paper stringlengths 9 16 | proof stringlengths 0 131k |
|---|---|
math/0010041 | Let MATH be the MATH-algebra generated by symbols MATH, MATH, MATH, and MATH subject to the relations MATH . Giving each MATH degree -REF and giving MATH degree REF makes MATH into a graded MATH-algebra. In fact, the natural quotient map MATH preserves this grading. Let MATH be the kernel of MATH. Then MATH is generate... |
math/0010041 | Let MATH denote the MATH-algebra generated by MATH and the set MATH. Given MATH and MATH in terms of monomials consisting of MATH, let MATH be such that MATH. We show that MATH. We can think of MATH as MATH, where MATH . By REF , for each MATH, we can find a 'MATH-integral' MATH of MATH in MATH. Since MATH we have MATH... |
math/0010041 | Let the algebra generated over MATH by the set mentioned in the statement of the lemma be MATH. Clearly, members of the set MATH map MATH to itself. Note that MATH . Thus, elements of MATH (degree REF endomorphisms in MATH) map MATH to itself. If MATH, then MATH and MATH and MATH. If MATH, then MATH, where MATH. Theref... |
math/0010041 | First we note that MATH as MATH-algebras. So, it suffices to show that if MATH then MATH for some finite number of commutators. Suppose MATH is such that MATH for some MATH. Then for any MATH, we have MATH. Thus, the MATH-commutators MATH. Now suppose that MATH. Then, MATH. Induction and the fact that MATH completes th... |
math/0010041 | We show by induction that MATH is surjective for MATH. When MATH, note that MATH is generated over MATH by MATH. The map MATH is clearly surjective. Let MATH. Consider MATH. Since MATH is surjective, there exists a MATH such that MATH. Consider MATH. Since MATH, we have MATH for MATH by REF . Consider MATH. Again by su... |
math/0010042 | In this proof, we let MATH and MATH. The MATH-bimodule MATH of first order MATH-differential operators is generated as a MATH-module by homogeneous MATH such that MATH for any MATH (see REF). Since MATH is generated by the two elements MATH and MATH, it is enough to consider those homogeneous MATH such that MATH and MA... |
math/0010042 | For any MATH we have, MATH for some integer MATH. Thus, MATH for any graded homogenous endomorphism MATH. |
math/0010042 | Since the given collection of operators are graded, it is enough to check for linear independence of homogeneous operators. This reduces to check that the set MATH is linearly independent over MATH which is clear. |
math/0010042 | By the corollary above there are MATH, MATH, MATH, MATH which are multiples of MATH and such that MATH . Put MATH and MATH. The operators MATH and MATH commute, so we have MATH for each MATH. We will prove the lemma by descending induction on MATH. To start, MATH, so MATH is in MATH. Hence MATH where MATH. However, MAT... |
math/0010042 | Let MATH be a homogeneous MATH-differential operator of order MATH. Then MATH where each MATH has the property that for some MATH, MATH is in MATH for any MATH. Then MATH will be in the algebra generated by MATH, MATH, MATH, MATH, and MATH if each MATH is so. Thus we will assume that there is a MATH such that for any M... |
math/0010042 | Let MATH be an ideal of MATH. Every element MATH of MATH can be written as MATH where MATH and MATH for all MATH. Since MATH is in MATH for every integer MATH, we conclude there is some MATH such that each MATH is in MATH. Similarly, we can assume that each MATH is in MATH. When we consider the commutators of MATH with... |
math/0010042 | Note that the elements of MATH and those of MATH commute with each other and that the algebras MATH and MATH generate MATH. Hence, we have a surjective map from MATH to MATH. This map is injective by the proposition above. Hence the theorem. |
math/0010042 | For each MATH, let MATH denote the set of all homogeneous operators in MATH of degree MATH. Then MATH where MATH and MATH are respectively the ring of operators in MATH and MATH of degree REF. By REF, MATH and MATH are localised polynomial rings. Hence, MATH is a localised polynomial ring and in particular a domain. Th... |
math/0010042 | Let MATH be a MATH-dimensional irreducible module of MATH, for MATH. Without loss of generality, we can assume that a MATH-basis of MATH is MATH where MATH is the highest weight vector, and that the action of MATH is given by the map MATH. By using the map MATH, we get the required MATH action on MATH. |
math/0010043 | Since a tight tree set is undirected there must exist a cut MATH in MATH that contains MATH. By REF every sequence MATH of cuts in MATH containing MATH must be finite. The last cut in such a sequence of maximal length must point at MATH. Suppose that there are two cuts MATH and MATH in MATH, such that MATH, MATH and MA... |
math/0010043 | By REF MATH is equivalent to MATH. |
