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math/0005176
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Let MATH denote the first NAME class of the given complex structure MATH, and, by a standard abuse of notation, let MATH also denote the pull-back class of this class to MATH. If MATH are the NAME duals of the exceptional divisors in MATH introduced by blowing up, the complex structure MATH has NAME class MATH . By a result of CITE, this is a monopole class of MATH. However, there are self-diffeomorphisms of MATH which act on MATH in a manner such that MATH for any choice of signs we like. Thus MATH is a monopole class on MATH for each choice of signs. We now fix our choice of signs so that MATH for each MATH, with respect to the decomposition induced by the given metric MATH. We then have MATH . This shows that MATH . If equality held, MATH would be almost-Kähler, with almost-Kähler class MATH proportional to MATH. On the other hand, we would also have MATH, so it would then follow that MATH for all MATH. However, the NAME invariant would be non-trivial for a spin-MATH structure with MATH, and REF would then force the homology class MATH to be represented by a pseudo-holomorphic MATH-sphere in the symplectic manifold MATH. But the (positive!) area of this sphere with respect to MATH would then be exactly MATH, contradicting the observation that MATH.
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math/0005176
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We may assume that MATH, since otherwise the result follows from the NAME inequality. Now MATH for any metric on MATH on MATH. If MATH is an NAME metric, the trace-free part MATH of the NAME curvature vanishes, and we then have MATH by REF . If MATH carries an NAME metric, it therefore follows that MATH . The claim thus follows by contraposition.
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math/0005179
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As MATH, MATH we get that MATH. Hence MATH.
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math/0005179
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We carry the proof inside MATH. The first point to observe is that MATH. So if MATH and MATH then MATH. Let MATH be a club of MATH. Hence MATH. For simplicity of notation let us assume MATH. The proposition follows from MATH .
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math/0005179
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We carry the proof inside MATH. The first point to observe is that MATH. So if MATH, MATH then MATH. Let MATH be a club of MATH. Then MATH. For simplicity of notation let us assume MATH. The proposition follows from MATH .
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math/0005179
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We carry the proof inside MATH. The point to observe is that if MATH, MATH then MATH. Let MATH be a club of MATH. Then MATH. For simplicity of notation let us assume MATH. The proposition follows from MATH .
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math/0005179
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We carry the proof inside MATH. The first point to observe is that MATH. The same technique as for the proof of REF is to be used. In MATH steps we can show that MATH, MATH, MATH.
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math/0005179
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Let MATH. As MATH we can assume without loss of generality that there is MATH such that MATH. Hence we can ignore these lower parts and assume that the set of conditions we start with is MATH. Let MATH. As MATH, MATH we can invoke the MATH-lemma. Hence, without loss of generality, there is MATH such that MATH. As MATH we can assume, without loss of generality, MATH. The MATH's, MATH's are always compatible. Hence we are left with handling of MATH. MATH and MATH satisfies MATH-c.c. Hence there are MATH such that MATH . Hence MATH.
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math/0005179
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Let MATH. Set MATH . We set MATH. It is clear that MATH, MATH, MATH. Let MATH. We define the condition MATH to be MATH with MATH substituted for MATH. Let MATH such that MATH, MATH. Of course the part below MATH poses no problem. So we are left to show that there is MATH such that MATH. By the definition of MATH there is MATH such that MATH. Let MATH be the last such that MATH. (MATH can be MATH.) If MATH we choose MATH. We split the handling according to where MATH is. CASE: MATH: MATH, MATH. As MATH we have MATH. Hence MATH. CASE: MATH: So MATH. In particular there is MATH. Let MATH. Then MATH. Once more, as MATH we have MATH, hence MATH. That is MATH.
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math/0005179
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Fix MATH. Then we can invoke REF on MATH, MATH, MATH to get MATH, MATH such that MATH is pre-dense below MATH. We do the above for all MATH. We let MATH. Of course, MATH is MATH with MATH substituted by MATH. We point out that MATH. We invoke MATH with MATH, MATH, MATH to get MATH, MATH such that MATH is pre-dense below MATH. We set MATH, MATH. We claim that MATH is pre-dense below MATH. Let MATH. Then there is MATH such that MATH. As MATH we have MATH. Choose MATH. There is MATH such that MATH. As MATH, MATH we get MATH. We complete the proof by noting that MATH was constructed such that MATH.
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math/0005179
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We give the proof for MATH. It is essentially the same for all MATH. Let MATH be large enough so that MATH catch `what interests us' and let MATH . Choose MATH such that MATH and set MATH where MATH. Let MATH be a well ordering of MATH such that MATH . We shrink MATH a bit so that the following is satisfied MATH . We start an induction on MATH in which we build MATH . Assume that we have constructed MATH . Set the following: CASE: MATH is MATH-minimal: MATH . CASE: MATH is the immediate MATH-successor of MATH: MATH . CASE: MATH is MATH-limit: Choose MATH such that MATH and set MATH . We start an induction on MATH. We construct in it MATH such that MATH is a maximal anti-chain in MATH below MATH. Assume we have constructed MATH and we do step MATH. CASE: MATH: MATH . CASE: MATH: If MATH is a maximal anti-chain below MATH then we finish the induction on MATH. Otherwise we choose MATH, MATH such that MATH for all MATH. We choose MATH such that MATH for each inaccessible MATH. MATH . CASE: MATH is limit: If MATH is a maximal anti-chain below MATH then we finish the induction on MATH. Otherwise we choose MATH, MATH such that MATH for all MATH. We choose MATH such that MATH for each inaccessible MATH. Choose MATH such that MATH and set MATH . Set MATH . If there is MATH such that MATH then set MATH otherwise we set MATH . When the induction on MATH terminates we have MATH . We complete step MATH by setting MATH . When the induction on MATH terminates we have MATH . We define the following function with domain MATH: MATH . By the construction MATH is a maximal anti-chain in MATH below MATH. Hence, MATH is a maximal anti-chain below MATH. By genericity of MATH over MATH, there are MATH's such that MATH and MATH is stronger than a condition in MATH. Let MATH be such that MATH. Note that we can use MATH here because the generic was built through the normal ultrafilter. If we would not have had this property we would have enlarged MATH to accommodate the intersection and we might have needed different MATH for each MATH. We combine the information gathered into MATH conditions. MATH . For each MATH we define MATH . We note that the construction ensures us that MATH, MATH when MATH and MATH and MATH do not contain contradictory information. Hence we can define for each MATH the following MATH . We consider the following sets MATH . There are MATH possibilities at this point CASE: There is MATH such that MATH: Of course, MATH satisfies the requested conclusion. Hence we set MATH and the theorem is proved. CASE: MATH: We claim that some shrinkage of MATH is enough to get us into clause REF. Let us assume, by contradiction, that a small shrinkage can not bring us to MATH. This means that there is MATH such that MATH . Let MATH and MATH be MATH with MATH substituted for MATH. Due to openness of MATH we still have MATH . Hence, by the construction, we have MATH . Invoking MATH on the above set yields MATH . Now, from the construction of MATH we see that MATH . Combining the above MATH formulas and recalling that MATH is open we get MATH . That is MATH. Contradiction. So, we have shown that MATH . By letting MATH be MATH with MATH substituted by MATH we get clause MATH.
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math/0005179
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Construct, by repeat invocation of REF for each MATH, a MATH-decreasing sequence MATH. Let MATH for all MATH. Choose MATH such that MATH. There is MATH such that MATH. By this we eliminated clause REF for MATH.
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math/0005179
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Of course this is completely trivial as MATH. In this case MATH and MATH are the same.
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math/0005179
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Let MATH. As MATH is MATH-closed we get immediately that MATH. We have much more than that. Namely, MATH are not collapsed. For this we remind the reader that the MATH we work with is a generic extension of MATH for a reverse NAME forcing. Let MATH be the reverse NAME forcing up to MATH and MATH be its generic. Let MATH be the forcing at stage MATH and MATH be its generic over MATH. Let MATH be the rest of the reverse NAME forcing up to MATH and MATH be its generic over MATH. Then we have MATH . Note that MATH is a reflection of the situation at REF and by this we see that MATH are not collapsed. In fact we see that nothing has changed as far as the definition of MATH in MATH is concerned. (We might have new anti-chains which is no obstacle to us).
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math/0005179
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In order to avoid excess of indices we give the proof of the case MATH. Let MATH be MATH-generic with MATH and let MATH be dense open in MATH. Then MATH is a dense open subset of MATH. By REF for MATH there are MATH and MATH, a MATH-fat tree, such that MATH and MATH . Hence there is MATH which forces the above. That is for each MATH there is a maximal anti-chain, MATH, of MATH below MATH such that MATH . By noting that MATH is MATH-closed and that MATH we see that there is MATH such that MATH .
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math/0005179
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Let MATH. Then MATH is dense open in MATH below MATH where MATH. By REF there are MATH, MATH, MATH such that MATH . By the definition of MATH we see that there is a function MATH with domain MATH such that MATH . As MATH there is MATH such that except on a measure MATH set we have MATH. By removing this measure MATH set from MATH we get MATH .
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math/0005179
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Our first observation is that MATH is MATH-c.c. (As opposed to the usual MATH-c.c. we have when MATH). We construct, by induction, the sequence MATH where MATH. Together with it we construct an auxiliary MATH-decreasing sequence MATH. CASE: MATH: By REF and openness of MATH there are MATH, MATH, MATH such that MATH and MATH is pre-dense below MATH. CASE: MATH: If MATH is a maximal anti-chain below MATH then the induction is finished. If it is not a maximal anti-chain we observe that MATH as MATH is MATH-c.c. Let MATH be such that MATH. As MATH is MATH-closed there is MATH such that MATH for all MATH. By REF starting from MATH there are MATH, MATH, MATH such that MATH and MATH is pre-dense below MATH. When the induction terminates we have a MATH-decreasing sequence MATH where MATH. By choosing MATH for all MATH we finish the proof.
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math/0005179
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In order to avoid too many indices we prove the lemma for the case MATH. Choose MATH large enough so that MATH contains everything we are interested in. Let MATH be such that MATH, MATH, MATH, MATH. Let MATH. Choose MATH such that MATH for all MATH. Let MATH. Note that MATH. Let MATH be a well ordering of MATH such that MATH. We shrink MATH a bit so that the following is satisfied: MATH. We start an induction on MATH in which we build MATH . Assume that we have constructed MATH. We start working in MATH. Set the following: CASE: MATH is MATH-minimal: MATH . CASE: MATH is the immediate MATH-successor of MATH: MATH . CASE: MATH is MATH-limit: Choose MATH such that MATH and set MATH . We make an induction on MATH which builds MATH. Assume we have constructed MATH and we do step MATH. CASE: MATH: MATH . CASE: MATH: If MATH is a maximal anti-chain below MATH we terminate the induction on MATH. Otherwise we choose MATH such that MATH and MATH. We make sure to choose the MATH such that if MATH is an inaccessible then MATH. We set MATH . CASE: MATH is limit: If MATH is a maximal anti-chain below MATH we terminate the induction on MATH. Otherwise we choose MATH such that MATH for all MATH and MATH. We make sure to choose the MATH such that if MATH is an inaccessible then MATH. Choose MATH such that MATH. We set MATH . Set MATH . Using the corollary of REF construct MATH and MATH a maximal anti-chain below MATH. So for each MATH there is MATH, a MATH-tree, such that MATH . We set MATH . When the induction on MATH terminates we have MATH . We complete step MATH by setting MATH . When the induction on MATH terminates we return to work in MATH and we have MATH . We define the following function with domain MATH: MATH . By the construction, MATH is a maximal anti-chain below MATH. We note that MATH as MATH is closed under MATH-sequences. Hence MATH as MATH. So MATH is a maximal anti-chain below MATH. As MATH is MATH-generic over MATH, there is MATH such that MATH and MATH is stronger than a condition in MATH. Note that we can use MATH here because the generic was build through the normal measure. If we would not have had this property we would have enlarged MATH to accommodate the intersection. We combine everything into one condition, MATH, as follows: MATH . We write what we have gained so far: For each MATH there is MATH, a maximal anti-chain below MATH, such that for each MATH there is MATH, a MATH-tree, such that MATH . We set MATH. Then MATH is a dense open subset of MATH. By invoking REF for MATH, MATH we find MATH, MATH such that MATH . Immediately we see that there are MATH, MATH such that by removing a measure MATH set from MATH we get MATH, MATH. So after the shrinkage of MATH we have for each MATH . We gather the additional information we have by setting MATH and letting the condition MATH be MATH with MATH substituted for MATH. So at this point we have the following MATH . Of course as MATH the above is just a convoluted form of MATH .
