paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9910063 | See REF. |
math/9910063 | For REF see REF. In fact, the assumptions of these propositions are formulated there for symplectic structures, but the proofs remain valid also for quasi-symplectic ones. REF treats the case of REF when MATH. We need to introduce some settings of this proof to explain the case when MATH is of finite order. For any MATH and for a sufficiently small neighborhood MATH of MATH, there exist local parameters MATH on MATH and a local section MATH of MATH identifying MATH with MATH in such a way that MATH becomes the section MATH of neutral elements of the fibers of MATH. One can realize MATH as the quotient MATH of the cotangent bundle of MATH by a local system MATH generically of rank MATH. Let MATH be the coordinates on MATH, dual to MATH. Then the restriction of MATH to MATH can be written in the form MATH . Here MATH does not depend on MATH, as was proved in CITE Let MATH. The invariance of MATH under the translations by sections from MATH is equivalent to MATH where MATH is any local section of MATH. Hence we have also the invariance under translations by sections from MATH, and the finite order case follows. REF is obtained by the successive application of REF . |
math/9910063 | We are in the algebraic situation, so we can apply REF to construct a symplectic structure MATH associated to the NAME - Markman form MATH. Define MATH, so that REF is verified. The definition of the relation ``associated" in REF imply also REF . |
math/9910063 | Take for MATH the locus of hyperplanes MATH such that MATH has at most one nondegenerate double point as singularity. Any cubic MATH with only one isolated singular point contains a normal elliptic quintic curve which does not pass through the singular point. To see this, take a hyperplane section of MATH which is a nonsingular cubic surface MATH. It contains a MATH-dimensional linear system of quintics of arithmetic genus REF, and its generic member is a non-singular space elliptic curve MATH. By REF, whose proof uses only the nonsingularity of MATH at points of MATH, we have MATH. Hence the NAME scheme of MATH is smooth of dimension REF at MATH. The family of quintics of genus REF in hypeplane sections of MATH being REF-dimensional, we see that MATH is deformable to a non-space curve, which is the wanted normal elliptic quintic. Define now MATH to be the subset in the moduli space of sheaves on MATH consisting of the vector bundles of rank REF obtained by NAME 's construction from all the normal elliptic quintics in MATH that do not pass through the singular point of MATH, and MATH the union of the MATH over all MATH with MATH. All our proofs concerning the dimension, the nonsingularity of MATH, MATH, as well as the definition and the proof of the nondegeneracy of MATH work without any modification with MATH in place of MATH for all MATH. Thus we obtain the NAME quasi-symplectic structure MATH on MATH. NAME - Markman CITE remark that their symplectic structure extends to a REF-form MATH on the generalized relative intermediate Jacobian MATH. They write that it is not difficult to see that it is nondegenerate over the boundary. This also follows easily from our setting. Indeed, the nondegeneracy of MATH implies that the vertical tangent bundle of MATH is isomorphic to MATH, hence its associated group scheme MATH is isomorphic to MATH, as in the proof of REF . The standard symplectic structure of the cotangent bundle of MATH descends to the quotient by MATH over MATH, hence also over MATH, as the descent REF extends by continuity. Thus we obtain a symplectic structure MATH on MATH. As the restrictions of MATH, MATH to MATH are associated to the same quasi-symplectic structure on MATH, they differ by a lift of a REF-form on MATH. Hence they define the same isomorphism between MATH and MATH over MATH, which is the identity one for MATH. By continuity, this holds also over MATH. Hence MATH is nondegenerate. As MATH, we have the uniqueness over MATH, hence MATH. The extension of REF follows trivially from the following observation: if the regular section MATH of the vector bundle MATH over MATH belongs to the vector subbundle MATH over MATH, then it does over the whole of MATH. |
math/9910068 | Without loss of generality we may suppose MATH, so MATH for any MATH. Let MATH. Then MATH for all MATH, and thus MATH, from which MATH. The opposite relation holds by symmetry. |
math/9910068 | Choose for MATH a finite generating set MATH, and define the weight MATH by MATH for all MATH. Then MATH for all MATH, so MATH. |
math/9910068 | Let MATH be a transversal of MATH in MATH, and set MATH. For every MATH of weight at most MATH, there is a unique MATH with MATH, and MATH. The map MATH is MATH-MATH, so its restriction to the set of elements of weight at most MATH is at most MATH-MATH, proving the first inequality. Considering all MATH of weight at most MATH and all MATH, we have MATH distinct elements MATH, with MATH. This proves the second inequality. |
math/9910068 | Let MATH be a minimal form for MATH. First, MATH may not contain two equal consecutive letters, since they could be cancelled, shortening the representation of MATH. Second, two unequal consecutive letters among MATH can be replaced by the third, and as MATH is triangular this operation will shorten the length of MATH while not enlarging its weight. Ultimately it will yield a word MATH of the required form, also representing MATH, and with smaller or equal weight. It can then be chosen as a minimal form of MATH. |
math/9910068 | Applying REF (with constants MATH and MATH) to the subgroup pairs MATH and MATH, we obtain from the previous proposition MATH for all MATH. At the cost of increasing MATH, we rewrite it as MATH, whence by iteration MATH for all integers MATH and MATH. Let MATH be such that MATH. Now given MATH, let MATH be maximal such that MATH be at least equal to MATH. We then have MATH, and since MATH, the corollary follows. |
math/9910068 | Consider the weight MATH for all MATH giving MATH, and consider the following two homomorphisms (that they are homomorphisms was proven by CITE): MATH . Then for any MATH we have MATH, and MATH is of length at most MATH, because MATH is a dihedral group of order MATH, and MATH. Now for any MATH, with MATH, we have MATH indeed applying MATH to the right-hand side gives MATH because MATH. Therefore we have MATH . |
math/9910068 | Let MATH be a path in MATH from MATH to MATH with MATH. Since the graph is finite, say with MATH vertices, MATH may be decomposed as a product of paths MATH, such that the MATH are all loops in MATH and MATH is of length at most MATH. Let MATH be the maximal weight of a word of length at most MATH. Then MATH . |
