lagrange s identity and minkowski s conjecture

... illustrate the strength as well as the limitation of this approach As we shall see below, the passage from lower estimates on balls to upper estimates on cylinders is simpler in this case Proposition ... true in dimensions n = 4, provided the solutions are also assumed to satisfy an anti-symmetry condition This is done by establishing (1.10) with q = under such an assumption More precisely, we have: ... classical result of Gidas-Ni-Nirenberg Indeed, the monotonicity in R may be shown as follows Suppose that for some R2 > R1 > R0 , 328 NASSIF GHOUSSOUB AND CHANGFENG GUI there is r0 ∈ (0, R1 ) such...

Ngày tải lên: 05/03/2014, 23:20

... this set Ξk is A-invariant, and has zero Hausdorff dimension transversally to the A-orbits For more details, see Section 10 and Section 11 Note that (1.3) is automatically satisfied if zero is attained ... assumption on K, we have that s s g ∈ L Choose a rational ˜ ∈ Σ close to s with α˜ ∈ A so that α˜x ∈ B1/ (x) s L s s Clearly g = α˜ s g satisfies α˜x = g x and so g ∈ B1/ Since x ∈ K ⊂ ˜ ˜ ˜ s ... a nullset and three compact sets As mentioned before we will work with two main assumptions: that µ satisfies the assump- 534 MANFRED EINSIEDLER, ANATOLE KATOK, AND ELON LINDENSTRAUSS tions of...

Ngày tải lên: 15/03/2014, 09:20

... second question is also yes Although the embedding assumption is partially needed, a weaker assumption on minors is the essential ingredient Besides considering extensions of 2-colored subgraphs we ... the set of vertices that are not in N i−1 (S) but are adjacent to at least one vertex in N i−1 (S) So N i (S) consists of those vertices in G whose distance from S is exactly i In Section we show ... = This means that no first neighbor of G[Pi ] is colored So we arbitrarily assign the colors and to the two color classes of G[Pi ] If the precoloring happened to assign these two classes the...

Ngày tải lên: 07/08/2014, 07:21

... Richard Stanley s 60th Birthday, Massachusetts Institute of Technology, Cambridge, Massachusetts, June 22–26, 2004, slide available at: http://www-math.mit.edu/~apost/talks/perm-slides.pdf [4] R P Stanley, ... ω uniformly at random, the probability that ω(v) is the smallest among the labels of the descendants of v is 1/h(v) So the number of possible labelings ω is n!/ v∈α h(v) Thus we have |Dn | = ... 1, we say that wi+1 is the right sibling of wi The set Gn can be viewed as a subset of Qn satisfying the following condition: Suppose that v is the right sibling of u in Q ∈ Qn Then m S( Q, u)...

Ngày tải lên: 07/08/2014, 08:22

... in some surface with edge-width at least w0 satisfies the Haj s Conjecture o K¨ hn and Osthus [5] proved that graphs whose girth is at least 186 satisfy the Haj s u o conjecture This excludes the ... (q) and two distinct vertices x, y are adjacent if and only if x − y = z for some z ∈ GF (q) Since p ≡ (mod 4), x − y is a square if and only if y − x is a square, and this assures that Pq is a ... triangulation of some surface may fail to satisfy the Haj s Conjecture In this sense, this note can be viewed as an echo to o the stimulating work of Thomassen [11] As noted in that paper, graphs embedded...

Ngày tải lên: 07/08/2014, 08:22

... detailed comments References [1] G E Andrews, q-Series: their development and application in Analysis, Number theory, Combinatorics,physics and computer algebra, NSF CBMS Regional Conf.Series, Vol.66, ... 3, (3.27)/(3.40)] for the details As far as we know, it is originally due to Carlitz [3] Main results In this section, we will present our main results At first, set f (x, y) = − xy, g(x, y) = ... order to find any possibly new or interesting results remained As an immediate consequence, we obtain the following bilateral summation from which two special cases of Ramanujan s ψ1 summation formula...

Ngày tải lên: 07/08/2014, 13:21

... its last e edges In fact, as c is odd, we can even assume that every path in G of length at least k − zigzags between L and S, except possibly on its first a and its last e edges As paths are symmetric, ... can shift its first (a + c + e) vertices at least m − (a + c + e) times So we may assume that every path in G of length at least k zigzags between L and S, except possibly on its first a and its ... particular classes of trees (such as paths [7]) have smaller Ramsey numbers, the bound k + m would be tight in the class of all trees In fact, the Ramsey number of two stars with k, resp m edges, is k...