math/0010043 | If these MATH pre-images are non-empty there exists a MATH for which MATH. The statement now is a consequence of MATH. |
math/0010043 | We can find an authomorphism MATH such that MATH and MATH. |
math/0010043 | Let MATH be a cut in a structure tree set MATH of the structure tree MATH. By REF there is a natural number MATH such that for every edge MATH there exist no more than MATH cuts in MATH which contain MATH but do not contain MATH. This constant is the same for all cuts in MATH. By this argument and by MATH we obtain MAT... |
math/0010043 | Since MATH is a subset of MATH we need only prove that MATH is finite if MATH has finite diameter. By the definition of a structure tree set the stabilizer MATH of the set MATH has at most two orbits MATH and MATH on MATH. For every cut MATH in MATH the set MATH has a maximal element. The same holds for the orbit MATH.... |
math/0010043 | Let MATH be the centre of MATH. In MATH there is a sequence MATH of vertices such that MATH. Since only finitely many vertices of the sequence lie in MATH and for all other elements of the sequence there exist components in MATH which contain them, MATH must be a star ball, too. |
math/0010043 | A graph that ramifies uniformly has infinite diameter. If MATH consisted only of components with finite diameter then MATH would be a star ball. If there were only connected components with finite diameter in MATH then MATH would be a star ball since MATH has infinite diameter. |
math/0010043 | For some MATH in MATH let MATH be a component of infinite diameter in MATH. By induction we now choose a sequence MATH of components having infinite diameter such that MATH . The existence of such a sequence is a consequence of REF . Let MATH be an arbitrary vertex in the inner vertex-boundary MATH. Since MATH is conne... |
math/0010043 | The set of automorphisms MATH acts transitively on both bipartite blocks of MATH. Every ball MATH in MATH with radius at least REF is a covering ball of MATH. Since MATH is a quasi-isometry the pre-image MATH, by REF , has a finite diameter. Every ball in MATH containing MATH is a covering ball. |
math/0010043 | We assume that there is a star ball MATH. Let MATH be a radius such that MATH is a covering ball. In MATH there is a component MATH containing a vertex MATH whose distance to MATH is greater then MATH. Let MATH denote a vertex of the MATH-orbit such that MATH is an element of the covering ball MATH. MATH is again a sta... |
math/0010043 | By connecting the MATH-images of adjacent vertices of a ray MATH of infinite diameter in MATH by geodesic paths whose lengths are at most the constant MATH in REF of quasi-isometry, we obtain a path MATH in MATH. By REF its diameter is infinite. Again by REF all MATH-balls in MATH, as a subgraph of MATH, contain at mos... |
math/0010043 | We have to prove that a ray MATH which is not equivalent to any vertex has infinite diameter. Let MATH be a cut containing MATH. Every vertex MATH in MATH can be separated from MATH by a cut MATH. The intersection MATH is again an edge-cut containing MATH. By induction we obtain a strictly decreasing sequence MATH with... |
math/0010043 | The implications REF are trivial. REF implies REF . We will now prove REF . CASE: If MATH is quasi-isometric to MATH then, by REF , there cannot exist a mixed end in MATH. Thick proper ends cannot be mapped onto MATH under MATH since thick ends cannot be described by a sequence of MATH-cuts. By REF they also cannot be ... |
math/0010043 | If MATH acts almost transitively on MATH, then also MATH must act almost transitively on MATH. Thus MATH must be an end in MATH. First we prove that all MATH-pre-images of vertices in MATH have finite diameter. If there is a constant MATH such that any two vertices in MATH with MATH-distance at least MATH have a differ... |
math/0010044 | Let MATH be any strongly fillable contact REF-manifold, MATH one of its strong fillings and MATH a Legendrian knot in MATH . It is not hard to find a neighborhood MATH of MATH in MATH and MATH of MATH in MATH and a diffeomorphism MATH such that MATH along MATH, and MATH . One may then use standard arguments (see REF) t... |
math/0010044 | The map MATH has order four with two fixed points and two points of order two (which are interchanged under MATH). One may thus conclude that the torus bundle is a NAME fibered space over MATH with NAME invariants MATH . To determine the sign of invariants, let MATH be a small disk about one of the fixed points with MA... |
math/0010044 | The proof can be found in CITE, but we reproduce it here for completeness. We first show the existence of a double cover MATH for which MATH is overtwisted. If MATH, then take two copies MATH and MATH of MATH and map MATH to MATH by MATH and MATH to MATH by MATH. In other words, MATH is the double cover of MATH given b... |
