paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0006094 | 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.... |
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 fam... |
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 clo... |
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 ... |
math/0006094 | This is a corollary of REF . |
math/0006094 | 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 obt... |
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. |
math/0006097 | 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 th... |
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 ... |
math/0006097 | Apply the previous result with MATH, MATH, and MATH. |
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 o... |
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 th... |
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 k... |
math/0006097 | Set MATH. Then the given determinant is MATH as required. |
math/0006097 | After specializing, MATH becomes MATH . Conjugating by MATH gives MATH since MATH unless MATH is even, the result follows. |
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 the... |
math/0006097 | We compute MATH so MATH . |
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... |
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 t... |
math/0006097 | We apply REF , with MATH and MATH . We find MATH . The theorem follows immediately. |
math/0006097 | 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 MAT... |
math/0006097 | We take MATH so MATH . We find MATH and thus obtain the stated kernel. |
math/0006097 | On the one hand, we have MATH on the other hand, we have MATH . The theorem follows. |
math/0006097 | 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, w... |
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 ... |
math/0006100 | Immediate from REF . |
math/0006100 | 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 ... |
math/0006102 | 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" expansi... |
math/0006102 | 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/0006102 | 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.... |
math/0006102 | 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. ... |
math/0006102 | REF allows us to repeat all the argument in REF , and the result follows immediately. |
math/0006102 | 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 ... |
math/0006103 | 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 . Usi... |
math/0006103 | 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 ... |
math/0006105 | 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, ... |
math/0006105 | 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... |
math/0006105 | 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. |
math/0006105 | 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. |
math/0006105 | 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 variet... |
math/0006105 | 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 cla... |
math/0006105 | 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 MA... |
math/0006105 | 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 th... |
math/0006105 | 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. |
math/0006105 | 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... |
math/0006105 | 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-modul... |
math/0006105 | 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 . |
math/0006105 | 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, ... |
math/0006105 | 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 tha... |
math/0006107 | The first formula is an immediate consequence of the theorem. It implies that MATH which yields the second formula. |
math/0006107 | Since uniruledness is a birational property, it is enough to observe that MATH is not uniruled because it has nonnegative NAME dimension. |
math/0006107 | 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. |
math/0006107 | 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 MAT... |
math/0006107 | 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 fr... |
math/0006107 | 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 permuta... |
math/0006107 | 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 . |
math/0006108 | 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 . |
math/0006108 | 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 . |
math/0006108 | 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. |
math/0006108 | 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 assum... |
math/0006108 | REF follows from REF follows from REF. |
math/0006108 | 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 . |
math/0006108 | 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 . |
math/0006108 | 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 dens... |
math/0006108 | 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. |
math/0006112 | 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... |
math/0006112 | 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 p... |
math/0006112 | 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 cas... |
math/0006112 | 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-... |
math/0006112 | 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 t... |
math/0006112 | 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 ... |
math/0006112 | 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 MA... |
math/0006112 | 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 numbe... |
math/0006112 | 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 M... |
math/0006112 | 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... |
math/0006112 | 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 bounda... |
math/0006112 | This is quite similar to the proof of REF and is left to the reader. |
math/0006112 | 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 ... |
math/0006112 | 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 tes... |
math/0006112 | 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... |
math/0006116 | 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,... |
math/0006122 | 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 th... |
math/0006122 | 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 th... |
math/0006122 | 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/0006122 | 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 , MA... |
math/0006122 | 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... |
math/0006122 | 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. |
math/0006122 | Follows from REF using REF . |
math/0006122 | MATH is already derivable intuitionistically. For MATH, use REF , and induction on MATH. |
math/0006122 | (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 al... |
math/0006122 | 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 a... |
math/0006122 | 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 , an... |
math/0006122 | 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. |
math/0006122 | 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 ordere... |
math/0006122 | 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... |
math/0006122 | 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... |
math/0006122 | 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. |
math/0006122 | If MATH, then MATH for some MATH. Since MATH for all MATH, MATH. |
math/0006122 | 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. |
math/0006131 | 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 l... |
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