Ngày tải lên: 07/08/2014, 21:20

... shortest (u, v)-paths, and for every vertex u of G and every subset B of size at most m, B |Ns (u)| ≥ (d/ 2s) s ≥ qds Moreover, qds /2 is an integer and qds /2 = ds ds ≥ > m 2( 2s) s 4( 2s) s Now ... choices for each of u2 , u3 , , us+1 , thus each v ∈ X is the first vertex of at most ds escaping paths This shows that v is on at most ds + msds−1 escaping paths Recall that since the robber was ... has almost the same number of zeros and ones (see [9]) (ii) This is clear as |S| = d (iii) This follows from the fact that each member of S has as its first coordinate, hence there is no odd-size...

Ngày tải lên: 08/08/2014, 12:23

... complexes of varieties whose differentials and morphisms are given by correspondences is homotopic to zero Since the functor associating the Gersten complex preserves homotopy, it is sufficient to show ... S, the restriction to H (S, Z)[−1] for s = s0 is expressed by the short exact sequence → Z → H (S, {s0 , s} ; Z) → H (S, {s0 }; Z) (= H (S, Z)) → 0, using the corresponding distinguished triangle ... Bernstein, and P Deligne, Faisceaux pervers, Analysis and Topology on Singular Spaces, I (Luminy, 1981), Ast´risque 100, 5–171, Soc Math France, Paris, e 1982 [6] J Biswas and V Srinivas, A Lefschetz...

Ngày tải lên: 14/02/2014, 17:20

... forms x0 = s0 + ω s1 + s4 , x1 = ω s0 + s1 + s4 , x4 = s0 + s1 + s4 , x2 = s2 + ω s3 + s5 , x3 = ω s2 + s3 + s5 , x5 = s2 + s3 + s5 ¯ ¯ lie in H (X ρ0 , ω ⊗2 ) This coordinate system yields ... over solvable extensions of Q This suggests that we consider the class of varieties X such that, if K is a number field, and SERRE S CONJECTURE OVER F9 1113 Σ is the set of all solvable Galois extensions ... q-expansion of s4 up to the trace-2/3 part is a b d 0 and that of s5 is c 0 Note that s0 = t1 s4 , s1 = t2 s4 , s2 = t1 s5 , s3 = t2 s5 , so it suffices for our 1 purposes to compute the q-expansions...

Ngày tải lên: 16/02/2014, 05:20

... classes of maps X −→ BDiff(F d ) where the disjoint union runs over a set of representatives of the diffeomorphism classes of closed, smooth and oriented d-manifolds Comparing these two classification ... this relies on transversality theorems; the other uses collapse maps to normal bundles of submanifolds in euclidean spaces See [43] and especially [35] Applied to our situation this identifies ... turns out to be a map between spaces with monoid structure (up to homotopy), and in this situation one easily passes between based and unbased homotopy classes Here are some details The monoid structure...

Ngày tải lên: 06/03/2014, 08:21

... and we conclude that λA ≤ i < j This case again splits into three sub-cases Sub-case (a) in which i ∈ S( α) and j ∈ S( α) and sub-case (b) in which i ∈ S( α) and j ∈ S( α) are similar to the ones ... conjugacy classes and P orbits In this section we describe the P = Pn (K) conjugacy classes of nilpotent elements in g = gln (K) We shall also describe certain Jacobson-Morosov triples that are associate ... process in that case for the set G × V ∗ (See Proposition 8.2 and Step B below.) Step B Induction on semisimple elements and their centralizers: As in Harish-Chandra s case we would like to use...

Ngày tải lên: 28/03/2014, 22:20

... Fermat s Last Theorem must match its charm and allure Kepler s sphere-packing conjecture is just such a problem—it looks simple at first sight, but reveals its subtle horrors to those who try to solve ... Mathematics and Statistics of the University of Saint Andrews in Scotland It stores a collection of biographies of about 1,500 mathematicians Friends and colleagues read parts of the manuscript and ... (Instructions on measuring with compass and ruler) was the first mathematics textbook in German In it he discussed perspective and proportions and showed how to construct figures with ruler and compass...

