paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0009236 | To check that the first pairing is well defined, let MATH be a coboundary in MATH. Then MATH is also a coboundary and we see that MATH since MATH is MATH-invariant and therefore MATH. Let MATH in MATH, where MATH and MATH. Then, by REF , MATH, so there exists a MATH-invariant invertible element MATH such that MATH. Then, by REF , we have MATH . Also, since f is equivariant, MATH. This finishes the proof of the first part. The proof of the second part is also similar to the non equivariant case as in CITE and is left to the reader. |
math/0009236 | We prove that the following maps define MATH-isomorphisms between MATH and MATH inverse to each other: MATH where MATH is the MATH-matrix of MATH. Since MATH and MATH and MATH therefore MATH, where MATH. Since MATH where MATH, we see that MATH is an algebra map. Also since MATH and MATH, we conclude that MATH which shows that MATH preserves the MATH-actions. |
math/0009236 | We prove that the following maps define a MATH-isomorphism between MATH and MATH inverse to each other: MATH . Since MATH and MATH we obtain MATH. Now since MATH is an algebra map. To check MATH preserves the MATH-actions, we see that MATH . |
math/0009239 | The sufficiency of the condition follows from the fact that MATH is a NAME subalgebra when MATH is graded. In order to check the necessity of the condition, notice that MATH gives a NAME system allowing to compute the homogeneous components of a vector field MATH. |
math/0009239 | Using NAME identity, we check that MATH and consequently MATH by induction on MATH. By definition, MATH for all MATH. Then, we check, by induction on MATH, that MATH. Therefore, MATH is a NAME subalgebra. It is trivially the smallest one to contain the subspaces MATH, MATH and MATH. |
math/0009239 | It is obvious that MATH. Furthermore, if MATH or MATH, MATH . Now, if MATH, then, for every polynomial function MATH, the field MATH belongs to MATH. |
math/0009239 | Notice that MATH cannot be made only of constant and linear vector fields. Indeed, it would then be included in the maximal subalgebra REF presented in the introduction, for instance. Therefore, MATH for some MATH. The conclusion follows from REF . |
math/0009239 | Let MATH be a stable subspace of MATH under the action of MATH. The space MATH satisfies the hypotheses of REF . Its algebraic closure is an infinite dimensional proper subalgebra containg MATH properly, hence a contradiction. Let now MATH be a non trivial ideal of MATH. It contains at least one constant vector field since MATH. It contains all of them since MATH is a stable subspace of MATH. |
math/0009239 | Indeed, MATH. The requirement for MATH to intertwine the action of MATH on MATH precisely means that MATH. If MATH and MATH, the equalities MATH show that MATH. The inclusions are strict because the dimension of MATH is infinite and because the dimension of MATH, for all MATH, is strictly less than that of MATH. |
math/0009239 | The sufficiency of the condition is obvious. Notice that a complex structure on MATH stabilizes the eigenspaces of MATH. Let MATH be a complex structure on MATH. Let MATH be the adjoint of MATH with respect to the MATH form MATH of MATH, that is, MATH . The so defined MATH intertwines the action of MATH on MATH. Moreover, MATH . Indeed, for all MATH and MATH, MATH . Define MATH . This map is actually valued in MATH since MATH for all MATH. The map MATH defined by its restrictions MATH to MATH is then a complex structure of MATH as a NAME algebra. |
math/0009240 | : This is an elementary exercise in modal reasoning. Scheme REF is actually a special case of scheme REF, obtained by replacing MATH with MATH in REF to obtain MATH. This is just MATH because MATH is equivalent to MATH. Conversely, scheme REF implies scheme REF because MATH implies MATH. Scheme REF is the contrapositive form of scheme REF (with MATH replacing MATH) and vice versa, because MATH is equivalent to MATH and MATH is equivalent to MATH. Scheme REF, similarly, is the contrapositive form of scheme REF. Finally, scheme REF is obtained from scheme REF (applied to MATH) by pushing the negation through to MATH, and conversely by pulling it out again. |
math/0009240 | : Let MATH simply be the statement, ``MATH is hereditarily countable." Of course, this is forceably necessary, since any set can be made countable by forcing, and once countable, it can never be made uncountable again by further forcing. But by the hypothesis on MATH, the statement MATH is not true. |
math/0009240 | : If a sentence MATH is forceably necessary over a forcing extension, then it was already forceably necessary over the ground model, and if necessary in the ground model, it remains necessary in any forcing extension. That is, MATH is downward absolute from a forcing extension, and MATH is upward absolute to any forcing extension. So MATH. |
math/0009240 | : In order to highlight various aspects of the theory, I will actually give two proofs of this theorem. The first proof is elementary, but the second proof will generalize to MATH. We begin with a simple observation. For no sentence MATH are both MATH and MATH forceably necessary. : Suppose that MATH is forced to be necessary by MATH, so that MATH holds in every forcing extension of MATH, and that MATH is forced necessary by MATH, so that MATH holds in every forcing extension of MATH. Consider now the forcing MATH. Since this is isomorphic to MATH, the extension MATH is a forcing extension both of MATH and of MATH. Thus, both MATH and MATH hold there, a contradiction. This argument generalizes to the following. Over any model of set theory, the collection of forceably necessary statements forms a consistent theory. : Consider any finite collection MATH of such sentences, each forceably necessary over a fixed model MATH. Suppose that MATH is forced necessary by MATH, and consider the partial order MATH. The resulting forcing extension MATH is a forcing extension of each MATH, and so each MATH holds there. Thus, any given finite collection of sentences forceably necessary over a fixed model of set theory is consistent. And so the whole collection is consistent. If MATH is forceably necessary, then this is necessary; and if MATH is not forceably necessary, then this also is necessary. In short, the collection of forceably necessary sentences is invariant by forcing. : Suppose that MATH is forced necessary by the forcing MATH, so that MATH holds in every forcing extension of MATH. Given any other forcing extension MATH, one can still force with MATH to obtain MATH. Since this is the same as MATH, it is a forcing extension of MATH. Thus, over MATH one could still force MATH to be necessary; and so MATH is forceably necessary in MATH, and hence necessarily forceably necessary in MATH. Conversely, suppose that MATH is not forceably necessary. This means that MATH is forceable over every forcing extension. Thus, over no forcing extension can MATH be forced necessary, so MATH is necessarily not forceably necessary. In summary, whether a statement is forceably necessary or not cannot be affected by forcing. To prove the theorem, now, suppose that MATH is any model of zfc. Let MATH be the collection of sentences that are forceably necessary in MATH. This includes every axiom of zfc, since these axioms hold in every forcing extension of MATH. By REF , the theory MATH is consistent, and so it has a model MATH. Suppose now that MATH is forceably necessary over MATH; I aim to show that MATH is true in MATH. First, I claim that MATH is actually in MATH. If not, then it is not forceably necessary in MATH, and so by REF the sentence MATH asserting that MATH is not forceably necessary is necessary and hence in MATH. In this case, since MATH is a model of MATH, the sentence MATH cannot be forceably necessary in MATH, contrary to our assumption. Thus, MATH must be in MATH, and so it holds in MATH, as desired. |
math/0009240 | A Second NAME Second Proof: I would like now to give an alternative proof of REF , relying on a more traditional iterated forcing construction. It is this iterated forcing argument that will generalize to the case of MATH. Let me motivate the first lemma by mentioning that if there is an inaccessible cardinal MATH, then by the proof of the downward NAME Theorem, one can find a closed unbounded set MATH of cardinals MATH with MATH. The structure MATH, therefore, is a model of zfc with a cardinal MATH such that MATH is an elementary substructure of the universe. Axiomatizing this situation, let the expression MATH represent the scheme, in the language with an additional constant symbol for MATH, which asserts of any statement MATH in the language of set theory that MATH . Each such assertion in this scheme is first order, since one need only refer to satisfaction in a set structure, MATH, and this is provided by NAME 's definition of the satisfaction relation. The little argument just given shows that if there is an inaccessible cardinal, then there is a model of zfc satisfying the scheme MATH; indeed, since the club MATH provides whole towers of such MATH, the consistency strength of MATH is easily seen to be strictly less than the consistency of the existence of an inaccessible cardinal. The really amazing thing, however, is that in fact MATH is equiconsistent with zfc. One might incorrectly guess that if zfc holds in MATH and the scheme MATH holds, then MATH knows that MATH is a model of zfc; but this conclusion would confuse the `external' zfc with the `internal' zfc of the model. What follows is only that MATH satisfies any particular instance of an axiom of zfc but not the formula asserting that MATH satisfies the entire scheme zfc. This subtle distinction is crucial, in view of the following elementary fact. If zfc is consistent, then so is MATH. : Assume that zfc is consistent; so it has a model MATH. By the NAME Reflection Theorem, every finite subcollection of the theory MATH is modeled in some rank initial segment of MATH, and therefore is consistent. So the whole theory is consistent. The theorem is now a consequence of the following lemma. Assume that MATH. Then there is a forcing extension of MATH, by forcing of size at most MATH, in which the Maximality Principle mp holds. : Since the Maximality Principle asserts, in a sense, that all possible switches have been turned permanently on that are possible to turn permanently on, the idea behind the proof of this lemma will be an iteration that forces all statements that are possible to force in a permanent way. The trick is to handle the meta-mathematical issues that arise on account of the non-definability of what is true or forceable. It is in doing this that the hypothesis MATH will be used. Let MATH enumerate all sentences in the language of set theory. I will define a certain forcing notion MATH, a MATH-iteration, all of whose initial segments MATH will be elements of MATH. So, suppose recursively that the iteration MATH has been defined up to stage MATH, and consider the model MATH. If MATH is forceably necessary over this model, then choose a poset MATH which forces it to be necessary, and let MATH using a suitable name for MATH. Let MATH be the finite support iteration of the MATH. I claim that if MATH is MATH-generic, then the Maximality Principle holds in MATH. To see this, suppose that MATH is forceably necessary over MATH. Necessarily, MATH is MATH for some MATH. Factor the forcing MATH at stage MATH as MATH, where MATH is the iteration after stage MATH. Since MATH is forceably necessary over MATH, it is also forceably necessary over MATH, because MATH is a forcing extension of MATH and so any forcing extension of MATH is also a forcing extension of MATH. Since MATH and MATH, whether a condition in MATH forces a given statement has the same answer in MATH as in MATH. Consequently, MATH. Thus, MATH is forceably necessary over MATH. In this case, the stage MATH forcing MATH forced it to be necessary, so MATH holds in MATH and all its forcing extensions. By elementarity again, MATH holds in MATH and all forcing extensions. In particular, since MATH is a forcing extension of MATH, it holds in MATH and all further extensions. That is, MATH is necessary in MATH. This completes the second proof of REF . |
