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math/0006094
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We give a sketch of the proof, for details one can see CITE. If MATH is the size of the shock MATH located in MATH, then we can apply REF to compute its shift MATH: by REF we obtain MATH . If MATH is sufficiently small, then we have MATH where MATH is the position of the shifted shock and MATH is its original position. Note that MATH is the shift rate of the shock MATH, after colliding with the shocks of the NAME problems MATH and MATH. Their total shift is proportional to MATH, and after the interaction with MATH, the shift of the latter is proportional to MATH. Thus taking the limit as MATH tends to MATH of REF, we obtain for MATH sufficiently small MATH which implies REF.
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math/0006094
|
As in the previous proposition, let MATH be the size of the shock MATH located in MATH in NAME coordinates. If MATH is its shift rate, then for MATH sufficiently small by REF we obtain MATH . In fact, by assumption, in the simplified wave patterns to compute the shift rate of MATH, there are no waves of the MATH-th family different from MATH. Dividing by MATH and taking the limit as MATH tends to MATH, we obtain MATH which implies MATH . We use the fact that MATH tends to MATH as MATH. Since MATH is compact, the conclusion REF follows easily.
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math/0006094
|
Consider two adjacent MATH-rarefaction fronts MATH and MATH, and let MATH, MATH, be the interaction times of MATH, MATH with other waves in the interval MATH. Fixed MATH for some MATH, let MATH be the characteristic line of the MATH-th genuinely nonlinear family starting in MATH (see REF ). Assume MATH sufficiently close to MATH such that MATH and MATH does not collide with shocks of other families for MATH. Let MATH be the characteristic curve starting in MATH. By the assumption of genuinely nonlinearity, at time MATH we have MATH for some constant MATH, depending only on MATH. Using REF , at time MATH we have MATH . Repeating the process, it is possible to find a countable number of times MATH such that MATH and using REF we get MATH . Repeating the process for MATH and for all intervals MATH, we obtain REF where MATH. The second equation follows noticing that the total amount of positive jumps in the interval MATH is bounded by MATH.
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math/0006094
|
The proof follows by REF . In fact, fixed a shock MATH, using REF , we have that at time MATH for a shock MATH of the MATH-th family there exist MATH if the shock MATH starts on both sides of MATH, or, using the same estimate of REF , MATH if MATH start on one side of MATH. Since there is at most MATH shocks such that REF holds, and the interval of influence is MATH, using REF together with REF we obtain MATH . The conclusion follows the linearity of the shift differential map.
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math/0006094
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This is a corollary of REF .
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math/0006094
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Consider two piecewise constant initial data MATH, MATH in MATH, and construct a pseudo polygonal path MATH, connecting MATH and MATH, such that MATH . We can assume that MATH has a finite number MATH of jumps. If we denote with MATH the path MATH, we have by REF MATH . If now MATH, since MATH converges to MATH, we obtain REF. Since this estimate does not depend on the number of initial jumps MATH, we can extend it uniformly on MATH. Using the same pseudo polygonal path, in a similar way we can prove that MATH . This shows that MATH converges uniformly to the solution MATH as MATH and MATH. It also implies that MATH . This concludes the proof.
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math/0006094
|
The statement follows easily, since we proved that MATH is the unique limit of wave front approximations, and for data with bounded total variation we can apply the results in CITE.
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math/0006097
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We first consider the case MATH. In that case, if the matrix MATH is singular, then the odd rows of MATH are linearly dependent and thus MATH. We may thus assume MATH and MATH is nonsingular. Then we can express MATH on MATH as a linear combination of the functions MATH. Using this we find that MATH where MATH . But then MATH as required. Now, suppose we know the theorem for sets of size MATH, and let MATH be a set of size MATH. Then MATH . It thus suffices to show MATH . Expand MATH along the bottom two rows and integrate, then simplify using the following integrals: MATH . We thus see that the MATH terms contribute nothing. For the MATH terms, MATH contributes MATH directly, while the terms associated to MATH give precisely the expansion of MATH along the first MATH column, up to an overall sign change. We thus obtain a total of MATH, as required.
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math/0006097
|
Define functions MATH on MATH by MATH and an antisymmetric function MATH on MATH by MATH . Then the function MATH on MATH is REF unless exactly half of the MATH are in MATH, in which case it equals MATH . Furthermore, the current matrix MATH is the same as the matrix associated to MATH and MATH. We thus find that MATH so we can apply REF ; we compute MATH thus obtaining the desired result, up to transformation by MATH .
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math/0006097
|
Apply the previous result with MATH, MATH, and MATH.
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math/0006097
|
We first observe that for any MATH, MATH is a unit in MATH; indeed, it agrees to valuation REF with the unit product MATH. Now, multiplication by a unit leaves the valuation unchanged, so MATH. This latter element is (up to sign) simply the determinant of the complementary minor to MATH; we easily see that every term of this determinant has valuation at least MATH. Now, let us consider how MATH is related to MATH. Recall that for a block matrix MATH with MATH invertible, the upper left block of MATH is given by MATH. In other words, the difference between the upper left block of MATH and the inverse of the upper left block of MATH is is MATH. Applying this to MATH, we find that MATH since MATH is a unit, we find MATH . By symmetry, we also find MATH . By induction on MATH, we find that MATH . In particular, defining an infinite matrix MATH by MATH we find MATH, and the lemma follows.
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math/0006097
|
The proof is essentially as above; the main difference is that the matrix MATH is now REF-dimensional, of the form MATH for some unit MATH. Then, since MATH, MATH is essentially just the determinant of a MATH submatrix of MATH. For the first equation, it is trivial to determine the valuation of this determinant; for the second equation, we simply relate the determinant of a MATH minor of MATH to the determinant of the complementary minor of MATH, and again the valuation is easy to determine.
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math/0006097
|
Since MATH we see that the theorem reduces formally to the symmetric function identity MATH . We first prove this formal identity, then consider the specific specialization of interest. If we restrict MATH so that MATH, then this only changes the left-hand-side by terms of order MATH; it will thus suffice to derive a kernel for each MATH such that the formal limit MATH of these kernels is MATH. When MATH, we find MATH . Thus we can apply REF above, with MATH . Defining MATH by MATH we find that MATH is the MATH-th principal minor of the infinite matrix MATH for MATH. Since MATH, we can restrict the second sum to MATH, and thus have MATH . (Recall MATH.) We thus find MATH . With respect to the natural valuation on the ring of symmetric functions in two variables, MATH satisfies the hypotheses of REF above; we thus find MATH for any fixed MATH. Since MATH, we find MATH . We compute MATH thus proving the desired formal result. For any complex number MATH and any parameter set MATH, we define a specialization MATH on the ring of symmetric functions in MATH by MATH . Now, specialize the formal identity by MATH and MATH. For MATH in a neighborhood of REF, both sides converge, and thus must agree in this neighborhood. Since both sides are analytic in a neighborhood of the interval MATH, it follows that they must agree at MATH, and the theorem is proved.
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math/0006097
|
Set MATH. Then the given determinant is MATH as required.
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math/0006097
|
After specializing, MATH becomes MATH . Conjugating by MATH gives MATH since MATH unless MATH is even, the result follows.
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math/0006097
|
We have MATH where MATH is the number of even parts of MATH, and thus MATH so the result reduces to showing the corresponding symmetric function identity. And again, we may take the limit MATH of the kernel corresponding to the restriction MATH. In that case, we have MATH with MATH . Now, if we define a kernel MATH then for nonincreasing sequences MATH and MATH, we find MATH if MATH; otherwise, the determinant is REF. We thus have MATH for MATH and MATH. Upon symmetrizing in MATH and MATH, we can apply REF , with MATH . We have MATH and MATH . Since MATH we can simplify the matrix resulting from REF by adding MATH times the second row/column to the first row/column and adding MATH times the third row/column to fourth row/column. Now, MATH and thus MATH . Taking MATH, MATH, we see that MATH satisfies the hypotheses of REF above. Thus if MATH are each either of MATH, or MATH, we find MATH . It thus remains to compute MATH . This gives the theorem, once we observe that MATH .
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math/0006097
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We compute MATH so MATH .
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math/0006097
|
The key step is to sum over the subsets of MATH. By the theorem, we have MATH where MATH is the kernel MATH . Subtract MATH times the second and third rows from the first and fourth rows (respectively), then subtract the first row from the fourth and the third from the second, then apply the same transformations to the columns. This transformation is symplectic (preserves MATH), and forces the last row of the MATH matrix to REF. We may thus expand along the bottom row, giving MATH since MATH . Thus MATH as required.
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math/0006097
|
As above, we reduce to an application of REF , with MATH and MATH . We compute MATH and MATH . Now, when MATH, we can simply shift the variables of summation to obtain MATH . When MATH, this gives MATH so we conclude that MATH . In particular, MATH satisfies the hypotheses of REF , so the kernels for finite MATH tend to a limit. We thus readily compute the kernel given above.