Ngày tải lên: 05/06/2014, 11:25

... Fermat s Last Theorem must match its charm and allure Kepler s sphere-packing conjecture is just such a problem—it looks simple at first sight, but reveals its subtle horrors to those who try to solve ... Mathematics and Statistics of the University of Saint Andrews in Scotland It stores a collection of biographies of about 1,500 mathematicians Friends and colleagues read parts of the manuscript and ... (Instructions on measuring with compass and ruler) was the first mathematics textbook in German In it he discussed perspective and proportions and showed how to construct figures with ruler and compass...

Ngày tải lên: 05/06/2014, 11:26

... of K2m It is easy to see that in the example the union of any t > colors consists of disjoint components of at most 2t vertices Also no open path can consist of edges colored (in this order) a, ... this note is the following question raised by P Erd s and A Gy´rf s [5] : Is it possible to have a proper edge coloring of Kn with cn o a a colors so that the union of any two color classes has ... the special case of n = 2m this example uses n − colors, proper coloring is a 1-factorization and if there is no 2-colored path with edges then this coloring is unique up to isomorphism [4] Closely...

Ngày tải lên: 07/08/2014, 06:20

... inequality (2) follows References [1] Bert Hartnell and Douglas F Rall, Domination in Cartesian Products: Vizing s Conjecture, in Domination in Graphs—Advanced Topics edited by Haynes, et al, Marcel ... That is, Pi = {v | (u, v) ∈ Di for some u ∈ Πi } Observe that for any i, Pi ∪ (V (H) − NH [Pi ]) is a dominating set of H, and hence the number of vertices in V (H) not dominated by Pi satisfies the ... |) < γ(G), and we have a contradiction This observation shows that |Rv | ≤ N= v∈V (H) |Qv | = |D| (5) v∈V (H) It follows from (4) and (5) that γ(G)γ(H) − |D| ≤ N ≤ |D|, and the desired inequality...

Ngày tải lên: 07/08/2014, 06:20

... 1.14, x, y admits a complete block system with blocks of size b for any b |b Suppose that b = rs We choose r, s in such a way that r is as large as possible so that Stab x,y (Cr )|C is faithful for ... xiB as = Hence b|iB s, say kB b = iB s But b = rs, so kB rs = iB s, meaning that kB r = iB for any B Thus h ∈ Stab x,y (Cs ) Since xiB a |B is nontrivial, this has shown that Stab x,y (Cs ) is ... the vertex gs is assigned the colour that has been associated with s All of the results of this paper also hold for Cayley colour digraphs This is not always made explicit, but is a simple matter...

Ngày tải lên: 07/08/2014, 07:21

... counterclockwise The quadrilateral obtained has a perimeter less than 2u sin β as well By Lemma 5, the perimeter of P1 is less than 2u sin β 1−cos β So one of the sides of P1 is less than assume that ... widths is still less than w, while the strips are in general position Consider the centrally symmetric convex polygon M of 2n sides which corresponds to the n strips as in Lemma 1, that is, each ... ⊂ T , then the annulus T \T0 is covered by the strips S1 , S2 , S3 , as shown in Fig.1 But the sum of the three strips’ widths is equal to (1 − ε)ω which is strictly less than ω the electronic...

Ngày tải lên: 07/08/2014, 15:22

... sets is ∅ This case is investigated in Proposition 2.3 Results for |X| = 11 All the proofs in this Section follow a similar pattern: we assume that certain sets are in F and F is not Frankl s ... cases: F contains three three-element subsets of a four-element set This case is considered in Proposition 2.1 F contains three three-element sets which all contain the same two elements The statement ... statement holds by Proposition 2.2 F contains three three-element sets whose union is a five-element set and whose intersection is a one-element set This case is solved in [12] The intersection of...

Ngày tải lên: 07/08/2014, 21:20

... the spaces of the East left newly arrived occupiers shaken Ober Ost s areas were separated from East Prussia by shallow, Xat lowland, with marshy woods and crossed by many rivers Rippled lines ... ‘‘empire and subjects’’ dissolved, as the newcomers now saw them as distinct and variegated Now they were understood as ‘‘Land und Leute,’’ ‘‘lands and peoples,’’ discrete unities of territory and ... articles and sketches searching for historical parallels, the search for traces of themselves in the region s history was consuming, yet ultimately without satisfying results The past in the present...

Ngày tải lên: 21/09/2012, 11:02