math/0009240 | : I claim, first, that if there is a model of zfc, then there is one in which the definable ordinals - and I mean ordinals that are definable without parameters - are unbounded. To see this, suppose that MATH is a model of MATH, not necessarily transitive, and let MATH be the cut determined by the definable elements of MATH. That is, MATH consists of those elements in MATH, where MATH ranges over the definable ordinals of MATH. I will show that MATH, by verifying the NAME criterion: suppose MATH for some MATH. By the definition of MATH, it must be that MATH is in some MATH in the sense of MATH for some definable ordinal MATH of MATH. In MATH, let MATH be the least ordinal such that for every MATH there is a MATH such that MATH, if there is any such MATH in MATH at all. Such a MATH exists by the Replacement Axiom. Moreover, MATH is definable in MATH, and consequently, MATH is included in MATH, and so MATH has all the witnesses it needs to verify the NAME criterion. Thus, MATH, and in particular, MATH is a model of MATH whose definable elements are unbounded. By beginning with a model MATH that also satisfies MATH, we may assume additionally that MATH satisfies MATH as well. Suppose now, towards contradiction, that MATH has an extension MATH, with the same ordinals as MATH, that satisfies MATH. Necessarily, MATH. Consider any definable ordinal MATH of MATH, defined by the formula MATH. Since MATH, the formula MATH also defines MATH in MATH, and since MATH is the MATH of MATH, we know further that MATH is definable in MATH as ``the ordinal satisfying MATH in MATH". Let MATH be the sentence expressing, ``the ordinal defined by MATH in MATH is countable". This sentence is forceably necessary over MATH, since if MATH is not countable in MATH, one can force to make it countable, and having done this, of course, it remains countable in any further extension. Thus, by MATH, the sentence MATH must be already true in MATH, and so MATH is countable in MATH. The key observation is now that since the definable ordinals MATH of MATH were unbounded in MATH, it must be that every ordinal of MATH is countable in MATH, a contradiction, since these two models of zfc have the same ordinals. |
math/0009240 | : REF shows that every model of MATH has a forcing extension that is a model of MATH. Conversely, suppose that MATH is a transitive model of MATH. Consider the MATH of MATH, which must have the form MATH, for MATH. The arguments of the previous theorem establish that every ordinal that is definable in MATH is countable in MATH. Let MATH be the supremum of such ordinals. Thus, MATH is at most MATH, and in particular, MATH. The initial claim of the previous theorem establishes that in the model MATH, the cut determined by the definable ordinals is an elementary substructure. In this case, this means that MATH in MATH, as desired. |
math/0009240 | : First, one shows that if there is a model of zfc, then there is a model of MATH is uncountable. This can be proved in the same way as REF , by simply adding the assertion MATH to the theory. The NAME Reflection Theorem shows that any finite collection of formulas is absolute from MATH to MATH for a closed unbounded class of cardinals, and so one may find such a MATH of uncountable cofinality. Given a model of MATH in which MATH is uncountable, one now proceeds with the argument of REF , using finite support in the iteration MATH of that theorem. Since on cofinality grounds MATH has size less than MATH, it follows that MATH in that argument. That is, the axiom MATH remains true in MATH, together with MATH. |
math/0009240 | : I will prove the theorem with a sequence of lemmas. Every model of theory REF has an inner model of theory REF. Specifically, if MATH holds, then MATH is inaccessible in MATH, and MATH. : Assume MATH and let MATH. I claim, first, that MATH is inaccessible to reals. To see this, suppose MATH is a real, and let MATH be the statement ``the MATH of MATH is countable". Since this is forceably necessary, by MATH it is true. Thus, MATH, as I claimed. It follows from this that MATH is inaccessible in MATH. Second, I claim that MATH is an elementary substructure of MATH (note: I prove this part of the Lemma only as a scheme). To see this, simply verify the NAME criterion: suppose that MATH satisfies MATH for a set MATH in MATH. Let MATH be least such that there is such a MATH in MATH. Consider the statement MATH asserting ``the least MATH, such that there is a MATH in MATH with MATH, is countable". This is expressed using the parameter MATH, which is coded with a real. Since it is forceably necessary, it must be true by MATH, and so MATH is countable. Thus, MATH can be found in MATH, and so the NAME criterion holds. In summary, MATH satisfies theory REF. If theory REF holds, then theory REF holds in a forcing extension. Indeed, if MATH and MATH is an inaccessible cardinal, then after the NAME collapse making MATH the MATH of the extension, the Maximality Principle MATH holds. : Assume that MATH and MATH is inaccessible. In order to obtain MATH in a forcing extension, I will define a certain finite support MATH-iteration MATH which at every stage forces with some poset of rank less than MATH. Let MATH enumerate, with unbounded repetition, all possible sentences in the forcing language for initial segments of such iterations with parameters having names in MATH. Suppose now that the iteration MATH is defined up to stage MATH; I would like to define the stage MATH forcing MATH. If MATH is a sentence with parameters naming objects in MATH, and if it is possible to force over MATH so as to make MATH necessary, then choose MATH to be (the name of) a poset accomplishing this. Otherwise, let MATH be trivial forcing. This defines the MATH-iteration MATH. Suppose that MATH is MATH-generic, and consider the model MATH. First observe that the cardinal MATH is not collapsed by this forcing, since it is a finite support iteration of MATH-c.c. forcing. Consequently, every real in MATH appears in some MATH. Thus, if MATH is a sentence with (names for) real parameters from MATH, it must be MATH for some MATH, with the names in MATH coming from MATH. So it suffices to consider the formulas MATH. Suppose that such a MATH is forceably necessary over MATH. Then it is also forceably necessary over MATH, since MATH is a forcing extension of MATH. Hence, by the elementarity MATH (as in REF ), the sentence MATH is forceably necessary over MATH. The stage MATH forcing MATH must then have forced it to be necessary, and so MATH became necessary in MATH, and hence also in MATH. Thus, since MATH is obtained by further forcing over this model, MATH remains necessary in MATH. So MATH satisfies MATH. Finally, let me consider the precise nature of the iteration MATH; in fact, we know MATH very well. Since the assertion ``MATH is countable" is always forceably necessary, it follows that every element MATH is countable in MATH. That is, unboundedly often, the forcing MATH collapses elements of MATH. Since MATH itself is not collapsed, it becomes the MATH of MATH. But up to isomorphism, the only finite support MATH-iteration of forcing notions of size less than MATH that collapses all cardinals below MATH to MATH is the NAME collapse of MATH to MATH. Thus, MATH is isomorphic to the NAME collapse, as I claimed. NAME has used a similar iteration to prove a similar conclusion relating to his generalized bounded forcing axioms (see REF). Theory REF implies theory REF. : Suppose that MATH and MATH is inaccessible. If MATH is a definable club, then of course MATH is unbounded in MATH. Thus, MATH, and so MATH contains a regular cardinal, as desired. If theory REF is consistent, then so is theory REF. : This argument is similar to REF . Suppose that MATH is a model of zfc plus the scheme asserting that MATH is NAME. By the NAME Reflection Theorem, any finite collection of formulas reflects from MATH to MATH for a closed unbounded class of cardinals MATH. Thus, since MATH is NAME in MATH, we can find in MATH a regular cardinal MATH such that the formulas are absolute between MATH and MATH in MATH. We may assume that the finite collection of formulas includes the assertion that for every MATH, the cardinal MATH exists, and from this it follows that MATH is a strong limit cardinal. Thus, MATH is actually inaccessible in MATH. So every finite subcollection of theory REF is consistent, and so the entire theory is consistent. This completes the proof of the theorem. |
math/0009240 | : Over any model one can force ch without adding reals, and this will preserve MATH. Alternatively, the model of REF satisfies MATH, since it is obtained via the NAME collapse of an inaccessible cardinal. To obtain a model of MATH, we will simply add NAME reals over this model. Specifically, I claim that the model MATH, where MATH is MATH-generic for the iteration of REF and MATH is MATH-generic for the forcing to add at least MATH many NAME reals, is a model of MATH. Certainly MATH is no problem, so consider MATH. The instances of the axiom scheme MATH, where MATH has real parameters from MATH, hold in MATH by REF , and each of these instances persists to every forcing extension, including MATH. At issue is whether the scheme holds of formulas using the new parameters available in MATH. Consider, therefore, a real MATH in MATH and the corresponding extension MATH. Since MATH is added by a countable forcing notion MATH, we may reorganize this forcing as MATH, where MATH is now MATH-generic for some countable forcing notion and MATH is MATH-generic for the NAME collapse of MATH in MATH (this is because the quotient forcing MATH has size MATH, is MATH-c.c. and collapses every cardinal below MATH to MATH; hence it is the NAME collapse). Since MATH, it follows by REF that MATH holds in MATH. Thus, instances of the MATH scheme using the parameter MATH hold in MATH. Since these persist to any forcing extension, they also hold in MATH. So MATH holds in MATH, as desired. |
math/0009240 | : Suppose that MATH is consistent with the existence of a proper class of, for example, measurable cardinals. The argument of REF produces a model of MATH with a proper class of measurable cardinals. Since the forcing of REF has size at most MATH, the measurable cardinals above MATH survive to the forcing extension in which MATH holds. For MATH, one should add the hypothesis that ``MATH is NAME to this argument in order to get a model as above for which MATH is inaccessible, and then carry on as in REF . This procedure shows that any large cardinal that is not destroyed by small forcing will be consistent with mp or MATH. To obtain the Maximality Principle in the absence of all large cardinals, one should simply apply the argument of REF using a model with no inaccessible cardinals. That is, if MATH is any model of zfc with no inaccessible cardinals, then no forcing extension of MATH will have inaccessible cardinals, and so the assertion that there are no inaccessible cardinals will be necessary in MATH and hence in the theory MATH of that proof. |
math/0009240 | : Suppose that the Maximality Principle holds, but that the inaccessible cardinals are bounded. Then the assertion ``there are no inaccessible cardinals" is forceably necessary, because one could force, in a permanent way, to make all the inaccessible cardinals countable. Thus, under mp, there must have been no inaccessible cardinals to begin with. |
