise and MATH is levelwise weakly equivalent to MATH. We prove the proposition by showing that for every strongly fibrant pro-space MATH, the map MATH is a weak equivalence. A retract argument then shows that the map MATH is a weak equivalence for all fibrant pro-spaces MATH. Assume that MATH is an arbitrary strongly fibrant pro-space. Note that each MATH is a fibrant simplicial set with only finitely many non-zero homotopy groups. Recall that MATH. Also recall that for every MATH, the map MATH is a fibration since MATH is fibrant. Therefore, the map MATH is a fibration. It follows that MATH is weakly equivalent to the homotopy limit MATH. Similarly, MATH is weakly equivalent to the homotopy limit MATH. Since homotopy limits are invariant under levelwise weak equivalence, we only need show that MATH is a weak equivalence of simplicial sets for each MATH. Therefore, we may assume that MATH is a fibrant simplicial set with only finitely many non-zero homotopy groups. By adjointness, to show that MATH is a weak equivalence, it suffices to find lifts in the diagrams of pro-spaces MATH . For simplicity, rewrite the fibration MATH as MATH. Our goal is to find a MATH and a commuting diagram of simplicial sets MATH . Note that MATH and MATH are fibrant simplicial sets, and MATH is a fibration of simplicial sets. Use NAME systems CITE to factor MATH as MATH where MATH, each MATH is a fibration, and the fibers of MATH (which may vary up to homotopy because MATH may not be connected) are of the form MATH for some fiber MATH of MATH. The fibers of MATH are of the form MATH, so they have nonzero homotopy groups only in finitely many dimensions. Therefore, there exists MATH such that MATH is a weak equivalence for MATH. We inductively construct partial liftings MATH . Since MATH, these lifts assemble to give us the desired lifting. Choose MATH such that the original square of pro-spaces is represented by the square of simplicial sets MATH . By the arguments of REF , there exists a MATH and a commutative diagram MATH . These arguments apply because MATH is a MATH-fibration and because MATH induces an isomorphism in homotopy of dimension MATH and MATH. For the moment, assume that MATH. Since the fibers of MATH are of the form MATH, MATH is abelian for every fiber MATH of MATH. By obstruction theory (for example, CITE), the only obstruction to finding a lift MATH is a class MATH in MATH, where MATH is the local system given by the first homotopy groups of the fibers of MATH. Note that MATH makes sense because these homotopy groups are abelian. But MATH since MATH induces a cohomology isomorphism. Therefore, there exists a MATH such that the image of MATH in MATH is MATH. Hence, there is a lift in the above diagram when MATH is replaced by MATH. The same obstruction theory argument applies inductively to give liftings MATH for each MATH. Now MATH is the local system given by MATH-th homotopy groups of the fibers of MATH. Again, MATH makes sense because MATH acts trivially on MATH for every fiber MATH of MATH. Let MATH. Recall that MATH is a trivial fibration for MATH. Therefore, lifts exist inductively in the diagrams MATH for MATH. Hence, the desired lifting exists when MATH. Now consider MATH. The argument given for MATH does not work. The trouble is that we cannot lift over MATH with obstruction theory because the first homotopy groups of the fiber are not necessarily abelian. When MATH, the map MATH is just the map MATH, so we need to find a MATH and a factorization of MATH through MATH. Note that such factorizations are the same as factorizations up to homotopy of MATH through MATH. CITE constructed such factorizations when MATH is connected. Their argument works even when MATH is not connected provided that MATH. Here we use the fact that MATH by REF . This proves the result. |
math/0106152 | Let MATH be a cofibrant object of MATH. Then MATH is a retract of another object MATH, where MATH is a transfinite composition of pushouts of maps in MATH CITE. Since there is a map from MATH to MATH, it suffices to find a map from some object of MATH to MATH. Since MATH is not the initial object, MATH is also not the initial object. Hence MATH is a non-trivial transfinite composition of pushouts of maps in MATH. Let MATH be a map in MATH occurring in the construction of MATH. Then there is a map from MATH to MATH. |
math/0106152 | Choose an infinite cardinal MATH larger than the size of any of the sets occurring in any of the objects of MATH. Let MATH be a set of size MATH. Define a pro-set MATH as follows. Consider the collection of all subsets MATH of MATH whose complements MATH are strictly smaller than MATH. Note that this implies that the size of MATH is MATH, but the converse is not true. These subsets form a pro-set, where the structure maps are inclusions. This system is cofiltered because MATH is strictly smaller than MATH when MATH and MATH are. Let MATH be an object of MATH. Suppose that there is a map MATH of pro-sets. Then there exists a MATH and a map MATH representing MATH. Let MATH be the image of MATH, so MATH is strictly smaller than MATH since MATH is strictly smaller than MATH. Consider the set MATH, which occurs as an object in the system MATH. Since MATH is a map of pro-sets, there exists a MATH such that the composition MATH factors through MATH. Since MATH and MATH have disjoint images in MATH, this is only possible if MATH is the empty set. However, MATH cannot be the empty set because MATH is not the empty pro-set. By contradiction, the map MATH cannot exist. |
math/0106152 | We argue by contradiction. Suppose that there exists a cofibrantly generated model structure for which every object is cofibrant. Let MATH be the set of generating cofibrations, and let MATH be the set of targets of maps in MATH. Apply MATH to MATH to obtain a small family of pro-sets MATH. Let MATH be the pro-set constructed in REF . We can think of MATH as a pro-space by identifying a set with a simplical set of dimension zero. By REF , there exists a non-empty MATH in MATH and a map MATH. This induces a map MATH. However, such a map cannot exist by REF because MATH belongs to MATH. |
math/0106153 | Let MATH, where MATH is even (the case of MATH odd is easier; see CITE). Let MATH. Let MATH be the selfmap of MATH defined by a degree MATH map on each factor. The map MATH induces a scalar homomorphism (namely, multiplication by a power of MATH) in each cohomology group of MATH. Namely, MATH . Let MATH be the sum of the standard generators MATH of the cohomology group MATH. Then MATH . Now the class MATH defines a map MATH such that MATH where MATH is a generator. Recall that MATH is a multiplicative generator of the polynomial ring MATH (MATH even). Our proof will rely on the following lemma. The map MATH extends to a map MATH which induces an isomorphism in rational homology through dimension MATH, where MATH is a finite CW complex obtained from MATH by attaching cells of dimension at most MATH. Furthermore, the maps MATH extend to MATH for MATH divisible by a sufficiently large MATH, in the following two situations: CASE: if MATH or MATH; CASE: if MATH. Since MATH, the map MATH induces an epimorphism MATH in each dimension less than MATH, and an isomorphism in dimension MATH. We will use the relative NAME theorem to eliminate the successive kernels of the homomorphisms REF, and construct MATH by skeleta MATH inductively on MATH (compare CITE). Let MATH and let MATH. Next, let MATH. Then MATH. Since the map MATH acts by MATH on MATH, it extends to MATH. We set MATH and note that, by patching the two pieces, MATH extends to MATH. Note that MATH, where MATH. We have MATH, MATH. Consider the exact sequence MATH . By NAME 's theorem, the map MATH induces multiplication by MATH on the group MATH . Now consider the diagram MATH where MATH are the NAME maps. By the relative NAME theorem, the homomorphism MATH is surjective. Thus MATH . Hence, every element in MATH is spherical. Let MATH, MATH, be a set of lifts via the NAME homomorphism MATH, of a set of generators of the kernel of MATH. From the exact sequence REF, we have MATH, where MATH is of finite order. By NAME 's result CITE on the nilpotence of the finite homotopy groups of spheres, there exists a MATH such that if MATH then MATH on MATH. Assume also that MATH is a multiple of the order of MATH. Then MATH that is, MATH is proportional to MATH. It follows that MATH extends to the next skeleton MATH and therefore also to MATH, proving REF. We can similarly define MATH, and use the NAME form of the relative NAME theorem (see CITE) to define MATH . Proceeding by induction, we extend MATH to the space MATH, which is the desired meromorphic extension of MATH (compare REF ). The above argument does not yield an extension of MATH to MATH (when MATH) because already the group MATH cannot be shown to be finite merely from NAME 's theorem, as in the calculation REF above. An alternative approach given below gives an easy proof only for MATH (compare REF ). For MATH, we can understand the effect of MATH on MATH easily using the NAME fibration over MATH. We argue inductively. To the extent that MATH already acts on MATH, the family MATH localizes homology of the space MATH. Therefore by CITE, the family also localizes homotopy, so that we have the same corollary as for the spheres: the map on MATH torsion of MATH induced by MATH is nilpotent. To show that the action of MATH is scalar on MATH, it therefore suffices to show that this group has rank precisely MATH . Consider the exact sequence of pair MATH: MATH by NAME. We need to show that the group MATH is finite. Now the map MATH, where MATH, has already been constructed. This map induces an isomorphism in rational homology in all dimensions. By the NAME theorem, the group MATH is finite for all MATH. The exact sequence of the pair shows that the groups MATH are all finite except MATH and MATH, since