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math/0005179
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The proof is done by induction on MATH. The case MATH is done in REF for MATH and REF for MATH. The case MATH is done in REF for MATH and REF for MATH.
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math/0005179
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Let MATH . Then MATH . Let MATH be the MATH, MATH such that MATH . We choose MATH such that for all MATH, for all MATH where MATH according to the selection of MATH. By REF, for each MATH there are MATH, MATH such that below MATH is pre-dense. Hence MATH . Let MATH. Let MATH be the condition MATH with its measure MATH set substituted by MATH. We get that for all MATH . Letting MATH be MATH restricted to MATH levels bring us to the beginning of the proof but with MATH instead of MATH. Hence, repeating another MATH steps as the above build MATH and MATH.
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math/0005179
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As usual we give the proof for the case MATH. Let MATH. MATH is a dense open subset in MATH. By REF there is MATH such that MATH . We use the above formula to fix MATH. Then for each MATH we fix MATH, MATH, MATH, MATH. In the same way we fix MATH, MATH, MATH for each MATH. By REF for each MATH there is MATH such that MATH . We choose MATH for all MATH. Hence we get MATH . Invoking REF again we get MATH such that MATH.
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math/0005179
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Let MATH be large enough, MATH, MATH, MATH, MATH, MATH, MATH. We find MATH such that MATH is MATH-generic. Let MATH be enumeration of all dense open subsets of MATH appearing in MATH. Note that for MATH we have MATH. Let MATH. Choose MATH such that MATH for all MATH. Let MATH. We shrink MATH a bit so that the following is satisfied: MATH. Let MATH. Elements of MATH are written in the form MATH. That is MATH. By MATH we mean MATH. Let MATH be well ordering of MATH such that MATH. We start an induction on MATH in which we build MATH . Assume that we have constructed MATH. Recall our convention: MATH. We start working in MATH. Set the following: CASE: MATH is MATH-minimal: MATH . CASE: MATH is the immediate MATH-successor of MATH: MATH . CASE: MATH is MATH-limit: Choose MATH such that MATH and set MATH . We make an induction on MATH which builds MATH. Assume we have constructed MATH and we do step MATH. CASE: MATH: MATH . CASE: MATH: If MATH is a maximal anti-chain below MATH we terminate the induction on MATH. Otherwise we pick MATH such that MATH and MATH. We make sure to choose the MATH such that if MATH is an inaccessible then MATH. We set MATH . CASE: MATH is limit: If MATH is a maximal anti-chain below MATH we terminate the induction on MATH. Otherwise we pick MATH such that MATH and MATH. We make sure to choose the MATH such that if MATH is an inaccessible then MATH. Choose MATH such that MATH . We set MATH . Set MATH . We construct MATH by invoking REF repeatedly for each MATH starting from MATH. We write explicitly what we have here: For each MATH there is MATH, a maximal anti-chain below MATH such that for each MATH there are MATH, MATH, a MATH-fat tree such that MATH and MATH is pre-dense below MATH. We set MATH . When the induction on MATH terminates we have MATH . We complete step MATH by setting MATH . When the induction on MATH terminates we return to work in MATH and we have MATH . For each MATH we define the following function with domain MATH . So MATH is a maximal anti-chain below MATH. We note that MATH as MATH is closed under MATH-sequence. As MATH we get MATH. So MATH is a maximal anti-chain below MATH. As MATH is MATH-generic over MATH, there is MATH such that MATH and MATH is stronger than a condition in MATH. Note that we can use MATH here because the generic was built through the normal measure. If we would not have had this property we would have enlarged MATH to accommodate the intersection. We combine everything into one condition, MATH, as follows: MATH . We claim that MATH is MATH-generic. So, let MATH be MATH-generic with MATH. Let MATH be a dense open subset of MATH. There is MATH such that MATH. Let MATH be such that MATH and MATH. For convenience let us set MATH. By the construction there is MATH, a maximal anti-chain below MATH, such that for each MATH there are MATH, MATH and MATH, a MATH-fat tree, such that MATH and MATH is pre-dense below MATH. Moreover, this pre-dense set is contained in MATH. We do the natural factoring MATH. By genericity there is MATH. Necessarily MATH. Hence MATH. So there is MATH such that MATH. So we finally got MATH.
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math/0005179
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If MATH then the claim is trivial. Hence we assume that MATH. Let MATH. Choose MATH large enough so that MATH contains everything we are interested in. By REF there are MATH, MATH such that CASE: MATH, CASE: MATH, CASE: MATH, CASE: MATH, CASE: MATH is MATH-generic. Let us set MATH. Note that MATH. Let MATH be MATH-generic with MATH. The MATH-genericity ensures us that for all MATH, MATH. Hence MATH. That is MATH.
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math/0005179
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MATH is not collapsed by REF. No cardinals MATH are collapsed as MATH satisfies MATH-c.c.
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math/0005179
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Let MATH be MATH-generic. For each MATH define MATH. It is routine to check that for MATH we have MATH. Hence MATH. For the other direction let MATH. By REF there are MATH large enough, MATH, MATH, MATH and MATH such that MATH is MATH-generic over N where MATH. Hence, MATH. That is MATH. As MATH when MATH and MATH the proof is completed.
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math/0005179
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This is immediate due to MATH and REF.
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math/0005179
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Let MATH be the natural factoring. By REF the fate of the cardinals in question is decided by MATH. We note that MATH where MATH. So we factor MATH. As it stands MATH is MATH-closed. So in order to prove the claim we make some finer analysis. We remind the reader that the MATH we work with is a generic extension of MATH for a reverse NAME forcing. Let MATH be the reverse NAME forcing up to MATH and MATH be its generic. Let MATH be the forcing at stage MATH and MATH be its generic over MATH. Let MATH be the rest of the reverse NAME forcing up to MATH and MATH be its generic over MATH. Then we have MATH . Comparison of the forcings used to construct MATH REF and MATH shows that the cardinal structure and power function of the model MATH in the range MATH behave in the same way as the cardinal structure and power function of the model MATH in the range MATH. From REF we see that in MATH: there are no cardinals in MATH, MATH are cardinals, MATH, MATH, MATH, MATH, MATH, MATH. Forcing with MATH does not change the power function and does not collapse cardinals above MATH by the previous claims adapted to the current context.
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math/0005179
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MATH is limit ordinal and by REF, there are unbounded number of cardinals below MATH which are preserved. Hence MATH is preserved.
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math/0005179
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Let MATH, MATH. Let MATH be large enough. By REF we have MATH, MATH, MATH, MATH, MATH, MATH such that MATH for each MATH. Choose MATH such that MATH, MATH. This is possible because MATH, MATH. Choose MATH such that MATH. Let MATH be MATH-generic with MATH. Of course, MATH also. Hence, MATH is MATH-generic over MATH. As MATH we have MATH. So there is MATH, MATH such that MATH.
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math/0005179
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Let MATH such that MATH is inaccessible and MATH be MATH-generic with MATH. (Forcing below an element of MATH eliminates a finite number of exceptions which we might otherwise have. That is if MATH and MATH then the interval MATH is untouched by the forcing). We set MATH . Note that MATH is a NAME generic sequence for the extender sequence MATH. Hence MATH is a club. The first ordinal in this club is MATH. We investigate the range MATH in MATH. We note that, by REF, for MATH it is enough to use MATH in order to understand MATH. So let MATH, MATH. CASE: MATH: Then there is MATH such that MATH and MATH. By REF, MATH remains a cardinal and by REF, MATH. CASE: MATH: Then there is MATH such that MATH and MATH. Let MATH be the MATH-immediate predecessor of MATH. By REF we have: MATH are cardinals, there are no cardinals in MATH, MATH, MATH, MATH, MATH, MATH, MATH. In fact due to all the cardinals collapsed we have MATH. Hence if MATH is a cardinal then MATH. By REF, MATH is an inaccessible cardinal. Let MATH be MATH-generic over MATH. In MATH remains inaccessible and MATH. So MATH is a model of ZFC satisfying MATH.
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math/0005180
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Let MATH be valleyless, then either MATH or MATH. Add one to every entry of MATH and then append REF to the beginning or end. The new permutation is valleyless of length MATH. The process is reversible so there are twice as many valleyless permutations of length MATH as there are of length MATH. Since MATH we have MATH and we are done.
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math/0005180
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Suppose that a permutation MATH with inversion table MATH is valleyless. If MATH for some MATH, MATH, then there is at least one symbol in MATH to the left of MATH that is greater than MATH. If there is any symbol in MATH greater than MATH to the right of MATH, then MATH would be a valley contradicting our assumption. Hence all numbers greater than MATH must be to the left of MATH and MATH. Conversely, suppose that MATH and MATH or MATH, MATH. We want to show that MATH belongs to MATH. Suppose not, then there is at least one symbol MATH in MATH such that MATH . Then MATH and MATH contradicting our assumption. Hence MATH belongs to MATH and we are done.
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math/0005181
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Pick MATH, MATH and MATH such that MATH is a MATH quasi-isometry, MATH, and MATH are uniformity data for MATH. Consider MATH such that MATH. We have MATH, and so MATH, from which it follows that MATH. Since MATH we obtain a bound MATH depending only on MATH. The usual ``rubber band" argument, using geodesics in MATH divided into subsegments of length MATH with a terminal subsegment of length MATH, suffices to prove that MATH is MATH coarsely lipschitz, with MATH depending only on MATH. For any MATH there is a point MATH such that MATH. For any MATH with MATH we have MATH and so MATH. As above we obtain an upper bound for MATH and the rubber band argument shows that MATH is coarsely lipschitz. For any MATH, setting MATH, we have MATH . It follows that MATH and so MATH, yielding an upper bound for MATH. Similarly, MATH is bounded for all MATH. Knowing that MATH and MATH are coarse lipschitz maps which are coarse inverses of each other, it easily follows that MATH is a quasi-isometry, with quasi-isometry constants depending only on the coarse lipschitz constants for MATH and MATH, and on the coarse inverse constants for MATH.
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math/0005181
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We start with the case when MATH is a REF-parameter NAME subgroup, and the proposition follows by examining each NAME block REF. The second case we consider is when MATH has all positive real eigenvalues. By REF we have MATH, and REF follows immediately from the first case applied to MATH, together with the fact that MATH takes the NAME decomposition of MATH to the NAME decomposition of MATH. In the general case, applying REF we have MATH. We can the apply the second case to MATH. Since MATH commutes with MATH it follows that MATH preserves the NAME decomposition of MATH. REF then follows from the boundedness of MATH.
