EMBEDDING INTO l n. 1 Notation and Lemmas

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1 EMBEDDING INTO l n We are looking at trying to embed a metric space into l n, our goal is to try and embed an n point metric space into as low a dimension l m as possible. We will show that, in fact, every n point metric space embeds into l n c for any integer c whenever n is sufficiently large. This proof hinges on the following (clever) observation: suppose we have an embedding from X to l n given by f. Then the individual co-ordinates f i gives us a 1-Lipschitz mapping to R with the property that for any x, y there is some i for which f i (x) f i (y) = ρ(x, y). So we can do the opposite, given a 1-Lipschitz function on X we can consider an oriented graph, join the edge x to y if f (x) f (y) = ρ(x, y). We then ask the question: given any space X, how many 1-Lipschitz functions do we need to cover the complete graph on X? It is evident that this is the same question. 1 Notation and Lemmas We will write (X, ρ) as our metric space on n points, and we will denote by m(x) the minimal k such that we can embed X into l k. Denote m(n) to be the largest m(x). Note that m(n) n 1 trivially. Write α(n) = n m(n). Our claim is that α(n). We will say that a metric space is generic if the distances are Q linearly independent. It is clear enough that if the result is true for generic metric spaces it is true for all metric spaces, we can approximate the distances in X with new ones that have been altered by at most ɛ. At each stage there are countably many forbidden choices, and thus we can do it at least n(n 1)/2 times (as required to alter all of them). Thus, without loss of generality we may consider generic metric spaces. Lemma 1.1. A 1-Lipschitz function f : Y R where Y X can be extended to a 1-Lipschitz function on X. Proof. Take f (x) = inf f (y) + d(x, y). This allows us to observe the following: if for Y X we have that m(y) Y c then we can extend all the 1-Lipschitz maps on Y and add into this collection ρ(z, ) for z X/Y. The first verify distances on Y and the second lot verify distances on X/Y and we thus get that m(x) X c. So we can do the work on a well behaved subset of X and then pass up. We now reduce the problem again. Let T be a tree on X. Orient the edges of T so that it contains no path of length 2 (there are two ways of doing this, pick any vertex and assign it a direction randomly and note that this forces the direction of all edges connected to the end vertices.) Define a function f on X by f (a) f (b) = ρ(a, b) for any edge ab in the graph. There are many ways of doing this, up to addition of a constant function, so choose any. We need one more lemma before getting to the combinatorics: Lemma 1.2. If T is a tree such that any two vertices are joined by a path of at most 4 edges, then a function of he form above is 1-Lipschitz if and only if we have that ρ(a, d) + ρ(b, c) ρ(a, b) + ρ(c, d). Proof. Fix any x, y X and we check that f (x) f (y) ρ(x, y). We just need to the cases, if they are connected by an edge then equality holds by definition. If they are connected x 1

2 y z then f (x) f (y) = ρ(x, z) ρ(z, y) ρ(x, y) as required. If they are joined by the path x z w y then we get f (x) f (y) = ρ(x, z) ρ(z, w) + ρ(w, y) The inequality ρ(x, z) ρ(z, w) + ρ(w, y) ρ(x, y) holds from the triangle inequality and ρ(x, z) ρ(z, w) + ρ(w, y) ρ(x, y) follows from the conditions in the lemma. Finally if x, y are joined by the path x z t w y then f (x) f (y) = ±(ρ(x, z) ρ(z, t) + ρ(t, w) ρ(w, y)). To check that ρ(x, z) ρ(z, t) + ρ(t, w) ρ(w, y) ρ(x, y) use that ρ(x, z) ρ(z, t) + ρ(t, w) ρ(x, w), as per the lemma, and then use the triangle inequality. We make some general remarks about trees of diameter at most 4 that are the graphs of Lipschitz functions (which we call admissible.) Choose the longest path in T and pick the vertex closest to the middle of this path as possible. This is called the centre of the tree T. Vertices to which this centre are called main vertices. Others are then called peripheral vertices. Then all vertices must be within 2 steps of the centre, otherwise we could create a longer path. If Y X we may extend an admissible tree to an admissible tree on X with the same main vertices, simply join a point x X/Y to the main vertex for which ρ(x, a) ρ(a, o) is minimized (where o is the centre). Thus we can denote T(o; a 1, a 2,..., a m ) the admissible tree with centre o and main vertices a 1,..., a m. 2 The Proof Consider the space as x 1,..., x n, and for any quadruple 1 a < b < c < d n consider the four points x a, x b, x c, x d. Consider the sums R 1 = ρ(x a, x b ) + ρ(x c, x d ), R 2 = ρ(x a, x c ) + ρ(x b, x d ) and R 3 = ρ(x a, x d ) + ρ(x b, x c ). These sums are different, since X is generic, and thus there are six possible orderings. Colour them by the symmetric group S 3, eg, colour (a, b, c, d) with 231 if R 2 > R 3 > R 1. By Ramseys theorem there is a monochromatic subset of size k whenever X is larger than some n(k). It thus suffices to consider a monochromatic subspace, eg, the proof transforms into Can we cover a monochromatic space by graphs of 1-Lispchitz functions?. Such a question does not depend on the metric, as per Lemma 1.2, so all we need to do is do it for a set of reals with some monochromatic structure. First of all, two of these are impossible for 5 point metric spaces, namely 213 and 312. Let x 1 < x 2 < x 3 < x 4 < x 5 be vertices of X. Then, in the 213 case, we get that ρ(x 1, x 2 ) + ρ(x 3, x 4 ) > ρ(x 1, x 4 ) + ρ(x 2, x 3 ), ρ(x 2, x 3 ) + ρ(x 4, x 5 ) > ρ(x 2, x 5 ) + ρ(x 3, x 4 ) and ρ(x 1, x 3 ) + ρ(x 2, x 5 ) > ρ(x 1, x 2 ) + ρ(x 4, x 5 ). Adding these together yields a contradiction. The same works for 312, except the inequalities are the other way around. So now all we need to do is go through the other 4 cases. The question we have asked is now independent of metric structure and only dependent upon whether monochromatic coverings exist, so we can focus on specific cases of reals. Each case is completely different, so we focus on them independently. Case 321 Let us look at the case 321. This tells us that if a < b < c < d then ρ(x a, x d ) + ρ(x b, x c ) > ρ(x a, x c ) + ρ(x b, x d ) > ρ(x a, x b ) + ρ(x c, x d ) 2