MATH from the exact sequence of the NAME fibration, proving REF. To generalize this proof to MATH, one could replace the spaces MATH by the skeleta of MATH (compare REF ). However, the calculation of the homotopy groups of such skeleta is not as immediate as that of complex projective spaces. CASE: REF provides a quick proof of REF , as follows. We appeal to rational homotopy theory, to conclude that the rationalisation MATH of MATH can be thought of as the MATH-skeleton of MATH. Due to the existence of the selfmaps MATH of MATH, according to CITE, the space MATH admits a model as an infinite telescope on MATH, whose MATH-th the cylinder of the self-map MATH of MATH, where MATH is induced by a degree MATH map on each of the factors in MATH. Now let MATH be a finite MATH-dimensional CW complex with MATH. Consider a map from MATH to MATH defined by any non-torsion MATH-dimensional cohomology class. The map may be assumed to have image in MATH, viewed as the skeleton of the classifying space. Being compact, the image of MATH lies in a finite piece of the telescope structure of MATH. Hence it can be retracted to the final stage, MATH, of the finite piece. Now MATH is just the space MATH, with some cells of dimension at most MATH attached. Hence MATH is a meromorphic extension of MATH. We thus obtain a MATH-meromorphic map MATH. The pullback REF completes the proof in REF cases mentioned in REF . In the general case MATH, we need a more delicate argument. We rely on the following property of maps of compact spaces into rationalisations. Let MATH be a CW complex admitting a telescope model (that is, there are self-maps which localize homology). Assume that a CW complex MATH is obtained from MATH by attaching cells of dimension at most MATH. Then the image of a compact space mapping into the rationalisation MATH of MATH may always be deformed into a subcomplex of MATH which is homotopy equivalent to a copy of MATH with cells of dimension at most MATH attached. Recall that MATH is obtained as the direct limit in the following construction: MATH . Here MATH is the identification of the MATH-th and MATH-th levels. A local CW complex is built inductively by attaching cones over the local sphere using maps of the local spheres into the lower `local skeletons'. For each cell of dimension MATH attached to MATH, we attach a corresponding local cell to MATH, which contains cells MATH of dimension MATH. Here the extra dimension is due to the presence of cylinders in REF . A map from a compact space MATH into MATH has image in a finite subcomplex MATH, which may be assumed to be of the following form: take a finite piece MATH of the infinite telescope MATH, and attach, to MATH, at most finitely many cells from among the MATH. Namely, MATH . From the homotopy equivalence MATH, we obtain the equivalence MATH, since each attaching map is an element of a homotopy group and hence a homotopy invariant, while homotopy groups commute with direct limits by a simple compactness argument. To complete the proof of REF , we argue as follows. Let MATH be the rationalisation of the space MATH of the lemma. By the universal property of a localisation CITE, the map from MATH to the local space MATH induced by MATH extends to a map MATH. Thus MATH may be thought of as the MATH-skeleton of MATH. Recall that MATH is obtained from MATH by adding cells of dimension at most MATH. If MATH is compact, by REF the image of the map can be deformed into a space MATH obtained from MATH by attaching cells of dimension at most MATH. The space MATH is thus a meromorphic extension (compare REF ) of MATH only if MATH. The case MATH is handled as above (following REF ), using the NAME fibration over MATH. Since we rely on an existence theorem for rationalisations, what we lose control of in this version of the proof is the exact form of the meromorphic extension of MATH, which admits a continuous map from MATH. |
math/0106153 | Recall that REF reduces the general case to that of the successive skeleta of the loop space MATH, arising from the cell decomposition of NAME, MATH with precisely one cell in each dimension divisible by MATH (see CITE). We now apply REF to the MATH-skeleton of MATH, to prove its MATH-systolic freedom modulo torsion, completing the proof of the Corollary. Our proof of REF can be simplified by using the telescope model for the localisation MATH of the loop space MATH. Such a model exists by CITE, since MATH admits self-maps defined by mapping a loop to its MATH-th iterate, which induce multiplication by MATH in all homology groups. Namely, let MATH be a finite MATH-dimensional CW complex, and let MATH be its MATH-th NAME number. A choice of a basis for a maximal lattice in MATH defines a map MATH. The composition of this map with the canonical map MATH, induces an isomorphism MATH. Now we argue as above. We use the compactness of MATH to construct a projection to the final level of a finite piece of the telescope. By the pullback REF , it remains to prove the MATH-systolic freedom of the MATH-skeleton of MATH. Next, we apply the carving up technique of CITE, to reduce the problem to the MATH-systolic freedom of the closures of the top-dimensional cells in the MATH-skeleton MATH of MATH. The cell decomposition of MATH induced by the NAME decomposition REF contains only cells of dimensions divisible by MATH. In particular, there are no cells of codimension REF. The absence of codimension REF cells in MATH is the crucial ingredient of the carving up technique. It allows us to isolate the different cell closures in MATH from each other. This is accomplished by inserting long cylinders based on the boundary, in MATH, of each top-dimensional cell. The absence of cells of codimension REF guarantees that these cylinders have positive codimension, and thus make no contribution to total MATH-dimensional volume. The effect of the long cylinders is to minimize the interaction between distinct top-dimensional cells, so that a MATH-cycle traveling from one to the other would have to pay a heavy price in terms of its MATH-volume. The formal argument, using coarea and isoperimetric inequalities, appears in CITE. Each of the top-dimensional cells in MATH is a product of skeleta MATH of MATH. Each MATH admits a meromorphic map to MATH by REF . Thus there exists a meromorphic map MATH . The MATH-systolic freedom of MATH now follows from the systolic freedom of products of spheres (compare REF ). |
math/0106157 | Let MATH and choose MATH such that MATH and MATH . Define MATH . We prove that MATH is independent of the choice of MATH. To see this, note that, since the cohomology of MATH is generated by classes of degree less than MATH, so is the quantum cohomology. This means that a quantum cohomology class MATH is zero if and only if MATH for every MATH and all MATH such that MATH for all MATH. Now suppose that the expression on the right of REF is nonzero. Then, by what we have just observed, there exist cohomology classes MATH of degrees less than MATH and a homology class MATH such that MATH . Since the homomorphism MATH is surjective (compare CITE), there exist classes MATH (of degrees less than MATH) such that MATH for every MATH. Hence, by Theorem A, MATH and hence MATH. This shows that MATH is well defined. The map MATH is obviously a ring homomorphism. The formula MATH follows immediately from Theorem A and the gluing REF for the NAME - NAME invariants. |
math/0106157 | Suppose, by contradiction, that there exists a sequence MATH such that MATH for every MATH and every MATH. By CITE there exists a subsequence, still denoted by MATH, such that the induced maps MATH converge in the MATH-topology to a smooth MATH-holomorphic curve. The limit curve represents the same homotopy class as the approximating curves and hence can be represented by a pair MATH. Since the sequence MATH converges to MATH in the MATH-topology, there exists a constant MATH such that, for every MATH, there exist a gauge transformation MATH and a section MATH such that MATH . The formulae MATH show that MATH converges to MATH in the MATH topology. This contradicts the choice of the sequence MATH and hence proves the lemma. |
math/0106157 | This is again a standard result for pseudoholomorphic curves and the proof is almost word by word the same as that of CITE. Here is a sketch. One argues by contradiction. If the result were false, there would be a sequence of smooth maps MATH that satisfies MATH but which does not satisfy the conclusion of the theorem for any constant MATH. This means that the MATH-distance of MATH to the space of MATH-holomorphic curves is not controlled uniformly by the MATH-norm of MATH. Now, by the NAME - NAME and NAME - NAME theorems, a suitable subsequence of MATH converges, strongly with respect to the sup-norm and weakly in MATH, to a MATH-holomorphic curve MATH. It follows from standard elliptic regularity for MATH-holomorphic curves that MATH then converges strongly with respect to the MATH-norm. To see this, write MATH and observe that MATH . Here the first inequality is the elliptic estimate for the NAME - NAME operator MATH and the second is the quadratic estimate for MATH. With this established it follows from REF and the implicit function theorem for the operator MATH that there exists a sequence of MATH-holomorphic curves MATH whose MATH-distance to MATH is bounded above by a fixed constant times the MATH-norm of MATH (see CITE). This shows that the sequence MATH does after all satisfy the conclusion of the theorem, in contradiction to our assumption. |
math/0106157 | We shall repeatedly use the identities MATH for MATH and MATH. To prove REF note that the triple MATH is given by MATH . The assertion now follows from the fact that MATH for MATH and MATH, and MATH for MATH. The first equation in REF follows from the fact that MATH for every MATH-form MATH on MATH (with values in any vector bundle). The second equation in REF follows from the first by duality. REF follows from the fact that the section MATH of the vector bundle over MATH with fibres MATH is MATH-equivariant. To prove REF note that the pair MATH is given by MATH . Here we have used the fact that MATH and hence MATH. The formula for the operator MATH follows by computing in local coordinates. |