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math/0005181
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We proceed in cases. CASE: Assume that MATH is the unique REF-parameter NAME subgroup such that MATH is conjugate to the absolute NAME form of MATH. Applying REF we have MATH where MATH and REF-parameter subgroup MATH is bounded. Choose MATH and MATH. We must show that the two numbers MATH and MATH have ratio bounded away from MATH and MATH, with bound independent of MATH. Setting MATH, it suffices to show that MATH and MATH have bounded ratio. But this is clearly true, with a bound of MATH since REF-parameter subgoup MATH is bounded. CASE: Assume that there exists MATH such that MATH for all MATH. Then the metrics MATH and MATH are identical. General case: Applying REF we may assume that MATH. Applying REF twice we may go from MATH to MATH to MATH, where MATH is conjugate to the absolute NAME form of MATH and of MATH.
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math/0005181
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We have MATH. The reverse inequality is similar, and so MATH is a quasi-isometric embedding. Since MATH is coarsely onto, an easy argument shows MATH is coarsely onto.
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math/0005181
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By construction, the foliations MATH and MATH are coordinate foliations in MATH; this shows that the flow MATH has a ``global product structure" in the language of hyperbolic dynamical systems. The lemma now follows the proof of the Shadowing Lemma in REF , page REF. A direct proof is also easy to work out, and is left to the reader.
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math/0005181
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Let MATH be some center leaf of MATH, of dimension MATH. From REF it follows that MATH is NAME close to some center leaf MATH of MATH, also of dimension MATH. By composition with nearest point projection (which moves points a uniformly bounded amount) we get an induced map MATH. By REF this map is a quasi-isometry. By REF , MATH and MATH are quasi-isometric to the nilpotent NAME groups MATH and MATH, respectively. As NAME and NAME have shown CITE, NAME 's invariant CITE may be used to prove that MATH.
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math/0005181
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We begin with: For each vertical flow line MATH in MATH, there exists a center leaf MATH in MATH such that MATH is contained in the MATH-neighborhood of MATH, where the constant MATH does not depend on MATH. Before proving the claim, we apply it to prove the proposition as follows. Consider any two vertical flow lines MATH in MATH. By the claim we have that MATH and MATH lie, respectively, in bounded neighborhoods of center leaves MATH and MATH of MATH. Since MATH as MATH, for each choice of sign MATH or MATH the following two statements are equivalent, and the second statement implies the third: CASE: The distance between the points MATH and MATH in MATH stays bounded as MATH. CASE: The distance between the points MATH and MATH in MATH stays bounded as MATH. CASE: The NAME distance between the sets MATH and MATH in MATH stays bounded as MATH. Using MATH signs, the first statement is equivalent to saying that MATH are contained in the same unstable leaf of MATH, and the third statement is equivalent to saying that MATH are contained in the same unstable leaf of MATH. It follows that MATH takes every unstable leaf of MATH into a bounded neighborhood of an unstable leaf of MATH. Applying the same argument to a coarse inverse MATH of MATH gives the opposite inclusion. Since MATH it follows that the image under MATH of any unstable leaf of MATH lies a bounded NAME distance from an unstable leaf of MATH, that is, MATH coarsely preserves the unstable foliations. A similar argument using MATH signs shows that MATH coarsely preserves stable foliations. By taking intersections of stable and unstable leaves it follows that MATH coarsely preserves center foliations. The final statements about dimensions follow from the fact that dimension is a quasi-isometry invariant, for leaves of the foliations in question; see CITE or CITE. It remains to prove the claim. Applying REF , we have an induced time change MATH which is a MATH-quasi-isometry with NAME constant MATH, where MATH depends only on MATH. Furthermore by REF and the comments following it, the map MATH is coarsely increasing: there exists MATH such that if MATH then MATH. We can furthermore increase MATH, depending only on MATH, so that: MATH . In fact taking MATH will do, for then we have MATH and, since MATH is MATH-Hausdorff close to MATH and MATH is MATH-Hausdorff close to MATH, it follows that MATH. To prove the claim, we first show that MATH is NAME close to some pseudo-orbit in MATH, and then we apply the Shadowing Lemma to show that the pseudo-orbit lies in a bounded neighborhood of some center leaf. To be more precise, fix a point MATH and consider the sequence MATH for MATH. Let MATH, and let MATH be such that MATH. From REF it follows that MATH. Let MATH. We claim that there exists MATH, depending ultimately only on MATH, so that MATH is a MATH-pseudo-orbit; in other words, MATH is bounded. To see why, first note that MATH and then MATH so we may take MATH. Applying the Shadowing Lemma, there exists MATH such that MATH is MATH-Hausdorff close to a MATH-pseudo-orbit MATH contained in some center leaf of MATH. On the other hand, since every point of MATH is within distance MATH of some MATH, it follows that MATH is uniformly NAME close to MATH, and so it is also uniformly close to the pseudo-orbit MATH.
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math/0005181
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This proof will define a sequence of constants which will depend on MATH and on the matrices MATH and MATH. We will indicate the dependence on MATH by writing, for example, MATH, but we will suppress the dependence on MATH. Although each constant in the sequence will depend on previous constants in the sequence, by induction it will ultimately depend only on MATH. For each fixed time MATH, and for each MATH, we have MATH for some MATH. Accepting this claim for the moment, we prove the proposition. The idea is simply that the conclusion of the claim, applied to both MATH and its coarse inverse MATH, with MATH, implies the proposition. Let MATH be a time parameter for MATH. Let MATH be a coarse inverse for MATH, also a quasi-isometry which coarsely respects the horizontal foliations and their transverse orientations, and with an induced time change MATH. The constants for MATH and MATH depend only on MATH. The claim therefore applies as well to MATH and we obtain, for each fixed time MATH and each MATH, MATH for some MATH. It is clear MATH is a coarse inverse for MATH, that is: MATH for some MATH. Also, by REF and the comments after it, the map MATH is coarsely increasing: there exists MATH such that if MATH then MATH. We reverse the inequality in the claim as follows. Fix MATH. Let MATH. Consider for the moment some MATH. Letting MATH it follows that MATH and so we have MATH . But MATH and MATH and so we obtain MATH . This has been derived only for MATH, but for MATH we obtain a similar inequality with another constant in place of MATH. Therefore, for all MATH we obtain MATH for some MATH. Note that this is true for all MATH, with MATH independent of MATH. In particular, taking MATH, for all MATH we obtain MATH . Now take any MATH, and since MATH we obtain MATH and so MATH . Taking MATH, this proves that MATH is an induced time change for MATH, with NAME constant MATH. Now we turn to the proof of REF . Let MATH, MATH be the real NAME forms. Let MATH (respectively, MATH) be the root space with eigenvalue MATH for MATH (respectively, MATH). Let MATH (respectively, MATH) be the direct sum of root spaces with eigenvalue MATH for MATH (respectively, MATH). Recall that MATH is the smallest eigenvalue MATH for MATH, and MATH is the smallest eigenvalue MATH for MATH. Let MATH be the direct sum of MATH and the eigenspace with eigenvalue MATH for MATH. We have MATH; let MATH be the corresponding foliations of MATH, whose leaves are parallel to MATH respectively. We also have MATH; let MATH be the corresponding foliations of MATH. Here is the idea for proving REF . Each leaf of MATH is foliated by leaves of MATH. Because MATH is the direct sum of MATH with the MATH eigenspace of MATH, it follows that as MATH distinct leaves of MATH in MATH diverge from each other exactly as MATH, measured in the time MATH horizontal plane of MATH. This is a consequence of the Exponential Lower Bound and the Exponential-MATH-Polynomial Upper Bound in REF ; notice that it is critical here that MATH not be the direct sum of MATH with the MATH root space, for then Exponential-MATH-Polynomial Upper Bound would be at best MATH times some polynomial, which would mess up the following calculations. Mapping over via the quasi-isometry MATH, distinct leaves of MATH in a single leaf of MATH must (coarsely) map to distinct leaves of MATH in a single leaf of MATH, which as MATH diverge from each other at least as fast as MATH, by the Exponential Lower Bound. The time change map MATH therefore cannot grow slower than MATH, as MATH. To make this precise, pick a leaf MATH of MATH, contained in some leaf MATH of MATH. We use the symbol MATH to denote a general leaf of MATH, which we will typically take to be a subset of MATH. By REF , there exists a leaf MATH of MATH such that MATH and for each leaf MATH of MATH there exists a leaf MATH of MATH such that MATH . Moreover, if MATH then MATH, because MATH and so MATH stays in a bounded neighborhood of MATH, but any leaf of MATH which is not a subset of MATH has points which are arbitrarily far from MATH. Let MATH be the horizontal subset of MATH at height MATH, and let MATH denote NAME distance in MATH between closed subsets of MATH. Let MATH be the horizontal subset of MATH at height MATH, and let MATH denote NAME distance in MATH. Since the NAME distance in MATH between MATH and MATH is at most MATH, the vertical projection from MATH to MATH induces a quasi-isometry between MATH and MATH; the multiplicative constant of this quasi-isometry is MATH, and its additive constant depends only on MATH. It follows that there exists a ``coarseness constant" MATH so that for any MATH, and for any MATH with MATH, if MATH are the vertical projections of MATH then MATH . To prove REF , fix a time MATH and let MATH. Let MATH be two leaves of MATH contained in MATH, and let MATH be the unique leaf of MATH within bounded NAME distance of MATH; this bound depends only on MATH, as shown in REF . In MATH, apply the Exponential Lower Bound and the Exponential-MATH-Polynomial Upper Bound of REF , and so for all MATH we have MATH where MATH depend only on MATH (note that MATH gives MATH). We want the distance between MATH and MATH in MATH to be greater than the coarseness constant MATH, for each MATH, in order that REF may be applied. We therefore impose a condition on MATH and MATH, namely that MATH which implies, for all MATH, that MATH and so MATH which implies MATH . Next, applying the Exponential Lower Bound of REF in MATH, for each MATH we have MATH . Taking MATH, and using the fact that MATH, this implies MATH . Therefore, MATH . Now divide both sides by MATH, and take logarithms, obtaining MATH and so MATH proving REF and therefore completing the proof of REF .
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math/0005181
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With what we know, the proof is mostly a matter of chasing through definitions. The quasi-isometry MATH is a bounded distance from a quasi-isometry MATH which takes the horizontal leaf MATH to the horizontal leaf MATH, and which simultaneously takes center leaves of MATH to center leaves of MATH. Now restrict the center foliations of MATH to MATH, MATH, and denote the respective leaf spaces as MATH. In order to apply REF , consider each horizontal leaf MATH of MATH as a geodesic metric space, with respect to the Riemannian metric induced by restriction from MATH. The inclusion map MATH is evidently MATH coarsely lipschitz, and it is uniformly proper, with a uniformity function MATH where MATH is larger than the maximum of the absolute values of all eigenvalues of MATH and their multiplicative inverses. Note in particular that the coarse NAME constants and the uniformity functions of the maps MATH depend only on MATH and on the matrix MATH, but not on MATH. Similar remarks apply to the inclusion map MATH. Applying REF , restricting MATH to MATH results in a map MATH which is a quasi-isometry. There is in turn an induced map MATH which is a quasi-isometry with respect to the associated NAME metric. The quasi-isometry constants of the maps MATH and MATH depend only on MATH. Now consider the coordinate identifications MATH, MATH. By construction of the left invariant metrics, for each MATH the space MATH is identified with MATH with the metric MATH, and the space MATH is identified with MATH with metric MATH, and so the maps MATH are uniform quasi-isometries from MATH to MATH for all MATH. Also, MATH is identified with MATH with the associated NAME metric MATH, and MATH is identified with MATH with the associated NAME metric MATH, and so the maps MATH are uniform quasi-isometries from MATH to MATH for all MATH. This implies that MATH is a quasi-isometry from MATH to MATH for all MATH. But for all MATH the map MATH is identical to MATH, proving the lemma.