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math/0006097
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We apply REF , with MATH and MATH . We find MATH . The theorem follows immediately.
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math/0006097
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For simplicity, we consider instead MATH which naturally differs only by rescaling MATH by MATH. We find that for MATH of the appropriate form with MATH, MATH where MATH and MATH. We then apply REF , with MATH in particular MATH satisfies the hypotheses of REF . We readily verify that MATH so MATH . Scaling MATH by MATH and simplifying gives the desired result.
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math/0006097
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We take MATH so MATH . We find MATH and thus obtain the stated kernel.
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math/0006097
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On the one hand, we have MATH on the other hand, we have MATH . The theorem follows.
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math/0006097
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This of course follows immediately from REF , but the following independent proof (based on the arguments of CITE) gives useful insight into how the kernel MATH can be derived. (The above proof, of course, has the advantage of using only finite methods.) From REF, we have MATH for any measure MATH. Thus, taking MATH, we find MATH where MATH . We thus find MATH and MATH . Thus MATH as required.
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math/0006097
|
Let MATH be the random partition associated to MATH, and set MATH, MATH, MATH. By the definition of the NAME pfaffian, MATH . Now, we have the following lemma: Let MATH be a decomposition as above. Then for any partition MATH with associated set MATH, MATH . Recall that for any partition, MATH . Setting MATH, MATH, we have MATH and MATH . Subtracting these two quantities, we conclude that MATH . We may thus replace MATH in the above sum with MATH. We thus have MATH . Similarly, MATH .
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math/0006100
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Immediate from REF .
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math/0006100
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Most of the details of the proof are contained in Refs. CITE and CITE, so we only sketch points not already covered there. The essential step (for the present applications) is REF , which shows that MATH intertwines the isometry MATH-with the restriction of the unitary operator MATH to the resolution subspace MATH. We have: MATH for all MATH, and all MATH. This proves REF . The rest of the proof will be given in a form slightly more general than needed. For later use, we record the following table of operators on the respective NAME spaces MATH and MATH, and the corresponding transformation rules with respect to the operator MATH. Let MATH be the scale number, and let MATH be given satisfying MATH and set MATH, MATH. Then we have the following transformation rules. MATH . Now the proof may be completed by use of the following sublemma, which we state just for MATH. Let MATH be a subspace of MATH which is invariant under multiplication by MATH. Then there is a MATH such that MATH and MATH, MATH. The proof follows from the NAME theorem CITE. Completion of proof of REF . Let MATH. Then we saw that MATH is isometric in MATH, and the complementary space MATH then satisfies the condition in NAME REF. Let MATH be the function determined from the sublemma. Then, after multiplication by a suitable MATH, we will get MATH, with the coefficients-MATH as in REF . Setting MATH, it follows then that REF - REF are satisfied.
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math/0006102
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The first assertion is known, see for instance CITE. For the second statement, we closely follow CITE. For MATH, of the form MATH, it turns out that MATH for any MATH. Let MATH, MATH, be orthonormal vectors such that MATH is a basis of MATH, and set MATH . Then, for MATH as before, we can write a ``NAME - type" expansion MATH . Assume now that MATH, that is, MATH . We plug REF into this relations, and we get the system MATH . Recalling that MATH and MATH are periodic, we find MATH . Therefore, MATH. This shows that MATH. Since the converse inclusion is is always true, the lemma follows.
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math/0006102
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This is proved as in CITE. We just sketch the argument. The idea is to use the contraction mapping principle to characterize the function MATH (see REF ). Indeed, define MATH . So MATH if and only if MATH and MATH. Now, MATH where MATH. Setting MATH one finds that MATH . By the NAME - NAME inequality, it turns out that MATH is a contraction mapping from some ball MATH into itself. If MATH is sufficiently small, we have proved the existence of MATH uniformly for MATH. We want to study the asymptotic behavior of MATH as MATH. We denote by MATH the functions MATH corresponding to the unperturbed energy functional MATH. It is easy to see REF that the function MATH found with the same argument as before satisfies MATH as MATH. Thus, by the continuous dependence of MATH on MATH and the characterization of MATH and MATH as fixed points of contractive mappings, we deduce as in CITE, proof of REF , that MATH. In conclusion, we have that MATH.
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math/0006102
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Observe that MATH, where MATH. According to REF , it suffices to look for critical points of MATH. From REF , it follows that either MATH everywhere, or has a critical point MATH. In any case such a critical point gives rise to a (non - trivial) closed geodesic of MATH. From REF , we know that MATH is MATH - invariant. This allows us to introduce the MATH - category MATH. One has MATH . Since MATH, (see CITE), then MATH. Finally, by the NAME - NAME theory, MATH carries at least MATH closed geodesics, distinct modulo the action MATH. This proves the first statement. Next, let MATH . Then REF immediately implies that MATH . Moreover, if REF holds, then MATH for MATH, and MATH for MATH. Since (recall REF ) MATH it follows that MATH . We can now exploit again the MATH invariance. By assumption, and a simple continuity argument, MATH, and similarly MATH, for a suitably large MATH. Hence MATH. The same argument applies to MATH. This proves the existence of at least MATH closed geodesics.
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math/0006102
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It is always true that MATH. By REF, we have that MATH. This implies that MATH. A generic element of MATH has the form MATH for MATH and MATH; then MATH and any two vector fields MATH and MATH along a curve on MATH can be decomposed into MATH . In addition, there results (see CITE) MATH and MATH where MATH, MATH, etc. stand for the curvature tensors of MATH, MATH, etc. By REF, as in the previous section, MATH is equivalent to the system MATH . As in the case of the sphere, the first equation implies that MATH is constant. The second equation in REF implies that MATH. Hence, MATH . This completes the proof.
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math/0006102
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REF allows us to repeat all the argument in REF , and the result follows immediately.
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math/0006102
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We wish to use REF . Since MATH, then MATH has a geodesic MATH such that MATH over some component MATH of MATH. See CITE. We consider the manifold MATH . Here we do not know, a priori, if MATH is non - degenerate in the sense of REF . But of course MATH has a minimum at the point MATH, where MATH. We now check that it is isolated for MATH. We still know that MATH. Take any point MATH, and observe that MATH. For all MATH sufficiently close to MATH, it holds in particular that MATH. Hence MATH since MATH and MATH (and hence MATH, due to MATH invariance) is an isolated minimum of MATH by assumption. Finally, thanks to REF , MATH. This concludes the proof.
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math/0006103
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REF follows from the fact that the MATH's have disjoint support on MATH. REF holds for NAME functions and REF follows from REF. By the density of step functions on MATH also REF follows. If MATH then MATH. Since MATH are orthonormal in MATH, we have MATH so by applying the NAME transform of order MATH we get MATH . Using the orthogonality of MATH we have MATH . By the NAME theorem, we then have MATH . The left-hand side can then be rewritten as MATH . Upon a change of variable letting MATH, the latter equals MATH . Comparing the previous two formulae for MATH we get MATH . Rewriting the above in terms of MATH we have MATH since MATH . From REF we get MATH . Thus MATH . Set MATH; thus we have MATH.
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math/0006103
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REF follows from the fact that the MATH's have disjoint support on MATH. REF holds as before and REF follows from REF. By the density of step functions on MATH also REF follows. If MATH then MATH. By applying the MATH-Hankel transform of order MATH we get MATH where we denote by MATH the MATH-Hankel transform to avoid confusion. The NAME Theorem for NAME transforms extends in a natural way to the case of MATH-Hankel transforms where it takes the following form: MATH where MATH is the MATH-measure REF . Then by using the orthogonality of MATH and the following fact: MATH we have MATH . By the NAME theorem, we then have MATH . The left-hand side can then be rewritten as MATH . Upon a change of variable letting MATH, the latter equals MATH . Comparing the previous two formulae for MATH we get MATH . Rewriting the above in terms of MATH we have MATH . Since from REF MATH thus MATH . Set MATH; thus we have MATH. Observe that the NAME functions have a ``multiplicative periodicity" on the unit circle in the following sense: MATH . This implies that MATH . Thus a MATH-analogue of REF holds. As in the previous section we construct representations of the NAME algebra in terms of the functions MATH whose existence is guaranteed from REF of Ref. CASE: As before we construct representations of the algebra MATH associated to the above multiresolution for the deformed case. The representations are realized on a NAME space MATH where the measure is given by MATH.
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math/0006105
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For MATH, let MATH be the number of MATH with MATH. The argument in the remark on p. REF (following REF ) shows that MATH for each MATH. Thus, we have MATH . If MATH, then MATH for each MATH and equality holds. This proves REF . For REF , note that under the given hypothesis we have MATH. Since MATH for all such MATH, REF follows immediately.