math/0106157 | Let MATH. Then, by REF , the formula MATH is equivalent to MATH . Take the MATH-inner product of the first equation with MATH and of the second equation with MATH. The sum of the resulting identities gives MATH . Here all norms are MATH-norms and all inner products are MATH-inner products. Choose MATH so small that MATH for all MATH. Then MATH . Now the required estimates follow from the inequalities MATH . The first inequality follows from REF below. In the second inequality the constant MATH can be chosen as an upper bound for the norms of the linear maps MATH over all MATH. The third inequality is the MATH-estimate for the NAME - NAME operator and it follows from the NAME formula. The constant MATH is gauge invariant and depends continuously on the pair MATH with respect to the MATH-norm and hence can be chosen independent of MATH. This proves the proposition in the case MATH. |
math/0106157 | Multiply the metric on MATH by MATH. Then the MATH-norm of MATH with respect to the rescaled metric is equal to MATH times the MATH-norm of MATH, and the MATH-norm with respect to the rescaled metric is equal to the MATH-norm. Hence the estimates follows from the NAME embedding theorem for the rescaled metric. The constant is gauge invariant and it depends continuously on MATH (with respect to the MATH-norm) and MATH (with respect to the MATH-norm). By REF , the estimate holds with a uniform constant MATH. |
math/0106157 | We prove first that, for every vector field MATH, there exists a constant MATH such that MATH for MATH, MATH and MATH. Here the MATH-norm labelled by MATH is understood as the (gauge invariant) MATH-norm for MATH. To prove REF we choose local holomorphic coordinates MATH on MATH. Thus MATH, and MATH, MATH, MATH, and MATH are as in REF . Write MATH where MATH. Define MATH and MATH by MATH . Then MATH and hence MATH . Since MATH and MATH we have MATH . This proves REF for the local vector field MATH. For MATH the proof is analogous. Hence the result follows for any linear combination of these vector fields supported in the given coordinate chart, and hence for every vector field on MATH. For MATH there are similar inequalities. By REF , there exists a constant MATH such that MATH for MATH, MATH, and MATH. Now the operator MATH is given by MATH and hence MATH for MATH, MATH, and MATH. Since MATH the required estimates for the operator MATH follow from REF . The proof for the adjoint operator is analogous. |
math/0106157 | We compute in local coordinates. Let MATH. Then MATH . Here MATH and MATH, MATH, MATH, MATH, MATH, and MATH are as in REF . It suffices to prove the estimate for the local operators MATH and MATH. Let MATH be defined by REF with MATH replaced by MATH. Since MATH we obtain MATH . Here we have used REF . For the vector field MATH, multiplied by any cutoff function, the estimates REF follow from these three identities. The proof for MATH is similar, and so is the proof for the adjoint operator. |
math/0106157 | By elliptic regularity, there exists a constant MATH such that MATH for every MATH and every MATH. Hence MATH . Here MATH is the constant of REF and MATH is the constant of REF . With MATH we obtain MATH where MATH. REF for MATH now follows from REF . To prove REF for MATH we use REF again to obtain MATH where MATH. Now let MATH. By definition of the MATH-norm and REF , there exists a constant MATH such that MATH for MATH, MATH and MATH. Let MATH be the constant of REF . Then, by REF with MATH and REF , we have MATH . The last inequality follows again from REF with MATH. The estimate REF for MATH now follows by taking the sum over finitely many suitably chosen vector fields MATH. To prove REF for MATH we observe that MATH and choose MATH such that MATH for every MATH. Let MATH be the constant of REF . Then, by REF with MATH and REF , we have MATH . The last inequality follows from REF with MATH. |
math/0106157 | The proof has nine steps. CASE: Let MATH such that MATH. Then there exists a constant MATH such that MATH for every MATH, every MATH, and every MATH. These are standard estimates for elliptic pdes. The first estimate uses MATH regularity for the operator MATH, the NAME embedding MATH, and the NAME inequality. The second estimate uses MATH regularity for MATH and the fact that MATH is injective. CASE: There exists a constant MATH such that MATH for every MATH and every MATH. Let MATH be a MATH-orthonormal basis of MATH. Given MATH choose MATH such that MATH . Let MATH be the unique section such that MATH . Then MATH . CASE: There exists a constant MATH such that MATH for every MATH, every MATH and every MATH. For every MATH we have MATH . Here the last inequality follows from REF . Now, by REF , MATH . CASE: There exist positive constants MATH and MATH such that MATH for every MATH, every MATH and every MATH. We apply REF to both operators MATH and MATH. Then, by REF , MATH . The last inequality follows from REF . We prove REF for MATH. By REF , MATH for all MATH and MATH and MATH. With MATH we obtain REF for MATH. CASE: We prove REF for MATH. By REF , MATH . Here the last inequality follows from REF . There exist positive constants MATH and MATH such that MATH for every MATH, every MATH and every MATH. The first estimate follows immediately from REF . To prove the second estimate, recall from REF that MATH where MATH is the NAME - NAME operator defined by REF and MATH is a zeroth order operator REF . Hence MATH . By REF in the proof of REF , the commutator MATH is a first order operator in MATH. Hence there exists a constant MATH such that MATH . Moreover, by REF , MATH . The last inequality follows from REF . Now the commutator MATH is a second order operator in MATH and a zeroth order operator in MATH. Hence the assertion follows from the last two inequalities. CASE: We prove REF for MATH. Let MATH be the constant in REF for MATH and MATH be the constant of REF Then, for every MATH, we have MATH . The penultimate inequality follows from REF , and the last step from REF and the definition of the MATH-norm. Now REF for MATH follows by taking the sum over finitely many vector fields MATH. CASE: We prove REF for MATH. By REF , suppose that REF holds with MATH and MATH, choose MATH such that MATH for every MATH, and let MATH be the constant of REF . Then MATH . The last inequality follows from REF . |
math/0106157 | In local holomorphic coordinates MATH on MATH the map MATH is given by MATH where MATH, MATH and MATH. Suppose that MATH . The second derivatives of MATH and MATH satisfy the following pointwise estimates for suitable constants MATH and MATH (that is, MATH does not depend on the derivatives of MATH): MATH . The estimate REF is obvious and REF will be proved below. Now consider the identities MATH . To prove REF replace MATH by MATH in REF , insert the resulting inequalities in REF , and integrate over MATH. Moreover, to derive the second assertion in REF from the first we use the inequality MATH of REF . To prove REF replace MATH by MATH in REF , insert the resulting inequalities in REF , and integrate over MATH. To prove REF we give an explicit formula for the second derivative of MATH in local coordinates on MATH: MATH . Here MATH for MATH (see REF), and MATH, MATH, MATH and MATH are defined as follows. The section MATH is given by MATH . Now we use the following notation for MATH: MATH . The section MATH is linear in MATH and is defined by MATH . The section MATH is bilinear in MATH and MATH, and is defined by MATH . REF now follow by a term by term inspection of MATH, MATH, MATH, and MATH, assuming MATH. |
math/0106157 | REF follow immediately from REF . To prove REF we observe that in estimating the quadratic terms in MATH we encounter products of the following forms CASE: MATH and MATH. Here the MATH-norms of MATH and MATH can be estimated by MATH and the MATH-norm of MATH by MATH. CASE: MATH, MATH, MATH, and MATH. The MATH-norms of these products can be estimated by MATH . CASE: MATH, MATH, and MATH. In these cases the MATH-norm of MATH is bounded by MATH and the MATH-norms of MATH and MATH are bounded by MATH . Similarly, in estimating the quadratic terms in MATH we encounter products of the following forms CASE: MATH and MATH. The MATH-norms of these products can be estimated by MATH . CASE: MATH. Here the MATH-norm of MATH bounded by MATH and the MATH-norm of MATH is bounded by MATH . This proves REF . The proof of REF is similar. |
math/0106157 | REF . |