3 Consider the space X = { k, (k 1),..., 1, 1,..., k} and the decomposition into the trees given by T i = T( i, k, (k 1),..., i 1, 1, 2,..., i). These trees have the property, upon drawing, that they are equal to T(i, k, k 1,..., i + 1, 1, 2,..., i). It is evident that these trees cover the complete graph on X, so all we need to do is check that the trees are admissible. Paths of length four in T have to go through both i and i, eg, they are of the form a i i b (or the reverse). So, what can we get? We have four separate cases, either a < i or a [1, i 1], and either b > i or b [ i + 1, 1]. If we have a < i and b > i then a < i < b < i, and the fact that R 3 > R 1 allows us to conclude that ρ(a, b) + ρ(i, i) ρ(a, i) + ρ(b, i) (the condition given in Lemma 1.2). If we have i < b < a < i we have that R 3 > R 1 allows us to give the condition in Lemma 1.2. If we have a < i < b < i the condition R 2 > R 1 gives Lemma 1.2. If we have i < a < i < b we have that R 2 > R 1 again gives Lemma 1.2. Case 132 Let us look at the case 132. This tells us that if a < b < c < d then ρ(x a, x b ) + ρ(x c, x d ) > ρ(x a, x d ) + ρ(x b, x c ) > ρ(x a, x c ) + ρ(x b, x d ) To solve this case we will use a lemma: Lemma 2.1. Let Y be a 132 monochromatic space, Y = n and that m admissible trees cover the complete graph on Y except for some k edges. Then we may add two vertices to Y and for the new space X add two admissible trees so that now the trees cover the complete graph on X apart from k 1 edges So we add two additional vertices and two additional trees but kill one edge. Proof. Consider the trees T(a, a + ɛ, b) and T(b, a + ɛ, b ɛ) (where ɛ is sufficiently small that we do not add in any crossings). We will show that there the unique path of length four in the first obeys the conditions of Lemma 1.2, the second tree following similarly. We have that a < a + ɛ < b ɛ < b, and thus (since we are in a 132 space) we have that ρ(b ɛ) ρ(a, b) > ρ(a + ɛ, b ɛ) ρ(a, a + ɛ), manipulating R 1 > R 3. So the vertex b ɛ connects to a + ɛ. Figure 1 contains images of the trees T(a; a + ɛ, b) and T(b, a + ɛ, b ɛ). We now apply this lemma starting with Y = c, m = 0 and k = c(c 1)/2. Applying the lemma k times we get an X such that X = c + 2k and X is covered by 2k trees, as desired. Since all 132 spaces are isomorphic we are done for any 132 space X. Case 123 We prove that given any integer c > 0 there exists a large N such that any 123-monochromatic space on N vertices may be covered by N c admissible tress. We use induction on c, for c = 0 then N = 1 suffices. We now show that if N works for c then 2N + 3 works for c + 1. Denote the vertices by 1, 0, 1,..., 2N + 1. Take trees T(0; 1, 2), T(1; 0, 2) and T(i + N + 1; i, i + 1). These trees are displayed in Figure 2. These N + 2 trees cover all the edges having at least one endpoint in { 1, 0,..., N + 1}. The inductive hypothesis says that we may cover using N c trees, so we use 2N + 2 c = (2N + 3) (c + 1). 3

4 Figure 1: The trees for 132 spaces Figure 2: The Trees for 123 spaces 4

5 Figure 3: The tree for 231 spaces Case 231 We show that a 231 monochromatic space on 4n + 1 vertices, given by X = {0, 1,..., 4n} can be covered by 3n trees, which is sufficient, whenever n > c. For i = 1,..., 2n take the trees T(0; i, 4n + 1 i), which are illustrated in figure 3. These cover all the edges except a b where 1 a 2n < b 4n + 1 a. So we have to cover the bipartite graph (with n vertices in each part) formed by such edges, using only n trees. Note that the tree T(1; 2n + 1,..., 4n) covers all the edges from vertices 1, 2n, 2n + 1, 4n), and after taking this tree we have an analagous bipartite graph with n 1 vertices in each part, which we can do by repeating this procedure (or induction.) 5