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math/0006105
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The rank REF situation leads to the item listed in the table. When the rank is at least REF, one applies REF to obtain MATH, whence MATH for some MATH; that is, MATH is minuscule. We handle the minuscule cases by classification. For each indecomposable root system MATH for which MATH, we list in the following table the NAME number, the set MATH, and the value MATH for each MATH. The simple roots are indexed as in the tables in CITE; the data recorded here, with the exception of the values MATH, may be verified by inspecting those tables as well. The values MATH are well known (and can anyway be computed from the formula, or by representation theoretic arguments). MATH . From this table, one can list all pairs MATH for which MATH has NAME number MATH and MATH is minuscule. It is a simple matter to see that MATH only when MATH is as claimed.
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math/0006105
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REF follows from the linkage principle for MATH CITE, and REF from the linkage principle for MATH CITE. REF follows from CITE. REF follows from CITE.
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math/0006105
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When MATH this follows since MATH is a simple MATH-module with restricted highest weight. When MATH, we have MATH. Since MATH is simply connected, we have MATH. Thus MATH is an indecomposable MATH-module with unique simple quotient MATH, and the lemma follows.
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math/0006105
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Let MATH be the (homogeneous) defining ideal of the variety MATH. We need to show that MATH. If not, then MATH for some proper MATH-submodule MATH. A look at the summary in CITE shows that, since MATH is simply connected, the only MATH-submodules of MATH have dimension REF or REF. On the other hand, by CITE, the variety MATH has codimension MATH in MATH and so clearly can't be contained in a REF dimensional linear subspace!
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math/0006105
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For REF see CITE. For REF , first suppose that MATH. By CITE, there is a MATH-equivariant isomorphism of graded rings MATH where MATH is again the graded coordinate ring of MATH, but with the linear functions on MATH given degree REF. The claim now follows from the lemma. When MATH, apply CITE to see that MATH; the claim follows again from the lemma in this case.
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math/0006105
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REF follows from CITE, and REF from CITE, see also CITE. For REF , note first that REF implies MATH by REF . If MATH, then REF imply that MATH, whence MATH follows from NAME 's degree formula. REF now follows since MATH is empty if MATH and MATH if MATH. In CITE, NAME made a list of all simple restricted modules for MATH with dimension MATH. Inspecting that list yields REF .
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math/0006105
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Write the highest weight of MATH as MATH with MATH restricted. Since MATH, we have MATH. Since MATH, REF implies that MATH and that MATH. We have in particular that MATH, hence the proposition will follow from REF if we show that MATH is REF. If MATH, NAME 's tensor product theorem gives MATH. If MATH then REF shows that MATH and MATH. If MATH for some MATH, then MATH by the linkage principle, whence MATH. Now REF applies; it shows that MATH whence we have MATH by another application of NAME 's theorem.
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math/0006105
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We verify REF , the argument for REF is the same. We must show that MATH; first note that MATH is the cohomology of the sequence MATH from which we get MATH. For any first quadrant spectral sequence one has (by similar reasoning) that MATH for MATH, so we get the desired isomorphism.
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math/0006105
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The NAME kernel MATH is a normal subgroup of MATH; thus there is a NAME spectral sequence computing MATH which in view of REF has the form MATH . If MATH, MATH by REF . There is an exact sequence of the form CITE MATH . Thus MATH. We get now REF by induction on MATH. REF shows now that MATH. Thus, the only possible non-REF MATH terms of total degree REF are MATH . For MATH, we apply REF Lemma B to see that MATH whence MATH by REF; thus REF will follow provided it holds for MATH. In that case, we have MATH by assumption, and the result just proved in REF shows that MATH. Thus REF applies; it shows that MATH as desired.
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math/0006105
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Let MATH be such that MATH for MATH, and such that MATH acts non-trivially on MATH. We have by REF that MATH. Also, we have by REF that MATH for MATH. If MATH, we are done. If MATH, then REF applies, and we get that MATH . We get by REF that MATH unless MATH and MATH with MATH. If MATH, then MATH is a simple MATH-module by REF . So if MATH then MATH whence MATH as claimed. In the remaining case, one must just note that weight considerations yield MATH for MATH, whence MATH.
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math/0006105
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Since MATH, REF shows that MATH is the simple module with highest weight MATH. It follows that MATH, and thus that MATH for MATH by REF . The proposition now follows from REF .
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math/0006105
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CASE: Follows from CITE. CASE: Since MATH is a pro-MATH group CITE, this follows from CITE. CASE: Choose a MATH vectorspace MATH and a non-trivial faithful MATH-rational representation MATH. For each extension MATH of MATH with integers MATH, the group MATH is a subgroup of (the group of MATH-points of) MATH. If MATH, the sequence in REF is split and MATH is a non-trivial MATH-module. Since MATH has characteristic REF, it is well known that the minimal dimension of a non-trivial MATH module is bounded below by the value MATH of a polynomial MATH, depending only on MATH, for which MATH as MATH. We may choose MATH such that MATH for each MATH, and REF follows at once. REF now follows from REF and CITE.
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math/0006105
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By CITE, MATH for some restricted dominant weight MATH. Thus MATH is a restricted, simple MATH module with dimension MATH. It follows from REF that MATH, that MATH, and that MATH as modules for MATH. Suppose that MATH for some MATH. By the linkage principle for MATH REF , we must have MATH, hence MATH. This implies that MATH. REF shows that MATH for some MATH, and REF yields MATH and lists the possible pairs MATH. For MATH, CITE have extended a result of CITE; this result implies in particular that the minimal degree for which MATH is non-REF is MATH, and that MATH . It is straightforward to compute for each pair MATH the length MATH; one gets in this way the result.
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math/0006107
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The first formula is an immediate consequence of the theorem. It implies that MATH which yields the second formula.
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math/0006107
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Since uniruledness is a birational property, it is enough to observe that MATH is not uniruled because it has nonnegative NAME dimension.
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math/0006107
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Assume the contrary that MATH, and let MATH be the corresponding family. Then each fiber MATH is a projective space bundle over the Jacobian MATH by NAME; in particular, it's uniruled. Therefore MATH and hence MATH are uniruled, but this contradicts the previous corollary.
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math/0006107
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We assume that MATH for some family MATH. By the previous corollary, we may suppose that MATH. Denote the maps MATH and MATH by MATH and MATH respectively. The map MATH is dominant and generically injective on the fibers of MATH. If a general fiber MATH has positive dimension, then an irreducible hyperplane section MATH meets MATH. Therefore MATH is still dominant, and we may replace MATH by MATH and MATH by the fiber product. By continuing in this way, we can assume that MATH is generically finite. Choose a desingularization MATH of a compactification of MATH, and a nonsingular compactification of MATH of MATH such that MATH extends to a morphism of MATH to a desingularization MATH of MATH. We then have MATH which implies that MATH has general type. On the other hand the general fiber of MATH is MATH is birational to an Abelian variety by REF and NAME. This implies that MATH by CITE, but this is impossible since MATH has general type.
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math/0006107
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Construct a commutative diagram MATH where MATH is a desingularization, and MATH is a MATH-equivariant desingularization of the fiber product. Then there are inclusions MATH . This implies the first part of the lemma. The inclusion MATH is an equality on the complement of the fixed point locus. Since MATH is locally free (hence reflexive) and MATH is torsion free, the second statement follows.
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math/0006107
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As the result is local analytic for MATH, we may replace it by MATH. Consider the action of the symmetric group MATH on MATH by permutation of factors, and let MATH be the corresponding homomorphism. This action is equivalent to a direct sum of MATH copies of the standard representation MATH where MATH acts via permutation matrices. Therefore MATH does not contain any quasi-reflections (because MATH) and MATH. When MATH is even, MATH. This implies that MATH is NAME by CITE, and therefore canonical of index one CITE. The case when MATH is odd is more laborious. For any element MATH of order MATH, define MATH as follows: choose a primitive MATH-th root of unity MATH and express the eigenvalues of MATH as MATH where MATH, set MATH. By CITE, to prove that MATH is canonical it will suffice to verify that MATH for every element MATH of order MATH. Let MATH be the permutation representation of MATH. If MATH is the standard basis then MATH. If MATH is a primitive MATH-th root of unity, then it is easy to see that the eigenvectors of the cycle MATH acting on MATH are MATH and MATH. Therefore the nonunit eigenvalues are MATH and these occur with multiplicity one. Hence MATH as required. The general case is similar. Let MATH be a permutation of order MATH and MATH as before. Write MATH as a product of disjoint cycles of length MATH. Therefore MATH is the least common multiple of the MATH, and let MATH. A list (with possible repetitions) of the nonunit eigenvalues of MATH acting on MATH is MATH . Therefore MATH and this proves that MATH is canonical. It remains to check that the index is at most MATH. For this it suffices to observe that if MATH are coordinates on MATH, then MATH is MATH invariant. This determines a generator of MATH, which shows that this module is free.
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math/0006107
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Let MATH be an integer such that MATH is even (hence a multiple of the index of MATH). Then MATH . By NAME 's formula, this equals MATH .