math/9910068 | Let MATH be a pair of words in minimal form; we are to construct a word MATH such that MATH satisfies REF . First we may suppose that neither MATH nor MATH starts by MATH; indeed we may add MATH's at their beginning (remembering that MATH and MATH are in MATH) and construct a word MATH such that MATH with MATH. The word MATH then satisfies MATH and is of weight at most MATH more than MATH, a fact that can be coped with by increasing the constant MATH. Second, we may suppose that MATH is shorter than MATH, and that their lengths differ by at most MATH. Indeed the shorter of these two can be extended by copies of MATH which is trivial in MATH, and MATH can be extended by an extra copy of MATH. Again the cost of this operation is at worst an increase of MATH by MATH. Now MATH satisfies REF for the constant MATH. This precisely means that there is an output word MATH satisfying REF . |
math/9910070 | The computations are as (or slightly more complicated than) the one for the expected value. We don't give more details. |
math/9910070 | Note that MATH and do some trivial rearrangements. |
math/9910070 | Obvious. |
math/9910078 | Use REF in the definition of NAME algebroid with MATH and arbitrary MATH and MATH, and then cyclically permute MATH, MATH and MATH: MATH . Now add the first two identities and subtract the third. Using NAME algebroid REF and the definition of MATH, we get: MATH . Using the definition of MATH again, we can rewrite this as: MATH . The statement follows from the nondegeneracy of MATH. |
math/9910078 | Starting with the NAME bracket on MATH, we shall, following CITE, extend it to a MATH on all of MATH satisfying REF for MATH. The extension will proceed, essentially, by induction on the degree of the argument: for each degree MATH will be a primitive of a certain cycle depending on the values of MATH on elements of lower degree. Higher MATH's will be introduced and extended in a similar fashion, as primitives of cycles (using the acyclicity of REF ). The main work will consist in calculating these cycles, in particular, showing that most of them vanish; these computations are mostly relegated to the technical lemmas of the next section. CASE: MATH. In degree REF, we are given MATH. Consider now an element MATH of degree REF. Then MATH is defined and is, in fact, a boundary by REF : MATH so we set MATH so that the SHLA identity REF for MATH, MATH holds in degree REF. Now, MATH is spanned by elements of the form MATH or MATH. As above, MATH is defined on elements of degree REF, and is, in fact, a cycle (compare CITE). We have MATH by NAME algebroid REF , whereas MATH so we set MATH. Now observe that, since MATH in degree REF, we can define MATH to be zero on elements of degree higher than REF as well and still have REF . We have thus defined a MATH that satisfies REF by construction. CASE: MATH. In degree REF, by NAME algebroid REF we have MATH where MATH is the NAME. So we set MATH, so that the homotopy NAME identity identity REF for MATH, MATH holds on MATH (as MATH). Consider now an element MATH . The expression MATH is defined and is a cycle in MATH (compare CITE), hence we can define MATH to be some primitive of this cycle, so that REF holds. But in our particular situation we in fact have (see the next section for a proof): MATH. Therefore, we can define MATH. Now observe that on elements of degree MATH has to be REF because deg-MATH, whereas MATH for MATH. We now have MATH defined on all of MATH and satisfying REF by construction. CASE: MATH and higher. Proceeding in a similar fashion, we look at the expression MATH (always a cycle in MATH) and define MATH to be its primitive in MATH, so as to satisfy REF . However, it turns out that (see the next section for a proof) MATH. Hence we can set MATH and observe that MATH has to vanish on elements of degree MATH as deg-MATH, while MATH for MATH. By similar degree counting, all MATH, MATH, have to vanish identically. This finishes the proof modulo REF . |
math/9910078 | Using NAME algebroid REF , we have MATH . |
math/9910078 | Using NAME algebroid REF , we can rewrite MATH as follows: MATH . Expressing the other summands of MATH in this form and collecting like terms in the parentheses, we find that the terms of the form MATH cancel out, terms of the form MATH add up to MATH, those of the form MATH add up to MATH, and finally, terms of the form MATH add up to MATH after we use NAME algebroid REF . Thus, MATH and the statement of the lemma follows immediately. |
math/9910078 | Let MATH be arbitrary. Then, using REF we have MATH . Hence, MATH . The first statement follows by the nondegeneracy of MATH . On the other hand, by REF , MATH . |
math/9910078 | We have to show that MATH vanishes if any two of the entries coincide. But MATH by REF . On the other hand, MATH where we have used REF . And finally, MATH just as in the previous calculation; we have used again REF . |
math/9910078 | It is not hard to show the equivalence of all of the properties, except for the NAME identity which will take a bit more work. So we shall first show the equivalence of REF for both definitions, and then prove the equivalence of the two versions of NAME using REF . Now, in view of REF it is obvious that REF is equivalent for the two definitions. REF in the new definition implies immediately that MATH for all MATH, hence MATH and we have the old REF . Moreover, by the new REF , MATH hence we have the old REF by the nondegeneracy of MATH. Conversely, if we start with the old definition, the new REF is immediate by REF , while the old REF combine to give MATH the new REF . As for the NAME rule, one has MATH for all MATH, MATH, hence it follows immediately that the new and old REF are equivalent. Now for the NAME identity. In view of REF , it is clear that one has MATH where MATH is as in REF , MATH is the NAME REF , and MATH . To show the equivalence of the old and new REF , we only need to show that MATH, where MATH is as in REF . But, by REF , MATH whereas MATH by REF . Therefore, MATH . However, since both MATH and MATH are completely antisymmetric REF , so is MATH; therefore, MATH is equal to its skew-symmetrization. But it is obvious that the skew-symmetrization of the first term on the right hand side of REF is MATH, whereas the skew-symmetrization of the second term is easily seen to be zero. Hence MATH, and we are done. |
math/9910078 | This is best done by direct computation in local coordinates, just as for ordinary manifolds. REF is obvious by REF and the definition of MATH, whereas for REF we can easily deduce from REF that MATH where MATH denotes the parity of the vector field MATH. |
math/9910078 | If locally MATH, MATH, then MATH, hence MATH by REF , and similarly for MATH. |
math/9910078 | Immediate from REF . |
math/9910078 | A straightforward computation, carried out in CITE. REF depends both on the NAME identity for MATH and MATH. |
math/9910078 | The skew-symmetry and derivation property are consequences of REF . Hence, we only need to consider fiberwise constant and fiberwise linear functions, that is, elements of MATH and MATH. We have: MATH for all MATH; MATH for all MATH, MATH. And finally, MATH for all MATH. We have made repeated use of the commutation relations REF . |