math/0010044 | To prove MATH is universally tight, we will first show that it can be made transverse to the MATH-fibers of the NAME fibration. We then show that this implies that MATH is covered by a tight contact structure on REF-torus. Since all of the tight contact structures on REF-torus are universally tight (see CITE), MATH mus... |
math/0010044 | Let MATH be a Legendrian knot isotopic to a regular fiber with MATH as in REF . Let MATH, MATH, be disjoint solid tori isotopic to tubular neighborhoods of MATH, for which MATH contains a contact-isotopic copy of MATH. By perturbing MATH we may assume MATH is convex with vertical (that is, parallel to the regular fiber... |
math/0010044 | From REF we have a tight contact structure MATH on MATH . Now assume that MATH is weakly symplectically semi-fillable. From REF we have a Legendrian knot MATH in MATH that is isotopic to, say, MATH with twist number REF. We may assume that the neighborhood MATH was chosen so that MATH . As discussed above, if we perfor... |
math/0010046 | Since MATH is abelian, every homomorphism MATH factors through MATH. By REF, we have MATH. Furthermore, every homomorphism MATH factors through MATH. Hence: MATH . By NAME 's enumeration principle REF, we have: MATH . Clearly, MATH, where MATH. The NAME function of a subgroup MATH of MATH can be computed from NAME 's R... |
math/0010046 | By NAME 's Lemma (compare CITE), MATH is isomorphic to MATH, the first homology of the chain complex REF, tensored over MATH with the module MATH. Under the identification MATH, the boundary map MATH is the integral MATH matrix MATH. Noting that MATH finishes the proof. |
math/0010046 | Let MATH be a finite presentation. Since MATH, we have MATH. As in the proof of NAME 's REF , MATH is the first homology of the chain complex MATH obtained by tensoring REF with MATH (viewed as a MATH-module via the representation MATH). In other words, MATH. Let MATH. In view of REF, the chain complex REF decomposes a... |
math/0010046 | Assume MATH. Let MATH be the cyclic group generated by MATH. Then there is an automorphism MATH such that MATH. The linear extension MATH is an isomorphism, taking MATH to MATH. Consequently, the MATH-modules presented by these two matrices are isomorphic. Hence, MATH, and so the contributions of MATH and MATH to the s... |
math/0010046 | CASE: The first inequality was already noted in REF. To prove the second inequality, pick a finite presentation of MATH with MATH generators, so that the Jacobian matrix MATH has MATH columns. For each representation MATH, the matrix MATH also has MATH columns, since MATH is one-dimensional. Hence, MATH, for all MATH. ... |
math/0010046 | By REF , we have MATH . Since MATH is sufficiently large, MATH. Let MATH be the order of a non-trivial representation MATH. It is readily seen that MATH. Thus, the assignment MATH establishes a bijection between MATH and the set of non-unit divisors of MATH. Moreover, MATH, where MATH. Now consider the representation M... |
math/0010046 | CASE: The cyclotomic polynomial MATH factors over the field MATH into MATH distinct, monic, irreducible polynomials of degree MATH. If MATH is any one of those factors, then the field MATH is isomorphic to MATH. Let MATH be the automorphism of MATH induced by multiplication by MATH in MATH. Clearly, MATH has order MATH... |
math/0010046 | The proof is by induction on the length of the word MATH. If MATH, the equality holds trivially. Suppose MATH, where MATH. Put MATH. We then have: MATH. Rewriting this last word in normal form (in the semidirect product MATH), we obtain the following equality (in the additive group MATH): MATH . Taking NAME derivatives... |
math/0010046 | Let MATH be a presentation for MATH. Let MATH be a representation, given by MATH. We want to lift it to a representation MATH. Such a representation is given by MATH, where MATH. In view of REF , we must solve the following system of equations over MATH: MATH where MATH is the NAME matrix of MATH. This system has MATH ... |
math/0010046 | Clearly, MATH, and the first identity follows from REF . NAME 's REF gives MATH. Recall that the subgroup lattice of the symmetric group MATH is MATH (see REF ). Thus, NAME 's REF gives MATH. Using REF , we get: MATH . The second identity now follows from REF . |
math/0010046 | The only group of order MATH is MATH; the only groups of order MATH are MATH and MATH; the only groups of order MATH are MATH (if MATH), and MATH and MATH (if MATH). The formulas follow from REF . |
math/0010046 | The proof is based on the computations displayed in REF . The first column lists the rigid isotopy classes (mirror pairs identified) of MATH-arrangements MATH (with MATH, or MATH and MATH horizontal), according to the classification by NAME, NAME\ĭ, and CITE. The next two columns list the NAME invariants MATH and MATH ... |