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math/0005181
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Consider MATH in the same leaf of MATH but not in the same leaf of MATH, and so MATH are in the same leaf of MATH but not in the same leaf of MATH. Define displacement vectors MATH, MATH, and so MATH and MATH. We know that MATH . We also know that REF is true for MATH sufficiently close to MATH, and so for MATH sufficiently close to MATH we have MATH . By induction on MATH, we shall prove that MATH if and only if MATH, or equivalently that MATH. For the basis step MATH, divide REF by MATH to obtain, for all MATH sufficiently close to MATH: MATH . By the Exponential Lower Bound and the Exponential-MATH-Polynomial Upper Bound of REF , the quantity MATH is bounded for MATH if and only if MATH; and the quantity MATH is bounded on MATH if and only if MATH. However by REF the boundedness of these two quantities on MATH are equivalent. For the induction step, assume that MATH, that is, MATH if and only if MATH. We must prove that MATH if and only if MATH. From REF, for MATH sufficiently close to MATH we have MATH . By the Exponential-MATH-Polynomial Upper and Lower Bounds of REF , the following two statements are equivalent: CASE: MATH CASE: For MATH, the quantity MATH is bounded, but the quantity MATH is not bounded. Similarly, the following two statements are equivalent: CASE: MATH. CASE: For MATH, the quantity MATH is bounded, but the quantity MATH is not bounded. But by REF are equivalent, and so REF are equivalent, completing the inductive proof of REF for all MATH. The foliation flag MATH must terminate at MATH for the same value of MATH for which the flag MATH terminates at MATH, proving that MATH, and completing the proof of REF .
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math/0005181
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To prove REF , consider MATH, let MATH, and let MATH be a geodesic connecting MATH and MATH. Let MATH be the MATH-neighborhood of MATH in MATH, so MATH. Applying REF iteratively, projecting inward starting from the edges of MATH furthest from MATH, it follows that vertical projection MATH distorts any distance MATH by at worst MATH, and so MATH. To prove REF , suppose that MATH and MATH and so MATH is MATH-homotopic to a constant in MATH where MATH depends on MATH but not on MATH. This homotopy may then be mapped back to MATH by vertical projection, distorting diameters of homotopy tracks by an amount bounded in terms of MATH as we saw above. The result is a MATH-homotopy of MATH to a constant in MATH, with MATH depending only on MATH and not on MATH.
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math/0005181
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Uniqueness of MATH follows obviously from the fact that distinct hyperplanes in MATH have infinite NAME distance. For existence of MATH we follow closely the proof of REF, concentrating on details needed to explicate the difference between the ``quasi-isometric" setting of CITE and the present ``uniformly proper" setting. Using the bounded geometry of MATH, uniform contractibility of MATH, and uniform properness of MATH, we may replace MATH by a continuous, uniformly proper map, moving values of MATH a bounded distance. Henceforth we shall assume MATH is continuous. Pick a topologically proper embedding of MATH in an open disc MATH. For each component MATH of MATH, the frontier of MATH in MATH is a bi-infinite line MATH in MATH. There is a homeomorphism of pairs MATH. Consider the topologically proper embedding MATH. Note that MATH is a contractible MATH-manifold. For each component MATH of MATH we have a homeomorphism MATH . The frontier of this set in MATH is MATH. Put a product metric and a product simplicial structure on MATH and glue to MATH. Doing this for each MATH, we impose a proper geodesic metric on MATH for which the inclusion MATH is an isometric embedding. The simplicial structure on MATH evidently has bounded geometry. Also, the metric space MATH is uniformly contractible. To see this, let MATH have diameter MATH. If MATH then homotoping along product lines of MATH for each MATH we obtain a MATH-homotopy of MATH into MATH, and then we use uniform contractibility of MATH. Whereas if MATH, then MATH for some component MATH of MATH; there is a MATH-homotopy of MATH into some MATH, and the latter is uniformly contractible by REF . We now plug this setup into the coarse separation and packing methods of NAME - CITE and CITE. NAME use a generalization of the Coarse Separation Theorem with more easily applied hypotheses, due to NAME - CITE. We denote the MATH-ball about a subset MATH of a metric space MATH by MATH. In a metric space MATH, a subset MATH is deep in MATH if for each MATH there exists MATH such that MATH. A subset MATH coarsely separates MATH if for some MATH there are at least two components of MATH which are deep in MATH; the constant MATH is called a coarse separation constant for MATH. Note that if subsets MATH and MATH of MATH have bounded NAME distance from each other, then MATH coarsely separates MATH if and only if MATH does. Here is an elementary consequence of the definitions: Let MATH be a quasi-isometry between geodesic metric spaces. If MATH coarsely separates MATH then MATH coarsely separates MATH, with separation constant depending only on the quasi-isometry constants of MATH and the separation constant for MATH. Here is the version of the Coarse Separation Theorem that we will use. Let MATH be a contractible MATH-manifold, MATH a contractible MATH-manifold, and suppose that MATH are uniformly contractible, bounded geometry, metric simplicial complexes. Let MATH be a uniformly proper map. Then MATH coarsely separates MATH, with coarse separation constant MATH depending only on the uniform contractibility and bounded geometry data for MATH and MATH and the uniform properness data for MATH. Moreover if MATH is continuous then we may take MATH, that is, MATH has at least two components which are deep in MATH. In fact there are exactly two components of MATH which are deep in MATH (see CITE). Following CITE we have a corollary: Let MATH be contractible MATH-manifolds, which are uniformly contractible, bounded geometry, metric simplicial complexes. Let MATH be a uniformly proper map. Then there exists MATH such that MATH. The constant MATH depends only on the uniform contractibility data and bounded geometry data for MATH and the uniform properness data for MATH. If no such MATH exists then the image of the map MATH does not coarsely separate MATH, violating REF . Continuing with the proof of REF , compose the continuous, uniformly proper map MATH with the isometric embedding MATH to obtain a continuous, uniformly proper map MATH. By the Coarse Separation Theorem it follows that MATH has at least two components which are deep in MATH. Now take the argument of REF , pages REF, and apply it verbatim, to produce a hyperplane MATH such that MATH. Next take the argument of REF , pages REF, and apply it verbatim, replacing ``quasi-isometric embeddings" with ``uniformly proper maps" and using the Packing Theorem above, to show the existence of MATH such that MATH, where MATH depends only on the metric fibration data for MATH, the uniform contractibility and bounded geometry data for MATH, and the uniform properness data for MATH. This finishes the proof of REF and of REF .
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math/0005181
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To prove REF , by REF the inclusion map MATH is uniformly proper and MATH is uniformly contractible, and clearly MATH is a contractible MATH-manifold. Composing with MATH we obtain a uniformly proper map MATH. Now apply REF . The idea of the proof of REF is that bushiness of the tree allows one to gain quasi-isometric control over horizontal leaves by considering them as ``coarse intersections" of hyperplanes. A subset MATH of a metric space MATH is a coarse intersection of subsets MATH, denoted MATH, if there exists MATH such that for every MATH there exists MATH so that MATH . Note that although such a set MATH may not exist, when it does exist then any two such sets are a bounded NAME distance from each other. The following fact is an elementary consequence of the definitions. For any quasi-isometry MATH of metric spaces, and MATH, if MATH exists then MATH is a coarse intersection of MATH, MATH, with constants depending only on the quasi-isometry constants for MATH and the coarse intersection constants for MATH and MATH. Consider now a metric fibration MATH. A subset of MATH of the form MATH, where MATH is an infinite ray in MATH, will be called a half-plane in MATH. The next lemma is an easy observation - see REF . Let MATH be a metric fibration over a tree MATH. Let MATH and MATH be distinct hyperplanes in MATH. Then MATH exists and is a bounded NAME distance from either a half-plane or a horizontal leaf in MATH. Moreover, MATH is a bounded NAME distance from a half-plane if and only if MATH is a half-plane. We remark that MATH can be an arbitrarily large finite NAME distance from a horizontal leaf; see REF . Let MATH, MATH be metric fibrations. Let MATH be a quasi-isometry. Suppose MATH and MATH are distinct hyperplanes in MATH which intersect in a half-plane. Then MATH and MATH are a uniformly bounded NAME distance from distinct hyperplanes MATH in MATH which intersect in a half-plane in MATH. By REF , there exists a constant MATH so that MATH is within NAME distance MATH of a unique hyperplane MATH in MATH. Since MATH are distinct they have infinite NAME distance, so MATH and MATH have infinite NAME distance and hence MATH. By REF , it is enough to prove that MATH is not a bounded NAME distance from a horizontal leaf in MATH. If MATH is a bounded NAME distance from a horizontal leaf, then since any horizontal leaf in MATH coarsely separates MATH it must be that MATH coarsely separates MATH. But MATH does not coarsely separate MATH. This contradicts REF . We now prove REF . Consider the quasi-isometry MATH. Since MATH is bushy, any horizontal leaf MATH in MATH can be realized as a coarse intersection of three hyperplanes MATH, such that the pairwise intersections MATH, MATH, MATH form three half-planes, any two of which have infinite NAME distance. Moreover, MATH where MATH is a bushiness constant for MATH (see REF ). Consider the unique hyperplane MATH which lies a NAME distance of at most MATH from MATH, MATH. By REF , the pairwise intersections MATH, MATH, MATH are all half-planes, any two of which have infinite NAME distance. The following elementary fact about trees, applied to MATH, now shows that MATH is a horizontal leaf MATH in MATH: Fact about trees: Let MATH be bi-infinite lines in a simplicial tree MATH, such that the pairwise intersections MATH, MATH, MATH are all infinite rays in MATH, any two of which have infinite NAME distance. Then MATH is a vertex of MATH. Since MATH it follows that MATH . But clearly we have MATH. To summarize, given a horizontal leaf MATH of MATH, we have found a horizontal leaf MATH of MATH such that MATH where MATH. A similar argument using a coarse inverse for MATH provides the desired bound for MATH. This completes the proofs of REF and of REF .