math/9910078 | We must show that MATH is a derivation of MATH if and only if REF holds. However, by REF , we have MATH . Since MATH is fiberwise quadratic, the second term in the last expression vanishes if and only if MATH. The statement follows by the injectivity of MATH. |
math/9910078 | The NAME transform MATH is a symplectomorphism. |
math/9910078 | REF is just a restatement of REF ; REF follows by computation: REF takes a bit more work. We have MATH for MATH; MATH for MATH; MATH for MATH, MATH; and finally, MATH . This proves REF . We have made extensive use of the commutation relations REF . |
math/9910078 | We need to verify REF . Since MATH is a NAME map, we can embed sections of MATH and functions on MATH into MATH using MATH as above and carry out all the computations up in MATH. We shall identify MATH and MATH with their images under MATH. Now, it follows that REF (the NAME identity) and REF (about the symmetric part) are just consequences of the properties of the derived bracket on a differential NAME superalgebra REF . On the other hand, REF , MATH translates, by REF , into MATH but this is obvious. REF , MATH translates into MATH . However, by the NAME identity for MATH, MATH since MATH because MATH is a NAME function for MATH. Finally, REF , MATH when both sides are applied to an arbitrary MATH translates into MATH for all MATH. Using NAME again, we have MATH since MATH by REF , while MATH since MATH is a NAME function on MATH. Thus all of the properties of a NAME algebroid are verified. Notice that REF was only needed to derive REF . |
math/9910078 | The modular vector field MATH rotates the necklace, hence cannot be Hamiltonian. |
math/9910078 | We will work in coordinates MATH on the unit disk in which MATH and MATH are given, respectively by REF . Locally MATH is a coboundary whose primitive is given by an NAME vector field MATH: it's easy to check that MATH. But MATH does not extend to a vector field on MATH since it does not behave well ``at infinity", that is, on the unit circle in the MATH-plane. Therefore, to prove that MATH is globally nontrivial it suffices to show that there does not exist a NAME vector field MATH such that MATH is tangent to the unit circle and the restriction is rotationally invariant. In fact, it suffices to show that there is no Hamiltonian vector field MATH such that MATH vanishes on the unit circle (since we can always add a multiple of the modular vector field to cancel the rotation). Assuming that such a MATH exists, we will have, in the polar coordinates MATH, MATH : MATH . Upon restriction to MATH this becomes MATH . In order for this to vanish it is necessary, in particular, that MATH be a nonzero constant which is impossible since MATH is periodic in MATH . |
math/9910078 | MATH is a nonzero multiple of MATH but MATH is nontrivial in MATH. |
math/9910080 | Assume on the contrary that MATH. Let MATH be the minimal cardinal such that MATH can be covered by MATH many ultrafilters, MATH. Clearly MATH. Let MATH be an injective function such that MATH for MATH. Then for each MATH and MATH there is MATH such that MATH. For each MATH pick an element MATH. By REF we can find MATH such that MATH for each MATH. Since MATH, there is MATH such that MATH is of size MATH. Then there is MATH such that MATH. Thus MATH by the choice of MATH. On the other hand, MATH, which contradicts MATH. |
math/9910080 | The length of a sequence MATH is denoted by MATH. If MATH then put MATH. Given sequences MATH and MATH we write MATH to mean that MATH is an initial segment of MATH. Denote by MATH the trivial Boolean algebra MATH. For each MATH choose a function MATH such that MATH is REF and MATH for MATH. Then for each MATH and MATH there is MATH such that MATH. We define, by induction on MATH, MATH as follows: CASE: MATH, CASE: if MATH is limit, then MATH, CASE: if MATH then let Let MATH. For MATH let MATH. For MATH we say that MATH is a sequence with nice tail iff we can write MATH, where MATH, MATH for some MATH and MATH for each MATH. Let MATH. For MATH put MATH. The sets MATH are pairwise disjoint. Define the function MATH as follows. Let MATH . For MATH write MATH and MATH and put MATH . For MATH let MATH and MATH. Let MATH be the Boolean algebra generated by MATH freely, except the relations in the following set MATH: MATH . MATH. Let MATH . It is enough to show that MATH. Assume on the contrary that MATH witnessed by MATH and ultrafilters MATH. We can assume that MATH. First observe that MATH because MATH can not be covered by finitely many ultrafilters either. Construct a sequence MATH of length MATH such that for each MATH: MATH . Assume we have constructed MATH for MATH satisfying REF . Then MATH by definition. Since MATH was minimal we have MATH and so we can choose a suitable MATH. Now MATH. For MATH write MATH. Since MATH there are MATH and MATH such that MATH for each MATH. Pick MATH such that MATH. Then MATH by REF . But MATH which contradicts MATH. We say that MATH is closed provided: CASE: if MATH and MATH then MATH, CASE: if MATH then MATH, CASE: if MATH and MATH then MATH. Every finite MATH is contained in a finite closed MATH. Observing MATH close MATH first for MATH and MATH, then close for MATH. Let MATH be closed and MATH. We say that MATH holds iff CASE: MATH for each MATH, CASE: MATH for each MATH. The following lemma is a special case of a well-known fact. Assume that MATH. Then there is a (unique) homomorphism MATH from MATH into MATH such that MATH iff MATH holds. For each MATH fix a homomorphism MATH with MATH, and finite, closed set MATH such that MATH is a Boolean combination of MATH. Define MATH by MATH. MATH has precaliber MATH for each MATH. Let MATH. It is enough to define a map MATH satisfying MATH such that MATH because MATH implies MATH. By thinning out MATH we can assume that MATH is a MATH-system with kernel MATH and that MATH. A pair MATH is called crossing pair if there are MATH such that MATH and MATH. The family of crossing pairs is denoted my MATH. For MATH let MATH . MATH. Let MATH, MATH, MATH, MATH. Then the pair MATH determines the pair MATH. Indeed, MATH is REF, so MATH is the unique pair MATH with MATH. We say that MATH and MATH are twins iff MATH but MATH. MATH . MATH. Let MATH, MATH, MATH. Fix MATH such that MATH and MATH are twins. Now the pair MATH determines the pair MATH. Indeed, MATH for some MATH, and there is at most one MATH such that MATH and MATH are twins. But MATH and MATH determine MATH because MATH is REF. Let MATH . MATH. Assume that MATH. Then MATH and MATH. Moreover MATH and MATH for some MATH and MATH. But for given MATH there is at most one pair MATH such that MATH and MATH are twins. Since there is at most one MATH with MATH and there is at most one MATH with MATH, we are done. So applying NAME 's free set mapping theorem we can thin out our sequence such that MATH for each MATH. Let MATH. Define MATH as follows. Let MATH iff either MATH or MATH for some MATH such that MATH. Since MATH we have MATH. We show that MATH holds. Assume that MATH and MATH are twins and MATH. Since MATH and MATH for each