math/0009240 | : It suffices, by REF, to find a violation of covering. For this, it suffices to show that MATH fails to compute the successors of singular cardinals correctly. In fact, I will show that MATH fails to compute the successor of any cardinal correctly. Consider any cardinal MATH and its successors MATH and MATH in the two models. Let MATH be MATH-generic for the canonical forcing that collapses MATH to MATH. This forcing is the same in MATH or MATH and has size MATH in either model, and so by the chain condition it preserves MATH over either model, making it the MATH of the extension. The key observation now is that the sentence MATH, asserting ``MATH is countable," is forceably necessary. Thus, by MATH in MATH, this sentence is true in MATH. Putting these conclusions together, we have MATH and so MATH does not compute the successor of MATH correctly. |
math/0009240 | : Clearly, REF is implied by REF, which is in turn implied by REF. Let me argue that REF is implied by REF. If MATH, then clearly MATH is a strong limit in MATH, and hence, if regular, MATH is inaccessible in MATH. Thus, the inaccessible cardinals are unbounded in MATH, and hence also in MATH, so statement MATH holds. REF easily follows from REF, because if MATH, then MATH has the correct MATH for MATH, and so MATH, giving statement MATH. Next, let me show that REF follows from REF, by verifying the NAME criterion. Suppose that MATH satisfies MATH where MATH. The least hereditary size in MATH of such a MATH is definable in MATH from MATH, and so by REF this size must be less than MATH. In particular, there will be such a MATH in MATH, and so the criterion is fulfilled. It remains only to prove statement MATH of the theorem. Suppose that MATH is definable in MATH from the object MATH. Let MATH be the hereditary size of MATH, so that MATH, and suppose that MATH is MATH-generic for the canonical forcing that collapses MATH to MATH. Let MATH be the sentence asserting ``MATH is hereditarily countable", using the definition of MATH in MATH from MATH, coded with the real MATH. Thus, MATH is a sentence with real parameters, and it is forceably necessary (please note that I am using here the invariance of MATH under under set forcing in order to know that MATH is still defined in the same way in any forcing extension). Consequently, since MATH implies MATH in MATH, the sentence MATH must be true in MATH. That is, MATH is hereditarily countable in MATH. Since MATH and MATH are preserved to MATH, it follows that MATH has hereditary size less than MATH in MATH, as desired. So the proof is complete. |
math/0009240 | : By this, I mean that such a class MATH will not compute successor cardinals above MATH correctly, that every cardinal above MATH will be a limit of inaccessible cardinals in MATH, and so on, and that every set MATH definable in MATH from an object in MATH for some MATH above MATH will be in MATH. Suppose, therefore, that MATH is definable from the parameters MATH, and that it is invariant by set forcing. In the forcing extension MATH that collapses MATH to MATH, the parameters MATH become hereditarily countable, and the class MATH therefore becomes definable from a real MATH. Thus, applying the previous proof to the class MATH, defined by the real MATH in MATH, the conclusion follows for the cardinals of MATH. Since these are precisely the cardinals above MATH in MATH, the corollary is proved. |
math/0009240 | : Fix any set MATH, and let MATH be a real collapsing cardinals so that MATH can be coded by a real MATH in MATH. Let MATH. This is definable from MATH, and is invariant in any further extension of MATH. Thus, by the previous corollary, if MATH holds, then MATH does not compute the successors of singular cardinals above MATH correctly. It follows that MATH exists in MATH, and so it also exists in MATH. |
math/0009240 | : Assume MATH. I will show, by induction on MATH, that every MATH formula MATH is absolute to any set forcing extension. This is trivial for MATH (and indeed, it holds automatically up to MATH by the NAME Absoluteness Theorem). Assume by induction that it is true for all MATH formulas. Further, since MATH holds in all set forcing extensions, suppose that this induction hypothesis also holds in all set forcing extensions, that is, that the MATH formulas are absolute from any set forcing extension to any further set forcing extension. Since the collection of MATH for which this hypothesis is true is clearly preserved under Boolean combinations, it suffices to consider an existential formula MATH, where MATH is absolute from any forcing extension to any further forcing extension. If MATH is true in such a model, then since MATH is absolute to any set forcing extension and both MATH and the witness MATH still exist there, MATH remains true in any set forcing extension. Conversely, suppose that MATH is true in a set forcing extension MATH, with MATH. Thus, the witness MATH to MATH was added by the forcing MATH. Since by the induction hypothesis the formula MATH is absolute to any further forcing extension, the formula MATH is necessary in MATH and therefore forceably necessary in MATH. Thus, by MATH, it is already true in MATH, as desired. So every projective formula is absolute by set forcing, and the theorem is proved. |
math/0009240 | : CITE has shown that projective absoluteness is equiconsistent with the existence of infinitely many strong cardinals. |
math/0009240 | : CITE has proved that if there is no inner model of a NAME cardinal and if the universe is closed under sharps and MATH is ineffable, then there is a definable class MATH, now widely known as the NAME core model, which is invariant under set forcing and which computes the successors of singular cardinals correctly. Since REF shows that if MATH holds, then the universe is closed under sharps but there can be no such class MATH, it follows under the hypothesis that there must be an inner model with a NAME cardinal. |
math/0009240 | : I will first prove that if there are a proper class of NAME cardinals, then the Maximality Principle MATH holds for local assertions using real parameters in the ground model. From this, since the hypothesis of a proper class of NAME cardinals is preserved by set forcing, it follows that the same conclusion holds in every forcing extension as well. That is, from the hypothesis of a proper class of NAME cardinals one can conclude the Necessary Maximality Principle MATH for local assertions as well. To begin, then, suppose that MATH is local and forceably necessary, forced necessary by the forcing MATH, with real parameters in MATH. Choose a NAME cardinal MATH above the rank of MATH and let MATH be the stationary tower forcing corresponding to MATH. Suppose that MATH is MATH-generic below a condition forcing that MATH becomes countable. By the basic properties of stationary tower forcing (see, for example, CITE, CITE), there is a generic elementary embedding MATH, with MATH in MATH. The model MATH therefore agrees with MATH well beyond the first inaccessible cardinal. Since furthermore MATH was collapsed, in MATH we may construct a MATH-generic filter for MATH, and therefore we may view MATH as a forcing extension of MATH. Since MATH is made necessary in MATH, it follows that MATH holds in MATH. Thus, since MATH agrees with MATH well beyond the first inaccessible cardinal, MATH holds also in MATH. Finally, since the real parameters of MATH are fixed by MATH, it follows by elementarity that MATH holds also in MATH, as desired. |
math/0009240 | : In the proof of REF , let us simply ensure that the forcing MATH at stage MATH comes from MATH. Then, define MATH to be the finite support or countable support iteration of the MATH, depending on the closure of MATH. Having done so, the iteration MATH, as well as the tail forcing MATH, will be in MATH, and so the argument of REF works to establish MATH in MATH. |
math/0009240 | : This argument follows the first proof of REF . Suppose that MATH is any model of zfc, and let MATH be the collection of sentences that are forceably necessary over MATH, referring throughout this argument only to forcing notions in MATH. I claim first that the theory MATH is consistent. Given any finite subcollection MATH, with MATH forced necessary by MATH, consider MATH. By the hypotheses of the theorem, this is in MATH. Further, the extension MATH is a forcing extension of each MATH, by forcing in MATH there. Thus, the extension MATH satisfies each MATH, and so this collection is consistent. Next, I claim that MATH is absolute to all forcing extensions of MATH. If a sentence is forceably necessary over MATH with respect to forcing in MATH, then the forcing that made it necessary will still be in MATH in any forcing extension, and serve to make it necessary there. So it will be forceably necessary over any forcing extension also. Conversely, if a sentence is forceably necessary over a forcing extension of MATH by forcing in MATH, then clearly it is forceably necessary in MATH. Finally, I claim that any model MATH of MATH is a model of MATH. Certainly MATH since these are necessary over MATH. If MATH is forceably necessary over MATH, then MATH must be in MATH, for if not, then the assertion that MATH is not forceably necessary would be necessary and hence in MATH, a contradiction. Thus, MATH and so MATH holds in MATH, as desired. |
math/0009240 | : The corresponding classes of posets have the closure properties of the theorem. |
math/0009240 | : Clearly, the first theory directly implies the second theory. Furthermore, I have already shown that the third and fourth theories are equiconsistent. Every model of theory REF contains an inner model of theory REF. This Lemma is a consequence of the following Lemma: If MATH holds (in the language with ordinal parameters below MATH) and MATH is regular, then MATH is inaccessible in MATH and MATH. : Suppose that the Maximality Principle MATH holds for c.c.c. forcing in the language with ordinal parameters below MATH, and that MATH is regular. Note that for any MATH, the assertion MATH is forceably necessary, since by c.c.c. forcing, one can pump up MATH as large as desired. Thus, this assertion must be true, and so MATH is weakly inaccessible and hence inaccessible in MATH. To prove MATH, I will simply verify the NAME criterion. Suppose that MATH for some MATH. Let MATH be large enough so that MATH. Now, by the Replacement Axiom, let MATH be least so that for every MATH with a MATH satisfying MATH, there is such a MATH in MATH. This property defines MATH in any forcing extension, and so the assertion MATH is forceably necessary. Thus, MATH, and so the witness MATH for MATH can be found in MATH, as desired. Any model of theory REF has a forcing extension that is a model of theory REF. : This argument follows REF above. Suppose that MATH and MATH is inaccessible. Enumerate MATH all sentences MATH (with unbounded repetition) in the language of set theory with (names for) parameters in MATH coming from a forcing extension by forcing of size less than MATH. Define a finite support MATH-iteration of c.c.c. forcing, so that at each stage MATH, the forcing MATH forces MATH to be necessary over MATH, if possible (that is, first, if this makes sense, if the parameters appearing in MATH are MATH-names, and second, if there is such a forcing in MATH). Let MATH be MATH-generic for MATH, and consider the model MATH. Note that periodically during the iteration, the posets will ensure that MATH is made arbitrarily large below MATH, and so MATH in MATH. Conversely, since the forcing has size MATH, it is easy to see that MATH. Further, since the iteration is c.c.c., MATH remains a regular limit cardinal. Finally, if MATH is a forceably necessary assertion over MATH with parameters in MATH, then MATH for some MATH such that the parameters are MATH-names. Since MATH was forceably necessary in MATH it must have been forceably necessary in MATH and hence, since MATH, it was forceably necessary in MATH. Thus, it was forced to be necessary in MATH, and so it was necessary in MATH. Thus, it is true in MATH, as desired. In summary, MATH is weakly inaccessible. This completes the proof of the theorem. |