math/0010048 | REF will be proven by using the compactness properties of MATH and by comparing MATH with a suitable family MATH-converging to MATH. Let MATH be such that MATH. Upon extracting a subsequence, we can suppose that this liminf is actually a limit, so that in particular MATH. We will modify MATH so as to obtain a compariso... |
math/0010051 | Only the ``if"-part remains to be shown. So let MATH be a leftinvariant subspace. Let MATH be an admissible vector for the regular representation, provided by REF . If MATH denotes the projection into MATH, then MATH is an admissible vector for MATH, since MATH, for all MATH. |
math/0010051 | Suppose MATH is an admissible vector. Then MATH, and the adjoint of MATH is given by MATH, where MATH, which defines a MATH function (here we use unimodularity). Hence, for every MATH, MATH, and the right-hand side is a continuous function. Hence for every MATH-function MATH there exists a continuous function which coi... |
math/0010051 | For MATH, consider the measurable field of operators MATH, defined by MATH. Then, by the boundedness condition, MATH. Moreover MATH exists and defines a field of trace class operators with integrable trace class norm. Hence we may apply the inversion formula and obtain for the inverse NAME transform MATH of MATH: MATH ... |
math/0010051 | The operator MATH commutes with left translation and hence has a decomposition MATH with a field MATH of essentially uniformly bounded operators. Of course the aim is to show that MATH-almost everywhere. For this purpose let MATH be a sequence with dense span in MATH. Then, since the NAME transform also intertwines the... |
math/0010051 | For the sufficiency we note that MATH, and we let MATH be the inverse NAME transform of that. Then REF shows that MATH, which means that MATH is the identity operator on MATH, and MATH is admissible. Now let MATH be an admissible vector for MATH, and define MATH. Then we have MATH and MATH (note that MATH is a bounded ... |
math/0010052 | We construct MATH locally; patching together the local constructions in order to obtain a globally defined almost-complex structure still satisfying the same type of bounds is an easy task left to the reader (recall that the space of MATH-compatible almost-complex structures is pointwise contractible). Let MATH be a lo... |
math/0010052 | The argument is very similar to that used by CITE, except that one needs to be slightly more careful in showing that the various bounds hold uniformly in MATH. Fix a point MATH : then we can find a neighborhood MATH of MATH and a local coordinate map MATH, such that MATH contains a ball of fixed uniform MATH-radius aro... |
math/0010052 | The argument is a direct adaptation of the proof of REF in NAME 's paper CITE. Pick a value of MATH and a point MATH. We work in the approximately MATH-holomorphic NAME coordinates given by REF , and use a trivialization of MATH in which the connection MATH-form becomes MATH. Then, we define a local section of MATH by ... |
math/0010052 | Let MATH be the stratification of MATH in which the only stratum is the zero section of MATH (these stratifications are obviously asymptotically holomorphic). By REF , starting from any asymptotically holomorphic sections of MATH (for example, the zero sections) we can obtain for large MATH asymptotically holomorphic s... |
math/0010052 | Let MATH, and let MATH be the stratification of MATH consisting of strata MATH, MATH, defined as follows : viewing the points of MATH as elements of MATH with coefficients in MATH, each MATH is the set of all elements in MATH whose rank is equal to MATH. By REF , the stratifications MATH are asymptotically holomorphic.... |
math/0010052 | By construction the NAME stratifications of MATH satisfy the assumptions of REF , as in every fiber of MATH they can be identified with the same holomorphic quasi-stratification of MATH. As a consequence, they are asymptotically holomorphic, and the existence of asymptotically holomorphic sections of MATH with the desi... |
math/0010055 | First we note the well-known facts that MATH . Recalling REF , we set MATH . This is a quadratic nonlinearity of the form REF. Thus, if MATH is null for each MATH, then the result follows by induction. In fact, if MATH, then MATH is the MATH-fold commutator MATH. A simple calculation shows that MATH . Thus, these commu... |
math/0010055 | Spatial derivatives have the decomposition MATH . So if we introduce the two operators MATH and the null vectors MATH we obtain MATH . On the other hand, if we write MATH REF can be transformed into MATH . Thus, we have MATH . Now, we may assume that MATH, for otherwise the estimates are trivial. But then it follows th... |