MATH, it follows that MATH. So MATH is impossible because MATH implies MATH. Thus MATH and MATH and so MATH by the construction of MATH. MATH. First fix a well-ordering MATH of MATH such that if MATH then MATH. Let MATH. Consider the product space MATH where MATH denotes the discrete topological space of size MATH whose underlying set is MATH instead of MATH. Applying MATH and MATH we can fix a dense family MATH. Write MATH. For MATH define MATH as follows: MATH and if MATH, then pick the unique MATH with MATH, MATH, and let MATH iff MATH. Let MATH . For MATH define MATH by recursion on MATH as follows. Let MATH iff MATH and for each MATH with MATH we have MATH and for each MATH with MATH we have MATH. By induction on MATH it is clear that MATH holds. Now let MATH. By construction of MATH we can find MATH such that MATH, moreover for each MATH if MATH then MATH. MATH. By induction on MATH. Assume that the claim holds for MATH provided MATH. We can assume MATH. If MATH, MATH then MATH so MATH as MATH holds. So, by the induction hypothesis, MATH. Assume MATH, MATH. If MATH then MATH and MATH so MATH and MATH. Thus MATH because MATH as MATH holds. If MATH then MATH by the assumption about MATH and so MATH. Thus MATH by the construction of MATH. Thus MATH, that is, MATH, which was to be proved, so the lemma holds. MATH is c.c.c by REF and MATH by REF so the theorem is proved. |
math/9910083 | Since we are interested only in facet and edge vectors meeting at the vertex MATH, we may renumerate the edge vectors MATH at MATH by the index set MATH of the facet vectors at MATH. To do this we just set MATH if MATH (that is, a facet vector and an edge vector have same index if the corresponding facet and edge of MATH span the whole MATH). Then for MATH one has MATH. Hence, MATH and MATH (see REF ). Now, since MATH is a primitive vector, it follows from REF that MATH. Changing the sign of MATH if necessary, we obtain MATH which together with REF gives MATH, as needed. |
math/9910083 | Let us consider the action of MATH near fixed point MATH corresponding to a vertex MATH of MATH. Let MATH be the edge vectors at MATH. The MATH-action induces a unitary MATH-representation in the tangent space MATH. We choose complex coordinates MATH in MATH such that the tangent space of two-dimensional submanifold MATH at MATH is given by REF (MATH is dropped). Then, the corresponding isotropy subgroup MATH is given by the equation MATH in MATH (see REF ). Hence, the weights of the MATH-representation are MATH, that is, an element MATH acts on MATH as MATH where MATH, MATH. A primitive vector MATH defines a one-parameter circle subgroup MATH. It follows from REF that this circle acts on MATH as MATH where MATH. The fixed point MATH is isolated if all the weights MATH of the MATH-action are non-zero. Thus, if MATH for all edge vectors, then the MATH-action on MATH defined by MATH has only isolated fixed points. |
math/9910083 | It follows from REF that for any vertex MATH one can write MATH . Then our lemma follows from the definition of MATH. |
math/9910083 | The NAME - NAME formula (CITE; see also CITE, where it was deduced within the cobordism theory) states that the MATH-genus of a stably complex manifold MATH with a MATH-action can be calculated as MATH where the sum is taken over all MATH-fixed submanifolds MATH, and MATH denotes the number of negative weights of the MATH-representation in the normal bundle of MATH. In our case all fixed submanifolds are isolated fixed points corresponding to vertices MATH. Hence, MATH depending on whether or not the orientation of MATH defined by MATH coincides with that defined by MATH. Thus, for quasitoric MATH we may substitute MATH for MATH in REF . REF shows that the weights of induced MATH-representation in MATH equal MATH, therefore MATH in REF is exactly MATH (see REF ), and the required formula follows. |
math/9910083 | For each vertex MATH of MATH the stably complex structure on MATH determined by MATH defines a complex structure on MATH via the isomorphism MATH (see REF ). The MATH-action on MATH determined by MATH induces the complex MATH-representation on MATH with weights MATH (signs of edge vectors are determined by REF ). NAME and NAME 's generalized NAME fixed point formula (CITE, see also CITE) gives the following expression for the equivariant MATH-genus of MATH where MATH, and MATH depending on whether or not the orientation of MATH defined by REF coincides with that defined by the original orientation of MATH. (We note again that this sign MATH is equal REF for all vertices if MATH is a true complex manifold, for example, a smooth toric variety.) NAME and NAME 's theorem CITE states that the above expression for MATH is independent of MATH and equals MATH. Taking the limit of the right hand side of REF as MATH, one obtains the NAME - NAME REF (since the MATH of each summand in REF is exactly MATH). In the case MATH corresponding to the NAME genus the same limit for the summand corresponding to a vertex MATH equals REF if there is at least one MATH and equals REF otherwise. This is exactly what is stated in the theorem. |
math/9910087 | Consider instead the coefficient of MATH in MATH. This coefficient is equal to MATH . The element MATH maps MATH to MATH. Consequently letting MATH be the number of cyclic descents of MATH, there are MATH values of MATH for which MATH has MATH descents, and MATH values of MATH for which MATH has MATH descents. Combining this with REF shows that the coefficient of MATH in MATH is MATH which simplifies to the formula in the statement of the theorem. |
math/9910087 | The first assertion is immediate from REF . The second assertion follows from the first together with the well-known facts that MATH and that MATH is the unique sequence satisfying NAME 's identity MATH. |
math/9910087 | Taking inverses and using the fact that MATH, it is enough to show that MATH. The coefficient of MATH in MATH is MATH . It is easy to see that MATH for all MATH. The result now follows from REF . |
math/9910087 | For the first assertion, observe that REF gives that MATH. Now use REF . Computations with the symmetric group MATH show that the inequality can be strict. For the second assertion, let MATH be the number of elements of MATH with MATH cyclic descents and let MATH be the number of elements of MATH with MATH descents. Observe that MATH . The second equality is the second part of REF and the final equality follows from REF . For the third assertion, the inequality is clear if MATH has no cuts. Otherwise, combining the fact that MATH for any MATH with REF shows that MATH is equivalent to a convolution of the form MATH (with MATH or MATH possibly MATH and MATH denoting the measure placing all mass the identity). Now observe that MATH . The first equality is REF , the second equality comes from REF , and the third equality is the second part of this theorem. |
math/9910087 | Given REF , this is an elementary combinatorial verification. |