math/0009240 | : Using the techniques of REF, one can show that if the first theory holds, then MATH is inaccessible in MATH and MATH. And this has already been proved equiconsistent with the second and third theories. Conversely, the second and third theories are equiconsistent with MATH for an inaccessible cardinal MATH. Given this, one enumerates the formulas MATH using (names for) parameters in MATH appearing in forcing extensions of size less than MATH, and then performs a countable support MATH-iteration MATH of proper forcing, which at stage MATH forces the necessity of MATH over MATH by proper forcing, provided that this make sense (that is, the names in MATH are MATH-names) and that this is possible (that is, there is a proper forcing poset accomplishing this). The resulting forcing extension MATH will be a model of MATH, just as in REF . |
math/0009242 | We will show that for MATH this small, on average MATH increases at each step. If MATH, then the size of MATH goes up by MATH, but if MATH, then the size of MATH may decrease by at most MATH [removing MATH (not previously included), MATH, and some neighbors]. Hence MATH which is positive precisely when MATH. Given an increase of MATH on average at each step, standard martingale stopping theorems (see, for example, CITE) show that after MATH expected time the value of MATH will be MATH, at which point MATH and the algorithm terminates. MATH . |
math/0009242 | Let MATH, and suppose that MATH. Let MATH be the event that MATH and MATH. Then MATH where the last step is exactly our assumption. Note that neither MATH nor MATH depends on MATH. Hence, summing over MATH, MATH . This completes the proof. MATH . |
math/0009242 | The acceptance probabilities were chosen precisely to match the requirements of REF . For instance, with heat bath RR, the left side of the equation in REF equals MATH which reduces to the right side of the equation with MATH. The calculation for arbitrary RR is entirely similar. MATH . |
math/0009242 | We use a potential function that rewards us for adding edges and penalizes us for connecting components. Let MATH where MATH will be determined later. When the edge MATH we attempt to add to MATH is between two vertices already connected in MATH, then MATH always goes up by REF, making this case uninteresting. It is when MATH would connect two previously unconnected components of MATH that the calculation becomes interesting. If the edge is chosen to be excluded from MATH, then MATH increases by MATH. If the edge is proposed to be included in MATH, then MATH changes by MATH if we accept. If we reject, we remove from MATH (and also from MATH) a component of size MATH and (from MATH) all of its adjacent edges. Not counting the edge MATH and making sure that we do not double-count, this totals at most MATH edges removed from MATH. However, we add exactly MATH new components to MATH by removing these edges. When we add the tree MATH back in, this produces MATH new edges for MATH, but for each such edge there is a MATH chance of including the edge in MATH and thereby reducing the number of components by MATH. Therefore, when we attempt to add MATH to MATH, but reject instead, the expected contribution to the change in MATH is at least MATH . Now MATH may be very large (nearly as large as MATH), so we choose MATH in such a way that the coefficient of MATH in this expression vanishes. That is, we set MATH and so the contribution in this case is bounded below by MATH. We try to put the edge in with probability MATH and to leave it out with probability MATH. We accept an inclusion with probability MATH. Putting everything together, we find that the expected change in MATH at any time step when MATH and MATH are not already connected in MATH is at least MATH which is positive exactly when MATH . |
math/0009244 | From REF and MATH we see that MATH are the eigenfunctions of MATH. Then the theorem is obtained from REF since it implies that the range of MATH is dense. |
math/0009244 | The symmetry of the operator MATH is trivial. Then we deduce that MATH is essentially self-adjoint on MATH. |
math/0009244 | For MATH, we have MATH due to MATH. Then MATH which implies that MATH is one-to-one and MATH. Similarly we have MATH and hence MATH. |
math/0009244 | We define the functions MATH and MATH by MATH . Then we see that MATH and MATH are real-holomorphic. Equivalently, MATH and MATH are real-holomorphic. By REF , MATH or the matrix MATH is invertible, which implies MATH is real-holomorphic. |
math/0009244 | Since the operator MATH is expressed in terms of MATH and MATH, it is enough to show the MATH cases. The NAME group of type MATH acts on the space of function on MATH by the permutation of the variable. We denote the action of MATH on MATH by MATH. Let us recall the definition of MATH, that is, MATH . Let MATH be a function in MATH. From the definition of the operators MATH, the function MATH satisfies the relations MATH, MATH. It remains to show the holomorphy of the function MATH on MATH. The function MATH is non-zero holomorphic function in MATH, because the function MATH is non-zero on MATH and does not admit any branching points on MATH. From the definition of the operators MATH, the function MATH does not have poles except for MATH on MATH. If the function MATH has a pole along MATH, the order of the pole is one, but it contradicts to the NAME group invariance of the function MATH. Therefore the function MATH is holomorphic along MATH. The holomorphy along MATH follows from the periodicity of MATH. |
math/0009244 | We prove MATH. For the other cases, the proofs are similar. From the definition, the function MATH is symmetric and rational with respect to the variables MATH. The possible poles of the rational function MATH are MATH and the degree of each pole is one, but it contradicts to the NAME group invariance of the function MATH. Therefore the rational function MATH does not have any poles, and we have MATH. |
math/0009244 | Since MATH and MATH, it is enough to show MATH. We have the equality MATH for some non-zero constant MATH, which follows from the correspondence between the integration of the MATH-invariant function and the one of the MATH. From this equality, REF , and the commutativity MATH, if we show MATH where MATH and MATH, then we obtain REF . If MATH then the functions MATH are MATH-class. From the definition of REF , we have the periodicity MATH for MATH. We set MATH and MATH. The functions MATH, MATH are smooth on MATH except for MATH. The behaviors of the functions MATH, MATH around MATH are MATH, that is, MATH and MATH are bounded around MATH. From the expression of REF , if we show MATH for all MATH such that MATH, then we obtain REF . The number MATH satisfies MATH. Though the function MATH have double pole along MATH, the integrands of REF are bounded around MATH from the properties MATH and MATH around MATH. Hence the singularities along MATH do not affect the integration. Since the integrands of REF are continuous, we can replace the range of integration of both sides of REF with MATH, where MATH. It is obvious that MATH . By applying the integration by parts repeatedly, we find that the right-hand side of REF is equal to MATH . Here we used the periodicities on MATH (MATH). Hence we obtain REF . |
math/0009244 | It is trivial for the MATH case. We assume MATH. Let MATH. Then MATH is a polynomial in the parameter MATH of degree at most MATH and MATH. We set MATH. Then MATH, because MATH for all MATH. For MATH and MATH, we set MATH and MATH. We fix the functions MATH. It is enough to show that the equations MATH hold for MATH and MATH. We set MATH . Then the equation MATH is equivalent to MATH, where MATH. From the equation MATH, it is sufficient to show MATH. We have MATH on the domain MATH, because MATH on MATH for MATH and the branch of the function MATH is chosen to be a positive real number. For MATH, the branch of the function MATH is canonically chosen by the relation MATH for MATH. Hence it is sufficient to show the equation MATH for MATH. From REF , MATH holds for MATH. From REF , we have MATH when MATH and MATH. Hence the integral MATH is well-defined if MATH. We fix MATH (MATH). Since the function MATH is holomorphic in MATH and the functions MATH and MATH are uniformly bounded in MATH for some MATH, the integral MATH is also holomorphic at MATH by the NAME 's theorem. By the identity theorem, the equation MATH holds for MATH such that MATH. Therefore we obtain the proposition. |
math/0009244 | We introduce variables MATH and set MATH . The numbers MATH and MATH are given in advance. We will investigate the conditions for the coefficients of the formal power series MATH and MATH satisfying the following relations MATH . We set MATH . By comparing the coefficients of MATH, we obtain that REF are equivalent to the following relations, MATH . We remark that the denominator of REF is non - zero by the non - degeneracy condition. The numbers MATH and MATH are determined recursively and they exist uniquely. We have recursively that for each MATH and MATH, MATH is finite and the summations in (REF - REF) on the parameters MATH and MATH are indeed the finite summations. At this stage, the apparent expression of the coefficients MATH may depend on MATH. We will show that the coefficients MATH do not depend on MATH. We denote MATH by MATH. From the commutativity of MATH and MATH we have MATH . Since the vector MATH admits the expansion MATH, we obtain the following relation from the uniqueness of the formal eigenvector. MATH where MATH and MATH is regarded as a formal power series on MATH from the formula MATH. Therefore we have MATH. On the other hand we have MATH. By the uniqueness of the formal eigenvector whose leading term is MATH, we have MATH. Therefore the coefficients MATH do not depend on MATH. From REF , we obtain recursively that the coefficients of the formal eigenvalue MATH are linear in MATH. Therefore the numbers MATH are determined appropriately. |
math/0009244 | The finiteness of the summation MATH follows from REF and the fact that the NAME polynomial forms a basis of MATH. The non-degeneracy of the joint eigenvalues MATH follows from the non-degeneracy of the joint eigenvalue of the NAME polynomial. |
math/0009244 | Let MATH be the sum of all divisors of MATH. By REF , we have MATH. Since the convergence radius of the series MATH is equal to MATH, the convergence radius of the series MATH is equal to or less than MATH. Therefore we have the lemma. |
math/0009244 | It follows from REF and the inequality MATH . |
math/0009244 | From REF , we have MATH where MATH is the highest root of the root system MATH. Since MATH and MATH, we have the lemma. |
math/0009244 | This follows from the NAME formula (CITE, p. REF.). |
math/0009244 | It is sufficient to show MATH for MATH and MATH. This inequality is equivalent to MATH. From the property MATH, we have MATH. |
math/0009244 | Immediate from the equality MATH. |
math/0009244 | First, we expand MATH by using NAME REF. Then MATH is expressed as the linear combination of MATH, where MATH and MATH. We set MATH . We repeatedly apply NAME REF for MATH. Then MATH is expressed as the linear combination of MATH, where MATH for some MATH. From NAME REF, we have MATH. Applying NAME REF for MATH and MATH, we obtain REF . |