math/0010055 | This result is essentially REF. |
math/0010055 | Recall that the weighted norm involves derivatives in the form MATH . In the case when MATH, the result was given in REF. Otherwise, if MATH, then the result is an immediate consequence of REF. |
math/0010055 | By REF , we must estimate terms of the form MATH but since MATH, we will consider MATH with MATH, and MATH. Let MATH. We separate two cases: either MATH or MATH. In the first case, REF is estimated as follows using REF: MATH . Otherwise, we use REF to estimate REF by: MATH . |
math/0010055 | Let MATH, MATH. Since MATH, we have MATH. Thus, by REF , we find using our assumption MATH . Thus, if MATH is small enough, the bound REF results. Again since MATH, we have MATH. From REF we now have MATH . If we apply REF and our assumption, then MATH from which REF follows. |
math/0010056 | Take MATH, MATH, and MATH. By REF , MATH . This proves the first part of the corollary, and the second is immediate. |
math/0010056 | The point MATH belongs to MATH. Since this point is nonconstant, it cannot be a torsion point. |
math/0010056 | The first statement is immediate from REF . The second statement follows without difficulty by applying REF first to the extension MATH, and then to the extension MATH. |
math/0010056 | This follows directly from REF (with MATH and MATH), REF , and REF . |
math/0010056 | This follows directly from REF , and REF . |
math/0010056 | Both sides have the same divisor, and evaluate to MATH at MATH. |
math/0010056 | By REF , MATH. As in the proof of REF , MATH and the identity of the proposition follows. |
math/0010056 | Suppose first that we are in REF . Let MATH be the linear fractional transformation which (after identifying the roots of MATH with the nonzero elements of MATH) agrees with the given automorphism of MATH on the roots of MATH. It follows from the NAME of the automorphism that MATH. If MATH, then (since MATH) we must ha... |
math/0010056 | That these points belong to MATH can be checked directly. Since they are nonconstant, both points have infinite order. The automorphism of MATH which sends MATH to MATH fixes the first point and sends the second point to its inverse, so they are independent in MATH. Since MATH, REF show that the rank cannot be greater ... |
math/0010056 | As with REF , the simplest proof is a direct calculation. |
math/0010056 | Let MATH, MATH, and MATH, and apply REF with MATH and MATH. |
math/0010056 | Take MATH to be the permutation of MATH which switches MATH and MATH, and MATH to be the permutation which switches MATH and MATH. Then the linear fractional transformations MATH act on MATH as MATH and MATH do, respectively. One computes in REF that MATH and MATH where MATH . If MATH, then MATH and MATH are MATH-linea... |
math/0010056 | Take MATH to be the permutation of MATH which switches MATH and MATH, MATH to be the permutation which switches MATH and MATH, and MATH to be the cyclic permutation MATH. Let MATH be the corresponding linear fractional transformation. Then in REF we have MATH where MATH . Now suppose MATH with MATH. Then MATH and MATH ... |
math/0010056 | We may write MATH as MATH where MATH has MATH rational roots. If MATH denotes the rational cyclic subgroup of order MATH, then MATH contains a rational point, and we may choose our model so that this point is MATH. Denote the other roots of MATH by MATH and MATH. If MATH is a generator of MATH and MATH is its MATH-coor... |
math/0010056 | The simplest proof is a direct computation. To construct this example one takes MATH to be MATH and proceeds exactly as in the proofs of REF , with MATH, MATH, which gives MATH . The curve defined by MATH has a rational point MATH, and using this one computes that MATH where MATH. REF with this input leads to the data ... |
math/0010056 | Without loss of generality we may assume that MATH, since if not, MATH by REF and there is nothing to prove. Let MATH, a homogeneous polynomial of degree MATH. REF are immediate from REF, respectively, applied to MATH. Suppose now that the Parity Conjecture holds, the irreducible factors of MATH all have degree at most... |
math/0010056 | If MATH has a real root then MATH contains both positive and negative values (MATH has no multiple roots because it was assumed to be squarefree). Thus by a result of NAME REF we have MATH . Now the corollary follows immediately from REF . |
math/0010056 | This is immediate from REF . |
math/0010056 | Suppose first that we are in REF . Translating the given rational root of MATH we may assume that MATH with MATH. Since MATH has MATH real roots we also have MATH. In particular, MATH. Let MATH be as in REF . Then MATH is divisible by MATH. We compute that MATH, but MATH is positive for large MATH, so MATH, and hence M... |