math/9910087 | Replacing MATH by MATH, it is enough to show that MATH . Using REF , one sees that MATH . To simplify things further, recall that MATH is MATH since the MATH's sum to the identity. The above then becomes MATH so the sought result follows from REF . |
math/9910087 | The case MATH is obvious. Multiplying MATH by MATH in the statement of REF shows that MATH . To get the generating function in MATH (for MATH) for the expected number of fixed points in a riffle shuffle followed by a cut, one multiplies the right hand side by MATH, sets MATH, differentiates with respect to MATH, and then sets MATH. Doing this yields the generating function MATH . The result now follows from the identity MATH which is equivalent to the assertion that a monic degree MATH polynomial over MATH has a unique factorization into irreducibles, since MATH is the number of irreducible polynomials of degree MATH over the field MATH. |
math/9910087 | REF and straightforward manipulations give that MATH . Setting all MATH gives the equation MATH . Taking reciprocals and multiplying by the first equation gives MATH proving the first assertion of the corollary. For the second assertion there is a technique simpler than that in CITE. Rearranging the last equation gives that MATH . Letting MATH be a generating function with a convergent NAME series, the limit coefficient of MATH in MATH is simply MATH. This proves the second assertion. |
math/9910087 | If MATH is relatively prime to MATH, then multiplication by MATH permutes the numbers MATH mod MATH. Thus MATH . |
math/9910087 | Observe that MATH. For suppose there is some MATH dividing MATH and MATH. Then MATH divides MATH and MATH, hence MATH. Since MATH divides MATH and MATH divides MATH, it follows that MATH divides MATH. This contradicts the assumption that MATH. |
math/9910087 | Suppose that REF is correct and recall the third definition of affine MATH-shuffles in REF. If MATH then both the affine MATH-shuffle and the MATH-riffle shuffle followed by a cut assign probability MATH to MATH. If MATH, then the affine MATH-shuffle assigns probability MATH to MATH if MATH, and REF otherwise. If MATH, then the MATH-riffle shuffle followed by a cut associates probability MATH to MATH. Since MATH, the MATH in REF is equal to MATH, which implies that for every conjugacy class MATH, the set of permutations in MATH with MATH cyclic descents has major index equidistributed mod MATH. Hence REF holds in this case. The third and final case is that MATH. Suppose that MATH divides MATH and MATH. REF implies that MATH divides MATH. Hence by REF , for any conjugacy class MATH, the set of permutations with MATH cyclic descents has its major index equidistributed mod MATH. Consequently (the second equality below holding by REF and the equidistribution property mod MATH), MATH . From REF (the formula for a MATH-riffle shuffle followed by a cut), REF follows. |
math/9910087 | The probability that an affine MATH shuffle yields MATH is MATH . Since MATH for any MATH in MATH, the assumptions on MATH and MATH imply that the only MATH dividing MATH and MATH is MATH. The result now follows from REF . |
math/9910087 | For the identity conjugacy class, use the third definition of affine MATH-shuffles in REF, together with the assumption that MATH. Next consider the case of simple transpositions. Suppose that MATH, the other cases being trivial. One checks that all simple transpositions MATH with MATH have either MATH or MATH cyclic descents. The easy case is that of MATH cyclic descents. The possible values of MATH are then MATH for MATH, MATH and MATH. The values of the major index so obtained are MATH and each value is hit once. Thus REF holds in this case. The harder case is that of REF cyclic descents. The relevant transpositions are MATH with MATH and MATH (having major index MATH) and MATH with MATH (having major index MATH). First suppose that MATH is odd. It suffices to prove that MATH is a multiple of MATH. Calculating gives MATH as desired. Next suppose that MATH with MATH. It suffices to prove that MATH is a polynomial multiple of MATH. Calculating as above (and omitting the steps analogous to the previous computation) gives that MATH . Since MATH is odd, it follows that MATH is divisible by MATH. |
math/9910087 | Both generating functions follow easily from NAME 's method of representing permutations by intercalations. The MATH's on the right hand-side index the letters of the alphabet. The point is that cycles are formed by fixing a smallest element MATH and specifying an ordered choice of elements larger than MATH; permutations are ordered multisets of such cycles. |
math/9910087 | For the first assertion, define MATH as an intercalation of cycles formed by entries in the bottom line of MATH, with cycles (from left-to-right) having lengths MATH. The assertion is then evident, and the following example may help to untangle the notation. The multiset permutation MATH has MATH with MATH, MATH, MATH, MATH, and MATH. Thus forming from MATH cycles of lengths MATH gives MATH as the intercalation MATH . For the second assertion, we give the argument for the first equality, the argument for the second assertion being analogous. The point is that the number of cards in pile REF of Solitaire with ties allowed applied to MATH is simply the number of left-to-right minima with ties allowed of MATH. The result now follows from the first assertion. |
math/9910087 | For the first assertion, note by NAME 's representation of multiset permutations as intercalations that each MATH-cycle is formed by fixing a smallest element MATH and specifying an ordered choice of MATH elements larger than MATH to occupy the first MATH positions of the cycle. Since multiset permutations are ordered multisets of such cycles, one concludes that MATH . Now replace each MATH by MATH. To prove the second assertion, replacing each MATH by MATH in the first yields the equation MATH . Setting all MATH and taking reciprocals shows that MATH . The result follows by multiplying the previous two equations. The arguments for the third and fourth assertions are analogous. |
math/9910087 | Write the NAME expansion MATH. Then observe that MATH. |
math/9910087 | By REF , MATH . Differentiating with respect to MATH and setting MATH implies that the sought expectation is MATH . Letting MATH, this simplifies to MATH . The second calculation is similar. |
math/9910087 | The proposition is proved by induction, the base case being trivial. Suppose that the proposition holds for MATH. Given MATH let MATH be the symbol with which MATH is switched. If MATH, then MATH is the same as MATH where MATH is obtained by crossing the symbols MATH and MATH out of MATH. If MATH, then MATH, where MATH is obtained by crossing the symbols MATH out of MATH. Consequently, MATH and the result follows by induction. |
math/9910088 | Set MATH; as in the proof of REF , the interesting case is when MATH. We split the integral in REF into two parts. First, consider MATH . The last inequality above holds because MATH is area preserving, and among all such transformations, the symmetric map MATH maximizes the integral over MATH. Next, we consider the case when MATH. Observe that MATH so that MATH . By combining the two estimates we complete the proof. |