math/0009244 | Since the normalized NAME polynomials form the complete orthonormal system with respect to the inner product MATH, we have MATH. We fix MATH. Let MATH be the smallest integer which is greater or equal to MATH. If MATH, then we have MATH by REF and the orthogonality. Therefore we have MATH . Similarly, we have MATH. Since the convergence radius of the series MATH is equal to MATH, we obtain that there exists a number MATH such that MATH and MATH for MATH. Hence we have the Proposition. |
math/0009244 | Let us recall that the operator MATH is defined by the NAME series REF . We fix the number MATH and set MATH. From the expansion REF , there exists a number MATH such that the inequality MATH holds for MATH REF and MATH. In this case, we have MATH. We write MATH. For the series MATH, write MATH. We have MATH for each MATH. Combining with REF , we obtain that for each MATH such that MATH and MATH. there exists MATH which does not depend on MATH but MATH such that MATH if MATH. To obtain REF , we use the method of majorants. We introduce the symbol MATH to avoid inaccuracies. We remark that MATH. We will apply the method of majorants for the formal series MATH instead of MATH . For the formal series MATH, we define the partial ordering MATH by the rule MATH . We will later consider the case that each MATH is expressed as the infinite sum. If one shows the absolute convergence of MATH for each MATH, one has the absolute convergence of MATH for each MATH by the majorant. We set MATH, where the coefficients MATH were defined by MATH. Our goal is to show REF for MATH such that MATH. Since MATH and MATH, it is enough to show that there exists MATH and MATH which do not depend on MATH (but depend on MATH) such that MATH is well-defined by MATH and satisfies MATH for all MATH such that MATH and MATH. We set MATH . MATH . We have the inequality MATH . Let MATH. If the coefficients of MATH with respect to the basis MATH converge absolutely, then the coefficients of the series MATH and MATH are well-defined and we have MATH . From the equality MATH and the property MATH, we have MATH . Set MATH, we have MATH where MATH . Remark that we used the inequality MATH for MATH and the formula MATH. The equality REF makes sense for MATH, where MATH is the positive number satisfying the inequalities MATH, MATH and MATH. Therefore each coefficient of MATH with respect to the basis MATH converges absolutely. Hence the following inequality makes sense, MATH . Let MATH be the solution of the equation MATH on MATH satisfying MATH. Then MATH is holomorphic in MATH near MATH and admits the expansion MATH. We have MATH where MATH is a holomorphic function defined near MATH. For the MATH case, we have the relation REF by calculating the residue around MATH. For the MATH case, we need to change the variable MATH and calculate the residue around MATH. The coefficient of MATH on the series MATH satisfying MATH has to be zero from the definition of MATH. By the inequality MATH, we have MATH for MATH and MATH: a sufficiently small positive number. Combining with the relation MATH, the inequality MATH, and the expansion MATH, we obtain REF and the proposition. |
math/0009244 | Since the spectrum MATH is discrete, there exists a positive number MATH such that MATH. We write MATH. From REF , we obtain that for each MATH, there exists MATH and MATH which does not depend on MATH such that MATH for all MATH such that MATH and MATH. Let MATH be the length of the circle MATH and write MATH. By integrating MATH over the circle MATH, we have MATH for all MATH such that MATH and MATH. Therefore we have REF . |
math/0009244 | Since MATH, it is enough to show that the function MATH is holomorphic when MATH is sufficiently small and MATH. We count roughly the number of the elements of MATH of a given length. The rough estimate is given by MATH. We will use this in REF . In the proof, we will use the notations and the results written in REF. In REF, there are parameters MATH and MATH. We fix MATH and MATH. There is another number MATH defined in REF. We have MATH by the formulae REF . If MATH then the bottom part of the inequality converges. We choose a positive number MATH which satisfies MATH. Then the series MATH is uniformly bounded and uniformly absolutely converges for MATH and MATH. Since the functions MATH are holomorphic, we have the holomorphy of the function MATH by the NAME 's theorem. |
math/0009245 | Let MATH be a small perturbation of a critical point MATH, such that MATH and MATH. The derivative of the function MATH, at MATH, is given by MATH where MATH. If for all MATH, we have that MATH then MATH . The second equation is obtained by considering a smooth curve MATH given by MATH, where MATH. It follows from the first order approximations MATH and MATH that MATH where MATH is the linear operator MATH . If MATH is a critical point of MATH, then for all MATH . Hence, MATH . |
math/0009245 | Since MATH we can perform the computation of MATH by fixing a class of MATH. For a class MATH, we fix a map MATH representing MATH and MATH. Thus, MATH . However, MATH . Let MATH. By NAME 's formula, MATH and MATH. Consequently, MATH . The group MATH is described below in REF ; (the symbol MATH stands whether the group is REF or not) |
math/0009245 | The Minimax Principle implies that the minimum value is attained in all connected component of MATH. REF shows that it is possible to construct non-contractible families of elements in MATH, consequently, by applying the Minimax Principle, it follows that there exist stable and non-stable critical points. In other words, there exist solutions to the equations REF . |
math/0009245 | The proof is splited in two easy claims; CASE: If MATH, then MATH. Let MATH; MATH CASE: If MATH, then there exits MATH and MATH such that MATH . Let MATH and MATH; MATH . |
math/0009245 | Let MATH; so MATH. NAME, MATH this implies that MATH and MATH . |
math/0009245 | Its follows from the discussion above, since it was concluded that MATH . |
math/0009246 | We prove this proposition by using the weak compactness REF . If the compactness fails, then there exists a finite number of bubble points MATH such that MATH has a non-trivial area concentration in each bubble point (we follow notations in REF above). Consider two cases: MATH or MATH is a torus or higher genus. In the first case, MATH . We can re-normalize the sequence so that MATH (by conformal transformations). According to REF , each bubble metric has a weak cusp singularity at its singular points. On the other hand, suppose that MATH is such a bubble point (MATH might be any of MATH), and MATH is a small disk centered at MATH (for convenience, we assume that it is a disk of radius REF). Now suppose the sequence of metrics can be re-written as: MATH . Here MATH is the coordinate variable for disk MATH . According to our assumption that there is a positive area concentration at MATH (or point MATH), thus MATH . Without loss of generality, we may assume that MATH . Now re-normalize this sequence of metrics by: MATH where MATH . Note that the metric MATH is just a re-normalization of MATH thus the scalar curvature and area is not changed! In other words, if we denote the scalar curvature of MATH by MATH then MATH . Thus, for any fixed MATH and MATH large enough we have MATH and MATH . By definition, we have MATH for any MATH . Therefore, it is not difficult to choose a subsequence of MATH which converges in every fixed disk MATH . Suppose the limit metric is MATH . Then MATH is a constant scalar curvature metric (REF or REF implies this) in the Euclidean plane with finite area. Thus, MATH is a smooth metric in MATH with positive constant curvature. This contradicts the earlier assertion that any bubble metric must only have weak cusp singularities (since the scalar curvature must be negative near a cusp singular point). Therefore, there is no bubble point after a possible re-normalization of conformal transformation. In other words, there is a subsequence of MATH which converges weakly in MATH (up to conformal transformation group) to a metric MATH . REF or REF implies that MATH has constant scalar curvature. In the second case, MATH is either a torus or surface of higher genus. We can not re-normalize like in MATH case. There are two cases to handle, the case when MATH and the case when MATH . In the first case when MATH one can argue like in MATH the bubble metric if existed, must be a round sphere on one hand; and must have one cusp singularity on the other hand. This is impossible since the scalar curvature near cusp singular point must be negative. Therefore there are no bubble point in the first case and the limiting metric must have constant scalar curvature metric (compare REF or REF ). In the second case, we have MATH . We will show that this is impossible by drawing a contradiction. First, it is a ghost vertex in the tree decomposition of the limit of MATH . REF imply that the limit of MATH can be decomposed in a tree structure where each vertex represents a limit of a subsequence of MATH under a local re-normalization (compare REF ). REF or REF ensures that each vertex (non-ghost) corresponds to a metric with constant scalar curvature and finite area. Therefore, each non-ghost vertex must be a sphere! Consider the total NAME character in the limit tree structure, it should be non-positive since this is torus or high genus surface. On the other hand, the contribution from each non vanishing vertex is always positive (since they are MATH); thus the total NAME character in the remaining ghost vertexes must be strictly negative (actually less than MATH if there is one MATH in the limit!). Moreover, the total area concentration is MATH at each ghost vertex. Therefore, the total energy concentration in the ghost vertex must be infinite by the NAME inequality: MATH where MATH denotes the total collection of "ghost vertexes." MATH because MATH . However, the total energy of the tree structure is finite, this is a contradiction! Thus, MATH must converge to some constant scalar curvature metric in both torus and surfaces of high genus. |
math/0009246 | Write MATH . Note that the NAME energy, NAME energy and area are uniformly bounded along this sequence of metrics. Then we have MATH and MATH . Here MATH is some uniform constant in this proof, and its value may change from line to line. Since area is fixed along the flow: MATH . It follows that MATH . Since the underlying surface is torus or surface of higher genus, we may assume (without loss of generality) that MATH is a non-positive constant. Combining the previous inequality and REF , we obtain MATH . Following from the NAME Inequality, we have MATH . Combining this with REF , we arrive at MATH . Note that MATH . It then follows that MATH . In particular, MATH . In view of REF , there is no concentration point for this sequences. It follows that MATH always converges by sequence. REF or REF implies that the limit metric has constant scalar curvature metric. |
math/0009246 | Suppose MATH . Then MATH in MATH . Thus MATH in MATH for any MATH . In other words, MATH uniformly. The later in turn implies that first eigenvalues of MATH must converge to that of MATH . The last statement of the proposition just follows from the NAME inequality. |
math/0009246 | Since the NAME flow exists for long time, REF holds for time MATH for some constant MATH . Therefore there exists at least a sequence of numbers MATH such that MATH . REF implies that there exists a subsequence (denoted again by MATH), a subsequence of conformal transformations MATH and a constant scalar curvature metric MATH such that MATH weakly in MATH . Now REF implies that MATH and MATH . However, the NAME energy decreases monotonely under the NAME flow. The above inequality implies that MATH . For any sequence MATH . REF again implies that there exists a subsequence (denoted again by MATH), a sequence of conformal transformation MATH and a constant scalar curvature metric MATH such that MATH weakly in MATH . |