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math/0006108
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Consider the following diagram MATH . The square on the right is commutative because MATH is a natural transformation; the square on the left is commutative since REF holds. One of the compositions from the left upper corner of REF to the right lower corner is MATH . The other composition equals MATH .
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math/0006108
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If MATH then by the NAME duality MATH, and hence MATH (compare REF ). Therefore in this case MATH coincides with MATH. The result now follows from REF .
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math/0006108
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We have a short exact sequence MATH of chain complexes in MATH. The corresponding long homological sequence provides the required isomorphism, since MATH is exact.
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math/0006108
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Applying REF to two manifolds MATH and MATH, we obtain two submodules MATH, described in REF . First, we observe that each of MATH is a metabolizer, that is, MATH. Indeed, by REF , the factor MATH is isomorphic to the cokernel of the intersection pairing MATH (we use the notations introduced in REF). However, our assumption MATH implies that MATH. Now we show that the metabolizers MATH and MATH are "disjoint", that is, MATH (here the intersection is understood as intersection of two subobjects of an object of an abelian category). Indeed, according to REF and the previous arguments, we may identify MATH with the kernel of the morphism MATH induced by the inclusion. Using the NAME sequence MATH and our assumption MATH, we obtain MATH. Now we want to show that MATH. We examine the NAME sequence REF again. Using the NAME duality we find that our assumption MATH implies MATH. Therefore we conclude that the direct sum MATH belongs to the image of MATH. Thus we get an isomorphism MATH which implies MATH. As a result, MATH are mutually complementary metabolizers in MATH, and this implies hyperbolicity of the linking form of MATH.
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math/0006108
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REF follows from REF follows from REF.
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math/0006108
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The proof is similar to the proof of REF described above in full detail. It is based on REF , which explicitly computes the linking form MATH. The distinction happens only at the very last stage of the proof, when one computes the torsion signatures of the elementary forms REF and uses additivity REF .
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math/0006108
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For any subobject MATH we have MATH because of the canonical isomorphism MATH, and MATH is isomorphic to MATH (not canonically). If MATH is the metabolizer, then MATH, (from REF ) and hence vanishing of excess REF is equivalent to REF .
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math/0006108
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From the proof of REF we know that without loss of generality we may assume that MATH is represented as MATH, where MATH is an object of MATH and MATH is a self-adjoint positive operator, MATH, and the metabolizer MATH is represented as MATH with MATH being the positive square root of MATH. If MATH is the spectral density function of MATH then MATH is the spectral density function of MATH. Therefore we obtain MATH . Now, REF shows that MATH which proves REF . If MATH then MATH contains the metabolizer of MATH (by the arguments in the proof of REF) and therefore MATH. This completes the proof.
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math/0006108
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If MATH then by REF MATH . Thus, if the excess REF vanishes, we get MATH which implies MATH. Conversely, if MATH, then REF shows that MATH. If REF holds, then MATH (again, because of REF ) and MATH . This completes the proof.
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math/0006112
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Begin by assuming MATH is stably simple. Let MATH and MATH be two transversal knots in the knot type MATH with the same self-linking numbers. Note MATH has a neighborhood MATH contactomorphic to MATH with the contact structure given by MATH . Now, for large integers MATH if MATH are the tori in MATH with MATH, then the characteristic foliation on MATH is by MATH curves. Let MATH be a leaf in this characteristic foliation. Note MATH is a Legendrian knot topologically isotopic to MATH and that MATH. Note also that if we have some MATH, then we have MATH for all MATH . Thus from MATH we have a well-defined stable isotopy class of Legendrian knot MATH and from MATH we similarly get MATH . Now since MATH we know that there is some MATH and MATH such that MATH is Legendrian isotopic to MATH . Finally, by observing that MATH and MATH, we see that MATH and MATH are transversally isotopic. Thus MATH is transversally simple. The other implication is proved at the end of the next section as it requires the theory of convex surfaces developed there.
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math/0006112
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Let MATH and MATH be any standardly embedded tori in MATH on which MATH and MATH respectively sit. We describe everything in terms of MATH and MATH but everything also holds for MATH and MATH . By REF we may make MATH convex without moving MATH, since the twisting of MATH with respect to MATH is MATH (recall MATH are positive). Now that MATH is convex and MATH is maximal, we may assume MATH is in standard form. This follows from observing that MATH implies the existence of a bypass along MATH, and hence a destabilization. Here MATH is the unsigned (actual) intersection number and MATH is the (minimum) geometric intersection number of the two isotopy classes. Let MATH REF be the slope of the dividing curves MATH and MATH be the number of dividing curves. According to REF , MATH . Thus for MATH to equal MATH we must have MATH and MATH . Let MATH and MATH be the NAME splittings associated to the tori MATH and MATH. Since the slopes of the dividing curves are the same, we may use the classification of tight contact structures on solid tori in REF to find a contactomorphism MATH (note that a meridional disk for MATH of MATH has exactly one dividing curve). Applying REF again, we may extend MATH to all of MATH, thus obtaining a contactomorphism of MATH which takes MATH to MATH . By NAME 's result REF we may find a contact isotopy of MATH taking MATH to MATH . So now we may assume that MATH and MATH are two Legendrian knots on the same convex torus MATH . Note MATH and MATH are both leaves in the ruling foliation of MATH . The other ruling curves exhibit a Legendrian isotopy from MATH to MATH .
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math/0006112
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Let MATH be a standardly embedded torus on which MATH sits. As in the proof of REF we may assume that MATH is convex and in standard form, since otherwise MATH can be destabilized. However, this time the slope MATH of the dividing curves is not MATH or the number of dividing curves is not REF. We first consider the case when MATH . Let MATH be the NAME splitting associated to the the torus MATH . Recall the ``slope" of the dividing curves is usually measured thinking of MATH as the boundary of MATH . As the boundary of MATH or MATH the slope of the dividing curves on MATH will be less than MATH . Assume MATH has this property. We know, by REF , that looking at concentric convex tori in MATH we will see dividing curves with any slope in MATH . In particular, there will be a torus MATH with two dividing curves having slope MATH . Let MATH be the region between MATH and MATH in MATH and MATH be an annulus lying between MATH and MATH in MATH . The boundary of MATH is convex and we may assume that the ruling curves on both boundary components have slope MATH . Thus we may assume that MATH is Legendrian and MATH is convex. The dividing curves will intersect MATH times and MATH times. As MATH and MATH are not both REF, MATH, so we can find a boundary-parallel arc along MATH among the dividing curves of MATH . This implies the existence of a bypass for MATH and hence a destabilization. Specifically, MATH for some Legendrian MATH. Repeating this argument we will eventually find a sequence of destabilizations which will exhibit MATH as MATH where MATH is the unique Legendrian MATH-torus knot with maximal NAME invariant. Now for the case when MATH and MATH . In this case we claim that there is a torus MATH in, say MATH parallel to MATH with two dividing curves having slope MATH . To see this just take a copy of MATH inside MATH, make the ruling curves meridional, and then look at a convex meridional disk MATH . We may use the bypasses on MATH to reduce the number of dividing curves on the copy of MATH until there are only two. (This follows easily from REF . Also see CITE.) Call the resulting torus MATH . Let MATH be the region between MATH and MATH . Now if we make the ruling curves on MATH have slope MATH then using the annulus MATH in MATH we may repeat the above argument to destabilize MATH .
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math/0006112
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First note that the example in REF shows that MATH . We show that MATH by contradiction. If MATH, then we can construct a NAME manifold MATH (with boundary) by adding a REF-handle to REF-ball along a MATH-torus knot with framing MATH . (Given a symplectic REF-manifold with convex boundary, one can add a symplectic REF-handle to a Legendrian knot MATH in the boundary with framing MATH and obtain a new symplectic REF-manifold with convex boundary, see CITE.) According to CITE, MATH is the connected sum of two lens spaces (neither of which is MATH or MATH and the knot is trivial). Now, REF says that MATH must be a boundary sum of two manifolds MATH and MATH . Using the NAME sequence we see that one of the MATH's, say MATH must be an integral homology ball, but this is impossible since MATH is a nontrivial lens space.
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math/0006112
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Since MATH we can use REF to show MATH lies on a convex standardly embedded torus MATH . We know the dividing curves MATH on MATH have slope MATH since MATH . Now as measured on either MATH or MATH the dividing curve have slope less than MATH . Assume MATH has such dividing curves. Now since REF says that we can find tori in MATH whose dividing curves have any slope in MATH we can find a torus MATH in MATH whose dividing curves have slope MATH . Now as in the proof of REF we can take an annulus MATH between MATH and MATH with one boundary on MATH and the other on a Legendrian divide of MATH and thus find a bypass for MATH since the dividing curves on MATH will intersect the boundary component containing MATH more than the one containing the Legendrian divide. Hence we may use this bypass to destabilize MATH .