math/9910088 | Note that on MATH, MATH, where MATH and MATH denotes the zero order modified NAME function of the second kind CITE. For simplicity, we take MATH and compute MATH as MATH. Set MATH and assume, without loss of generality as MATH is bounded, that MATH. If MATH, then MATH, so that MATH . Since MATH is continuous and decays at infinity, this implies a bound of the form MATH . If, on the other hand, MATH, we use the mean value theorem to estimate MATH which again implies a bound of the form REF. In the last step we have used REF in conjunction with MATH. To recover the scaling of the estimate in MATH, divide REF by MATH, rescale MATH, MATH, and MATH by MATH, and note that MATH. |
math/9910088 | Due to the quasi-Lipschitz condition for MATH, we can adopt the method that NAME developed for the NAME equations in CITE, by simply replacing the kernel estimates in MATH by the corresponding estimates in MATH. Our presentation follows to some extent that of CITE. For simplicity, we assume MATH throughout this proof. We introduce a sequence of approximate solutions MATH for MATH. The proof now proceeds in several steps. Prove that MATH for every MATH. We proceed inductively. Notice that for every MATH the vector field MATH is quasi-Lipschitz in space and continuous in time. This is a consequence of REF as MATH . This implies uniform continuity in time, because MATH is bounded for every MATH: MATH . Since MATH is continuous in time and quasi-Lipschitz in space, the vector field generates a local flow MATH for some MATH - see, for example, REF. Because of the global bound REF, the right side of REF is bounded and the flow exists globally in time. Show that there exists a limiting flow map MATH. We first prove that the sequence MATH is NAME in MATH for some MATH. To simplify notation, we shall drop the explicit time dependence of MATH and MATH, and estimate MATH . By taking the supremum over MATH on both sides, we obtain MATH . Defining MATH we can simplify the previous estimate, and obtain MATH . It is well known that this implies MATH uniformly on MATH for MATH sufficiently small. Since MATH depends only on MATH and the MATH-norm of MATH, this result can be extended to arbitrarily large times. Thus, the contraction mapping theorem implies the assertion of REF . Show that the Lagrangian flow REF is satisfied in the limit, and that MATH. We define the limiting potential vorticity MATH and the limiting velocity MATH in the obvious way, and check by direct estimation that MATH weakly in MATH, and MATH in MATH; both limits are uniform over finite intervals of time. To prove that MATH, MATH, and MATH solve the limit REF , we consider its integrated version MATH . Thus, the left side must be zero. Since MATH is continuous in MATH, we can differentiate with respect to MATH, and find that MATH satisfies REF and that MATH is in fact continuous. Due to the time-reversibility of the equation, the result extends to negative times as well. Moreover, one can show - first by formal calculation for smooth function, and then extending by the usual density argument - that the weak solution MATH defined through REF satisfies MATH for every MATH. This shows that solutions of the vortex method, and hence the equations of second-grade non-Newtonian fluids, preserve the (weak) co-adjoint action. Prove that the solution is unique. Uniqueness is shown by a direct estimate on the difference of two flow maps. This leads to another log-Gronwall inequality, which can be treated in the same way as the previous ones; we omit all details. |
math/9910088 | The result immediately follows from the fact that the vanishing of the weak co-adjoint action is equivalent to the weak formulation of REF . REF , giving global well-posedness of weak solutions, then concludes the argument. |
math/9910088 | We estimate the difference of the NAME and NAME-MATH flow maps: MATH . To estimate MATH, we note that on MATH, the difference of the kernels is explicitly given by MATH, so that MATH . The other two integrals can be estimated by using the quasi-Lipschitz conditions, REF , respectively. One finds that MATH and MATH . By inserting the bounds for MATH to MATH back into REF and taking the supremum on both sides, we obtain the log-Gronwall inequality MATH . To obtain explicit bounds that are valid on any finite interval of time MATH, we set MATH and use the tangent approximation of the concave function MATH; namely, for any MATH, MATH . This makes the right-hand-side of REF linear in MATH. For notational simplicity, we also rescale MATH and MATH such that MATH. We substitute REF into REF and obtain the usual NAME inequality; it follows that MATH must satisfy the differential inequality MATH . Setting MATH and integrating REF with this choice of MATH, we find that MATH . Thus, MATH uniformly on MATH. |
math/9910088 | As in the proof of REF , we estimate MATH . We find, after a change of variables, that MATH . By REF below with MATH and MATH, this expression converges to zero as MATH. Moreover, by REF , MATH . By inserting these estimates back into REF and taking the supremum in MATH on both sides, we obtain an integral inequality that can be solved with the log-Gronwall inequality exactly as in the proof of REF . The result then follows. |
math/9910088 | Set MATH and MATH. Let MATH be fixed. By assumption on the MATH, there exists a MATH such that for every MATH, MATH . Moreover, there exists a MATH such that MATH for MATH. Since MATH is uniformly continuous on compact sets, there exists MATH such that MATH for all MATH with MATH. NAME MATH with finitely many balls of radius MATH and denote the centers of these balls by MATH, MATH. Choose MATH large enough such that for MATH, MATH . Then for MATH there exists a MATH such that MATH, and MATH . On the other hand, if MATH, then MATH . This completes the proof. |
math/9910089 | Since MATH, where MATH denotes the NAME operator MATH in MATH with a NAME boundary condition at MATH (and the same MATH at MATH as MATH, if any), there is a MATH such that for all MATH, MATH is analytic in MATH. Using MATH, the estimate REF holds on the boundary of a sector with vertex at MATH, symmetry axis MATH, and some opening angle MATH. An application of the NAME principle (compare CITE) then extends REF to all of the interior of that sector and hence in particular along the ray MATH. Since REF results from REF upon integrating (compare REF), MATH from MATH to MATH, the extension of REF to MATH with MATH just proven, allows one to estimate MATH uniformly with respect to MATH, and hence to extend REF to MATH. |