math/0106157 | The proof is similar to that of REF. However, in the present case the nonlinearities (in the quadratic estimates) appear in the highest order terms, and we establish estimates for the MATH-norms and not just the MATH-norms (as in CITE). We assume throughout that the exponential map at each point in MATH is defined in a ball of radius one. Abbreviate MATH and let MATH be defined by REF . Then MATH . Hence, by REF , there exists a constant MATH, depending only on MATH and MATH, such that MATH . We use NAME iteration to obtain a zero of MATH, and hence a solution of REF . Let MATH be the sequence defined recursively by MATH and MATH . We prove by induction over MATH that there exist positive constants MATH, depending only on MATH and MATH, such that MATH for MATH and MATH. The constants are chosen such that the linear estimates of REF hold for MATH with MATH, the quadratic estimates of REF hold for MATH with MATH, the MATH estimates of REF hold for MATH with MATH, and MATH . For MATH the estimates REF follow from REF . Namely, by REF with MATH, we have MATH . The estimate REF for MATH follows from the identity MATH and REF . Namely, since MATH we have MATH . Hence the hypotheses of REF are satisfied with MATH and MATH, and hence MATH . Since MATH this proves REF for MATH. Now assume that the sequences MATH and MATH have been constructed up to some integer MATH and that the estimates REF have been established for all integers up to MATH. Then, by REF , MATH and hence MATH . This shows that MATH lies in the domain of the exponential map at MATH for every MATH and so MATH lies in the domain of MATH. Let MATH and MATH be defined by REF . Then, by REF , MATH satisfies the estimate REF . To prove REF we observe that, by the induction hypothesis, MATH . To prove REF we observe that, by REF , MATH . Thus the hypotheses of REF are satisfied with MATH and MATH. Hence MATH . This completes the induction. By REF , the sequence MATH is NAME in the MATH-norm. Moreover, by examining the second component of MATH we find that MATH satisfies REF for every MATH and hence so does its limit MATH . By REF , this limit also satisfies REF with MATH. Moreover, by REF , the sequence MATH converges to zero in the MATH-norm and hence MATH. Hence MATH satisfies REF and it follows from elliptic regularity that MATH is smooth. This proves existence. We prove uniqueness. Suppose MATH satisfies REF , and MATH. Then, by REF , MATH . If MATH then MATH. This proves uniqueness. Since REF , and REF are gauge invariant, it follows that the map MATH is MATH-equivariant. |
math/0106157 | In this proof we drop the subscript MATH. Fix two pairs MATH and MATH that satisfy REF , and MATH . We prove that MATH satisfies REF , provided that MATH and MATH are sufficiently small. By ellipticity of the operator MATH, there exists a constant MATH such that MATH . Now let MATH be the constant of REF and MATH be the constant of REF . Then MATH where MATH. Hence MATH where MATH is the constant of REF . Since MATH we obtain MATH where MATH. Now we use the refined quadratic estimate of REF with MATH. By REF , we have MATH (provided that MATH). Thus the hypotheses of REF are satisfied with MATH and MATH replaced by MATH. Hence, by REF , MATH . Here the last two inequalities follow from REF . With MATH we have MATH where MATH. Since MATH satisfies REF we can apply REF (with MATH) to obtain MATH . By REF with MATH and REF , we now have MATH . Here we have used the fact that MATH. Moreover, the penultimate inequality follows from REF and the last inequality, with a suitable constant MATH, follows from REF . By REF with MATH, we have MATH . Here the penultimate inequality follows from REF and the last follows from REF . Combining these two estimates with REF we obtain MATH . If MATH and MATH are sufficiently small, we obtain MATH . Hence the result follows from the uniqueness argument at the end of the proof of REF . |
math/0106157 | For a fixed connection MATH and MATH the result is obvious. Choose MATH such that MATH for MATH, whenever MATH are sufficiently small in the MATH-norm and satisfy MATH. Here the MATH norms are understood with respect to the connection MATH. It follows that MATH and hence MATH for MATH. The other three inequalities follow by similar arguments. This proves the lemma for a fixed connection MATH. Moreover, the constant MATH depends continuously on MATH with respect to the MATH-norm, and is gauge invariant (with respect to the action of MATH on MATH by conjugation). Hence, by REF , it can be chosen independent of MATH as long as MATH for some MATH. |
math/0106157 | The operator MATH is given by MATH . By our standing hypotheses, MATH is compact and MATH is injective for every MATH. Hence there exists a constant MATH such that MATH for every MATH, every MATH, and every MATH. (Here MATH denotes the metric on MATH induced by MATH and MATH.) Hence the operator MATH is injective and hence, by elliptic regularity, it is bijective. Next we prove that there exists a constant MATH such that, for every pair MATH, every MATH, and every MATH, we have MATH . For a fixed connection MATH this follows directly from the interpolation inequality in CITE and the MATH-estimate for the operator MATH. Now the identity MATH shows that the constant in REF depends continuously on MATH with respect to the MATH-norm. Moreover, REF is gauge invariant. Hence it follows from REF (with MATH) and the NAME theorem that the estimate REF holds with a uniform constant MATH for all MATH such that MATH for some MATH. Using the identity MATH for MATH and integration by parts we obtain MATH . The last term on the right is negative. Now REF is equivalent to MATH . Hence, by the previous identity and REF , we have MATH . Therefore, by NAME 's inequality, MATH . Hence MATH and hence MATH . Thus we have proved the inequalities MATH . By REF , and REF , MATH . This proves the second estimate in REF . To prove the first estimate in REF we use a rescaling argument in local holomorphic coordinates on MATH. NAME MATH by finitely many open sets, each of which is holomorphically diffeomorphic to the unit square in MATH, suppose that the coordinate charts extend to a closed square of side length two, and choose trivializations of the bundle MATH over each of these (extended) open sets. In these coordinates we write the metric in the form MATH, and we write MATH, MATH. Moreover, MATH, MATH is a vector field along MATH, and MATH. In this notation REF has the form MATH where MATH and MATH . Now we introduce new functions, defined on the square MATH, by MATH . Then REF is equivalent to MATH where MATH and MATH . This equation can be written in the form MATH where MATH and MATH are given by MATH . Hence there exists a constant MATH such that, for all real numbers MATH such that MATH, we have MATH . Here the constant MATH is independent of MATH and MATH. It follows that MATH where the constant MATH depends on the metric and on the MATH-norms of MATH and MATH. With MATH, MATH, and MATH we obtain MATH . Hence, by taking the sum over the coordinate charts, MATH . Here MATH is the number of open sets in the cover and the constant MATH depends continuously on MATH with respect to the MATH-norm. Hence, by REF , MATH can be chosen independent of the pair MATH. Combining the last inequality with REF we obtain the first estimate in REF as claimed. |
math/0106157 | Throughout the proof we denote by MATH positive constants depending only on MATH and MATH (and not on the pair MATH). Fix a pair MATH and choose a positive constant MATH that is smaller than the injectivity radius of MATH on the compact set MATH. Suppose that MATH and MATH satisfy the hypotheses of REF with a sufficiently small constant MATH. Let MATH be the constant of REF . Then, by REF , MATH . Hence, if MATH is sufficiently small, it follows that the MATH-distance between MATH and MATH is smaller than MATH for every MATH. Hence there exists a unique smooth path MATH starting at MATH such that MATH where MATH . The endpoint MATH of this path obviously satisfies REF . We prove the inequalities MATH for MATH, where the constant MATH depends only on MATH and MATH. Then REF follow by integrating the function MATH over the interval MATH. For every MATH whose MATH-distance to MATH is less than MATH we define the linear operator MATH by MATH for MATH. Then MATH where MATH and MATH are as in REF . We prove the first inequality in REF . Since MATH we have MATH and we must estimate the three terms on the right with MATH. Since MATH it follows from REF , with MATH replaced by MATH, that MATH . Moreover, MATH . Hence, by REF , MATH and hence the first inequality in REF follows from REF . Next we prove the second inequality in REF . Using the identity MATH and REF we obtain MATH . Similarly, using the identity MATH and REF we obtain MATH . Hence, by REF , MATH . Moreover, since MATH, we have MATH . Here we have used the inequality MATH from REF . The constants MATH and MATH in the last two estimates depend continuously on the pair MATH with respect to the MATH-norm and are gauge invariant. Hence, by REF , they can be chosen independent of MATH. Hence the second inequality in REF follows from REF . To prove REF we observe that MATH where MATH and MATH are as in REF . The terms on the right have been estimated above and this proves REF . Thus we have proved the existence of MATH. REF with MATH sufficiently small guarantees that the MATH distance between MATH and MATH is smaller than the injectivity radius. This proves uniqueness. |
math/0106157 | The proof is based on a NAME type iteration. Let MATH . For MATH we define MATH inductively by MATH where MATH, and MATH is the unique solution of the equation MATH . To construct these sequences we must ensure that in each step MATH and MATH satisfy the hypotheses of REF so that MATH can be chosen as in the assertion of REF . We shall prove this below. And we shall also prove that these sequences satisfy the following estimates. MATH . The constants MATH and MATH are chosen as follows. Suppose that the constants MATH and MATH satisfy the following conditions. CASE: The injectivity radius of MATH on MATH is bigger than MATH. CASE: REF holds with MATH for every MATH. CASE: The assertion of REF holds for MATH with MATH replaced by MATH. CASE: The assertion of REF holds for MATH with MATH replaced by MATH and MATH replaced by MATH. CASE: The assertion of REF holds for MATH with MATH replaced by MATH and MATH replaced by MATH. CASE: The assertion of REF holds for MATH with MATH replaced by MATH. Now choose positive constants MATH and MATH such that MATH . We prove that the estimates REF hold for MATH. Since MATH, REF with MATH follows from REF . Since MATH, REF holds for MATH. REF is vacuous for MATH and REF is obvious. Now suppose that the sequences have been constructed and the inequalities REF have been established up to some integer MATH. Then MATH . Hence the hypotheses of REF are satisfied with MATH replaced by MATH and MATH replaced by MATH. Choose MATH