math/0009246 | Let MATH and let MATH denote any generic constant. Then MATH as MATH . We want to show that MATH decays exponentially fast (note that MATH in the following calculation): MATH . Now we need to consider two cases: MATH or MATH . The first case is easier, while the second case is more delicate. Consider the first case. If MATH then there exists some positive constant MATH and MATH such that MATH for some positive constant MATH . Thus, we have MATH . Next we consider the second case. If MATH then our normalization yields that MATH . We can rewrite REF : MATH . Since the first eigenvalue of MATH converges to MATH the above inequality seems to give us little control on the decay rate of MATH . However, by using NAME condition, we can still prove the above estimate in MATH . Notice that NAME condition implies MATH where MATH is any holomorphic vector field in MATH and MATH is the potential of the vector field MATH with respect to the metric MATH. The NAME condition can be re-written as MATH . Note that MATH converges to some constant scalar curvature metric in MATH in the MATH sense. This should be enough for our purpose. For MATH large enough, we can choose a constant scalar curvature metric MATH (which may depends on MATH) such that MATH and MATH as MATH . This means that the eigenvalue of MATH converges to those of some constant scalar curvature metric (whose first, second eigenvalues are MATH respectively). As REF implies, the eigenspace MATH of MATH corresponds to eigenvalues between MATH and MATH converges to the first eigenspace of MATH . If we decompose MATH into two components: MATH where MATH . Then the NAME REF condition implies that : MATH where MATH as MATH . Plugging this into REF , we have (note that MATH in the following calculation): MATH where MATH . Choose MATH small enough and MATH large enough (note that MATH), we have MATH . It follows that we can easily see the exponential decay of MATH norm of MATH: MATH . |
math/0009246 | We already know that there exists a family of conformal transformations MATH such that MATH converges to a constant scalar curvature metric MATH . Then the set MATH must be compact . Otherwise, suppose that MATH is a non-compact family of conformal transformations. By a direct calculation, one should yield: MATH . On the other hand, MATH . By a triangle inequality for distance function, we have MATH . However, MATH is invariant under the NAME flow, MATH is in the image of MATH under NAME flow at time MATH . By Theorem C, the distance in MATH decreases under the NAME flow. Consequently, we have MATH . This is a contradiction! Thus the set of conformal transformations MATH must be compact and the conformal factor must be bounded. |
math/0009249 | We shall write MATH and MATH; we are going to view MATH as the pre-Hilbert MATH-module described above, and also as a dense subspace of the NAME MATH-module MATH. For MATH, we define MATH by MATH . We shall prove that MATH extends to a unitary operator of MATH onto MATH which intertwines the given representations. We first prove that MATH is well-defined and isometric: for both, it suffices to show that MATH for MATH, MATH and MATH. (Inserting the function MATH allows us to deduce from the properness of the actions that the integrands in the following calculations have compact support.) To prove REF , we note that MATH . From this and an application of NAME 's Theorem, we deduce that MATH as desired. We next prove that MATH is surjective. We begin by observing that, with the pointwise action MATH and the inner product MATH is a pre-Hilbert MATH-module; we denote its completion by MATH. (Although we shall not need this, it might help to observe that MATH is naturally isomorphic to the external NAME tensor product MATH of, for example, CITE.) Writing out the formulas shows that the map MATH is inner-product preserving from MATH to MATH, and we can see by considering elementary tensors MATH that it has dense range. So we can view MATH as MATH. Functions MATH act as multipliers on MATH and hence also on MATH and MATH, where the action is given by MATH. Since MATH and since an isometry with dense range is surjective, it suffices for us to show that MATH is dense in MATH. If MATH has MATH, then using the MATH-norm to estimate the MATH-norm gives MATH . Thus it suffices to show that we can approximate MATH uniformly on a compact neighbourhood of MATH by functions in MATH with support in that neighbourhood. Because MATH acts freely and properly on MATH, the map MATH is a homeomorphism of MATH onto MATH. (The inverse is given by MATH, where MATH is the translation function characterised by MATH; a routine compactness argument shows that MATH is continuous.) Thus the map MATH defined by MATH is a linear isomorphism which preserves the kind of approximation we want. For MATH, we choose an extension MATH of MATH to a function of a compact support on MATH, and now standard arguments show that we can approximate MATH in MATH by functions MATH in MATH; then MATH is the required approximation to MATH. Thus MATH is surjective. It remains to check that MATH intertwines the given representations as claimed. Let MATH, and for this calculation denote the action MATH of MATH on MATH by MATH. Then for MATH, we have MATH . On the other hand, MATH . Since we have MATH the only difference between REF is the location of the integral with respect to MATH. But the integrands in both formulas are continuous and compactly supported, so there is no difficulty verifying that they have the same inner product with every vector of the form MATH, and hence must be equal. We deduce that MATH and we have proved the Theorem. |
math/0009249 | Every regular representation MATH induced from a faithful representation MATH of MATH has the same kernel MATH, and MATH has kernel MATH. We choose MATH to be the restriction MATH of a faithful nondegenerate representation of MATH. Then MATH is the restriction of the regular representation MATH of MATH, and we can apply REF with MATH. We deduce that, for this MATH, MATH is equivalent to the right-regular representation of MATH induced from the representation MATH of MATH. Thus MATH . (If we knew that MATH is faithful then we would have equality in REF above; instead we show that equality holds and deduce that MATH is faithful.) By symmetry MATH, and now applying MATH we see that MATH. It follows from standard properties of the NAME correspondence CITE that the reduced crossed products are NAME equivalent. Finally, if MATH is amenable, then MATH, so MATH and the system MATH is amenable; the last part follows by symmetry. |
math/0009249 | This is the special case of REF in which MATH; the dynamical system MATH is then trivially amenable. |
math/0009249 | REF immediately gives the ``if" direction. So suppose MATH is equivalent to the regular representation MATH for some representation MATH of MATH. Let MATH, and note that MATH is equivalent to MATH. REF implies that MATH is equivalent to MATH, and applying MATH shows that MATH is equivalent to MATH. |
math/0009249 | This follows from REF by taking advantage of the symmetry of the situation. Suppose that MATH is a representation of MATH instead. Let MATH and note that MATH is regular because it is equivalent to MATH. By REF MATH is the restriction of a covariant representation MATH of MATH. |
math/0009249 | We can use a partition of unity on MATH to write every function in the dense subalgebra MATH as a sum of functions supported on MATH-saturated open subsets of MATH which are trivial as MATH-bundles. Since MATH is nondegenerate and in particular nonzero, MATH must be nonzero on one of these sets. More formally, there is a MATH-saturated open set MATH such that there is a bundle isomorphism MATH, and such that MATH is not identically zero on MATH. Because MATH is MATH-saturated, MATH is invariant, and thus MATH is a nonzero MATH-invariant subspace of MATH. We now let MATH denote the composition of MATH with the projection onto the second factor MATH, and define MATH by MATH where MATH. Because MATH is nondegenerate on MATH it extends to a representation MATH of MATH. Since MATH is nondegenerate as a homomorphism into MATH it follows that MATH is a nondegenerate representation of MATH on MATH whose range commutes with every MATH. Since MATH is MATH-equivariant, MATH is equivariant for the action MATH of MATH on MATH and the action MATH of MATH on MATH; thus the covariance of MATH implies that MATH is a covariant representation of MATH. |
math/0009249 | From REF and a NAME 's Lemma argument, we obtain a decomposition MATH into MATH-invariant subspaces, each of which admits a suitable nondegenerate representation MATH. Now we just take MATH, and apply NAME 's imprimitivity theorem as described above. |
math/0009249 | To avoid having to write out the opposite version of REF , we instead prove the equivalent assertion that the regular representation MATH of the system MATH satisfies MATH. To do this, we apply REF with MATH absent. Then MATH is the bimodule MATH of CITE, MATH is the trivial bimodule MATH, and REF says that MATH . Since MATH, we have MATH, and CITE implies that MATH. Thus our choice of MATH implies that MATH, and MATH as required. |
math/0009249 | Suppose that MATH is amenable. Applying REF to MATH shows that MATH . Since MATH, MATH factors through a faithful representation MATH of MATH; the amenability of MATH implies that MATH is faithful, or, equivalently, that MATH. REF says that the NAME correspondence MATH carries MATH to MATH. Thus MATH . On the other hand, another application of CITE, with MATH missing this time, shows that the NAME correspondence MATH takes MATH to MATH. Thus REF gives MATH . So MATH factors through a faithful representation MATH of MATH, and the representation MATH appearing in REF factors through the regular representation MATH. Since we know from REF that MATH has kernel MATH, MATH must be faithful on MATH. Thus MATH is amenable. The result follows by symmetry. |
math/0009250 | We use induction on MATH; the case MATH follows directly from hypothesis PL REF . Suppose the result is true for MATH, and fix MATH, MATH with property MATH and let MATH be a tree on MATH isomorphic to MATH such that every MATH-subsequence of MATH satisfies MATH. Let MATH be the sequence of initial nodes of MATH with MATH for some MATH. Using PL REF we may find a subsequence MATH of MATH such that MATH has property MATH. Let MATH and set MATH. This tree is still isomorphic to MATH. Now for each MATH let MATH so that MATH and every MATH-subsequence of MATH satisfies MATH. By PL REF MATH has MATH and we may apply the induction hypothesis to obtain a subtree MATH of MATH isomorphic to MATH and having property MATH for MATH. It is easy to see that the tree MATH is the required subtree of MATH isomorphic to MATH with property MATH for MATH. Let MATH be a limit ordinal and suppose the result is true for every ordinal MATH. Let MATH be the sequence of ordinals increasing to MATH so that MATH. Let MATH, let MATH have MATH and let MATH be a tree isomorphic to MATH satisfying the requirements of the lemma. Let MATH be the subtree of MATH isomorphic to MATH and let MATH be the subtree of MATH isomorphic to MATH with property MATH for MATH. Let MATH be the sequence of initial nodes of MATH and let MATH for MATH. We have that MATH has property MATH for MATH, for each MATH, and so by condition PL REF we can find sequences MATH and MATH with MATH and such that MATH has property MATH. Let MATH. Now MATH has MATH for MATH and is still isomorphic to MATH. We now set MATH, then MATH and has property MATH for MATH. |
math/0009250 | Let MATH, and consider the following two possibilities: CASE: MATH; CASE: MATH. In REF so that MATH . For REF MATH and REF now follows from REF . |
