By Vitali D. Milman, Gideon Schechtman

The Israeli GAFA seminar (on Geometric element of sensible research) throughout the years 2002-2003 follows the lengthy culture of the former volumes. It displays the overall tendencies of the speculation. lots of the papers take care of diverse points of the Asymptotic Geometric research. furthermore the quantity includes papers on similar elements of chance, classical Convexity and in addition Partial Differential Equations and Banach Algebras. There also are expository papers on issues which proved to be a great deal on the topic of the most subject of the seminar. One is Statistical studying concept and the opposite is versions of Statistical Physics. all of the papers of this assortment are unique study papers.

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Fig. 1. A simple transport problem The problem is to ﬁnd a partition of R2 into sets A and B with µ(A) = µ(B) = 1/2 so as to minimise the cost of transporting A to (−1, 0) and B to (1, 0): x − (−1, 0) A 2 x − (1, 0) dµ + 2 dµ. B I claim that the best thing to do is to divide the measure µ using a line in the direction (0, 1), as shown in Figure 1. To establish the claim, we need to check that given two points (a, u) and (b, v) with a < b, it is better to move the leftmost point to (−1, 0) and the rightmost point to (1, 0), than it is to swap the order.

Am )) whose coordinates are the measures of the sets on which φ is linear. As ti approaches 0, the ith linear function drops to negative inﬁnity and so µ(Ai ) decreases to zero. In view of the hypothesis on µ, H can be deﬁned continuously on 46 K. Ball the simplex and maps each face of the simplex into itself. It is a well-known consequence (or reformulation) of Brouwer’s ﬁxed point theorem, that such a map is surjective. (If the map omits a point, say in the interior of the simplex, then we can follow it by a projection of the punctured simplex onto its boundary, to obtain a continuous map from the simplex to its boundary which ﬁxes each face.

Let us denote by T the circle {z ∈ C| |z| = 1} ⊂ Q. + Proposition 2. (1) The stabilizer of E0 ∈ Gr2,4 in Q × Q is equal to T × T . (2) The stabilizer of span{1, i} in non-oriented Grassmannian Gr2,4 is equal to T × T ∪ j(T × T ). Proof. Clearly T × T ⊂ StabE0 . Hence we have a covering + (Q × Q)/(T × T ) → Gr2,4 . + Since Gr2,4 is simply connected, this covering is an isomorphism. This proves part (1). Let us prove part (2). Clearly (T × T ) ∪ j(T × T ) is a subgroup contained in the stabilizer of span{1, i} ∈ Gr2,4 .