as in the assertion of REF . By REF , we have MATH . Moreover, MATH, and hence, by REF , MATH . Since MATH, this proves REF with MATH replaced by MATH. Moreover, by REF , MATH . Since MATH, this proves REF with MATH replaced by MATH. Now let MATH be the unique solution of MATH . Then, by REF , MATH . Since MATH this implies REF with MATH replaced by MATH. It remains to prove REF with MATH replaced by MATH. By REF , we have MATH for MATH. Hence MATH . Since MATH this proves REF with MATH replaced by MATH. This completes the induction. By REF , MATH is a NAME sequence in the MATH-norm. Moreover, MATH where MATH. We prove by induction that there exists a sequence MATH such that MATH . For MATH we set MATH and MATH. Suppose that the sequence has been constructed for all integers up to MATH. Then MATH . Hence, by REF , and REF , MATH . By REF , there exists a section MATH such that MATH . Applying REF to MATH and MATH we find MATH . The last inequality follows from REF . This completes the induction. Thus we have proved that MATH satisfies REF and hence is a NAME sequence in MATH. Denote MATH . Then MATH . The last equation follows from REF . Moreover, by REF , we have MATH . Hence REF holds with MATH. To complete the existence proof we must show that MATH and MATH are smooth. We shall prove that the sequence MATH is bounded on MATH for every MATH. Here it suffices to obtain rather crude estimates with constants which depend on MATH and are allowed to diverge as MATH tends to zero. We fix a constant MATH and prove by induction that for every integer MATH there exists a constant MATH such that, for every MATH, MATH . For MATH this follows from REF . Now let MATH and assume that these estimates have been established with MATH replaced by MATH. Observe that there exists a constant MATH such that, for every MATH, MATH . The first two inequalities are obvious, and the last follows by inspecting REF in the proof of REF . Combining these inequalities with the induction hypothesis, we obtain MATH . Abbreviate MATH and choose MATH so large that MATH. Then MATH for all MATH and hence the sequence MATH is bounded. It follows that the sequences MATH and MATH are bounded. Thus we have proved that MATH and MATH satisfy REF . This completes the induction. It follows that MATH is smooth and hence, so is MATH. This proves existence. We prove uniqueness. Choose MATH so small that MATH . Assume that MATH and MATH satisfy the requirements of the proposition. Then MATH for MATH, where MATH. By REF , we have MATH for MATH. Hence, by REF , there exists a unique element MATH such that MATH . The gauge transformation MATH satisfies MATH . Moreover, MATH . Hence MATH and MATH satisfy the hypotheses of REF . We use REF and the estimate REF to obtain MATH . Since MATH we have MATH and hence MATH. Hence MATH and MATH. By REF , and REF , we have MATH for MATH. Hence MATH. |
math/0106157 | Let MATH where MATH and denote MATH, MATH, MATH. Let MATH and MATH be the constants of REF . Then MATH . Differentiating these identities we obtain MATH where MATH and MATH (see REF), and MATH where MATH is given by MATH (see REF ). Inserting the expressions for MATH and MATH in REF into REF gives MATH . Since MATH and MATH we have, by REF , MATH . Hence MATH . By REF , there exists a constant MATH such that MATH . The last inequality follows from REF . With MATH it follows that MATH . Hence REF follows from REF . |
math/0106157 | We denote MATH . Moreover, we drop the subscript MATH and write MATH. Then, in local holomorphic coordinates on MATH and a local frame of MATH, MATH is given by MATH . Here we use the notation of REF . Differentiating these formulae with respect to MATH we obtain MATH . Here MATH is defined by REF . The required estimates follow from these three identities via a term by term inspection. |
math/0106157 | Let MATH for MATH and MATH. Then MATH . Hence, by REF , there exist constants MATH (depending only on MATH) such that MATH . The last inequality follows from REF and the basic elliptic estimates for the operator MATH. Thus we have proved REF . To prove REF let MATH so that MATH . Then, by REF with MATH, MATH and hence, by REF (with MATH and MATH replaced by MATH), MATH . Now it follows from REF (with MATH) and REF (with MATH) that MATH . Hence MATH . This proves REF . |
math/0106157 | Abbreviate MATH. Consider the identity MATH . By REF , the MATH-norm of MATH is bounded above by a constant times MATH. By REF , the MATH-norm of MATH is bounded above by a constant times MATH. Hence the first term satisfies the required bound. For the second term the estimate follows from REF and the fact that the MATH-norm of MATH is bounded above by a constant times MATH. For the third term we use REF and for the last the estimate follows from REF . |
math/0106157 | Let MATH be an open set containing zero and MATH be the composition of the map MATH defined in REF with a NAME space isomorphism MATH. Then MATH for MATH; in particular, MATH . Now choose MATH smooth sections MATH so that MATH and MATH for MATH and MATH. Given MATH we abbreviate MATH . If MATH is sufficiently small then, by REF , this operator is surjective for every MATH. In this case we define MATH by MATH for MATH. By REF , these vectors form a basis of MATH for MATH sufficiently small. Now let MATH and MATH be the constants of REF . Choose MATH so small that MATH for MATH and MATH. Let MATH be the map of REF as introduced above. Define MATH by MATH where MATH denotes the MATH-inner product on MATH. Then MATH . We shall establish the existence of a zero of MATH with the inverse function theorem. We must prove that MATH on a ball of radius MATH and that MATH is less than MATH. To see this, we first observe that MATH . Here we have used the fact that the MATH-norm of MATH is less than or equal to the MATH-norm of MATH, that the MATH-norm of MATH is controlled by MATH, that the MATH-norm of MATH is controlled by its MATH-norm, and that, by REF , the latter is bounded above by MATH . Thus we have proved that MATH . Now let MATH be the unique smooth section defined by MATH for MATH sufficiently small. Then there exists a constant MATH such that MATH for MATH sufficiently small and MATH. Hence, by REF , we have that, for MATH and MATH, MATH . Moreover, there exists a constant MATH such that MATH for MATH sufficiently small. Now suppose that MATH and MATH have been chosen so small that the assertion of REF holds, with MATH replaced by MATH, for the paths MATH, MATH, MATH, and MATH. Then MATH for MATH and MATH. Thus the Jacobian MATH satisfies MATH . Choose MATH and MATH so small that MATH. Then MATH . Hence the inverse function theorem asserts that MATH whenever MATH. Now suppose that MATH. Then, by REF , we have MATH and hence we can apply the inverse function theorem with MATH. Then MATH contains zero and, by the inverse function theorem, there exists a point MATH such that MATH . The last inequality follows from REF . Now define MATH . Then REF are satisfied by definition. The estimate REF follows from REF : MATH . Moreover, the vector MATH in the assertion of REF is the image of MATH under our NAME space isomorphism MATH. Hence, by elliptic regularity for the NAME - NAME operator, its MATH-norm is bounded by MATH and hence by the MATH-norm of MATH. |
math/0106157 | Let MATH and MATH be given. Choose positive constants MATH, MATH, MATH, and MATH such that REF holds with MATH replaced by MATH and MATH replaced by MATH, REF holds with MATH replaced by MATH, and both results hold for MATH. Now choose MATH so small that MATH . Let MATH and suppose that MATH and MATH satisfy the hypotheses of Theorem B, namely MATH and MATH where MATH satisfies MATH . By REF , there exist MATH and MATH satisfying REF , and REF , with MATH replaced by MATH. Hence MATH . This estimate together with REF shows that MATH and MATH satisfy the hypotheses of REF . Hence, by REF , MATH . Moreover, again by REF , MATH and, by REF , MATH . This proves the theorem. |
math/0106157 | Let MATH denote the standard Laplacian. For MATH denote MATH, MATH, and MATH . By REF , we have MATH and hence MATH . Here MATH is as in REF . Thus MATH . Now choose a constant MATH such that MATH for all MATH and MATH. Then MATH and hence MATH where MATH and MATH. Fix a constant MATH such that MATH . Then, by REF below, we have MATH for every MATH. Applying REF again we obtain MATH for every MATH, where MATH . The MATH-estimate now follows by taking the sum over finitely many balls of radius MATH that cover the compact set MATH. Moreover, by REF , the function MATH is subharmonic in MATH for every MATH. Hence, by the mean value inequality, MATH for MATH and MATH. Assume MATH. Then we can choose MATH and this proves MATH estimate. |
math/0106157 | For MATH we have MATH and hence MATH . Integrate this inequality over the interval MATH to obtain MATH for MATH. The first inequality follows by integrating this inequality over the interval MATH. The second inequality was proved in CITE. |
math/0106157 | Consider the functions MATH given by MATH . We prove that there exists a constant MATH such that MATH . To see this, recall from the proof of REF that MATH and hence MATH . Moreover, by REF , MATH . Hence MATH . Combining this with the formula for MATH we obtain MATH . The first row on the right is bounded below by MATH. Moreover, by assumption, the image of MATH is contained in the compact set MATH. Hence the last six terms can be estimated from below by MATH for some constant MATH whenever MATH is sufficiently small. Thus we have proved REF . Hence, by REF , there exist constants MATH and MATH such that MATH for every MATH. Since MATH and zero is a regular value of MATH there is an inequality MATH whenever MATH is sufficiently small. Thus we have proved REF for MATH as well as MATH . Now let us define MATH by MATH . We shall prove that there exist positive constants MATH, MATH, and MATH such that MATH for MATH. We consider the equation MATH and use the formula MATH where MATH . Here we abbreviate MATH. The last equality for MATH follows from REF . Now consider the tensors MATH and MATH defined by MATH for MATH and MATH. Then MATH . Hence, by a term by term inspection, we obtain an inequality MATH for MATH sufficiently small. Note, in particular, that the term MATH contains the two positive summands MATH and MATH. Since MATH the last inequality implies REF with MATH. Now it follows from REF , and the formula MATH that REF holds for MATH. The estimate REF for MATH follows by interpolation. |
