Banach Spaces from Barriers in High Dimensional Ellentuck Spaces

Transcription

1 Banach Spaces from Barriers in High Dimensional Ellentuck Spaces A. ARIAS N. DOBRINEN G. GIRÓN-GARNICA J. G. MIJARES Two new hierarchies of Banach spaces T k (d, θ) and T(A k d, θ), k any positive integer, are constructed using barriers in high dimensional Ellentuck spaces [10] following the classical framework under which a Tsirelson type norm is defined from a barrier in the Ellentuck space [3]. The following structural properties of these spaces are proved. Each of these spaces contains arbitrarily large copies of l n, with the bound constant for all n. For each fixed pair d and θ, the spaces T k (d, θ) and T(A k d, θ), k 1, are l p -saturated, forming natural extensions of the l p space, where p satisfies dθ = d 1/p. Moreover, the T k (d, θ) spaces form a strict hierarchy over the l p space: For any j < k, the space T j (d, θ) embeds isometrically into T k (d, θ) as a subspace which is non-isomorphic to T k (d, θ). The analogous result for the spaces T(A k d, θ) is open. 46B03, 46B07, 03E05; 03E02, 05D10 1 Introduction Banach space theory is rich with applications of fronts and barriers within the framework of the Ellentuck space (see for example [9], [23], and Part B of [3]). Infinitedimensional Ramsey theory is a branch of Ramsey Theory initiated by Nash-Williams in the course of developing his theory of better-quasi-ordered sets in the early 1960 s. This theory introduced the notions of fronts and barriers that turned out to be important in the context of Tsirelson type norms. During the 1970 s, Nash-Williams theory was reformulated and strengthened by the work of Silver [26], Galvin and Prikry [17], Louveau [21], and Mathias [22], culminating in Ellentuck s work in [15] introducing topological Ramsey theory on what is now called the Ellentuck space. Building on seminal work of Carlson and Simpson in [7], Todorcevic distilled key properties of the Ellentuck space into four axioms which guarantee that a topological space satisfies infinite-dimensional Ramsey theory analogously to [15] (see Chapter

2 2 A. Arias, N. Dobrinen, G. Girón-Garnica and J. G. Mijares 5 of [27]). These abstractions of the Ellentuck space are called topological Ramsey spaces. In particular, Todorcevic has shown that the theory of fronts and barriers in the Ellentuck space extends to the context of general topological Ramsey spaces. This general theory has already found applications finding exact initial segments of the Tukey structure of ultrafilters in [24], [13], [14], [12], and [10]. In this context, the second author constructed a new hierarchy of topological Ramsey spaces in [10] and [11] which form dense subsets of the Boolean algebras P(ω α )/Fin α, for α any countable ordinal. Those constructions were motivated by the following. The Boolean algebra P(ω)/Fin, the Ellentuck space, and Ramsey ultrafilters are closely connected. A Ramsey ultrafilter is the strongest type of ultrafilter, satisfying the following partition relation: For each partition of the pairs of natural numbers into two pieces, there is a member of the ultrafilter such that all pairsets coming from that member are in the same piece of the partition. Ramsey ultrafilters can be constructed via standard methods using P(ω)/Fin and the Continuum Hypothesis or other settheoretic techniques such as forcing. The Boolean algebra P(ω 2 )/Fin 2 is the next step in complexity above P(ω)/Fin. This Boolean algebra can be used to generate an ultrafilter U 2 on base set ω ω satisfying a weaker partition relation: For each partition of the pairsets on ω into five or more pieces, there is a member of U 2 such that the pairsets on that member are all contained in four pieces of the partition. Moreover, the projection of U 2 to the first copy of ω recovers a Ramsey ultrafilter. In [6], many aspects of the ultrafilter U 2 were investigated, but the exact structure of the Tukey, equivalently cofinal, types below U 2 remained open. The topological Ramsey space E 2 and more generally the spaces E k, k 2, were constructed to produce dense subsets of P(ω k )/Fin k which form topological Ramsey spaces, thus setting the stage for finding the exact structure of the cofinal types of all ultrafilters Tukey reducible to the ultrafilter generated by P(ω k )/Fin k in [10]. Once constructed, it became clear that these new spaces are the natural generalizations of the Ellentuck space to higher dimensions, and hence are called high-dimensional Ellentuck spaces. This, in conjunction with the multitude of results on Banach spaces constructed using barriers on the original Ellentuck space, led the second author to infer that the general theory of barriers on these high-dimensional Ellentuck spaces would be a natural starting point for answering the following question. Question 1.1 What Banach spaces can be constructed by extending Tsirelson s construction method by using barriers in general topological Ramsey spaces? The work in this paper is a first step toward the broader goal of finding structural properties of Tsirelson analogues using infinite dimensional barriers on high and infinite