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math/0005181
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The idea of the proof is to compare the growth types of the filling area functions for ``quasivertical bigons" in MATH and in MATH. In MATH this growth type will be quadratic, while in MATH it will be exponential. Let MATH, MATH, MATH, or MATH. There is a quotient map MATH whose point pre-images give the horizontal foliation of MATH, and such that the NAME distance between two horizontal leaves equals the distance between the corresponding points in MATH. A path MATH in MATH is said to be MATH-quasivertical if its projection to MATH is a MATH-quasigeodesic. Define a MATH-quasivertical bigon in MATH to be a pair of MATH-quasivertical paths MATH which begin and end at the same point. If MATH are fixed, we define a filling area function MATH for MATH-quasivertical bigons in MATH. Given a MATH-quasivertical bigon MATH, its filling area is the infimal area of a NAME map MATH whose boundary is a reparameterization of the closed curve MATH; such a map MATH is called a filling disc for MATH. For each MATH define MATH to be the supremal filling area over all MATH-quasivertical bigons MATH in MATH such that MATH. Suppose there is a quasi-isometry MATH which coarsely respects horizontal foliations. Let MATH be a coarse inverse for MATH, also coarsely respecting horizontal foliations. Clearly MATH takes any MATH-quasivertical bigon in MATH to a MATH-quasivertical bigon in MATH, distorting lengths by at worst an affine function; this affine function, and the constants MATH, depend only on MATH, the quasi-isometry constants for MATH, and the NAME constant for the induced height function. Fill the resulting bigon in MATH as efficiently as possible, and map back to MATH via MATH, distorting area by at worst an affine function which again has the same dependencies. We thereby obtain a filling of the original bigon in MATH. If MATH denotes the filling area function for MATH-quasivertical bigons in MATH, and if MATH denotes the filling area function for MATH-quasivertical bigons in MATH, it follows that the growth type of MATH is dominated by the growth type of MATH, that is, MATH for some positive constants MATH independent of MATH. However, we shall now show that MATH has a quadratic upper bound while MATH has an exponential lower bound, contradicting the above inequality. Consider a MATH-quasivertical bigon MATH in MATH. Applying the argument of REF , there are center leaves MATH in MATH and quasivertical paths MATH which stay uniformly close to MATH, respectively. The initial points of MATH are at a uniformly bounded distance, as are the terminal points, and it follows that MATH stays uniformly close to a quasivertical path MATH. Connecting initial and terminal endpoints with short paths MATH we thus obtain a closed curve MATH, contained in a center leaf of MATH, which stays uniformly close to MATH. Since center leaves of MATH are isometric to Euclidean space, in which the filling function is quadratic, it follows that MATH has a quadratic upper bound. To show that MATH has an exponential lower bound, we now construct quasivertical bigons in MATH which can be filled only by discs of exponential area. In the case where MATH is a MATH matrix such loops are given explicitly in REF; examples for general MATH are simple modifications of this example. To be explicit, choose an eigenvalue of MATH of absolute value MATH; such an eigenvalue exists because MATH. Choose an affine subspace MATH parallel to the MATH-eigenspace of MATH. Consider the subspace MATH. For each fixed MATH, choose two vertical segments MATH in MATH whose upper endpoints are in MATH and whose lower endpoints are in MATH, and so that the distance in MATH between the upper endpoints, measured using the Riemannian metric on MATH, is equal to MATH; it follows that the distance in MATH between the lower endpoints, measured using the Riemannian metric on MATH, is within a constant multiple of MATH. Now double this picture, in the doubled MATH horoball MATH, to get a closed loop in MATH, that is: in one horoball go up MATH, across MATH unit, and down MATH, and then in the other horoball go up MATH, across MATH unit, and down MATH; let MATH be the resulting closed curve in MATH. We have MATH. To see that the filling area of MATH is exponential in MATH, note that any filling disc for MATH must contain a path in MATH connecting the lower endpoints of MATH, because MATH separates the two halves of MATH in MATH. This path has length exponential in MATH; and a neighborhood of this path in the filling disc has area exponential in MATH.
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math/0005183
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By induction on the length of the proof. It will suffice to show that the axioms are valid, and that the quantifier rules and (tt) preserve validity. The soundness of the quantifier rules is established by observing that corresponding quantifier shifting rules are intuitionistically valid. For instance, since MATH are intuitionistically valid, it is easily seen that MATH is a sound rule. The only problematic rules are MATH and MATH. Suppose MATH is derivable in MATH. By induction hypothesis, MATH is valid. Then certainly MATH is NAME. Since MATH did not occur in MATH or MATH, we may now assume that MATH does not either. Since the quantifier shift MATH, that is, MATH is valid in IF, we see that MATH is valid. The result follows since MATH is intuitionistically valid, and hence NAME. The communication rule is sound as well. Suppose the interpretation MATH satisfies the premises of (cm). The only case where the conclusion is not obviously also satisfied is if MATH and MATH. If the left lower sequent is not satisfied, we have MATH, and hence MATH, and thus the right lower sequent is satisfied. Similarly if the right lower sequent is not satisfied. For (tt) we may argue as follows: Suppose that the hypersequent MATH is NAME. Let MATH be an interpretation, and let MATH be just like MATH except that MATH. Since MATH does not occur in the conclusion hypersequent MATH we have MATH and MATH. If MATH we are done. Otherwise, assume that MATH, that is, MATH . Let MATH. Now consider MATH: MATH by assumption; MATH, since MATH; and MATH, since MATH. Hence, MATH, a contradiction.
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math/0005183
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Observe that a hypersequent MATH and its canonical translation MATH are interderivable using the cut rule and the following derivable hypersequents MATH . Thus it suffices to show that the characteristic axioms of IF are derivable; a simple induction on the length of proofs shows that proofs in intuitionistic predicate calculus together with REF and MATH can be simulated in MATH. REF is easily derivable using the communication rule. MATH . The formula MATH can be obtained thus: MATH . The last line is obtained from the preceding by two REF inferences, followed by an external contraction. We indicate this with the double inference line.
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math/0005183
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Let MATH and MATH be the cut-free proofs of MATH and MATH, respectively. We may assume, renaming variables if necessary, that the eigenvariables in MATH and MATH are distinct. The proof follows NAME 's original NAME. Define the following measures on the pair MATH: the rank MATH, the degree MATH, and the order MATH is the number of applications of the (ec) rule in MATH, MATH. We proceed by induction on the lexicographical order of MATH. If either MATH or MATH is an axiom, then MATH can be derived from MATH or MATH, respectively, using only weakenings. (This includes the case where MATH). Otherwise, we distinguish cases according to the last inferences in MATH and MATH. The induction hypothesis is that the claim of the lemma is true whenever the degree is MATH or is MATH and either the order MATH, or the order MATH and the rank MATH. CASE: MATH or MATH ends in an inference which acts on a sequent in MATH. We may invoke the induction hypothesis on the premises of MATH or MATH, and MATH or MATH, respectively. CASE: MATH or MATH ends in (MATH). For instance, MATH ends in MATH . Apply the induction hypothesis to MATH and MATH. The resulting proof MATH of MATH has one less REF than MATH (although it may be much longer), and so the induction hypothesis applies again to MATH and MATH. CASE: MATH or MATH end in another structural inference, (tt), or (cm): These cases are unproblematic applications of the induction hypothesis to the premises, followed by applications of structural inferences. For example, assume MATH ends in (cm), that is, MATH where MATH. Apply the deduction hypothesis to the right premise and MATH to obtain a cut-free proof of MATH . Using applications of (ew) and (cm), we obtain the desired result. The case of REF may be of special interest. Suppose MATH ends in(tt), with MATH . Apply the induction hypothesis to the premises of MATH and MATH, and apply (tt) to obtain the desired proof: MATH . The case of MATH ending in REF is handled similarly. CASE: MATH ends in a logical inference not involving the cut formula, or MATH ends in a logical inference not involving the cut formula. These cases are easily handled by appeal to the induction hypothesis and application of appropriate logical and structural inferences. We outline the case where MATH ends in MATH: MATH . We apply the induction hypothesis to the left premise and MATH, and apply CASE : CASE: Both MATH and MATH end in logical inferences acting on a cut formula. For instance, if MATH we have MATH . First we find proofs MATH and MATH of MATH either by applying the induction hypothesis to MATH and MATH or MATH if MATH or MATH, respectively, contain MATH, or otherwise by adding (ic)-inferences to MATH and MATH. Now apply the induction hypothesis based on the reduced degree of the cut formulas twice: first to MATH and MATH to obtain MATH, and then to the resulting proof and MATH to obtain MATH . The desired result follows by several applications of (ic). The other cases are similar and are left to the reader.
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math/0005183
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This is proved exactly as for the classical and intuitionistic case (see CITE). First, observe that all axioms are cut-free derivable from atomic axioms. The cut-elimination theorem thus provides us with a cut-free proof MATH of MATH from atomic axioms. Next, observe that REF rule can be simulated without using cuts by the rule MATH . The rule can be derived as follows (we omit side sequents): MATH . Of course, REF together with (ec) simulates (MATH). We replace all applications of REF by applications of REF in our cut-free proof. Define the order of a quantifier inference in MATH to be the number of propositional inferences under it, and the order of MATH as the sum of the orders of its quantifier inferences. The proof is by induction on the order of MATH. The only interesting case is of MATH occurring below a quantifier inference, since this case does not work for intuitionistic logic. Suppose MATH contains a REF inference above a REF inference, and so that all the inferences in between are structural. We have the following situation: MATH where MATH contains only structural inferences. We reduce the order of MATH by replacing this part of MATH by: MATH .
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math/0005183
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Easy induction on MATH. Every occurrence of MATH must arise from a weakening, simply delete all these weakenings.
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math/0005183
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By induction on the length of MATH. We distinguish cases according to the last inference MATH in MATH. For simplicity, we will write MATH in what follows below instead of MATH or MATH with the understanding that it denotes an arbitrary multiset of MATH's. CASE: The conclusion of of MATH is so that MATH only occurs on the right side of sequents, or only on the left side. Then REF applies, and the desired hypersequent can be derived without (tt). CASE: MATH applies to sequents in MATH. Then the induction hypothesis can be applied to the premise REF of MATH and appropriate inferences added below. CASE: MATH is structural inference other than (cut) and (cm), or a logical inference with only one premise, or a logical inference which applies to a MATH. These cases are likewise handled in an obvious manner and are unproblematic. One instructive example might be the case of (MATH). Here the premises would be of the form, say, MATH . Let MATH. The induction hypothesis provides us with MATH . We obtain the desired hypersequent by applying REF successively MATH times, together with some contractions. CASE: MATH is a cut. There are several cases to consider, most of which are routine. The only tricky case is when the cut formula is MATH and MATH occurs both on the left and the right side of sequents in both premises of the cut. For simplicity, let us consider the cut rule in its multiplicate formulation MATH . We want to find a derivation of MATH where MATH. The induction hypothesis applied to the premises of the cut gives us MATH . We obtain the desired hypersequent by MATH successive applications of (cm). CASE: MATH is (MATH), or REF applying to MATH or MATH. Consider the case of (MATH), the others are treated similarly. The premises of MATH are, for example, MATH . By induction hypothesis, we obtain MATH . It is not straightforwardly possible to derive the desired hypersequent from these. If MATH, let MATH. Then we do easily obtain, however, the following by repeated application of (MATH), REF and CASE : MATH . Now a single application of (MATH), plus (ec) gives us MATH . Then we derive, using MATH cuts: MATH where MATH is the derivation MATH . The desired hypersequent is obtained by MATH cuts with REF MATH is a communication rule. This is the most involved case, as several subcases have to be distinguished according to which of the two communicated sequents contains MATH. Neither of these cases are problematic. We present two examples: CASE: One of the communicated sequents contains MATH on the right. Then the premises of MATH are MATH where. The induction hypothesis applies to these two hypersequents. If we write MATH, we have MATH . We obtain the desired result by applying MATH instances of (cm), internal weakenings and external contractions as necessary, to obtain, in sequence MATH . The sequents participating in the application of (cm) are marked by boxes. The original end hypersequent follows from the last one by internal weakenings. CASE: The communicated sequents both contain MATH, once on the right, once on the left. The premises of MATH are MATH . We have proofs of MATH . Again, a sequence of MATH applications of (cm), together with internal weakenings and external contractions produces the desired end sequent.