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math/0006112
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Let MATH and MATH be tori on which MATH and MATH respectively sit. We assume that MATH and MATH have been arranged as described above. The proof is finished as we finished the proof of REF . We only need to recall that REF says that the contactomorphism type of a tight contact structure on a solid torus (with standard convex boundary) is determined by the number of positive bypasses on a meridional disk and the number of positive bypasses on the meridional disks for MATH and MATH where MATH are determined by the rotation number of MATH .
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math/0006112
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We assume MATH and leave the similar case to the reader. In this situation there must be a MATH so that MATH and MATH . Thus MATH and MATH . In the notation set up above, let MATH we the convex torus outside MATH with boundary slope MATH and MATH be the corresponding one for MATH . From REF stated above we know that MATH and MATH . By examining dividing curves on the meridional disks, we also have MATH (from REF ). Now arrange for the Legendrian ruling curves on MATH to have slope MATH and consider the annulus MATH between MATH and MATH with slope MATH . We can take one boundary of MATH to be MATH and the other a Legendrian ruling curve on MATH . Once we make MATH convex, the dividing curves do not intersect the boundary component touching MATH and intersect the other boundary component MATH times. Since the rotation number (values of MATH and MATH) of the boundary components of MATH differ by MATH, we may use REF (or a slight generalization of it, see CITE) to see there are MATH boundary-parallel dividing curves separating off MATH negative disks from MATH . This shows that MATH . Similarly, we see that MATH is MATH . We may now use the argument in the proof of REF to find a contact isotopy taking MATH to MATH . A further contact isotopy takes MATH to MATH .
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math/0006112
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Let MATH and MATH be two MATH-torus knots with the same invariants. Let MATH destabilize to MATH and MATH to MATH . If MATH and MATH have the same invariants, then they are Legendrian isotopic and MATH and MATH are the same stabilization of the same knot and hence are Legendrian isotopic. Now suppose the rotation numbers of MATH and MATH differ by MATH as in REF . Then we can realize MATH as a stabilization of MATH and MATH as a stabilization of MATH . Thus they both destabilize to MATH and hence they both destabilize to MATH and MATH implying they are Legendrian isotopic. Finally, if the rotation numbers of MATH and MATH differ by something else, then a similar argument will show that MATH and MATH will destabilize to the same Legendrian knot, finishing the proof.
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math/0006112
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Arrange for all the Legendrian ruling curves on MATH, and MATH to be meridional and let MATH be a meridional disk for MATH that intersects these three tori in Legendrian curves. We may now isotop MATH (relative to its intersection with the tori) so that it is convex. Let MATH where MATH and MATH . Orient MATH so that MATH is the oriented boundary of MATH . Now the dividing curves on MATH intersect MATH times (since that is the number of times the dividing curves on MATH intersect MATH). Moreover, they intersect each of MATH and MATH REF times. We claim that all the dividing curves on MATH separate off disks that contain no dividing curves. Note this implies that all the bypasses on MATH are of the same sign. If this were not the case, then we would have bypasses on MATH of both signs and we would be able to glue this to a bypass of the same sign on MATH, creating an overtwisted disk. To match a bypass on MATH with any bypass on MATH we might have to ``add copies of MATH to MATH," that is, take a copy of MATH, cut it along MATH to obtain an annulus, and glue one of its boundary components to MATH and the other to MATH. This has the effect of shifting the dividing curves on MATH relative to those on MATH by MATH (for details see the section on Sliding Maneuvers in CITE). Since all the bypasses have the same sign, REF implies that MATH or MATH .
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math/0006112
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Make the Legendrian ruling curves on MATH be meridional and let MATH be the meridional disk for MATH with Legendrian boundary. Note MATH (or at least a translate of it). Now the dividing curves intersect the boundary of MATH times. And since there are no closed homotopically trivial dividing curves we can conclude that there are exactly MATH dividing curves. By examining the possible configurations and using REF , one may readily conclude that MATH . Moreover, by recalling how MATH is related to the rotation number of MATH and noting the values of MATH realized in REF , we see that all the possible values of MATH are actually realized.
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math/0006112
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Make the Legendrian ruling curves on MATH and MATH longitudinal and let MATH be a longitudinal annulus spanning between MATH and MATH with Legendrian boundary. After making MATH convex, the dividing curves will intersect MATH in MATH points and MATH in MATH points. We claim that MATH dividing curves run from one boundary component of MATH to the other boundary component and there is one boundary-parallel dividing curve with endpoints on MATH . If this is not the case, then there must be boundary-parallel dividing curves for MATH and hence a bypasses. Carefully applying REF , we can then find a convex torus between MATH and MATH with slope not lying between MATH. This, by REF , implies that there is a convex torus with boundary slope REF and the Legendrian divides on this torus will be the boundaries of overtwisted disks. Therefore, the dividing curve configuration on MATH is as claimed. Now make the Legendrian ruling curves on MATH longitudinal and make MATH intersect MATH in one of these longitudinal curves, say MATH . The curve MATH separates MATH into two annuli - MATH which touches MATH and MATH which touches MATH . Note the dividing curves are still as described above. From this we can deduce the structure of the dividing curves on MATH . To this end, note the dividing curves on MATH intersect the boundary component touching MATH times and the boundary component touching MATH times. Due to the structure of the dividing curves on MATH, we know that the MATH dividing curves emanating from MATH run across MATH to the other boundary component. Thus we know there are MATH other dividing curves, all of whose boundaries lie on MATH . As argued in the proof of REF , we can conclude that these all separate off disks that contain no dividing curves and thus give MATH bypasses of the same sign. Let MATH be the relative NAME class for the region MATH between MATH and MATH. Since MATH, REF yields MATH . If MATH, then we claim that MATH . Assuming the contrary, we obtain a contradiction to the tightness of MATH as follows: Using REF one may easily check that MATH . Thus, if MATH then MATH . However, looking at the convex annulus MATH of slope MATH in MATH, we see that the there are only MATH non-closed dividing curves, all of which have their boundary on MATH . This means that for MATH there must be some dividing curves bounding disks, violating the tightness of MATH . Therefore, MATH .
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math/0006112
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This is quite similar to the proof of REF and is left to the reader.
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math/0006112
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Note that any claim we will be making concerning MATH should be interpreted up to a MATH-small isotopy - in particular, we will make extensive use of the Legendrian Realization Principle REF without explicitly mentioning each time that a MATH-small perturbation is taking place first. Let MATH be a convex fiber in MATH with Legendrian boundary, and MATH an open convex neighborhood of MATH. Then in MATH we look for an annulus MATH with MATH, where the MATH are Legendrian curves in (a MATH-isotoped copy of) MATH, and MATH contains a bypass. Denote the dividing set of MATH by MATH. Let us consider Case I. Observe that MATH is always even. Here we may always find a closed Legendrian curve MATH on MATH which does not intersect MATH. We take MATH to be the annulus MATH . Note that MATH and the geometric intersection number MATH, because an NAME map cannot fix a curve isotopy class. Realize MATH as a Legendrian curve which has minimal intersection number with MATH in its isotopy class. Then MATH will have at least MATH dividing curves emanating from MATH and none from MATH, from which we can conclude using the Imbalance Principle REF that there is a bypass along MATH . If MATH, then we may always find an embedded bypass, and we apply REF . Let MATH be the attaching Legendrian arc MATH for the bypass, and MATH be the intersections with MATH in consecutive order along MATH. Since MATH, MATH lie on distinct dividing curves. If MATH both lie on the MATH's (or the same for MATH), then there will exist a boundary-parallel dividing curve on the new MATH, after bypass attachment (and thus a bypass for the new MATH). If MATH both lie on the MATH's (or the same for MATH), then we will be able to reduce MATH by two. Hence, if MATH and MATH, every bypass will be as above, and we may either reduce MATH by REF or obtain a bypass for MATH along MATH. Now there are two other cases: MATH, MATH, and MATH lie on MATH's and MATH lies on a MATH; or MATH, MATH, and MATH lie on MATH's and MATH lies on a MATH. In the latter case, we reduce MATH by REF. (Note that, in the former case, the bypass attachment yields a dividing set which puts us in Case III, where two of the isotopy classes of arcs are represented by a single MATH.) We eventually have REF MATH and MATH, REF MATH and MATH, REF MATH and MATH, or REF a bypass along MATH. We can actually do better, and remove REF MATH, MATH. If MATH intersects MATH exactly twice, then we have a degenerate bypass along MATH (one with same endpoints), and attaching this degenerate bypass gives rise to a dividing set which has a boundary-parallel dividing curve. If the intersection number is greater than REF, then keep MATH but repeatedly attach bypasses until we arrive at boundary-parallel components of (the new) MATH. Observe that, once we have reduced to REF or REF , there will still exist bypasses that can be attached onto MATH. However, they do not necessarily yield boundary-parallel dividing curves after attachment and cannot be used to destabilize MATH - at least not immediately. In Cases II and III, MATH by REF . Consider Case II. Let MATH and MATH be the two isotopy classes of curves of MATH, and MATH, MATH be the number of arcs in each. Note that both MATH and MATH are even. For convenience, we will identify MATH, and MATH, MATH with minimal (shortest-length) integral vectors MATH. It is easy to find a curve MATH on MATH so that MATH intersects MATH and MATH in a different number of points. Indeed consider the subset MATH of the set MATH of (isotopy classes of) closed (connected) curves MATH on MATH, consisting of MATH which have minimal geometric intersection MATH among curves in MATH. Since the NAME map MATH cannot preserve a finite set of closed curves, MATH. Noting that MATH consists of curves MATH with minimal intersection number MATH among curves in MATH, we pick MATH. Using this MATH we may argue as in REF to find a bypass along MATH. As long as MATH, MATH (as above) lie on three distinct dividing curves, and MATH (or MATH) lie on `consecutive' dividing curves of the same type MATH (or MATH). Thus we can produce a boundary-parallel component of the new MATH after attaching the bypass to MATH. Therefore, at least one of the MATH must equal MATH. We will eliminate MATH, MATH (or MATH, MATH). In this case, MATH consists of one curve MATH parallel to MATH. As above, MATH, so MATH, since MATH intersects both MATH and MATH. We then let MATH, and find a bypass along MATH. The only nontrivial case is when our bypass along MATH involves only the two dividing curves isotopic to MATH, MATH lie on the same curve, and the bypass attachment changes the dividing curves MATH, MATH to MATH, MATH. In this case we will have a sequence of `nested bypasses'. More precisely, attach the (outermost) bypass onto MATH and take a new MATH such that MATH has fewer intersections with the new MATH. Note that MATH intersects MATH twice, but MATH still intersects MATH at least four times. In any event, if we reduce this intersection number to four, there can no longer exist any bypasses of the nontrivial type, and we produce a destabilization. The only case left is when MATH. In Case III, if we crush the boundary MATH to a point, then homologically one of the isotopy classes of MATH is the sum of the other two isotopy classes. Therefore we write the three classes as MATH, MATH, MATH, and the number of dividing curves as MATH, MATH, MATH (these must have the same parity). Using the same argument as in Case II (with MATH), we produce an annulus MATH, together with a bypass along MATH. As long as three of the isotopy classes have more than one dividing curve, any bypass along MATH gives rise to a boundary-parallel component of MATH on MATH. Thus, one of the isotopy classes will have one dividing curve.