3 Banach Spaces from Barriers in High Dimensional Ellentuck Spaces 3 dimensional Ellentuck spaces. The second author s motivation was the hope that this approach can shed new light on distortion problems. The constructions presented in this paper have as their starting point Tsirelson s groundbreaking example of a reflexive Banach space T with an unconditional basis not containing c 0 or l p with 1 p < [28]. The idea of Tsirelson s construction became apparent after Figiel and Johnson [16] showed that the norm of the dual of the Tsirelson space satisfies the following equation: (1) { a ne n = max sup n n a n, 1 m } 2 sup ) E i ( n a ne n, where the sup is taken over all sequences (E i ) m of successive finite subsets of integers ( with the property that m min(e 1 ) and E i n a ) ne n = n E i a n e n. The first systematic abstract study of Tsirelson s construction was achieved by Argyros and Deliyanni [1]. Their construction starts with a real number 0 < θ < 1 and an arbitrary family F of finite subsets of N that is the downwards closure of a barrier in Ellentuck space. Then, one defines the Tsirelson type space T(F, θ) as the completion of c 00 (N) with the implicitly given norm (1) replacing 1/2 by θ and using sequences (E i ) m of finite subsets of positive integers which are F -admissible, i.e., there is some {k 1, k 2,..., k m } F such that k 1 min(e 1 ) max(e 1 ) < k 2 < k m min(e m ) max(e m ). It became customary in the literature to identify Tsirelson s original space with its dual. So, in the above notation, Tsirelson s original space is denoted by T(S, 1/2), where S = {F N : F min(f)} is the Schreier family. In addition to S, the low complexity hierarchy {A d } d=1 with A d := {F N : F d} is noteworthy in the realm of Tsirelson type spaces. In fact, Bellenot proved in [4] the following remarkable theorem: Theorem 1.2 (Bellenot [4]) If dθ > 1, then for every x T(A d, θ), 1 2d x p x T(A d,θ) x p, where dθ = d 1/p and p denotes the l p -norm. We refer the reader to [8, 20] for a systematic study of Tsirelson s space and Tsirelson type spaces.

4 4 A. Arias, N. Dobrinen, G. Girón-Garnica and J. G. Mijares There are three non-trivial directions in which this construction may be extended using high-dimensional Ellentuck spaces, two of which are considered in this paper. We construct two new hierarchies of Banach spaces extending the low complexity hierarchy on the Ellentuck space to the low complexity hierarchy on the finite dimensional Ellentuck spaces. This produces various structured extensions of the l p spaces. In Section 2 we review the construction of the finite-dimensional Ellentuck spaces in [10], introducing new notation and representations more suitable to the context of this paper. Our new Banach spaces T k (d, θ) and T(A k d, θ) are constructed in Section 3, using finite rank barriers on the k-dimensional Ellentuck spaces. These spaces may be thought of as structured generalizations of l p, where dθ = d 1/p, as they extend the construction of the Tsirelson type space T 1 (d, θ), which by Bellenot s Theorem 1.2 is exactly l p. We prove the following structural results about the spaces T k (d, θ). They contain arbitrarily large copies of l n, where the bound is fixed for all n (Section 4), and that there are many natural block subspaces isomorphic to l p (Section 5). Moreover, they are l p -saturated (Section 6). The spaces T k (d, θ) are not isomorphic to each other (Section 7), but for each j < k, there are subspaces of T k (d, θ) which are isometric to T j (d, θ) (Section 8). Thus, for fixed d, θ, the spaces T k (d, θ), k 1, form a natural hierarchy in complexity over l p, where dθ = d 1/p. The second class of spaces we consider involves the most stringent definition of norm. These spaces, T(A k d, θ), are constructed in 3 using admissible sets which are required to be separated by sets which are finite approximations to members of the E k. The norms on these spaces are thus bounded by the norms on the T k (d, θ) spaces. In Section 9, the spaces T(A k d, θ) are shown to have the same properties as the as shown for T k(d, θ), the only exception being that we do not know whether T(A j d, θ) embeds as an isometric subspace of T(A k d, θ) for j < k. The paper concludes with open problems for further research into the properties of these spaces. Acknowledgement. Dobrinen was partially supported by National Science Foundation Grants DMS and DMS High Dimensional Ellentuck Spaces In [10], the second author constructed a new hierarchy (E k ) 2 k<ω of topological Ramsey spaces which generalize the Ellentuck space in a natural manner. In this section, we