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math/0005192
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Let MATH be a MATH - colored Y - link with MATH components. We call MATH the defect of the coloring. If MATH, the coloring is simple. Suppose MATH; then there are at least two components of the same color MATH. Therefore, the MATH - colored Y - sublink of MATH can be split into two disjoint non-empty parts MATH and MATH. It is easy to see that MATH where MATH is obtained from MATH by recoloring all components of MATH in a new, MATH-th, color. The defect of the coloring for each of MATH, MATH and MATH is less than MATH. The statement follows by the induction on MATH.
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math/0005192
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It suffices to prove, that there exist two Y - links MATH and MATH in MATH, each colored in at least MATH colors, such that MATH. Since MATH is a MATH-HS, it may be obtained from MATH by Y - surgery on a Y - link MATH in MATH. Let MATH be a MATH - colored Y - link in MATH, such that its image under Y - surgery on MATH is isotopic to MATH. Let MATH be the MATH - colored Y - link, obtained from MATH by an addition of MATH, with all components of MATH colored in a new color. Then MATH and hence MATH.
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math/0005192
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Follows from the construction of MATH by an application of MATH.
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math/0005192
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Follows from the construction of MATH by an application of MATH: MATH .
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math/0005192
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The equality MATH follows from the construction of MATH by an application of MATH. Topologically, this corresponds to cutting the corresponding solid handlebody introducing a pair of complimentary handles as shown below: MATH . Let MATH be the component of MATH containing this edge, and MATH, MATH be the components of MATH replacing MATH. Note that each subclover of MATH which contains exactly one of the components MATH and MATH, has a trivial leaf. Thus by the definition of MATH and REF MATH if MATH has the same number of components as MATH, and MATH otherwise.
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math/0005192
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It suffices to prove the surgery equivalence of Y - links obtained by the moves MATH - MATH in the standard handlebody. Instead of drawing the handlebodies we will draw thick lines passing through the handles (encoding a set of surgery and clover components), similarly to REF . By REF , MATH is surgery equivalence. To verify MATH, depict the Borromean linking with one component passing on the boundary of an embedded surface with two handles, and then twist one of the handles along the other: MATH . To verify MATH, use an isotopy and REF : MATH . The verification of MATH is similar: MATH .
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math/0005192
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Let us first verify the statement for a Y - graph. By an isotopy and a subsequent application of MATH and MATH, we get MATH . The general case now follows by REF . Finally, MATH by MATH.
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math/0005192
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Let MATH and MATH be the Y - graphs depicted in REF b. Note that MATH looks exactly like MATH, except for the way its lower leaf links the thick line. Turning this leaf to the same position changes the framing of the adjacent edge by a half twist: MATH . By REF , MATH; also, MATH and MATH by REF . Thus MATH and the lemma follows. Alternatively, one can show that MATH for a Y - graph MATH depicted in REF a and pull its edge off the lower thick line by REF to obtain once again MATH.
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math/0005192
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Rotating the special leaf we can change the framing of the adjacent edge while preserving the isotopy class of MATH. Thus by REF MATH and the lemma follows.
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math/0005192
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Cutting the neighboring internal edge we obtain two new leaves MATH and MATH. Without a loss of generality suppose that the framing of MATH is MATH. Then slide MATH along MATH by MATH as shown below: MATH . Notice that MATH changes the framing of a leaf by MATH, so the new leaf MATH is REF - framed. Splitting MATH as shown above by REF , we obtain two clovers each of which contains a trivial leaf (either MATH or MATH), and so can be removed by MATH.
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math/0005192
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We use MATH to pass from MATH to MATH and MATH with MATH as in the proof of REF . Then we split the leaf MATH of MATH by REF introducing new Y - graphs MATH and MATH with MATH: MATH . But MATH by REF , and MATH by REF . Hence MATH. On the other hand, MATH . The comparison of two above expressions for MATH proves the theorem.
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math/0005192
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It suffices to prove the statement for MATH; the general case then follows by REF . Consider a standard Y - graph MATH in a handlebody MATH of genus REF and attach to MATH an additional handle MATH. We are to slide all three edges of MATH along MATH in a genus REF handlebody MATH as indicated in REF . Each time we will use REF to compute the corresponding change of MATH. Sliding the first edge of MATH, we obtain a new Y - graph MATH with MATH. Sliding the next edge of MATH (or rather of MATH), we obtain a new Y - graph MATH with MATH. After sliding the third edge we return back to MATH and get MATH. Here a degree REF clovers MATH, MATH, and MATH are shown in REF . Summing up these three equalities we get MATH . But MATH differs from MATH only by an edge slide, and MATH differs from MATH by an edge slide and a cyclic ordering of edges in a vertex, see REF . Hence MATH by REF , and MATH by REF . A substitution of two these expressions into REF proves the theorem.
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math/0005192
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The map MATH factors through AS and IHX relations by REF . The independence on the choice of MATH follows from REF . The surjectivity follows from the results of REF . The torsion result follows from REF .
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math/0005192
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Any Y - link in MATH, in particular MATH, can be made into a trivial Y - link by framing twists and crossing changes. Moreover, it suffices to use crossing changes which involve only the leaves of Y - graphs. Indeed, instead of a crossing change which involves an edge of a Y - graph, one can do two subsequent crossing changes with the neighboring leaf of this graph (by sliding first the other branch towards the leaf). Using MATH we can realize each of these framing and crossing changes by surgery on a trivial unimodular surgery link, as illustrated in REF . The resulting collection of these surgery components comprises MATH.
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math/0005192
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By REF , MATH is generated by all MATH, with MATH and MATH as above. It remains to notice that MATH and that any sublink MATH of MATH also laces MATH.
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math/0005192
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Let MATH be a trivial Y - link in MATH of degree at least MATH and MATH be an arbitrary link lacing MATH. In view of REF , it suffices to prove that MATH belongs to MATH. Suppose that some leaf of MATH is not linked with MATH, that is, bounds a disc which does not intersect MATH; then MATH by REF . Otherwise, all (that is, at least MATH) leaves of MATH are linked with MATH. But each component of MATH is linked with at most two leaves of MATH; hence the number of components of MATH is at least MATH. Therefore MATH.
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math/0005192
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By REF , MATH is generated by all MATH, with MATH and MATH as above. It remains to notice that MATH and that any Y - sublink MATH of MATH also laces MATH.
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math/0005192
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Proceeding similarly to the proof of REF , we rotate MATH together with the trivial leaf linked with it. This adds a half twist to the framing of the adjacent edge of MATH, while preserving the isotopy class of MATH and MATH. Thus MATH by REF .
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math/0005192
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Let MATH be a lacing pair in MATH with MATH having at least MATH components. In view of REF , it suffices to prove that MATH. We proceed by downward induction on the degree MATH of MATH. If MATH, then obviously MATH and the theorem follows. Suppose now that the inclusion holds for all Y - links of degree higher than MATH and let us prove it for a Y - link MATH of degree MATH. By a repeated use of REF we can reduce the problem to the case when none of the edges of MATH pass through the discs MATH bounding the components of MATH. If for some MATH the disc MATH does not intersect MATH, then MATH by REF , and we are done. If some disc MATH intersects MATH in exactly one point belonging to a leaf MATH of MATH, the statement follows from REF (applicable by the induction assumption). We are left with the case when each component of MATH is linked with at least two leaves of MATH. But each leaf of MATH can be linked with at most one component of MATH. Therefore, MATH should have at least MATH leaves, that is, at least MATH components. Hence MATH, and the theorem follows.
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math/0005192
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Cut out a tubular neighborhood MATH of MATH; this is a disjoint union of genus REF handlebodies. Components of MATH bound non-intersecting discs in MATH, which intersect the boundary MATH by either one or two circles, which are meridians of MATH. Perform surgery on MATH by cutting, twisting by an appropriate element of the mapping class group, and gluing back the handlebodies MATH. This transforms the meridians into some other curves on MATH. It suffices to show that for each handlebody MATH of MATH these curves on MATH are bounding inside MATH. Let MATH be a meridian on MATH. The surgery on this Y - graph corresponds to the action of an element of the NAME subgroup (which acts trivially on MATH), so the image of MATH still bounds a surface MATH. Now, recall that MATH is a good link, so there are only two possible configurations of the meridians. Firstly, MATH may contain just two meridians MATH and MATH, corresponding to the same component MATH of MATH. Then, smoothing the intersections of MATH with MATH, we obtain a surface bounding MATH. Secondly, these meridians may appear only on one of the three handles of MATH. Pick one of the meridians MATH; all other meridians on MATH may be considered as small push-offs of MATH in the normal direction. Thus their images under the Y - surgery, together with the corresponding surfaces, can be obtained by a similar push-offs. This finishes the proof of the lemma.
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math/0005192
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By REF , the image of B under surgery on MATH bounds in the complement of MATH. Thus it will bound also in MATH, for any MATH.
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math/0005192
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Suppose that there is a leaf of MATH, which has at least MATH neighbors, apart from the other leaves of the same Y - graph. Then the corresponding components of MATH comprise MATH. It remains to consider the case when each leaf of MATH has less than MATH neighbors belonging to other Y - graphs. Pick an arbitrary component MATH of MATH and remove from MATH all neighboring components of MATH. Repeat this step, each time picking a new component MATH of the remaining link, until there are no more components left. Finally, take MATH. Let us establish a lower bound for the number of these steps. Each MATH is linked with at most two Y - graphs, leaves of which have altogether at most MATH other neighboring leaves. Thus the removal of MATH and of the link components neighboring MATH may unlink at most MATH leaves of MATH. In the beginning, MATH was linked with all MATH leaves of MATH. Therefore, the number of steps is at least MATH, thus MATH has more than MATH components.
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math/0005192
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Let MATH be a trivial MATH - component Y - link in MATH and MATH be an arbitrary link lacing MATH. In view of REF , it suffices to prove that MATH. If some leaf of MATH is not linked with MATH, then MATH. Otherwise, all leaves of MATH are linked with MATH and we can use REF to find a good sublink MATH of MATH with at least MATH components. It remains to apply REF for MATH and MATH.
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math/0005192
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Let MATH be a clover of degree MATH in MATH. We proceed by the downward induction on MATH. If MATH, then MATH by REF , and the theorem follows. Suppose that the statement holds for any clover of degree higher than MATH, and let us prove it for a clover MATH of degree MATH. Note that by the induction assumption we can use REF . Thus it suffices to prove the statement for a clover MATH which is a disjoint union of a degree MATH clover MATH with no leaves and MATH copies MATH of a Y - graph with three special leaves. We call a path in a connected graph maximal, if it is connected, passes along each edge at most once, and contains the maximal number of edges. A path in MATH is maximal, if it is maximal in each of its connected components. The number MATH of vertices of MATH which do not belong to a maximal path is called the length-defect of MATH. If MATH is positive, pick a vertex of MATH which does not belong to a maximal path, but is connected to it by an edge. Applying to this edge the IHX relation of REF , we obtain two clovers, each of which has the length defect MATH. Hence it suffices to prove the theorem for MATH. Notice that if MATH, there are at least MATH edges of MATH without common ends. Thus, including MATH's, there are at least MATH edges of MATH without common ends. Cut all edges of MATH by REF . Now, unlink all pairs of newly created leaves and change the framing of all leaves of MATH's to REF by MATH, as shown below: MATH . Denote by MATH the resulting link and by MATH the MATH - component Y - link obtained from MATH. Clearly the link MATH laces MATH and MATH . We should show that MATH for each MATH. If some leaf of MATH is unlinked with MATH, then this leaf is trivial and MATH. Otherwise, by the construction of MATH, there are at least MATH non-neighboring components of MATH (since there were at least MATH edges of MATH without common ends). These components comprise a good sublink of MATH. The theorem now follows from REF .