math/0106157 | Suppose the assertion is false. Then there exist a constant MATH and sequences MATH and MATH such that MATH . Here MATH is chosen smaller than the number MATH required for the definition of the map MATH. We prove in four steps that there exist an integer MATH, positive constants MATH and MATH, and sequences MATH such that MATH for every MATH. For MATH sufficiently large it then follows from REF that MATH in contradiction to our assumption. CASE: There exist constants MATH and MATH such that MATH for MATH and MATH. By the graph construction in REF, it suffices to establish the estimate under the hypothesis that MATH is independent of MATH. Namely, MATH where MATH, MATH is the connection induced by MATH on MATH, and MATH. Hence we can use the results of REF. Since MATH and MATH, the pair MATH satisfies the hypotheses of REF and so the sequence MATH is uniformly bounded. Hence there exists a constant MATH such that MATH for every MATH. This implies that, in local holomorphic coordinates on MATH, the pair MATH satisfies the hypotheses of REF for MATH sufficiently large. Hence the estimate holds in local holomorphic coordinates on MATH with MATH replaced by MATH. Hence, by a partition of unity argument, it holds globally. CASE: There exists an integer MATH and a constant MATH such that, for every MATH, there exists a unique MATH such that MATH . Define MATH and MATH by MATH so that MATH, and let MATH. Then there exists a constant MATH such that MATH for every MATH. The existence of MATH for large MATH follows from the implicit function theorem for the map MATH. This sequence satisfies an estimate of the form MATH for every MATH and every MATH. Here the constants MATH and MATH are independent of MATH and MATH, and the second inequality follows from REF . For MATH there is actually a better estimate (by MATH instead of MATH), but we shall not use this here. In the following we suppress the subscript MATH and write MATH instead of MATH, respectively. We establish the required estimates in local holomorphic coordinates on MATH. As in REF , we write MATH for some NAME algebra valued functiona MATH and MATH, and denote MATH . Then MATH . We assume that the functions MATH are defined on an open set MATH and fix any compact subset MATH. We must prove the estimates MATH on the subset MATH, where MATH . Abbreviate MATH, MATH. Then MATH . Hence, by REF , MATH where MATH . Hence MATH . Since MATH we have MATH and, since MATH, MATH . It follows that MATH . If MATH is sufficiently small this gives MATH . Here the second inequality follows from a similar argument as the first. Combining these inequalities with REF we obtain MATH . In order to estimate MATH we apply the operator MATH to REF and use the formula MATH to obtain MATH . Combining this with REF and the estimate for MATH we obtain MATH . Hence, by REF , MATH . Similarly, MATH and MATH . Now use REF again to obtain MATH . The second equality uses the fact that MATH and that REF-form MATH takes values in MATH. It follows that MATH . It remains to show that MATH . To estimate the term MATH differentiate REF with respect to MATH. Then apply the operator MATH to the resulting expression to eliminate MATH and obtain an estimate of the form MATH . Then apply the operator MATH to the equation obtained from differentiating REF , and estimate MATH using the upper bound found for MATH. The estimate for MATH is obtained in a similar manner. To estimate MATH and MATH, we begin with the identity MATH instead of REF and then follow the same procedure. CASE: There exist an integer MATH, a constant MATH, and a sequence MATH such that MATH and MATH for MATH. By REF , MATH . Hence the induced maps MATH form a sequence of approximate MATH-holomorphic curves which satisfy a uniform MATH-bound on their first derivatives. Hence, by MATH and REF , there is nearby a true MATH-holomorphic curve MATH whose MATH-distance to MATH is controlled by the MATH-norm of MATH. Now this MATH-holomorphic curve has a unique lift MATH of the form MATH . Let MATH be the connection determined by MATH via MATH. Then MATH . Here the last inequality follows from REF . Since MATH we obtain MATH . In particular, these inequalities together give a uniform MATH-bound on the MATH-holomorphic curves MATH. Hence, by the elliptic bootstrapping techniques for MATH-holomorphic curves, the sequence MATH satisfies a uniform MATH bound on the first derivatives. This proves REF . Unfortunately, the estimate on MATH in REF is only in the MATH-norm and not in the MATH-norm. A further modification of the pair MATH is required to improve this estimate. CASE: There exist an integer MATH, a constant MATH, and a sequence of gauge transformations MATH such that the sequence MATH satisfies the following. For MATH the original sequence MATH has the form MATH where MATH satisfies REF . The idea is to choose MATH for large MATH such that MATH . This can be done by using pointwise, for every MATH, the implicit function theorem to obtain the local slice condition. This suffices to obtain the missing estimates for the first derivatives of MATH. We sketch a proof of this estimate below. By REF , the distance between MATH and MATH is unformly bounded by a constant times MATH while the distance in the MATH-norm is bounded by a constant times MATH. By REF , the distance between MATH and MATH is bounded in the MATH-norm by a constant times MATH. Hence there exists a sequence of smooth sections MATH and a constant MATH such that MATH . Moreover, the sequence MATH is uniformly bounded in the MATH-norm and MATH . Here the last inequality follows from the identity MATH which in turn follows from REF . Now, by the inverse function theorem for the map MATH there exists a constant MATH and (unique) sequences MATH and MATH such that MATH and MATH . Define MATH . We shall prove that the pair MATH satisfies REF . To see this note first that MATH . The endomorphism MATH of MATH is MATH-close to the identity, MATH, and MATH . Hence, by REF , there is an estimate MATH for all MATH. Hence, by REF , there exists a constant MATH such that MATH . Next observe that, by REF , MATH . Hence there exists a constant MATH such that MATH for all MATH. Thus we have proved that MATH for all MATH. It remains to estimate the MATH-norm of the first derivatives of MATH. For this we shall drop the subscript MATH and write MATH instead of MATH. Moreover, we use local coordinates on MATH as in REF and write MATH and MATH and MATH . Consider the formula MATH . By REF , we have MATH and hence, by elliptic regularity for the NAME operator, MATH . Moreover, since MATH, it follows from REF that MATH and hence MATH . Here we use the fact that, by elliptic bootstrapping for MATH-holomorphic curves, there is a uniform MATH-bound on MATH. Now consider the pointwise inequality MATH . Since the operator MATH is small, we obtain MATH . Now use REF and the estimates REF , and REF to obtain MATH . The terms MATH, MATH, and MATH are estimated similarly. This proves REF . It follows from REF that MATH for some constant MATH and MATH sufficiently large. This contradicts our assumption and hence proves the theorem. |
math/0106157 | Fix a MATH-invariant and MATH-compatible almost complex structure MATH. If MATH is sufficiently small then there exist unique loops MATH and MATH such that MATH . By REF , we have MATH . Since the image of MATH is the orthogonal complement of the kernel of MATH we deduce that there exists a constant MATH such that MATH pointwise for every MATH. Define MATH and MATH by MATH . Then MATH . Hence MATH and so MATH . This implies that there exists a path MATH such that MATH . Hence MATH is a loop and MATH . Moreover, with MATH, we obtain MATH . This proves the lemma. |
math/0106157 | Let MATH and MATH be as above. Then the local equivariant symplectic action can be expressed in the form MATH . By REF , we have the pointwise inequality MATH . Moreover, by REF , MATH where MATH for MATH. With MATH we obtain, by REF , MATH and this implies MATH . Hence MATH . This proves the lemma. |