5 Banach Spaces from Barriers in High Dimensional Ellentuck Spaces 5 reproduce the construction, though with slightly different notation more suited to the context of Banach spaces. Recall that the Ellentuck space [15] is the triple ([ω] ω,, r), where the finitization map r : ω [ω] ω [ω] <ω is defined as follows: For each X [ω] ω and n < ω, r(n, X) is the set of the least n elements of X. Usually r(n, X) is denoted by r n (X). We shall let E 1 denote the Ellentuck space. We now begin the process of defining the high dimensional Ellentuck spaces E k, k 2. The presentation here is slightly different than, but equivalent to, the one in [10]. We have chosen to do so in order to simplify the construction of the Banach spaces. In logic, the set of natural numbers {0, 1, 2,... } is denoted by the symbol ω. In keeping with the logic influence in [10], we shall use this notation. We start by defining a wellordering on the collection of all non-decreasing sequences of members of ω which forms the backbone for the structure of the members in the spaces. Definition 2.1 For k 2, denote by ω k the collection of all non-decreasing sequences of members of ω of length less than or equal to k. Definition 2.2 (The well-order < lex ) Let (s 1,..., s i ) and (t 1,..., t j ), with i, j 1, be in ω k. We say that (s 1,..., s i ) is lexicographically below (t 1,..., t j ), written (s 1,..., s i ) < lex (t 1,..., t j ), if and only if there is a non-negative integer m with the following properties: (i) m i and m j; (ii) for every positive integer n m, s n = t n ; and (iii) either s m+1 < t m+1, or m = i and m < j. This is just a generalization of the way the alphabetical order of words is based on the alphabetical order of their component letters. Example 2.3 Consider the sequences (1, 2), (2), and (2, 2) in ω 2. Following the preceding definition we have (1, 2) < lex (2) < lex (2, 2). We conclude that (1, 2) < lex (2) by setting m = 0 in Definition 2.2; similarly, (2) < lex (2, 2) follows by setting m = 1, as any proper initial segment of a sequence is lexicographically below that sequence. Definition 2.4 (The well-ordered set (ω k, )) Set the empty sequence () to be the -minimum element of ω k ; so, for all nonempty sequences s in ω k, we have () s. In general, given (s 1,..., s i ) and (t 1,..., t j ) in ω k with i, j 1, define (s 1,..., s i ) (t 1,..., t j ) if and only if either

6 6 A. Arias, N. Dobrinen, G. Girón-Garnica and J. G. Mijares (1) s i < t j, or (2) s i = t j and (s 1,..., s i ) < lex (t 1,..., t j ). Notation. Since well-orders ω k in order-type ω, we fix the notation of letting s m denote the m-th member of (ω k, ). Let ω k denote the collection of all nondecreasing sequences of length k of members of ω. Note that also well-orders ω k in order type ω. Fix the notation of letting u n denote the n-th member of (ω k, ). For s, t ω k, we say that s is an initial segment of t and write s t if s = (s 1,..., s i ), t = (t 1,..., t j ), i < j, and for all m i, s m = t m. Recall the concatenation operation: Given sequences s = (s 1,..., s i ) and t = (t 1,..., t j ), s t denotes the concatenation of s and t, which is the sequence (s 1,..., s i, t 1,..., t j ) of length i + j. As is standard, for a natural number n, s n denotes the sequence (s 1,..., s i, n). Definition 2.5 (The spaces (E k,, r), k 2, Dobrinen [10]) An E k -tree is a function X from ω k into ω k that preserves the well-order and initial segments. For X an E k -tree, let X denote the restriction of X to ω k. The space E k is defined to be the collection of all X such that X is an E k -tree. We identify X with its range and usually will write X = {v 1, v 2,...}, where v 1 = X( u 1 ) v 2 = X( u 2 ). The partial ordering on E k is defined to be simply inclusion; that is, given X, Y E k, X Y if and only if (the range of) X is a subset of (the range of) Y. For each n < ω, the n-th restriction function r n on E k is defined by r n (X) = {v 1, v 2,..., v n } that is, the -least n members of X. When necessary for clarity, we write rn(x) k to highlight that X is a member of E k. We set AR k n := {r n (X) : X E k } and AR k := {r n (X) : n < ω, X E k } to denote the set of all n-th approximations to members of E k, and the set of all finite approximations to members of E k, respectively. Remark Let s ω k and denote its length by s. Since X preserves initial segments, it follows that X(s) = s. Thus, E k is the space of all functions X from ω k into ω k which induce an E k -tree. Notice that the identity function is identified with ω k and therefore ω k is regarded as a member of E k. It is of course the maximal member of E k : Every X E k satisfies X ω k. Notice also that ω k is an E k -tree, and this tree is induced by ω k. The set ω k is the prototype for all members of E k in the sense that every member X of E k will be a subset of ω k which has the same structure as ω k, according to the interaction between the two orders and. Definition 2.5 essentially is generalizing the key points about the structure, according to and, of the identity function on ω k.