math/9910089 | By REF we may assume, without loss of generality, that MATH and MATH are compactly supported such that MATH and by REF we may suppose that REF holds along the ray MATH, that is, MATH . Denoting by MATH and MATH the MATH-functions and NAME solutions associated with MATH, MATH, integrating the elementary identity MATH from MATH to MATH, taking into account REF, yields MATH . By REF, and REF, the right-hand side of REF is MATH as MATH, that is, MATH . Denoting by MATH the transformation kernels associated with MATH, MATH, REF implies MATH where MATH . Insertion of REF into REF, interchanging the order of integration in the double integral, then yields MATH . An application of REF then yields MATH . Since REF is a homogeneous NAME integral equation with a continuous integral kernel MATH, one concludes MATH a.e. on MATH. |
math/9910089 | We denote by MATH the half-line REF MATH-functions associated with MATH on MATH, MATH. Then a straightforward combination of REF yields MATH and hence REF, applying REF separately to the two half-lines MATH and MATH and using the argument following REF. |
math/9910089 | As in the scalar case, we may assume without loss of generality that MATH . The fundamental identity REF, in the present non-commutative case, needs to be replaced by MATH where MATH denote the MATH matrix-valued NAME solutions associated with MATH, MATH, and MATH the matrix-valued Wronskian of MATH matrices MATH and MATH. Identity REF then becomes MATH utilizing the fact MATH . MATH obeys a transformation kernel representation MATH . From this, REF, and the hypothesis of REF, one concludes by REF that MATH . Now let MATH be right multiplication by MATH on MATH matrices and MATH be left multiplication by MATH. Then MATH where MATH is an operator on MATH matrices which is a sum of a left multiplication (by MATH), a right multiplication (by MATH), and a convolution of a left and right multiplication. It follows by REF , and REF that MATH . This is a NAME equation and the same argument based on MATH that a NAME operator has zero spectral radius applies to operator-valued NAME equations. Thus, REF mplies MATH for a.e. MATH. |
math/9910093 | It suffices to prove REF under the assumtpion that MATH - the general case follows by rescaling. This case, however is trivial, and follows from the observation that if MATH, then MATH implies that MATH, with equality if and only if MATH. |
math/9910093 | Since, by REF , MATH, this follows immediately from REF . |
math/9910093 | We will prove the theorem by induction. Let MATH be a vertex in MATH of maximal degree MATH. Such a vertex is contained in, at most, MATH triangles. This is because there is at most one triangle per edge connecting two vertices adjacent to MATH. and removing it together with the edges incident to it, leaves a graph MATH with MATH triangles, MATH edges, and MATH vertices. Note, first of all, that if the two endpoints of an edge in MATH have valences MATH and MATH, then, if MATH, MATH is contained in at most MATH triangles. So if the degree of MATH (assumed to be maximal) was smaller than MATH, no edge of MATH was contained in as many as MATH triangles, so we are done. If MATH, then MATH has MATH edges, and so each edge incident to MATH is contained, on the average, in at most MATH triangles. Now, MATH so the edges incident to MATH are contained, on the average, in fewer than MATH triangles. The number of edges of MATH is smaller than MATH (by a simple calculation), so each of them is contained, on the average, in at most MATH triangles. Since, at best, each of them was contained in one more triangle containing MATH, this tells us that the average was smaller than MATH. If MATH, repeating the argument as above shows us that for the equality to hold MATH has to be a complete graph on MATH vertices, and so MATH is a complete graph on MATH vertices. |
math/9910093 | As before, we set up the NAME multiplier problem, which has MATH equations of the form: MATH . Adding all of the equations together, we find that MATH while multiplying MATH by MATH and adding the results together we get MATH so that that sought-after sum is equal to MATH, as before. Further, note that the derivative of MATH is equal to MATH, which has exactly MATH real zero whatever the value of MATH. Therefore, the equation MATH has at most two real roots. The specifics of our problem are such that we know that there are exactly two roots, one positive, the other negative. Call the positive root MATH, and the negative root MATH, and suppose that MATH of the MATH are equal to MATH, while MATH of the MATH are equal to MATH. It follows that MATH . By REF it follows that MATH . From REF , we have the following equation for MATH where we have substituted for MATH from REF : MATH . Dividing through by MATH get MATH from where, rearranging terms, and replacing MATH by MATH, we get MATH since, amazingly, everything cancels after clearing denominators. So, finally, we see that MATH while MATH thus showing the first part of the theorem. Now, the sum MATH which we seek is given by MATH . This is obviously maximal when MATH is as large as possible, to wit MATH, from which the second part of the theorem follows immediately. |
math/9910094 | Let us denote by MATH the adjoint of MATH with respect to the pairing REF . Consider the symmetric operator MATH which exists whenever MATH. Notice that it has the same principal symbol as MATH. Now, the map MATH is obviously MATH-equivariant. The result follows then from the uniqueness of MATH-module isomorphism - see REF . |
math/9910096 | Since the technique is the same for all the instances, it is enough to discuss e. g. the MATH case. Adding a new slice means adding a pair MATH with MATH, MATH, replacing MATH by REF and providing the factor MATH. But MATH which explains the recursion. The starting value is just MATH . |
math/9910096 | The proofs of the first REF relations are very similar, and we only sketch the first instance. Summing up we find MATH . Iterating that we find for MATH: MATH from which the announced formula follows by solving for MATH. REF others are trickier, because of a term MATH. Again, let us discuss one case. Observe that MATH because one more ``up" step should replace MATH by MATH. Now the generating function MATH of the quantities MATH (upcoming) is obtained independently, whence we get MATH . Now iteration as usual derives the desired result. |
math/9910096 | The proof works as in the easy cases of the tangent recursions and is omitted. For the starting value, we must consider the first pair of numbers. |
math/9910096 | The proofs are quite similar as before; however, iteration must be done for the function MATH, and REF must be added at the end. |
math/9910096 | The desired relation gives us more and more restrictions when we look at the coefficients of MATH. By a tedious search that will not be reported here we find these REF possibilities, and all others can be excluded. The proof that this indeed works is very similar for all of them, so we give just one, namely the instance MATH. Note the following expansions: MATH . So we must prove that for MATH or, reversing the order of summation in the second sum, MATH . We rewrite this again as MATH . Therefore we have to prove that MATH . We use REF, see also CITE, MATH . The desired result now follows by replacing MATH by MATH and plugging in MATH. |