math/0106157 | Let MATH be as in MATH and let MATH be a finite energy solution of REF in radial gauge. We prove in seven steps that MATH has the properties asserted in the proposition. CASE: MATH uniformly in MATH. Abbreviate MATH and MATH as in REF. Let MATH . Then REF with MATH has the form MATH . Since MATH is contained in a compact subset of MATH this gives an inequality MATH . Namely, choose MATH such that MATH is injective whenever MATH. Then the first term in the secomd row can be estimated from below by MATH whenever MATH. In case MATH we can use the inequalities MATH and MATH. Now it follows from CITE that there is a constant MATH such that MATH . With MATH this implies MATH . CASE: For MATH sufficiently large, we have MATH where MATH and MATH . The energy identity on MATH has the form MATH . For MATH sufficiently large denote by MATH the function used in the definition of the local symplectic action of MATH. Then MATH and MATH for some point MATH and some loop MATH. The homotopy class of the connected sum MATH (the orientation of MATH is reversed) is independent of MATH. Hence the number MATH is independent of MATH. Since MATH tends to zero as MATH it follows that MATH for every sufficiently large number MATH. This proves REF . MATH. Suppose, by contradiction, that MATH. Then there exists a regular value MATH of MATH such that MATH . Hence the set MATH is a smooth submanifold of MATH with boundary. Since MATH it follows from REF that there exists a number MATH such that MATH . Hence MATH is compact and has a nonempty boundary. By MATH, MATH in MATH (see CITE). Hence MATH . This contradiction proves REF . Consider REF in polar coordinates MATH. Define MATH and MATH by MATH . Then MATH, MATH, and REF is equivalent to MATH where MATH . The radial gauge condition has the form MATH for large MATH. The energy of the triple MATH is given by MATH . CASE: There exist positive constants MATH and MATH such that, for every MATH, MATH . By REF , we have MATH for some constant MATH and every sufficiently large real number MATH. Hence there exists a real number MATH such that MATH . CASE: There exist positive constants MATH and MATH such that, for every MATH, MATH . The MATH estimate for MATH and MATH follows from REF . The MATH-estimate for MATH follows from REF . There exists a MATH-function MATH and a MATH-function MATH such that MATH and MATH. Moreover, MATH, and if MATH holds then MATH. By NAME 's inequality and the radial gauge assumption, we have, for MATH, MATH . Hence the existence of the MATH-limit of MATH follows from REF . That MATH converges uniformly as MATH tends to infinity follows from the exponential decay of MATH in REF . That the limit is a MATH-function and satisfies MATH follows from the fact that MATH converges (exponentially) to zero as MATH tends to infinity. That MATH follows from the energy identity in the proof of REF and the MATH-convergence of MATH. That MATH is an integer multiple of MATH (when MATH holds) follows from the proof of REF . |
math/0106157 | NAME MATH by finitely many distinct balls MATH, MATH, and choose points MATH such that MATH. Then, for every MATH, there exists a MATH such that MATH. Thus the open set MATH is contained in the finite union of the following open sets MATH, MATH. Choose MATH so small that MATH extends to an equivariant function (still denoted by MATH) from MATH to MATH and define MATH . For every smooth path MATH and any two points MATH and MATH the holonomy MATH of the connection MATH is defined by MATH where MATH is the unique horizontal lift of MATH with MATH. It satisfies MATH for MATH and MATH. Hence the map MATH is MATH-invariant and MATH-equivariant. Choose a finite sequence of smooth functions MATH such that MATH . Then the functions MATH and MATH, defined by MATH for MATH, satisfy REF . Now choose a MATH-invariant smooth function MATH such that MATH for MATH and MATH for MATH. Define MATH and MATH by MATH for MATH. Then MATH is smooth with respect to the MATH . NAME manifold structure on (the completion of) MATH. Moreover, MATH and MATH whenever MATH. Now choose a finite sequence of smooth functions MATH, MATH, such that MATH and MATH . Then the function MATH, defined by MATH is the required classifying map. |
math/0106157 | Suppose the result is false. Then there exist a constant MATH and sequences MATH such that MATH for every MATH. This means that, for every MATH, there exists a sequence MATH such that MATH . This contradicts the bubbling argument in REF of the proof of Theorem A below. |
math/0106157 | Let MATH be the smooth path defined by MATH . Then, by REF , and REF, we have MATH . The last inequality follows as in REF in the proof of REF . Since MATH the lemma is proved. |
math/0106157 | Let MATH be as in REF and MATH be as in REF so that MATH . Let MATH be defined by REF . Then MATH and hence MATH . Differentiating this identity with respect to MATH we find MATH . (See REF .) Hence, by REF , MATH . Since MATH this shows that MATH is an orientation preserving embedding. Indeed, it follows that the restriction of MATH to every ball of radius MATH is an embedding for MATH and MATH sufficiently small and hence, by REF , MATH is an embedding for MATH sufficiently small. For MATH denote MATH and consider the map MATH defined by MATH . Then it follows from REF and the inequality MATH of REF that MATH . For MATH sufficiently small we have MATH, where MATH is given by MATH. This proves the proposition. |
math/0106157 | The result is obvious when MATH. Moreover, both moduli spaces are empty when MATH is a nonzero torsion class. Hence assume that MATH is a nontorsion homology class, denote by MATH the corresponding equivariant homology class, fix a compact NAME surface MATH, and let MATH be a principal MATH-bundle whose characteristic class MATH is the pushforward of MATH. In the course of the proof it will be necessary to also consider other bundles MATH with corresponding equivariant homology clsses MATH. By MATH, there exists a constant MATH such that MATH for every solution MATH of REF over any NAME surface. Note that MATH can be chosen to be a regular value of the function MATH. Let MATH such that MATH for every MATH and let MATH and MATH be as in REF . Fix MATH points MATH such that the points MATH are pairwise distinct. Choose an integer MATH, a MATH-equivariant smooth map MATH, and MATH smooth classifying maps MATH defined by MATH, where MATH is as in REF . Then MATH and MATH for MATH. For MATH consider the evaluation maps MATH given by MATH . Let MATH denote the product map defined by MATH . For any subset MATH such that MATH and any class MATH that descends to MATH we consider the evaluation map MATH given by MATH . Now fix equivariant cohomology classes MATH such that MATH . There is a natural embedding MATH and we denote by MATH the pullback of MATH under this embedding. Note that MATH is a compact manifold with boundary. Replacing MATH by some integer multiple of MATH, if necessary, we may assume without loss of generality that, for every MATH, there exists a compact oriented manifold with boundary MATH of dimension MATH and a smooth map MATH such that the homology class in MATH represented by MATH is NAME dual to MATH. For MATH such that MATH we denote the corresponding product map by MATH . For MATH we abbreviate MATH and MATH . The functions MATH can be chosen such that the following holds. CASE: MATH is transverse to MATH for every MATH and MATH is transverse to MATH for every subset MATH and every equivariant homology class MATH. Now the notation has been set up and we shall prove Theorem A in five steps. For MATH and MATH consider the set MATH . CASE: The map MATH is a pseudo-cycle. The map MATH is the composition MATH, where the evaluation map MATH is given by MATH and the embedding MATH is given by MATH . That MATH is a pseudo-cycle was proven in CITE. Hence MATH is a pseudo-cycle. (see CITE for the definitions). CASE: MATH is a finite set and the number of elements of MATH, counted with appropriate signs, is the NAME - NAME invariant: MATH . Here the function MATH denotes the intersection index of the maps MATH and MATH. Consider the functions MATH given by MATH . Since MATH is transverse to MATH, MATH is a smooth submanifold of MATH and MATH is dual to the cohomology class MATH obtained from MATH by pullback under the obvious inclusion MATH. The class MATH agrees with the image of the class MATH under the homomorphism MATH . Hence another representative of the class MATH can be obtained as follows. Let MATH be a smooth function, defined on a compact manifold MATH that is dual to MATH (replace MATH by an integer multiple of MATH, if necessary). Lift MATH to a MATH-equivariant map MATH, defined on the total space of a principal MATH-bundle MATH, and consider the induced map MATH . It is homologous to MATH. Let MATH and MATH. Then MATH . The first equality follows from the definition of MATH, the second from the fact that MATH is a pseudo-cycle REF and MATH is homologous to MATH, and the third equality follows from the definition of the NAME - NAME invariants (see CITE for example). CASE: The invariant MATH can be expressed as the intersection number MATH for MATH sufficiently small. The map MATH is dual to the class MATH, where MATH denotes the projection onto the MATH-th factor. Moreover, MATH. Hence MATH . Here MATH denotes the obvious inclusion. The last equality follows from the fact that MATH is homotopic to the evaluation map MATH in the definition of MATH. CASE: For MATH sufficiently small there is an injective map MATH such that MATH satisfies MATH . Here MATH denotes the intersection index of the maps MATH and MATH (in the transverse case). Choose MATH such that MATH and consider the map MATH . By REF , this map converges to MATH in the MATH-topology as MATH tends to zero. By MATH the map MATH is transverse to the diagonal MATH. Hence MATH is transverse to MATH for MATH sufficiently small. Moreover, by REF , the image of MATH under MATH is MATH-close to MATH. Hence, by the implicit function theorem, there is, for MATH sufficiently small, a unique injective map MATH such that the distance between each point and its image is bounded above by a constant times MATH. Composing this map with MATH we obtain the required map MATH. By REF , the map MATH identifies the two intersection indices. CASE: Assume MATH. Then there exists a constant MATH such that the map MATH of REF is surjective for MATH. Suppose, by contradiction, that there exist sequences MATH and MATH such that MATH . Consider the sequence MATH . We prove that MATH diverges to MATH. Assume otherwise that MATH is bounded. Then, by Theorem D, there exists a constant MATH such that MATH belongs to the image of the map MATH for MATH sufficiently large. Write MATH . Since MATH is compact we may assume that the limit MATH exists. Moreover, since MATH is compact, we may assume, by passing to a further subsequence if necessary, that the limit MATH exists. Since MATH converges to MATH in the MATH-topology, and MATH we deduce that MATH and, for MATH sufficiently large, MATH . The last assertion follows from the uniqueness part of the implicit function theorem used in the definition