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math/0006112
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If MATH has dividing set MATH, then act via MATH to get MATH on MATH, which is a convex surface which is isotopic to MATH. Hence, we may freely modify MATH for any MATH, via an isotopy. Let MATH be the slopes corresponding to the two eigendirections of MATH, with MATH. On the circle at infinity MATH of the standard tessellation of MATH, MATH is an attracting fixed point and MATH is a repelling fixed point under the action of MATH. Suppose that the initial configuration MATH corresponds to slopes MATH respectively. Then we may act repeatedly via MATH so that MATH, and we have three possibilities: CASE: MATH. CASE: MATH. CASE: MATH, MATH, or MATH, MATH. Assume that in all the cases, bypass attachments do not give rise to boundary-parallel dividing curves (and hence a destabilization), since that would contradict MATH. This means that, for dividing curve configurations of type III REF , we only use moves of type A through I in REF . For configurations of type I REF , we only use the moves in REF . To simplify notation, write MATH for the shortest integral vector with nonnegative entries, corresponding to a slope MATH. CASE: We may assume in addition that MATH . Indeed, if MATH then we are done, and if MATH then MATH since any triangle in the tessellation whose vertices are between MATH and MATH and whose clockwise-most vertex MATH has all its vertices MATH . Now since MATH we may use MATH to achieve the desired slopes. We will show that using bypasses we can perform a sequence consisting of the following moves to our MATH and MATH (hence affecting MATH): CASE: replace MATH with MATH, where MATH and MATH (corresponds to right drawing in REF ) CASE: replace MATH with MATH, where MATH and MATH (corresponds to left drawing in REF ) or CASE: replace MATH with MATH, where MATH and MATH and MATH are determined by MATH, and the triangle corresponding to MATH is the innermost triangle in the tessellation with MATH and all vertices counterclockwise of MATH . Here, `innermost' means closest to the center of the disk. Note that, if we start with a triangle MATH corresponding to MATH and MATH, then there is a finite number MATH of triangles that can be obtained from MATH by the sequence of moves listed above. Moreover, the triangle with vertices at MATH is one of the triangles that can be so obtained. Now, if MATH is obtained from MATH by one of the moves above, then MATH . Thus any (sufficiently long) finite sequence of these moves will take us from MATH to the triangle with vertices at MATH . We now show how to find the bypasses that allow us to perform the above moves. Since MATH, MATH, and, by using an annulus MATH with MATH parallel to MATH, we find a bypass of type A, F, or H in REF . The bypasses are of the stated type, since the slope of the bypass is smaller than MATH or MATH . Suppose we do not already have MATH. Moves of type A and H replace the ideal triangle with vertices MATH by another triangle with vertices MATH (see REF ). Here MATH corresponds to the slope of MATH. Thus, MATH and MATH are affected by applying one of the first two moves mentioned above. A move of type F will give rise to a configuration of type I REF with MATH and MATH. Just as we denoted a configuration of type III REF by MATH, we will denote a configuration of type I REF by MATH. We deal with the case MATH at the end of the paragraph. Otherwise, in order to modify MATH, take an annulus MATH, where MATH has slope MATH. Note that MATH. Since bypass attachments cannot give rise to boundary-parallel dividing curves on MATH, there are two possibilities, given in REF . Here MATH is the smallest slope for which there exists an edge of the tessellation from MATH to MATH. (Note: in general there are other possibilities for the slope of the bypass but with this choice of basis for the torus and our choice of MATH only those shown in REF are possible.) After the bypass attachment, MATH has dividing curves of type III REF and two of the dividing curves have slopes MATH respectively. Let MATH and MATH be the minimal integral vectors corresponding to these slopes. Then the third dividing curve on MATH is spanned by MATH or MATH . In the first case we have performed a move of type REF listed above and in the second case we have performed a move of type REF followed by one of type REF. Thus we may perform a sequence of the above moves eventually obtain MATH or at some point we obtain a configuration of type I REF with curves of slope MATH. In this case use MATH to modify MATH. Let this be MATH. Then, using the annulus MATH where MATH has slope MATH, we obtain a bypass which modifies MATH. CASE: Assume MATH. Without loss of generality, assume MATH (this argument is the same as the one at the beginning of REF ). Use the annulus MATH, where MATH has slope MATH, to obtain a bypass of type A, F, or H. The same argument as in REF allows us to isotop MATH so that the dividing curves are of type I REF with slope REF or of type III REF with slopes MATH . In the first case, attaching the only possible bypass along MATH which does not yield a boundary-parallel component on MATH, we obtain MATH. Next suppose we have MATH. If we use the annulus MATH where MATH has slope MATH, then MATH, and F cannot happen (for F to happen, we need MATH). Therefore, MATH, which is dealt with in REF . CASE: Here we start with a triangle MATH in the NAME tessellation whose vertices straddle MATH . Below we show that attaching bypasses will either put us back in REF , which we have already dealt with, or we perform one of the following moves on MATH and MATH CASE: replace MATH with MATH and MATH (if MATH and MATH) or CASE: replace MATH with MATH and MATH (if MATH and MATH). Assuming this is true for the moment. There is some MATH such that the vertices MATH span a triangle MATH in the NAME tessellation that is disjoint from MATH and separates MATH from MATH . Each time we perform a move of type REF or REF to MATH, we obtain a new triangle MATH lying between MATH and MATH. Since there are only a finite number of triangles in the tessellation lying between MATH and MATH, this process must stop. It can only stop if MATH (or a bypass attachment puts us in REF ), in which case we are done. Suppose that MATH and MATH. Use the annulus MATH, where MATH has slope MATH, to obtain a bypass of type C, E, or G. C gives MATH, with MATH, which was already treated under REF . Bypasses of type E or G modify MATH to MATH - and we are still in REF (though we might have MATH and MATH). Now suppose that MATH and MATH. Similarly, we have bypasses of type B, D, or I. I puts us in REF , and B, D modify MATH to MATH (still in REF ).