math/0009250 | We may assume that MATH for some sequence MATH. We may also assume that MATH with MATH isomorphic to the replacement tree MATH. It is sufficient to find a sequence MATH and trees MATH of order MATH such that MATH satisfies the conditions of the lemma for each MATH. Let MATH and let MATH and sets MATH satisfy for MATH: MATH . Let MATH and choose MATH. Consider a subtree MATH of MATH isomorphic to MATH and let MATH, with MATH, be the defining map for the replacement tree. Then MATH is isomorphic to MATH. If MATH for all MATH, and every MATH, then MATH is the subtree we seek. If not, then there exists a terminal node MATH of MATH and MATH with MATH and MATH. Let MATH and MATH, so that the restricted tree MATH is isomorphic to MATH. Let MATH be the restriction of MATH to MATH so that MATH. Again, either MATH is the required tree, or else there is a terminal node MATH in MATH and MATH with MATH and MATH. Continuing in this way we obtain either a subtree MATH isomorphic to MATH such that MATH for each MATH, and every MATH, or else there is a branch MATH of MATH and normalized vectors MATH such that MATH for MATH. But then there exist MATH such that MATH, and hence MATH a contradiction. The last condition on MATH is clear from the proof. |
math/0009250 | Let MATH be the given MATH tree on MATH of order MATH and let MATH be a MATH tree on MATH of order MATH. Let MATH be the sequence of terminal nodes of MATH, so that MATH. Choose MATH, MATH and for each MATH find MATH such that MATH whenever MATH. Apply the previous lemma to MATH for MATH and MATH with MATH to obtain a tree MATH of order MATH such that for each node MATH we have MATH whenever MATH. From REF and the construction of MATH there exists MATH such that MATH is a MATH tree and hence by REF for property MATH, there exists MATH such that the tree MATH is a MATH tree on MATH. The tree MATH is the required MATH tree on MATH of order MATH. |
math/0009250 | If MATH, then, since the closure of a MATH tree is a MATH tree for some MATH by REF , it follows from REF that there exists an infinite sequence as in the statement of the theorem. Otherwise we assume the index is countable and let T be a MATH-tree on MATH of order MATH. By the previous lemma there exist numbers MATH and MATH-trees MATH on MATH of order MATH for MATH. Therefore the MATH-index is at least MATH. It follows that the MATH-index is MATH for some MATH. |
math/0009250 | We only need make a couple of modifications to the proof of REF . Instead of embedding MATH into MATH and using a basis there we use the basis MATH of MATH. Then in the proof of REF we must ensure that we construct a block basis tree. But this is easy. In REF , for each terminal node MATH we choose MATH so that MATH, and once we find the subtree MATH of MATH, we take a further subtree MATH which also has order MATH. Since we started with a block basis tree MATH, the subtrees MATH and MATH will also be block basis trees, as will the final tree. |
math/0009250 | REF are results of CITE and CITE stated in terms of the MATH-index. In particular REF follows from REF or CITE Corollary of REF , while REF follows from CITE and CITE. Before we can give the proof of REF we need some results on the relationship between MATH and MATH sequences, so we shall postpone the proof until we have these. For REF , if MATH is not shrinking, then it has a normalized block basis which is a MATH sequence, giving a MATH-block basis tree of order MATH, and the other direction follows from REF , since MATH is separable and the closure of a MATH-block basis tree is again a MATH-block basis tree. If MATH, then there exist MATH and a normalized block basis MATH of MATH so that MATH is a MATH sequence for each MATH. By a compactness argument it is easy to find MATH such that MATH, where MATH is the basis constant of MATH. Thus MATH is not boundedly complete. The converse is clear and REF follows. The proof of REF requires more work. First, if MATH is separable, then by NAME, CITE MATH embeds into a NAME space with a shrinking basis. Thus it is sufficient to prove that if MATH has a shrinking basis MATH, then MATH. If we show that MATH, then this would follow from REF . To show this we will take a MATH-weakly null tree on MATH of order MATH, and apply the Pruning Lemma to obtain a perturbation of a MATH-block basis tree on MATH of order MATH. Let MATH be a MATH-weakly null tree on MATH of order MATH, so that MATH is isomorphic to MATH. Define MATH to have property MATH if it is weakly null, and define MATH to have property MATH if it is a MATH-perturbation of a block basis of MATH. Using that MATH is shrinking it is standard work to show that MATH and MATH satisfy the requirements of the Pruning Lemma, and hence we may obtain a subtree MATH of order MATH which is a perturbation of a MATH-block basis tree of order MATH on MATH, that is, each terminal node is a MATH-perturbation of a MATH block basis of MATH. Next suppose that MATH is not separable; we shall show that MATH. Let MATH denote the NAME set, and let MATH be a sequence of subsets of MATH for MATH and MATH such that MATH and each MATH is the union of the disjoint, non-empty, clopen sets MATH and MATH, with MATH. NAME system on MATH (relative to MATH) is a sequence of continuous functions, MATH with MATH where MATH is the characteristic function of the set MATH. A sequence of continuous functions, MATH is a NAME system on MATH up to MATH if there exists a NAME system MATH on MATH such that for each MATH, MATH, MATH and MATH for every MATH. Given a NAME space MATH, and a subset MATH which is weak-MATH homeomorphic to the NAME set, a sequence MATH is a NAME system up to MATH, relative to MATH if the restrictions MATH form a NAME system up to MATH on MATH. By a result of CITE, since MATH is not separable, we have that given MATH there exists a set MATH which is weak-MATH homeomorphic to the NAME set, and a dyadic tree MATH of elements in MATH which form a NAME system up to MATH, relative to MATH. The dyadic tree is the natural one, with MATH if and only if MATH. Let MATH be the countably branching tree with MATH levels, ordered by MATH if and only if MATH and MATH for MATH. Choose MATH isomorphic to MATH so that if MATH has immediate successors MATH (that is, if MATH is equivalent to the node MATH of MATH, then MATH corresponds to MATH for MATH), then MATH . Since MATH does not contain MATH, it follows by CITE that we may prune MATH to obtain a further subtree MATH isomorphic to MATH so that if MATH is the sequence of immediate successors of MATH as above (or the sequence of initial nodes), then MATH is weak-Cauchy. Furthermore they will still satisfy the property above and we may assume that if we set MATH then MATH is REF-basic and weakly null. From this we can create a tree MATH which is isomorphic to MATH with nodes MATH as above and letting the immediate successors of such a node MATH be formed in the same manner from the successors in MATH of MATH. The resulting tree has the property that if MATH is a branch of MATH, then MATH for each MATH. Hence for each such branch there exists MATH such that MATH for every MATH so that MATH is a MATH sequence. Now, for all MATH, since MATH is isomorphic to MATH, it follows that we may find a subtree of MATH which is a weakly null tree isomorphic to MATH and so MATH as claimed. |
math/0009250 | We prove this by induction on MATH. There is nothing to prove for MATH, and if the result has been proven for every MATH, then it is also clear when MATH is a limit ordinal. Thus suppose the result has been proven for MATH, let MATH, and let MATH be a tree of order MATH. By taking a subtree we may assume that MATH has a unique initial node, MATH, and let MATH, a tree of order MATH. The terminal nodes of MATH are MATH, and let MATH be the sequence from the induction hypothesis on MATH for MATH and the sequence MATH. Now let MATH be the subtree of MATH, MATH. We have that MATH whenever MATH, MATH and MATH for MATH. We must now stabilize the maps on MATH for each MATH. Let MATH and for MATH let MATH . This forms a partition of the terminal nodes of MATH and by REF one of the trees MATH has order MATH for some MATH. The sequence MATH is now the required sequence. |
math/0009250 | We shall show that for each MATH there exists a MATH-tree on MATH of order MATH if and only if there exists a MATH-tree on MATH of order MATH. Let MATH and let MATH be a MATH-tree on MATH of order MATH for some MATH. Choose MATH and a sequence MATH decreasing rapidly to zero. Let MATH be the tree obtained when REF is applied to MATH with MATH where the functions MATH on MATH are in MATH with MATH for each terminal node MATH. We can find these functions since the terminal nodes are MATH sequences. Let MATH be the map MATH where MATH and let MATH . Clearly MATH has order MATH and is a MATH-wide-MATH tree by REF . Now let MATH, then MATH is a wide-MATH tree by REF above and clearly MATH is isomorphic to MATH so that MATH. Finally let MATH, then MATH is a MATH-tree by REF and isomorphic to MATH so it is of order MATH. This completes the proof of one implication. The other is similar. |
math/0009250 | From REF we know that MATH, and so it suffices to show (see for example, Monk CITE) that if MATH, then MATH. We may regard MATH as a subspace of MATH, and let MATH be a monotone basis for MATH. Let MATH be a MATH-weakly null tree on MATH of order MATH. To make a tree of order MATH we want to add a tree of order MATH after each terminal node of MATH. Let a sequence MATH in MATH have property MATH for MATH if it is normalized and weakly null. Let MATH have property MATH if it is a MATH perturbation of a normalized block basis of MATH. REF and MATH clearly satisfy conditions PL REF - REF of the Pruning Lemma. Note also that if we apply the Pruning Lemma to a tree MATH for MATH and a sequence MATH satisfying MATH, then the resulting sequences MATH are MATH-basic. We apply the Pruning Lemma to MATH with MATH and the empty sequence, so we may assume that for every MATH-node MATH of MATH with MATH, the sequence MATH is normalized, weakly null and a MATH perturbation of a normalized block basis of MATH. Now let MATH be the sequence of terminal nodes of MATH with MATH, and apply the Pruning Lemma to MATH for MATH with MATH and MATH, for each MATH, to obtain a tree MATH which has MATH for the sequence MATH. To complete the proof we put the trees together as follows. Let MATH . The ordering on MATH is that inherited from MATH and the trees MATH. It is easy to see that MATH is a MATH-weakly null tree. Indeed, if MATH, then the sequence is REF-basic and both MATH and MATH are MATH sequences so that MATH is a MATH sequence. Since we have extended only the terminal nodes it is clear that the new tree is weakly null. Finally, MATH, since MATH, and hence MATH. |