math/0010056 | This is immediate from REF , and REF (the last to handle the excluded value MATH in REF ). |
math/0010056 | REF follows directly from REF . The polynomial MATH of REF has degree MATH, and hence it has a real root, so REF follows from REF . |
math/0010056 | If MATH and MATH, let MATH denote the squarefree part of MATH, that is, the unique squarefree integer MATH such that MATH for some integer MATH. For every squarefree integer MATH let MATH denote the hyperelliptic curve MATH of genus MATH. The map MATH defines an injection MATH (where MATH denotes the hyperelliptic invo... |
math/0010066 | Suppose the ring has length REF. Then the component is a rooted tree with a loop at the root. The loop has no effect on the permutation generated. Ignore it, and suppress the corresponding transposition. This leaves us with a permutation of the form MATH whose associated graph is a tree MATH with vertex set, MATH say, ... |
math/0010066 | Any representation of MATH as an exchange shuffle will determine a partition of the set MATH into one- and two-element blocks, the one-element blocks being those cycles represented as trees, and the two-element blocks those pairs of cycles represented as unicycles. Think of such a partition as an involution, by letting... |
math/0010066 | We have MATH . For MATH, an involution on MATH can either fix the first element, leaving MATH ways of moving the remaining elements, or swap it with a member of MATH, leaving MATH ways of moving the remaining elements. Hence MATH . If we set MATH then induction on MATH proves that, for all MATH, MATH . From the above, ... |
math/0010066 | See NAME and NAME. |
math/0010066 | We have just seen that MATH. Since MATH, the first inequality is true for MATH. We have the following form of NAME 's formula CITE: MATH . For MATH, this yields the estimate MATH, which implies the result. For the second inequality, let MATH be the maximum of MATH over all permutations MATH of MATH. We will induce on M... |
math/0010066 | Let MATH be an arbitrary permutation of MATH, and let MATH be its decomposition into disjoint cycles. By REF , MATH . Plugging in the bounds from REF above, MATH that is, MATH . From REF , it follows that for MATH, MATH is maximized at MATH, so MATH unless MATH, that is, unless MATH is the identity permutation. However... |
math/0010066 | First, let the component be a tree. We will prove the induced permutation is a cycle, but now without using conjugation. As before, let us suppose we have a permutation of the form REF : MATH . We wish to prove by induction on MATH that REF is a MATH-cycle. We will need a stronger induction hypothesis. Let MATH, , MATH... |
math/0010066 | First we comment on what is to be proved. Start with the left-hand side above. Suppose we are given a representation of MATH as a unicyclic exchange shuffle, and suppose that in the associated directed graph MATH is the upper and MATH the lower cycle. Break up the ring into maximal segments of consecutive vertices that... |
math/0010066 | If every member of MATH is less than every member of MATH, then MATH must always vanish, yielding MATH. On the other hand, if MATH is a tree with root REF representing MATH, then the condition MATH is vacuous, so MATH . Similarly, MATH and substituting into REF yields the desired result. Of course, the first equation i... |
math/0010066 | Represent MATH as a product of transpositions, MATH . We will exhibit a sequence MATH with MATH . Let MATH be the permutation of MATH that moves the first MATH terms past the last MATH. Then by REF and similarly MATH . Using MATH, REF becomes MATH . Rearrange terms to get: MATH . Now we use repeatedly the conjugation r... |
math/0010066 | Without loss of generality, let MATH end in its smallest element, which we call MATH. We claim that MATH and MATH where in these sums MATH, but not MATH, MATH, or the MATH's, may be empty. The lemma will follow from these claims. To see where MATH is maximized, use REF . If MATH is decreasing, then MATH, MATH, , MATH, ... |
math/0010066 | Let MATH and MATH be sequences of distinct positive integers with disjoint ranges, and let MATH. We wish to show that, for fixed MATH and MATH, MATH is maximized iff we can write MATH . We will induce on MATH. Without loss of generality, let REF be the smallest element in the domain of MATH. Exchange MATH and MATH, if ... |
math/0010066 | Let MATH be an arbitrary permutation of MATH, and let MATH be its decomposition into disjoint cycles, where MATH is a MATH-cycle, MATH is a MATH-cycle, , and MATH is a MATH-cycle. By REF , MATH . Using the bounds in REF then shows that MATH is no bigger than REF and that the equality condition is as claimed. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.