of the maps MATH. This contradicts our assumption. Thus we have proved that MATH diverges to MATH as claimed. Now choose a sequence MATH such that MATH . Passing to a subsequence, if necessary, we may assume that MATH converges. Denote MATH . Moreover, by applying NAME 's trick (see CITE for example) we may assume that MATH . We distinguish three cases. CASE: MATH. CASE: There exists a MATH such that MATH for all MATH. CASE: MATH. We shall prove that in REF a nonconstant MATH-holomorphic sphere in MATH bubbles off at the point MATH, in REF a nontrivial solution of the vortex REF bubbles off, and in REF a nonconstant MATH-holomorphic sphere in MATH bubbles off. To see this, we choose a local holomorphic coordinate chart MATH on MATH that maps MATH to zero, identifies a neighbourhood of MATH with the ball MATH, and identifies the volume form MATH with the form MATH, where MATH. Moreover, we choose a local frame of the bundle MATH along this coordinate chart. We use the notation of REF . Then the sequences MATH and MATH satisfy MATH . Moreover, there is a sequence MATH such that MATH . Let us define MATH and MATH by MATH and MATH and MATH by MATH . Then MATH converges to MATH in the MATH-topology and MATH converges to MATH in the MATH-topology. Moreover, MATH . CASE: Suppose that MATH diverges to infinity. Then, by REF , the curvature MATH converges uniformly to zero. Hence, by NAME 's weak compactness theorem CITE, we may assume that MATH and MATH converge in the sup-norm and weakly in MATH. This implies that the sequence MATH is bounded in MATH. Hence, by the usual elliptic bootstrapping argument for pseudoholomorphic curves, it is bounded in MATH (the lower order terms in the equation have the form MATH and hence are bounded in MATH). Hence there exists a subsequence, still denoted by MATH, that converges strongly in MATH to a MATH-holomorphic curve MATH with finite energy. Since the sequence MATH is bounded it follows that MATH and hence MATH extends to a nonconstant holomorphic sphere in MATH. This contradicts MATH. CASE: Suppose that the sequence MATH is bounded and does not converge to zero. Let us assume, without loss of generality, that MATH . Then we can use the compactness result of CITE to deduce that, after a suitable gauge transformation and after passing to a further subsequence, the triple MATH converges to a solution MATH of the vortex REF with finite energy. Moreover, MATH and hence the energy is nonzero. Hence, by REF , we have MATH . CASE: Suppose that MATH . Then, by REF , MATH for every compact set MATH. It follows that the sequence MATH is uniformly bounded in MATH. Hence, by NAME 's weak compactness theorem, we may assume that MATH and MATH converge weakly in MATH and strongly in MATH, on every compact subset of MATH. Here MATH is any fixed real number, say MATH. Hence, the sequence MATH is bounded in MATH. Now it follows again from the elliptic bootstrapping analysis for pseudoholomorphic curves that MATH is bounded in MATH and hence has a subsequence that converges strongly in MATH on every compact subset of MATH. The limit MATH is a finite energy solution of REF on MATH. This solution represents a MATH-holomorphic sphere in the quotient MATH. Moreover, since MATH it follows that MATH and hence the resulting holomorphic sphere in MATH is nonconstant. Hence MATH . Thus we have proved in all three cases that MATH for every MATH. This shows that, after passing to a suitable subsequence, bubbling can only take place at finitely many points MATH. On every compact subset of MATH the sequence MATH is uniformly bounded. (As an aside: this is used in the proof of REF .) Hence it follows as in REF , that a suitable subsequence in a suitable gauge converges on this complement to a finite energy solution of REF . The limit MATH descends to a holomorphic curve MATH with finite energy. Hence, by the removable singularity theorem for MATH-holomorphic curves, it extends to a holomorphic curve on all of MATH, still denoted by MATH. The energy of this MATH-holomorphic curve satisfies MATH . By REF , the dimension of the moduli space reduces by at least MATH at each bubble. Thus the limit MATH belongs to a moduli space MATH of dimension MATH . If MATH then the limit curve MATH still satisfies MATH for every MATH and hence cannot exist, by the transversality condition MATH. In general, denote MATH . Then the limit MATH satisfies MATH . Since the points MATH are pairwise distinct we have MATH and so MATH . Here we have used the fact that MATH for each MATH. It follows again from MATH that such a limit curve cannot exist. Hence our assumption that the map MATH were not surjective for every MATH must have been wrong. This proves the theorem. |
math/0106157 | By definition of MATH we have MATH. Alternatively, we can compute in local holomorphic coordinates MATH on MATH. In such coordinates the Hamiltonian perturbation, the connection MATH, and the volume form on MATH have the form MATH and REF have the form MATH . Moreover, the almost complex structure MATH is given by MATH . This proves the lemma. |
math/0106157 | Continue the notation of the proof of REF . Then the curvature MATH is given by MATH where MATH denotes the NAME bracket on MATH, and MATH where MATH and MATH denote the differential on MATH. Abbreviate MATH and MATH for MATH. Then MATH . The last identity follows from the fact that MATH and MATH. |
math/0106157 | Since MATH is MATH-invariant we have MATH for every MATH. This formula can be expressed in the form MATH . Using this formula we obtain MATH as claimed. |
math/0106157 | We shall use the identity MATH . By REF , we have MATH for MATH and MATH. Here the penultimate equality follows from the fact that MATH is a gradient vector field and that MATH is skew-adjoint. This shows that MATH is a MATH-form on MATH with values in the bundle of skew-Hermitian endomorphisms of MATH, and so is MATH. Moreover, since MATH the operator MATH is a Riemannian connection. By REF , this shows that MATH is a Riemannian connection. It follows directly from the definition that MATH preserves the complex structure MATH. |
math/0106157 | Since the metric MATH is MATH-invariant for every MATH we have MATH . Hence the first identity follows from the fact that MATH and that MATH . The second identity follows from the first and the fact that MATH is MATH-invariant for every MATH. |
math/0106157 | Given a map MATH, a vector field MATH along MATH, and a MATH-connection MATH where MATH, we denote MATH . Then the assertion can be restated in the form MATH . To prove this we use the identities MATH . The first and second identities are the definition of the connection MATH and the curvature tensor MATH. The other identities use the equations MATH for every MATH-invariant vector field MATH and all MATH. |
math/0106157 | By definition of MATH, we have MATH . Hence the formula for MATH follows from the relation between the NAME tensor and MATH. Now this equation shows that the operator MATH is complex anti-linear. |
math/0106157 | Since the group action preserves geodesics, MATH . Differentiate this identity with respect to MATH to obtain the first identity. To prove the second differentiate the first identity covariantly and use the definition of MATH and MATH. For more details see CITE. |
math/0106157 | See CITE. |
math/0106157 | We compute MATH . Here the first equation follows from REF , the second from the definition of MATH, and the last from the definitions of MATH and MATH. |
math/0106157 | REF . |
math/0106157 | We only prove the first identity. The proof of the second is similar. By REF , MATH . Hence, by the definition of MATH and REF , MATH and MATH . Inserting these two identities into the previous formula we obtain MATH . Now the result follows from REF . |
math/0106158 | By general nonsense from CITE, we need only show that MATH takes relations REF described above to weak equivalences of pro-spaces. NAME replacements are no problem because the targets and sources of every map in question are already projective cofibrant. To show that MATH is projective cofibrant for every hypercover MATH, use REF to conclude that MATH is a split simplicial scheme. For relations of type REF , note that MATH commutes with coproducts of schemes CITE. For relations of type REF , see REF . |
math/0106158 | The first claim follows from the formal machinery of NAME adjoint functors CITE. The last claim follows from general nonsense and the fact that every representable presheaf is projective cofibrant. |
math/0106158 | The proof is entirely analogous to the proof of the main theorem of CITE. We colocalize with respect to the objects MATH for all MATH. More precisely, a pro-map MATH is a weak equivalence if the induced map MATH is a weak equivalence of simplicial sets for every MATH. NAME have sufficiently good properties that this kind of colocalization always exists CITE. |
math/0106158 | The argument is basically the same as in the proof of REF . The only significantly different part is in showing that MATH is a MATH-cohomological weak equivalence for every scheme MATH in MATH. We need to show that this map induces an isomorphism in cohomology with coefficients in MATH. In order to understand these cohomology maps, CITE allows us to consider the map on étale cohomology induced by the projection MATH . The projection induces an isomorphism in étale cohomology by CITE. |
math/0106158 | By REF , it suffices to show that the square MATH is a homotopy pushout square. Let MATH be the homotopy pushout of the diagram MATH . From the paragraph preceding this theorem, we know that MATH is a local weak equivalence of presheaves. The functor MATH preserves weak equivalences by REF , so MATH is also a weak equivalence. Left derived functors commute with homotopy colimits, so the homotopy pushout of the diagram MATH is weakly equivalent to MATH. |
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