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math/0006112
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Note that MATH is a genus two handlebody. In general, when analyzing handlebodies with convex boundary, we use compressing disks MATH and MATH so that the cut-open manifold is a REF-ball. To obtain a smooth convex boundary of MATH, we need to round the edges MATH. This is done using the NAME Lemma REF . When using NAME, we must be careful to remember that, since MATH was the convex boundary of a neighborhood of a Legendrian curve, there is holonomy (as we go from MATH to MATH). We then arrange for MATH, MATH to have Legendrian boundary - for this we use the Legendrian Realization Principle. Then, MATH, MATH with Legendrian boundary are perturbed, fixing the boundary, so they become convex. What remains is to study the configuration of dividing curves on each compressing disk. We have MATH where MATH is the annulus MATH . Below we discuss the dividing curves on MATH and MATH but first we consider the dividing curves on then annulus MATH and how they are related to the dividing curves on the MATH's. To this end consider REF . In this figure we identify the right and left edges, then the center rectangle forms the annulus MATH . The top and bottom parts of the picture form neighborhoods of MATH in MATH and the shaded region is obtained using the NAME Lemma. The dividing curves on MATH divide the MATH into REF intervals. We label the intervals REF so we can identify them with the corresponding intervals on the annulus MATH . CASE: MATH and MATH. See REF . There exists a disk MATH where MATH is an arc with slope REF, with geometric intersection MATH. Note that MATH are represented by solid lines in REF , and MATH lies on MATH . Though MATH is not drawn in REF , one may easily determine that MATH. By the Legendrian Realization Principle, we may realize MATH as a zero-twisting (rel MATH) Legendrian curve. This implies that there is an overtwisted disk and that there cannot be a tight contact structure on MATH with this boundary condition. CASE: MATH and MATH. Let MATH be an arc on MATH with slope REF and let MATH be an arc with infinite slope. Choose the compressing disks MATH and MATH for MATH to be isotopic to MATH and MATH, drawn as in REF . After using the Legendrian Realization Principle, we take the boundaries to be Legendrian and MATH and MATH to intersect the dividing curves on MATH (minimal geometric intersection number) REF times, respectively. There is only one possibility for the dividing curves on MATH. However, there are several possible dividing curve configurations for MATH. Consider the intersections between MATH and MATH - count the intersections from bottom to top along MATH shown in REF . One may easily check that there cannot exist a bypass along MATH which straddles the second intersection or the fourth intersection, without immediately yielding a bypass for MATH. Thus, the only two possibilities for MATH are shown in REF . Now, if MATH had the right-hand configuration shown in REF , then a bypass straddles the third intersection between MATH and MATH. Note this bypass is nested inside another bypass on MATH . If we added both these bypasses in succession, the resulting copy of MATH would have dividing curves MATH . Hence we could find an overtwisted disk as above. Thus, the dividing curves on MATH are shown on the left-hand side of REF . Now, since we have normalized each of MATH, MATH to have a unique dividing curve configuration, we have a unique tight contact structure on MATH with given configurations MATH, MATH, up to isotopy. This is because any two tight contact structures with these boundary conditions can first be matched up along MATH, MATH, and then matched up inside MATH using REF .
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math/0006116
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We will use an argument in CITE together with standard facts about refined NAME maps from CITE. The rough idea of the proof is that the conjecture is obviously true for a NAME MATH, and then the results of CITE and CITE will show that it remains true when we pull back from MATH to MATH. To see how this works in detail, we first recall some notation. Suppose we have REF stacks: MATH where MATH is a regular embedding. In this situation, we have the refined NAME map MATH defined in CITE, where we use CITE to extend from schemes to stacks. Now assume that there is also a vector bundle MATH on MATH such that MATH is the inclusion of the zero locus of a regular section of MATH. By CITE, it follows that MATH for all MATH, where MATH and MATH are from REF. To apply this to our situation, suppose that MATH is an embedding bundle on MATH and MATH is the zero locus of a section MATH of MATH. By REF , we can assume that MATH is embedded in the NAME MATH. Furthermore, MATH induces a section MATH of the universal quotient bundle MATH on MATH via the tautological quotient mapping MATH . Let MATH be the zero locus of MATH. It follows from this description that MATH is a regular section of MATH. This description also implies that MATH and that MATH. Now fix MATH and suppose that MATH maps to MATH. Then we have a cartesian diagram: MATH where MATH and MATH are the natural inclusions. Since MATH and MATH are homogeneous spaces, it follows that MATH is a regular embedding of smooth stacks. Applying the above construction, we get the class MATH . Using the argument of CITE, we see that MATH . NAME 's lemma uses MATH and MATH rather than MATH and MATH, but the proof still applies since MATH and MATH are convex. Also, while the statement of REF is different from REF, the final sentence of his proof shows that REF follows from his argument. In REF, we discussed how MATH induces the bundle MATH on MATH. In a similar way, the universal quotient bundle MATH induces a bundle MATH on MATH. The map MATH is the zero locus of a section of MATH by CITE. Using REF, it follows that MATH for all MATH, where MATH and MATH are from REF. Since MATH, we obtain MATH . Combining this with REF, the proposition follows.
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math/0006122
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It is easily seen that the rules of inference preserve validity. For instance, if MATH is valid, then, for any valuation MATH, MATH where MATH. If MATH does not occur in MATH, then MATH and we have MATH. That MATH is sound for arbitrary NAME logics was shown in CITE. The tedious but straightforward verification that the remaining axioms REF - REF are valid is left to the reader.
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math/0006122
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By induction on the complexity of MATH. The claim is obvious for atomic formulas, conjunction and disjunction. If MATH we have to distinguish two cases. Suppose first that MATH. By induction hypothesis, MATH, and hence the first disjunct in the definition of MATH is true. Thus MATH defines MATH and MATH. Now suppose that MATH. Then MATH, MATH and MATH, and thus MATH. If MATH, let MATH be the valuation which is just like MATH except that MATH, and let MATH be the corresponding interpretation which is like MATH except that it assigns MATH to MATH. By induction hypothesis, MATH. We again have two cases. Suppose first that MATH. For all MATH, MATH, since MATH by induction hypothesis. On the other hand, MATH for all MATH, and so MATH. Now consider the case where MATH. Here there is no bound MATH on the the members of sets defined by MATH where MATH. Hence, MATH and MATH. The case MATH is similar.
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math/0006122
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By induction on the complexity of MATH. The claim is again trivial for atomic formulas, conjunctions or disjunctions. If MATH, two cases occur. If MATH, then MATH. By induction hypothesis, MATH, and hence MATH. Otherwise, for some MATH we have MATH but MATH. So MATH must be true and the predicate defined is the same as MATH. Now for REF If MATH, then there is a prefix closed witness MATH so that MATH. By induction hypothesis, MATH, and hence MATH for all MATH, and thus MATH as well. Consider MATH. First, suppose that MATH. That means that for some MATH, MATH, and for no MATH and no MATH, MATH. By induction hypothesis, MATH and for all MATH, MATH. Hence MATH. If MATH does not exist, for each MATH there is a witness MATH with MATH. By induction hypothesis, for each MATH we have MATH, and so MATH. The case MATH is similar.
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math/0006122
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If there is a valuation MATH such that MATH, then by REF there is a MATH with MATH and MATH so that MATH, and hence MATH. Conversely, suppose MATH. We may assume, without loss of generality, that all propositional variables in MATH are bound. Then there is an interpretation MATH with MATH so that some MATH. By REF , MATH. Hence, if MATH, then MATH for all MATH, and, also by REF , MATH. Thus a formula MATH is a tautology in MATH iff MATH. The claim follows by the decidability of MATH.
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math/0006122
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CASE: From REF we have MATH, which, together with the left-to-right direction of REF yields the result. CASE: The left-to-right implication immediately follows from REF together with REF . For the converse, replace MATH by MATH in REF and use REF to derive MATH. Then, using REF , one has MATH. The claim follows by MATH. CASE: In MATH, we have MATH. Replacing MATH by MATH and MATH by MATH, we have MATH. The result follows using REF and MATH.
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math/0006122
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By induction on the complexity of MATH. Cases for MATH, MATH, and MATH are easy. If MATH, we use the induction hypothesis and REF . If MATH, we argue: MATH . The case of MATH is handled similarly.
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math/0006122
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Follows from REF using REF .
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math/0006122
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MATH is already derivable intuitionistically. For MATH, use REF , and induction on MATH.
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math/0006122
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(See also REF.) First note that MATH is a tautology and provable in MATH. Since for each MATH we have MATH, the right-to-left implication MATH follows by case distinction. For the left-to-right implication, consider MATH. This is provable, since MATH is provable. By distributivity of MATH over MATH, we have MATH. We also have MATH for each MATH from MATH. Together we get MATH .