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math/0005192
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By REF , MATH is generated by all MATH, with MATH and MATH as above. It remains to notice that MATH, and that any Y - sublink MATH of MATH also MATH - laces MATH.
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math/0005192
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Let MATH be a MATH - lacing pair in MATH such that MATH has at least MATH components. We will show by downward induction on the degree MATH of MATH that MATH. If MATH, then obviously MATH. Inductively suppose that the statement holds for any MATH - lacing pair MATH in MATH with MATH of degree greater than MATH. If for some MATH the disc MATH bounding a component of MATH does not intersect MATH, then obviously MATH and we are done. If for some MATH the disc MATH intersects MATH in exactly one point, the statement follows from REF (applicable by the induction assumption). We are left with the case when each disc MATH intersects MATH in at least two points. Since MATH - laces MATH, these points should belong to the leaves of different Y - graphs, which do not intersect any other disc MATH, MATH. Therefore, MATH should have at least MATH components, and the theorem follows.
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math/0005193
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MATH makes up the core of the cylinder that is left over if the n-holed sphere is compressed along disks parallel to all of the uninvolved vertices. If the core of the cylinder were knotted (see REF ), then by standard satellite knot theory so is any arc running through the cylinder, but the edge between these two vertices in our graph is unknotted and can be assumed to be inside the cylinder, so this cannot be the case.
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math/0005193
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If MATH and MATH were linked, then MATH and MATH (edges of the the original unlinked n-graph running from vertex MATH to vertex MATH and MATH to MATH respectively) would have to be linked.
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math/0005193
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It is well known that if MATH, then MATH produces a MATH REF-bridge link, which is trivial if and only if q=REF. Similarly, of course, MATH is an unknot if and only if MATH so in that case MATH (the picture is just rotated ninety degrees). REF have expositions on the these facts. The classification was first done in REF .
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math/0005193
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By REF the star core is not knotted, so we may assume it consists of one straight edge MATH running from MATH the north pole to MATH at the south pole and another edge MATH meeting this edge at MATH at the origin and winding around in some manner before ending up at a MATH somewhere on the Southern hemi-sphere (as in REF ). Act upon MATH by an ambient isotopy leaving MATH fixed until MATH lies entirely below the equatorial disk except for at MATH REF . Now if we examine the two edges of the core in MATH the Southern hemi-ball (the ball we get by cutting the original ball in half along the equatorial disk) and pull MATH and MATH apart slightly so that they no longer intersect to get MATH and MATH, the core's edges cannot be linked by REF this is the same as contracting MATH and looking at MATH and MATH. This together with the fact that neither MATH nor MATH can be knotted in MATH by REF , means that MATH is just a rational tangle in MATH. Now by REF we know that the rational tangle to which they correspond, must be an unknot if vertices on the equatorial disk are connected as are the vertices on the sphere. By REF we may assume that the rational tangle is obtained merely by twisting the two vertices on the southern hemisphere around each other, and therefore that it may be untangled without affecting the rest of the core REF . Therefore the core is standard, as must be the enveloping n-holed sphere.
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math/0005193
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Assume there is a counterexample for MATH and examine what it must look like. MATH minus the interior of the non-standard ball, bounded by the non-standard enveloping REF-holed sphere is homeomorphic to MATH minus an open neighborhood of four arcs MATH which run from the inner sphere to the outer sphere. Since the enveloping REF-holed sphere is non-standard, the arcs cannot each simultaneously be isotoped in MATH to be MATH. Since the original edges of MATH were unknotted and pairwise unlinked, they could be thought of as rational tangles. Examine the three pairs of edges of MATH corresponding to the three possible pairings of the vertices, MATH, MATH, MATH. Pick one pair, say MATH. Because the pair is a rational tangle, there is a disk MATH in MATH that separates the edges. Assume MATH intersects MATH minimally. An innermost curve of intersection on MATH yields a subdisk MATH which compresses MATH into two annuli MATH and MATH which are just the boundary of a regular neighborhood of MATH and MATH respectively. The same arguments can be made for the other two pairs. Thus, there are three disks that can be added to MATH that separate the end points of the MATH into pairs, each yielding a rational tangle. If we take the branched double cover of MATH over the four arcs MATH, we get a manifold MATH with boundary two tori MATH, MATH because the branched double cover of a sphere over four points is a torus. Since adding MATH to MATH minus the ``inside" of MATH, thought of as MATH minus a neighborhood of the four branching arcs yields the exterior of a rational tangle, and the branched double cover of a ball over a rational tangle is, of course, just a solid torus. Thus, MATH lifts to a disk that gives a filling of MATH that turns MATH into a solid torus, so MATH is just a solid torus minus a knot. Note that the three pairings of the vertices gives three different fillings of MATH each of which yields a solid torus. [B REF ] If MATH is a nontrivial knot in MATH such that MATH is not parallel into MATH and there exists more than one nontrivial surgery on MATH yielding MATH, then MATH is equivalent either to MATH or its mirror image MATH. MATH is the MATH pretzel knot embedded as shown in REF . We refer to the this as the NAME knot. Since the arcs were not standard, we must lift to the NAME knot or a knot MATH parallel into MATH, the boundary of MATH. If MATH is parallel into the boundary of MATH then MATH fails to produce a counter example to REF . The proof of the proposition requires two steps. First we prove that MATH can be assumed to have the standard embedding in MATH by showing there is a unique strong inversion on MATH. Second we prove that the punctured sphere produced by quotienting out by the MATH symmetry is either boundary parallel or else violates REF and therefore is not a counterexample. To prove that MATH must be embedded in the standard manner we examine an annulus MATH that runs from MATH, the torus boundary component corresponding to MATH in the exterior of MATH, to MATH, the boundary of the solid torus. MATH can be assumed to be embedded (see REF ). Let MATH be the involution of MATH. Let MATH. There is a unique strong inversion of MATH, a torus knot parallel to the boundary of a solid torus. Our first goal is to show that MATH can be chosen such that MATH. We choose MATH to have a minimal number of intersections with MATH. It is clear that MATH can be assumed to be fixed by MATH, so perturbing MATH slightly we can assume that MATH. Now MATH must consist of simple closed curves. An innermost disk argument suffices to show that each of the simple closed curves is essential on MATH and MATH, so the intersection consists of disjoint simple closed curves MATH parallel to MATH on MATH. Let MATH be the curve of intersection closest to the boundary component of MATH on MATH, MATH be the second, and so on increasing the index as the curves move towards the boundary component on MATH. Likewise on MATH the intersection consists of parallel essential circles MATH labeled in the same manner. Let MATH be the sub-annulus of MATH running from the boundary component on MATH to MATH, and MATH be the sub-annulus of MATH running from the boundary component on MATH to MATH. Let MATH be the curve of intersection on MATH that corresponds to the intersection with MATH at MATH. If we cut and paste MATH replacing MATH by a push off of MATH we reduce the number of intersections of MATH and MATH by at least one. In order to preserve the property that MATH we must also replace MATH by a push off of MATH. This, however, cannot increase the number of intersections, so we have a new annulus running from MATH to MATH that has fewer intersections with its image under MATH, contradicting minimality. Thus, we can assume that MATH. This, however, implies that restricting to the solid torus between MATH and MATH, MATH takes MATH to itself, exchanging MATH with MATH. This in turn implies that there must be an annulus in MATH that is fixed by MATH. Now that we know that MATH is fixed, we use it to show that we have a standard inversion of MATH. Let MATH be a meridional disk for MATH that is fixed by MATH. Examine MATH. In general the intersection pattern on MATH will look something like REF with a collection of arcs running from MATH (which punctures MATH several times) to MATH and a collection that run from one of the punctures from MATH to another. We can remove the arcs running from MATH to MATH by picking an outermost arc on MATH that runs from one component of MATH to itself. This small disk gives an isotopy of MATH together with MATH that reduces the number of intersections of MATH with MATH. because MATH we can simultaneously do a second isotopy of MATH that also reduces the number of intersections of MATH with MATH and preserves the symmetry of MATH. Thus, we can assume that MATH consists solely of arcs running from MATH to MATH. This, however, shows that we have the standard symmetry for MATH because cutting the solid torus along MATH turns it into a cylinder and MATH becomes bands running from the top of the cylinder to the bottom in the unique way possible. Now we need only argue that torus knots with standard embeddings fail to give a counterexample. Let MATH be a torus knot embedded in MATH, fixed by an involution MATH of MATH. Let MATH be the fixed annulus above. Let MATH be MATH with four branching arcs MATH, the quotient of the exterior of MATH in MATH by MATH. In the quotient, MATH maps down to MATH which we will designate MATH and MATH maps to MATH designated MATH. Either MATH violates REF or it is standard and therefore in either case is not a counterexample to REF . Because MATH, each run from MATH to MATH the two arcs MATH, MATH that are fixed in MATH by MATH must each intersect MATH in exactly one point. In the knot exterior MATH and MATH are then broken each into two arcs with one end point of each arc on MATH and one end point on MATH. In MATH the MATH then become the four branching arcs MATH. Under the quotient, the annulus MATH becomes a rectangular disk MATH which without loss of generality runs down MATH along MATH up MATH and back along MATH and is disjoint from MATH and MATH. This is clear because MATH consisted of the preimage of MATH and MATH but was disjoint from the preimage of MATH and MATH (recall that MATH, like the MATH is fixed by MATH and runs from MATH to MATH). This means that MATH looks exactly like REF , where MATH and MATH form a rational tangle inside the ball designated MATH. By REF we see that MATH must be the tangle MATH that results from two horizontal arcs whose eastern vertices are twisted MATH times around each other or MATH will be knotted violating REF . On the other hand, if MATH is the tangle MATH above, then twisting the eastern portion of MATH, MATH times shows MATH is homeomorphic to the standard picture and therefore again fails to be a counterexample. Note: One can in fact prove that (if MATH is a torus knot other than the unknot) MATH never produces a standard enveloping sphere, but instead always creates one that violates REF completes the proof that MATH cannot be parallel to MATH and therefore must be a NAME knot if it is to produce a counterexample. We may transform the traditional picture of the NAME Knot to a symmetric one as in REF . NAME REF tells us that this knot (entered as a link) has exactly one MATH symmetry, which we can now see. We quotient out by the symmetry to either get a counterexample, or proof that there is none. We get MATH minus four arcs MATH. We will show that three of the arcs violate REF (See REF ). Label the edge omitted from the picture MATH. Label the horizontal edge that has one of its vertices on the eastern side of the sphere MATH. Contraction of MATH takes edges MATH and MATH to MATH and MATH as pictured in REF . To see that MATH and MATH are in fact linked we call on the following well known facts about rational tangles. (See, for example, REF .) CASE: A REF-bridge knot (or link) is the denominator of some rational tangle CASE: Conversely, the denominator of a rational tangle is a REF-bridge knot (or link). If MATH is the denominator of a rational tangle, then MATH is prime. The denominator of the tangle in REF is the connect sum of a trefoil and a figure eight knot and therefore it is not a rational tangle by REF . Since neither of the arcs is knotted, they must be linked violating REF . Since this was the only possible counterexample, REF must be true.