math/0009250 | We first show by induction on MATH that if MATH is a MATH-block basis tree on MATH of order MATH, then we can extract a certain subtree MATH isomorphic to MATH. The tree MATH will have the property that each MATH-subsequence MATH is a normalized block basis of the shrinking basis MATH and hence is weakly null. Thus MATH will be a MATH-weakly null tree, and so MATH. We prove this by induction on MATH. For MATH we may assume that MATH so that MATH consists of disjoint branches MATH of length at least MATH for each MATH. Since each branch is a block basis of MATH, there must exist a sequence MATH and a vector MATH in one of the nodes of branch MATH such that MATH is a normalized block basis of MATH. Set MATH for each MATH, then MATH, a sequence of incomparable nodes, is the required subtree. If the result has been proven for MATH, then we let MATH be a MATH-block basis tree with order MATH and assume MATH, since this is a minimal tree of order MATH. Recall that MATH is constructed by taking MATH, and then after each terminal node putting a tree isomorphic to MATH. Applying the above argument for MATH to the initial part of the tree, which is isomorphic to MATH, we may construct a sequence of incomparable nodes MATH, with MATH and MATH a normalized block basis of MATH. After each node MATH we have a tree isomorphic to MATH from which we may construct a MATH-weakly null subtree of order MATH with the required properties, using the induction hypothesis. Putting these trees together with the nodes MATH we obtain the desired MATH-weakly null tree. If MATH is a limit ordinal and the result has been proven for any ordinal smaller than MATH, then let MATH be a MATH-block basis tree on MATH isomorphic to the minimal tree MATH. Recall that MATH is constructed by taking a certain sequence of ordinals MATH increasing to MATH and letting MATH be the disjoint union of the trees MATH isomorphic to MATH. By the induction hypothesis we may find a subtree MATH of MATH for each MATH which is isomorphic to MATH and has the required properties. We must now be a little careful when putting the trees MATH together. We cannot just take their union since we will need the initial nodes to form a weakly null sequence. Let MATH be the sequence of initial nodes of MATH, with MATH. Since for each MATH the sequence MATH is a normalized block basis of MATH, it follows that we may find a sequence MATH and MATH such that MATH is a normalized block basis of MATH and MATH. Let MATH and set MATH. Then MATH is the required tree. This completes the first part of the proof and shows that MATH. We noted in the proof of REF that MATH, and since we know that both indices are of the form MATH for some MATH the result follows from the two inequalities. |
math/0009250 | As usual we use induction on MATH. For the initial case MATH, let MATH and let MATH be a weak-MATH relatively open neighborhood of MATH. We may find MATH such that MATH and MATH. Set MATH, then MATH is the required tree. We next suppose the result has been proven for MATH; let MATH, and let MATH be a weak-MATH relatively open neighborhood of MATH. We may find MATH with MATH and MATH. From the induction hypothesis, for each MATH there exists a tree MATH isomorphic to MATH satisfying the requirements for MATH. Let MATH so that the trees MATH are disjoint. Define MATH, then MATH and satisfies the requirements of the lemma for MATH. If MATH is a limit ordinal and the result has been proven for each MATH, let MATH be the sequence of successor ordinals increasing to MATH so that MATH. Let MATH (so that MATH for each MATH) and let MATH be a decreasing collection of weak-MATH relatively open neighborhoods of MATH, so that MATH, which may be chosen since MATH is separable. By the induction hypothesis we may find a tree MATH isomorphic to MATH satisfying the lemma for MATH and MATH for each MATH. Let MATH be the sequence of initial nodes of MATH with MATH, so that MATH and MATH for each MATH and every MATH. Since the sets MATH are decreasing we may find a subsequence MATH of MATH and numbers MATH with MATH for each MATH (where MATH is the function from the Pruning Lemma) such that MATH and MATH. It is at this limit ordinal stage that we use the relatively open neighborhoods MATH to ensure that we choose MATH, so that the order of the tree MATH below will be MATH. The tree MATH satisfies the requirements of the lemma. |
math/0009250 | Let MATH and let MATH be the tree for MATH from the previous lemma for MATH. Replace each node MATH with the node MATH to obtain the tree MATH which is still isomorphic to MATH. Clearly, if MATH is any MATH-subsequence, then MATH and MATH. Let MATH have property MATH if it is weak-MATH null with MATH, and property MATH if it is weak-MATH null with MATH for every MATH. It is clear that MATH and MATH satisfy condition PL REF of the Pruning Lemma, as modified by REF after it and we obtain condition PL REF using the fact that MATH is separable. Thus we may apply the Pruning Lemma and prune MATH to obtain a tree MATH with property MATH satisfying REF above. To see that REF holds, note that each node MATH is of the form MATH so that MATH. |
math/0009250 | Let MATH, to be chosen later. We first choose a sequence MATH with MATH for each MATH. Since MATH is separable we may assume MATH is weakly NAME (by taking a subsequence of MATH and then the same subsequence of MATH). Again, by taking subsequences and using that MATH, we may assume that MATH if MATH (where MATH is small, to be chosen later). Now set MATH so that, since MATH, it follows that MATH, and hence MATH and MATH. Further, for MATH, MATH . If MATH and MATH, then MATH and MATH when MATH. Next, since MATH, we may pass to subsequences of MATH and MATH to obtain MATH when MATH. We now pass to one last pair of subsequences MATH and MATH so that MATH is MATH basic for every MATH as required. |
math/0009250 | We have that MATH and we want to construct a MATH-weakly null tree on MATH of order MATH and constant MATH. Let MATH be the tree on MATH for some MATH from REF . We want to construct a tree MATH in MATH, isomorphic to MATH, so that if MATH has immediate successors MATH etc., and MATH are the corresponding nodes of MATH, then MATH is weakly null, MATH, MATH whenever MATH and MATH is MATH basic for every MATH. The proof is very similar to that of the Pruning Lemma, although a little stronger as we must keep track of two trees MATH and MATH, so we will not give it here. We claim that MATH is the required MATH-weakly null tree. We already know that MATH is a weakly null tree. We must show that if MATH, then MATH is a MATH-sequence. We know that MATH is MATH basic, so we seek MATH such that MATH for MATH. Let MATH be the corresponding node in MATH to MATH and recall that MATH. Now, MATH for MATH. Finally, setting MATH we still have MATH for each MATH, and hence MATH is a MATH-sequence with MATH. |
math/0009250 | As usual we proceed by induction on MATH. If MATH, then MATH where MATH and MATH is a normalized weakly null sequence. For each MATH pick MATH with MATH, then choose a subsequence MATH which converges weak-MATH to some MATH. Choose a sequence MATH; since MATH is weakly null, it follows that for each MATH there exists MATH such that MATH for every MATH. But now MATH so that MATH and hence MATH. The result for the case MATH follows easily from this. If the result has been proven for MATH, let MATH be a MATH-weakly null tree of order MATH and let MATH be a weak-MATH closed subset of MATH which contains a full set of branch functionals of MATH. Let MATH be the sequence of initial nodes of MATH with MATH and MATH a normalized weakly null sequence. For each MATH let MATH be the weak-MATH closure of the set of branch functionals of MATH in MATH for branches whose initial node is MATH. Thus MATH for every MATH. Further, let MATH, so that MATH is a MATH-weakly null tree of order MATH, and MATH is a weak-MATH closed subset of MATH which contains a full set of branch functionals of MATH. Hence, by the induction hypothesis, there exists MATH with MATH. Now, MATH for each MATH, so we may now proceed as in the case MATH to obtain MATH with MATH. Thus MATH. Then, since MATH is weak-MATH closed, and since MATH for each MATH, it follows that MATH as required. For the case where MATH is a limit ordinal we simply note that if the result has been proven for each MATH, and if we have MATH and MATH as in the statement of the lemma, then MATH for each MATH. This forms a countable decreasing sequence of non-empty weak-MATH closed sets in the weak-MATH compact set MATH. Thus MATH, which completes the proof. |
math/0009250 | If MATH is a MATH-weakly null tree on MATH of order MATH, then there exists a branch functional for each branch of MATH. Thus we may take MATH in the above lemma, to obtain MATH. |
math/0009250 | We first show that MATH is a monotone basic sequence, and then apply the NAME theorem to obtain that its span is all of MATH. Let MATH, then MATH . Since MATH is hereditary and the enumeration of MATH is admissible, it follows that there is an index MATH and MATH such that MATH. Next, observe that for all MATH with MATH, if MATH, then for MATH, MATH if and only if MATH and MATH (since MATH). Thus MATH and so MATH is a monotone basic sequence. To see that MATH we shall apply the NAME theorem. Since MATH for each MATH, it follows that MATH contains the constant function. It is easy to see that MATH separates the points of MATH, so it remains to show that the set contains the algebra generated by MATH. If MATH and MATH is not identically zero, then there exists MATH such that MATH, that is, MATH and MATH so that both MATH, and MATH. Hence either MATH, or MATH which gives MATH is MATH or MATH respectively. In either case we have that the algebra is contained in the linear span, as required, which completes the proof. |
math/0009250 | The heart of the proof lies in choosing an appropriate block basis of MATH. To do this we shall construct a tree isomorphism MATH from MATH into MATH by induction, and then the map from the basis MATH of MATH to a block basis of MATH will be given by MATH . This immediately gives the ordering requirement on MATH that if MATH and MATH, then MATH precedes MATH, that is, if MATH and MATH, then MATH. To help us write down the construction of MATH more explicitly we define subtrees MATH of MATH, for MATH, by MATH, with the order MATH, and as usual MATH is the order of the tree. Clearly MATH and MATH is a successor ordinal for each MATH, since MATH is the unique initial node. Let MATH be a non-terminal node of the tree MATH, so that there exists MATH with MATH. Then there are infinitely many sets MATH with MATH and MATH. Thus, if MATH, then MATH for every such set MATH. To simplify the notation for the induction we shall use an enumeration MATH of MATH, the domain of MATH, which satisfies: MATH, MATH, and if MATH, then MATH. Now let us inductively define MATH. Let MATH, MATH, and set MATH. Suppose that MATH has been defined for MATH such that if MATH, MATH and MATH, then MATH, MATH if and only if MATH and for MATH, MATH. We next define MATH. From our enumeration MATH of MATH there exists MATH such that MATH, or MATH for some MATH. In the first case let MATH be the least integer such that MATH and MATH. We can achieve this last condition because MATH. In the second case let MATH be least with MATH, MATH, and MATH. The existence of MATH is guaranteed by the conditions on MATH. Indeed, since MATH, it follows that MATH is not a terminal node of MATH so that MATH. We also have MATH, so that neither is MATH a terminal node of MATH and hence MATH has an infinite sequence of successor nodes MATH such that MATH, MATH, MATH for each MATH and MATH, where MATH. If MATH is a successor ordinal, then choose MATH so that MATH . Otherwise MATH, and we may choose MATH so that MATH. We then set MATH. Clearly MATH satisfies all the requirements of the induction. It remains to show that this is sufficient to ensure that the map MATH is an isometry. We must show that MATH. Now, MATH . First note that for MATH fixed, if there exist MATH such that MATH, and MATH, then both MATH and MATH so that we may assume MATH. By the conditions on MATH this forces MATH, but MATH so that MATH, and hence MATH. Thus, for each MATH, MATH we have MATH or MATH. To complete the proof we define a map MATH from MATH into the collection of finite subsets on MATH by MATH and show that the range of MATH is MATH, for then MATH as required. First note that if MATH, then MATH. Indeed, MATH . Now fix MATH and let MATH satisfy MATH and MATH. Then MATH, so that MATH, that is, MATH. On the other hand, if MATH, then MATH which gives MATH for some MATH, so that MATH by REF , and hence the two sets are equal. Thus MATH is contained in the range of MATH. Finally let MATH, set MATH and find MATH such that MATH and MATH. But then MATH, while MATH. Hence MATH and so the range of MATH is exactly MATH as required. This completes the proof. |
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