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math/0006122
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By REF , MATH where MATH is a MATH-chain normal form over MATH. Consider a disjunct of MATH of the form MATH, where MATH, , MATH is the ordered partition of MATH corresponding to MATH. If MATH, then MATH, since MATH. Otherwise, MATH with MATH. Then the sequence MATH, , MATH corresponds to a conjunction MATH where for at least one MATH, MATH, and MATH, where MATH is the part of MATH corresponding to MATH, , MATH. Since MATH, we have MATH . As is easily seen, the right-hand side of REF is provably equivalent to MATH . In sum, MATH, and MATH is a MATH-chain. By induction on the number of disjuncts in MATH one shows that there is MATH which is a disjunction of MATH-chains such that MATH. Now we have to prove that there exists a disjunction of MATH-chains MATH satisfying REF so that MATH. Suppose that for some disjunct MATH in MATH we have MATH and MATH where MATH and MATH. Then, since MATH we have MATH where MATH is the MATH-chain corresponding to MATH. Consider a disjunct MATH of MATH where for some MATH, both MATH and MATH where MATH. Then MATH. To see this, recall that MATH if MATH. By definition of MATH, that means that MATH . Since MATH, we have MATH which together with the left conjunct of REF gives MATH. Thus, as before, MATH is provably equivalent to the MATH-chain corresponding to MATH, , MATH. Lastly, suppose that for a disjunct MATH of MATH we have both MATH and MATH for some MATH, MATH such that MATH. Then by REF together with transitivity we get MATH, and since MATH we have MATH where MATH is the MATH-chain corresponding to MATH. By induction on the number of disjuncts in MATH we obtain the desired MATH.
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math/0006122
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CASE: The left-to-right implication follows easily from the two instances of REF MATH . For right-to-left, consider MATH which are derived easily from REF using MATH. Use REF to introduce the existential quantifier in the antecedent of REF , and then REF to obtain MATH . The antecedent of REF is an instance of REF , and so MATH from which the right-to-left direction of REF follows by (MATH). CASE: The argument is analogous to the derivation of REF .
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math/0006122
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Suppose MATH. Let MATH, MATH. At stage MATH, pick the non-innermost quantified subformula MATH or MATH of MATH corresponding to MATH and replace MATH to obtain MATH. The procedure terminates with MATH. At each stage MATH follows by induction on MATH from REF . The lower bounds are obvious from the construction of MATH.
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math/0006122
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Let MATH be the maximal exponent of a subformula MATH and let MATH. REF provides us with MATH in minimal normal form over MATH so that MATH. Since MATH distributes over MATH, we only have to consider formulas of the form MATH where MATH is a MATH-chain and satisfies the conditions of REF . MATH corresponds to an ordered partition MATH, , MATH over MATH. We prove that MATH for some quantifier-free MATH by induction on MATH. If MATH, then either MATH or MATH. In the first case, MATH, in the second one, MATH. Now suppose MATH. NAME cases arise, according to how the equivalence classes containing MATH are distributed. CASE: The partition corresponding to MATH is of the form MATH . Then MATH is of the form MATH . Since MATH is provable, MATH. CASE: The partition corresponding to MATH is of the form MATH and MATH. Then MATH is of the form MATH . We first show that MATH. For the right-to-left direction, observe that MATH from which the claim follows by (MATH). The left-to-right direction is proved by induction on MATH, using REF . In sum, we have REF The partition corresponding to MATH is of the form MATH with MATH, MATH. Because of the condition on MATH we can assume that MATH with MATH. We proceed by induction on MATH. If MATH, then we have a conjunct MATH, and MATH. Otherwise, we have a conjunct MATH with MATH. Using REF , this conjunct is provably equivalent to MATH. Hence, MATH is equivalent to the disjunction of two MATH-chains corresponding to MATH . For the first MATH-chain, the maximum exponent of MATH is smaller and hence the induction hypothesis of the present subcase applies. The second MATH-chain is shorter overall, and hence the induction hypothesis based on number of equivalence classes applies.
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math/0006122
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Let MATH be the minimal normal form of MATH. It is provably equivalent to the formula obtained from MATH by replacing each element of a chain MATH by MATH. By distributivity then, MATH where MATH is a conjunction of disjunctions of implications of the form MATH. Any such disjunct of the form MATH is provably equivalent to MATH if MATH (in which case the entire disjunction can be deleted), or to MATH if MATH. The part of a disjunction in MATH containing MATH thus can be assumed to be of the form MATH where MATH. This, in turn, is equivalent to a conjunction of disjunctions of the form MATH . This can again be simplified by taking MATH and MATH, since MATH if MATH. Since MATH and MATH if MATH, it suffices to show that a formula of the form MATH is equivalent to a quantifier free formula. We distinguish three cases: CASE: MATH, MATH. Then MATH and hence MATH. CASE: MATH, MATH. Then MATH, and hence MATH. CASE: Since MATH by assumption, this leaves only the case MATH. Then MATH. The left-to-right implication is obvious by (MATH), instantiating MATH by MATH. For the right-to-left implication two cases arise: CASE: MATH. By REF , we have MATH. Furthermore, MATH. In sum, we have MATH . Since MATH, we have MATH. CASE: MATH. By REF , MATH, and so MATH. Using induction and REF , it is easy to show that MATH . Each of the disjuncts MATH implies MATH, which in turn implies MATH, so MATH. In sum, we have again MATH. The bound on MATH follows by inspection.
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math/0006122
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We may assume, renaming variables if necessary, that each variable in MATH is bound by only one quantifier occurrence. By induction on MATH. If MATH, there is nothing to prove. If MATH, let MATH be as in REF . Replace each innermost quantified formula MATH, MATH by MATH or MATH, respectively. The resulting formula MATH satisfies MATH and MATH.
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math/0006122
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Consider the minimal normal form MATH of MATH over MATH. Each chain in MATH is of one of two forms MATH is provable, so MATH, and MATH. So if MATH contains MATH, then MATH, otherwise MATH, where MATH is the maximum of MATH occurring in MATH.
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math/0006122
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If MATH, then MATH for some MATH. Since MATH for all MATH, MATH.
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math/0006122
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MATH is sound for each finite-valued NAME logic, so MATH for each MATH. Conversely, if MATH, then MATH for some MATH. Since MATH is sound for MATH, we have MATH as obviously MATH.
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math/0006131
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We prove this by induction. First, we make a reduction. Suppose that some rank other than the top and bottom rank contains a single vertex MATH. Then the lattices MATH are dismantlable and rank-connected. Any lexicographic shelling of MATH and separately of MATH will be a lexicographic shelling of MATH as long as all labels in MATH are greater than all labels of MATH. By induction, therefore, we may assume that every rank except the very top and the very bottom contains at least two elements. Since MATH is dismantlable, every non-trivial sublattice of MATH contains a doubly irreducible element. We show that the covers of doubly irreducible elements and the vertices covered by doubly irreducible elements cannot themselves be doubly irreducible. Let MATH be doubly irreducible in MATH. Suppose MATH. Let the unique vertex that MATH covers be MATH and the unique vertex that covers MATH be MATH. Let MATH be the induced bipartite graph of the NAME diagram with vertices MATH. Then both MATH and MATH are in MATH, and MATH is a connected graph because MATH is rank-connected. Since MATH is doubly irreducible, MATH is a degree REF vertex in MATH, hence MATH is still connected. Since MATH contains at least two vertices, MATH must contain MATH and at least one vertex in MATH, hence MATH has an edge to another vertex of rank MATH. Therefore MATH (and similarly MATH) cannot be doubly irreducible in MATH. For rest of the proof, we rely on the following definition and previous result. We say that MATH is a corner of MATH in MATH if there exist MATH and MATH such that MATH both cover MATH and are covered by MATH, and MATH is doubly irreducible. REF guarantees that MATH is lexicographically shellable if MATH is lexicographically shellable. We prove that there must always exist a doubly irreducible element of MATH that has a corner, hence MATH is lexicographically shellable by induction. Let the doubly irreducible vertices of MATH be MATH. Suppose that no element of MATH has a corner. Let MATH cover MATH and be covered by MATH in MATH for MATH. As seen before, rank connectedness guarantees that MATH and MATH are not doubly irreducible in MATH. Consider the relation MATH (which is true in MATH) in the sublattice MATH. There must be a chain from MATH to MATH in MATH. However in MATH, the rank of MATH and MATH differs by exactly REF. Therefore, any chain between MATH and MATH in MATH must have length less than or equal to REF, since removing vertices cannot make chains longer. If the length of the chain is REF, then the middle vertex of the chain will be a corner of MATH. By our assumption, MATH has no corners, hence MATH must cover MATH in MATH. Suppose that MATH covers both MATH and MATH for some MATH. Now MATH is not a corner of MATH, so MATH. Therefore, MATH covers both MATH and MATH in MATH. Thus if element MATH covers MATH vertices in MATH, after removing the doubly irreducible vertices, MATH still covers MATH vertices in MATH. Each doubly irreducible vertex MATH is replaced by the unique vertex MATH that it covers. Similarly, if element MATH is covered by MATH vertices in MATH, it is still covered by MATH elements in MATH. Hence the sublattice MATH is composed entirely of elements that are not doubly irreducible. It is not empty, since MATH must contain at least one doubly irreducible element MATH and therefore contains both the unique element that covers and the unique element that is covered by MATH. Since MATH is dismantlable, MATH must be the trivial lattice that contains only a top and bottom element. Thus, MATH must have rank REF and all its doubly irreducible elements have rank REF. Each of these rank REF elements is a corner of every other, contradicting our assumption that no doubly irreducible element has a corner.
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