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math/0005193
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REF shows a MATH curve in a ball. Such a graph is well known not to be standard, but every subgraph is standard. Let a MATH curve be the star core of an enveloping n-holed sphere. Although it is not standard, it supports a pairwise unlinked n-graph MATH. If we think of the enveloping sphere as bounding a central (round) ball with n tentacles running to the boundary, we can picture the pairwise unlinked graph as being a standard unlinked graph in the central ball which is extended by a product down each of the tentacles. Now since MATH any two edges of MATH miss at least one vertex, but there is an isotopy of any MATH arcs of a MATH curve that makes those arcs appear standard. Likewise since the edges of MATH completely miss one of the tentacles, they may be pictured as being embedded in a ball bounded by a standard enveloping MATH-holed sphere. The arcs remain standard within the central ball and are extended by a product down the tentacles throughout the entire process, so clearly the edges are not linked pairwise. (See REF ).
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math/0005193
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Assume MATH is chosen with a minimal number of intersections with MATH. Examine an innermost curve MATH on MATH. If MATH is not essential on MATH, then there is an obvious isotopy through which this intersection could have been eliminated, so we may assume it is essential on MATH. Therefore the innermost loop gives us a compressing disk for MATH. Homology is enough to assure us that the disk must be on the inside of MATH (the component of MATH containing MATH ). Since MATH is assumed to be essential in MATH, it must separate the vertices of MATH into two non-empty sets. With no loss of generality, let MATH be in one set and MATH be in the other. Since MATH separates the vertices (as in REF ), the disk it bounds does too, and MATH must intersect it. This, however, is a contradiction since the interior of MATH is disjoint from the edges of MATH.
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math/0005193
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Choose a point on MATH to be the origin of MATH. Let MATH be the sphere of radius MATH centered at the origin in MATH, and let MATH be the ball that it bounds. We may alter MATH slightly if necessary so that we may assume that it intersects each MATH transversally. Fix MATH and examine MATH. The pieces of MATH in MATH may be broken into two sets. The first set consists of the connected piece containing the origin MATH, and the second set, all the other pieces, MATH. We shall isotope MATH until MATH contains one piece of the first type and none of the second on the ``outside" of MATH in MATH. The latter does not prevent us from claiming that the former is boundary parallel in MATH, so we do not worry about them. See REF We examine the pre-images of the MATH in MATH. We now make an argument to show that we may assume that none of them have a boundary curve which is trivial in the fundamental group of MATH. If there were a trivial boundary curve, we could choose an innermost one (on MATH). This would bound an embedded disk MATH on MATH that meets MATH in a simple closed curve. The boundary curve splits MATH into two disks, each of which bounds a ball with MATH. If MATH is on the interior of MATH, then we choose the ball that does not contain the surface MATH (MATH, is connected and disjoint from MATH, so it can only be in one of the two balls). If MATH is on the exterior of MATH then we choose the smaller of the two balls (it is contained in the other ball). Either way we push MATH across the ball and through MATH, eliminating at least its intersection with MATH and possibly more extraneous intersections that were contained in the ball through which our isotopy was done. Note that MATH is unchanged away from its boundary and its boundary can only be changed by capping off trivial components. We continue this process until there are no more trivial components in the entire collection of MATH. At the risk of sloppy notation we shall continue throughout to call the new surfaces MATH carefully noting at each step that we still have done only a finite number of isotopies to a finite number of pieces. We now notice that MATH fits all of the criterion of the standard finite case. Since the geodesics are dense within the NAME Set, at any finite stage there will be a complete graph in MATH on the vertices that are given by the intersection of MATH and MATH. Thus, by the previously proven finite case, MATH is standard in MATH. We can use its inherited product structure to isotope MATH to make sure that it does not intersect MATH on the outside of MATH. Now we repeat the process for some MATH. We might worry that this process results in our pushing some piece of MATH an infinite number of times, but this is not the case, as every point in MATH is in some MATH for large enough MATH and our isotopies never affect points of MATH that are not near the boundary of MATH. Thus, for each point there is some MATH such that the point is left alone for good after MATH steps. Since each step involved only a finite number of isotopies, each point is moved only a finite number of times. The only thing left for us to check is that the product structure for MATH can be chosen to correspond exactly with the product structure we already chose for MATH. MATH may be left fixed as we do our operations for MATH, so naturally MATH remains fixed, too. Since MATH is boundary parallel in MATH, we may substitute part of MATH for it in MATH and the resulting surface still contains a complete graph on one side and must be boundary parallel. If we concatenate its product lines in MATH with the product lines of MATH in MATH we see a product structure that suits our desires as in REF .
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math/0005195
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Because MATH, any self-diffeomorphism of MATH preserves orientation. Now MATH is a minimal complex surface of general type, and hence, for the standard `complex' orientation of MATH, the only NAME - NAME basic classes CITE are MATH. Thus any self-diffeomorphism MATH of MATH satisfies MATH . Letting MATH be the NAME dual of the fiber of MATH, and letting MATH denote the genus of MATH, we have MATH for MATH. The adjunction formula therefore tells us that MATH where the intersection form is computed with respect to the `complex' orientation of MATH. If we had a diffeomorphism MATH with MATH, this computation would tell us that that MATH and that MATH . In either case, we would then have MATH . On the other hand, MATH, so intersecting the previous formula with MATH yields MATH and hence MATH in contradiction to our hypotheses. The assumption that MATH is therefore false, and the claim follows.
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math/0005196
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To specify a connected reductive group, it is enough to specify a compact torus together with the collection of characters of that torus which will serve as the weights for the semisimple part of the group. The torus in turn can be described as MATH for some lattice MATH, which is the form used in the statement of the lemma. (This choice of torus is dictated by physical considerations.) To complete the specification, the weight spaces must be given. The NAME model is singular along a (reducible) curve MATH; the general singularity over each irreducible component of MATH is a rational double point CITE. Let us consider the intersection configuration of the exceptional curves and the exceptional divisors on MATH. In most cases, the intersection matrix is (up to a sign) the unique NAME matrix of a NAME algebra MATH; here we also find the non-simply laced algebras, as the exceptional curves might undergo a monodromy transformation as they move in the exceptional divisors along the curve MATH. In some cases a more delicate argument is needed CITE.
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math/0005196
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Following the algorithm provided in CITE one can in fact show that MATH . This statement is also buried in the proof of CITE; even though the author claims it only for MATH.
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math/0005196
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It follows from CITE.
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math/0005196
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It can be verified by inspection and explicit computations.
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math/0005196
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We compute the NAME characteristic of MATH via the structure of elliptic fibration (the NAME characteristic of the general fiber is zero) and NAME - NAME 's sequence.
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math/0005196
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From the previous corollary we have: MATH . Note that MATH, which gives: MATH . Substituting this in the above equation we obtain the statement of the proposition.
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math/0005196
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MATH is the number of cusps away from MATH; our assumptions in REF imply that the cusps are determined by the common zeroes of the polynomials MATH away from MATH (these are ordinary vanishing, see REF ). MATH and MATH might also vanish along MATH, of orders MATH and MATH; MATH and MATH might have a common zero along MATH. The multiplicities of these latter zeros are measured by MATH. (See REF.)
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math/0005196
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CASE: MATH is regular along MATH (simply laced groups in NAME 's exceptional series.) In this case MATH, which (except for the trivial case) are precisely the simply laced groups in NAME 's exceptional series. Here the singular fibers are of types MATH, and MATH. By assumptions, MATH and MATH becomes REF : MATH . Now we use the geometry of the singularities of MATH: MATH . By solving the system we have CASE : MATH . By substituting the above equation in the right hand side of MATH, we see that every term is a multiple of MATH. Then: MATH in fact, by the definition of MATH, MATH (see REF). In this case we also have MATH (see REF ) and thus: MATH . Now, for these groups MATH where MATH is the NAME number (see the following REF ). This is in agreement with the expectations from physics (see REF) together with REF . CASE: MATH has a pole along MATH . Since MATH is well defined around MATH, MATH. This together with the assumption MATH, implies MATH, that is, MATH . The substitution in REF gives MATH. This is again consistent with the expectations in REF.
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math/0005196
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Case by case checking.
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math/0005196
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As we have already seen in REF, the intersection numbers of the various parts of the discriminant in MATH determine the geometry of MATH and the choice of the group MATH and vice versa. Following REF, we write all the terms in MATH in REF , as coefficients of MATH, the genus of the curve of singularities, the number of points where the singularities are non-generic, and MATH, when the groups are non-simply laced, and then interpret the results. The coefficients in REF are determined by the group and the local geometry (the degeneration of the general rational double point) and are listed in REF. We divide the proof in REF steps. MATH . CASE: We show how the geometry suggests the appropriate substitutions for MATH, MATH, MATH and also MATH if the group has monodromy branched at MATH points. If MATH, then the substitutions are uniquely determined (see REF). In REF we show how these substitutions are equivalent to certain representation-theoretic facts. If MATH or MATH, then after the substitutions we obtain the data in REF . That is, the resulting formula for MATH can be written as a sum of local terms, associated to various points MATH, which can be collected into a formula of the form MATH . The local contributions MATH are recorded in REF . In the cases MATH, MATH, there are choices in making the substitutions but if a careful choice is made we can again write things in the form REF (see also REF for a better interpretation). As we will point out in REF below, the substitutions can be formulated in a very general way which allows them to be applied in cases beyond the specific ones considered here CITE. MATH . CASE: We show how we can naturally interpret the entries in REF as charged dimensions of certain representations (multiplied by the coefficient MATH), given in REF . That is, MATH. If MATH is not a branch point, then the (resolution of the) general elliptic surface through MATH can be associated to a group MATH containing MATH, and the representation is obtained via the branching rules for the adjoint representation of MATH. If MATH is non-simply laced, then we consider MATH, MATH simply laced, and we use again the branching rules. (This gives the representation-theoretic interpretation of the number ``MATH " from REF .) MATH . CASE: Finally we show how the number MATH can be derived from the geometry of the degeneration of the general double point to the singularity over MATH.
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math/0005196
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The statement follows from NAME 's formula.
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math/0005196
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CASE: When MATH is finite and there is no monodromy, that is, cases MATH, MATH, MATH, MATH, MATH, MATH, MATH corresponding to the simply laced groups MATH, MATH, MATH, MATH, MATH, MATH, MATH, in NAME 's exceptional series, then the local geometry is given by the following equations: MATH which can be solved since MATH: MATH . CASE: When MATH is finite and there is monodromy, that is, cases MATH, MATH, MATH, MATH corresponding to groups MATH, MATH, MATH, MATH which includes the remainder of NAME 's exceptional series, the local geometry is the same but we also use REF to eliminate MATH in favor of MATH: MATH . CASE: If MATH, MATH, REF tells us that MATH where MATH is the number of non-simple normal crossings intersections. The genus formula then says that MATH . (The case of MATH is similar, using MATH.) CASE: If MATH, MATH, coming from MATH with monodromy, then MATH so that MATH . Combining this with the genus formula yields MATH . CASE: Finally, if MATH or MATH coming from MATH, MATH, then MATH which can be combined with the genus formula and solved to give: MATH .
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math/0005197
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It is sufficient to define a composition for elements MATH of the special type MATH, where MATH, MATH, MATH, MATH is a bijection, and MATH. Composition of elements of this type is defined using the composition in the modular operad MATH.
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