math/9910096 | For the proof by induction we must do the following: Set MATH and MATH . We must show that MATH . Now look at MATH . By the induction hypothesis we only have to show that MATH . However, we can easily show by induction that MATH and MATH holds (the hard part is to find these formul). We have to prove that MATH or MATH . Thus we must prove MATH or MATH or MATH . Now MATH . Similarly, MATH . This finishes the proof. The continued fraction for MATH follows by replacing MATH by MATH. |
math/9910096 | The proof follows the same lines; this time the polynomials (continuants) are MATH and MATH . Hence we have to prove that MATH from here on we can use the previous proof. An alternative proof is by noting that MATH and using the previous result. |
math/9910096 | This follows from the previous theorem by replacing MATH by MATH. |
math/9910096 | The proof of CITE covers the first REF instances, since we note that MATH . The only open case is thus MATH, as the remaining one would follow from duality. Thus, let us now consider MATH and MATH. We need the following computation that is akin to the one in REF . MATH . Although we do not need it, we also mention the dual formula MATH . A similar computation gives the result REF MATH . Now we write MATH and thus MATH . Comparing coefficients, we find MATH . The induction argument is as in CITE; MATH has a factor MATH and, according again to CITE, MATH is still a polynomial. The two factors MATH and MATH mean that everything in REF must be divisible by MATH, and this finishes the proof. It is likely that stronger results as in CITE hold, but we have not investigated that. |
math/9910098 | The inequality is well known in the case of MATH - see Sect. REF (easily adapted to the semi-classical setting), CITE and CITE. The localization argument presented in the Appendix of CITE then gives the lemma. |
math/9910102 | Assume MATH is a regular branch group on its subgroup MATH. Define MATH for all MATH. Clearly MATH contains the direct product MATH, and it is of finite index in MATH, so all the more of finite index in MATH. The second implication holds because branch groups are infinite, and `finite index in infinite group' is stronger than `infinite'. |
math/9910102 | Consider the `level' function MATH . Then MATH is increasing, and by REF, MATH. It then follows that MATH let us take for MATH a natural number satisfying MATH, for instance MATH where MATH is the least integer greater than MATH. The result follows, because then MATH and MATH. |
math/9910102 | Suppose that MATH. Then MATH can be identified with the MATH-orbit of MATH, and, by REF , with the set of all infinite sequences over MATH that eventually coincide with MATH. If MATH, it sends the infinite sequence MATH to one of the MATH sequences in MATH; thus the image of MATH under the set of elements of depth at most MATH is of cardinality bounded by MATH. The image of MATH under the set of elements of length at most MATH is then asymptotically bounded by MATH, by REF . |
math/9910102 | If MATH is amenable, then the trivial one-dimensional representation MATH of MATH is weakly contained in MATH. Inducing up, MATH is weakly contained in MATH. Conversely, if MATH is weakly contained in MATH, we obtain by restricting to MATH that MATH is weakly contained in MATH. Now MATH is a subrepresentation of MATH, because the NAME mass at MATH is MATH-fixed; and MATH by NAME reciprocity. It follows that MATH is weakly contained in MATH, and therefore that MATH is amenable. |
math/9910102 | The first statement follows from NAME 's REF . The second holds because MATH splits as the direct sum of the MATH's, and is weakly equivalent to the direct sum of the MATH's. It was mentioned that MATH and MATH are equivalent: they are both finite-dimensional and act on MATH-equivalent sets, namely MATH and MATH. The third statement then follows from REF below. It is obvious that MATH is an eigenvalue of MATH with constant eigenfunction. Now the fourth follows from REF. |
math/9910102 | Since MATH is amenable, MATH is in its spectrum. By the existence of MATH, the NAME graph of MATH with respect to MATH is two-colourable (that is, bipartite), so its spectrum is symmetrical, and therefore contains MATH. By the NAME conjecture, proved for this case by NAME and NAME, the MATH-algebra MATH contains no idempotents. It follows by functional integration that MATH's spectrum is connected, so is MATH as claimed. |
math/9910102 | Let MATH and MATH be the NAME graphs MATH and MATH respectively, and mark in them the vertices MATH and MATH. Then MATH in the sense of CITE, that is, coincide with MATH in ever increasing balls centered at MATH; therefore, for their associated spectral measures, MATH, in the sense of weak convergence. Therefore the support of MATH is contained in the closure of the union of the supports of MATH, and the proposition follows. |
math/9910102 | The quasi-regular representation MATH is weakly contained in MATH, by ``approximation of coefficients" CITE. Indeed, as MATH, the NAME function at MATH, is a cyclic vector for MATH, it is enough to see that the function MATH can be pointwise approximated on MATH be coefficients of the MATH's. But MATH so MATH . We then have a surjection MATH, where MATH is the MATH-algebra generated by the image of MATH. The spectrum inclusions follow. |
math/9910102 | Let MATH be the NAME graph MATH. Choose MATH, with a corresponding eigenvector MATH. As MATH is amenable, there is a NAME sequence MATH in MATH, that is, a family of subsets of MATH with MATH as MATH. Define now functions MATH on MATH, MATH . Then MATH. If we let MATH be a fundamental domain of MATH in MATH, of diameter MATH, and let MATH denote the MATH-neighbourhood of MATH, then we have MATH, whence MATH so MATH. |
math/9910102 | MATH now MATH so the lemma follows. |
math/9910102 | By direct computation, MATH so MATH and MATH. Let us temporarily agree to write MATH . We will prove by recurrence that MATH . Indeed MATH . Now using REF we have for MATH proving the claim. |
math/9910102 | We prove by recurrence that for all MATH with MATH we have MATH . Indeed for MATH equality holds trivially, and if MATH we combine the terms for MATH and MATH, with MATH: letting MATH designate the MATH-th term, MATH so MATH. Letting MATH in REF proves the proposition, in light of REF . |
math/9910102 | Consider the NAME graph MATH described in Subsection REF. It has vertex set MATH, a simple edge between MATH and MATH for all MATH, a double edge between MATH and MATH for all MATH, a loop at each MATH, and a triple loop at MATH and MATH. The homogeneous space MATH is isomorphic to MATH graph through the correspondence MATH. The choices of MATH clearly define an eigenvector for MATH, because MATH if MATH is an eigenvalue for MATH. |
math/9910102 | The proof follows by recurrence on MATH: first it is readily checked that MATH; then for MATH we compute MATH . |
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