7 Banach Spaces from Barriers in High Dimensional Ellentuck Spaces 7 The family of all non-empty finite subsets of ω k will be denoted by FIN(ω k ). Clearly, AR k FIN(ω k ). If E FIN(ω k ), then we denote the minimal and maximal elements of E with respect to by min (E) and max (E), respectively. Example 2.6 (The space E 2 ) The members of E 2 look like ω many copies of the Ellentuck space. The well-order (ω 2, ) begins as follows: () (0) (0, 0) (0, 1) (1) (1, 1) (0, 2) (1, 2) (2) (2, 2) The tree structure of ω 2, under lexicographic order, looks like ω copies of ω, and has order type the countable ordinal ω ω. Here, we picture the finite tree { s m : 1 m 21}, which indicates how the rest of the tree ω 2 is formed. This is exactly the tree formed by taking all initial segments of the set { u n : 1 n 15}. (0,0) (0,1) (0,2) (0,3) (0,4) (1,1) (1,2) (1,3) (1,4) (2,2) (2,3) (2,4) (3,3) (3,4) (4,4) (0) (1) (2) (3) (4) () Figure 1: Initial structure of ω 2. Next we present the specifics of the structure of the space E 3. Example 2.7 (The space E 3 ) The well-order (ω 3, ) begins as follows: () (0) (0, 0) (0, 0, 0) (0, 0, 1) (0, 1) (0, 1, 1) (1) (1, 1) (1, 1, 1) (0, 0, 2) (0, 1, 2) (0, 2) (0, 2, 2) (1, 1, 2) (1, 2) (1, 2, 2) (2) (2, 2) (2, 2, 2) (0, 0, 3) The set ω 3 is a tree of height three with each non-maximal node branching into ω many nodes. The maximal nodes in the following figure are technically the set { u n : 1 n 20}, which indicates the structure of ω 3. (0,0,0) (0,0,1) (0,0,2) (0,0,3) (0,1,1) (0,1,2) (0,1,3) (0,2,2) (0,2,3) (0,3,3) (1,1,1) (1,1,2) (1,1,3) (1,2,2) (1,2,3) (1,3,3) (2,2,2) (2,2,3) (2,3,3) (3,3,3) (0, 0) (0, 1) (0, 2) (0, 3) (1, 1) (1, 2) (1, 3) (2, 2) (2, 3) (3, 3) (0) (1) (2) (3) () Figure 2: Initial structure of ω 3.

8 8 A. Arias, N. Dobrinen, G. Girón-Garnica and J. G. Mijares 2.1 Upper Triangular Representation We now present an alternative and very useful way to visualize elements of E 2. This turned out to be fundamental to developing more intuition and to understanding the Banach spaces that we define in the following section. We refer to it as the upper triangular representation of ω 2 : Figure 3: Upper triangular representation of ω 2. The well-order (ω 2, ) begins as follows: (0, 0) (0, 1) (1, 1) (0, 2) (1, 2) (2, 2). In comparison with the tree representation shown in Figure 1, the upper triangular representation makes it simpler to visualize this well-order: Starting at (0, 0) we move from top to bottom throughout each column, and then to the right to the next column. The following figure shows the initial part of an E 2 -tree X. The highlighted pieces represent the restriction of X to ω 2. Figure 4: Initial part of an E 2 -tree. Under the identification discussed after Definition 2.5, we have that

9 Banach Spaces from Barriers in High Dimensional Ellentuck Spaces 9 X = {(2, 4), (2, 6), (6, 6), (2, 8), (6, 8), (9, 9), (2, 10),...} is an element of E 2. Using the upper triangular representation of ω 2 we can visualize r 10 (X): Figure 5: r 10(X) for a typical X E Special Maximal Elements of E k There are special elements in E k that are useful for describing the structure of some subspaces of the Banach spaces that we define in the following section. Given v ω k we want to construct a special Xv max E k that has v as its first element and that is maximal in the sense that every other Y E 2 with v as its -minimum member is a subset of X. For example, if v = (0, 4) ω 2, then a finite approximation of Xv max this: E 2 looks like

10 10 A. Arias, N. Dobrinen, G. Girón-Garnica and J. G. Mijares Figure 6: A finite approximation of X max (0,4) E 2. Now let us illustrate this with k = 3 and v = (0, 2, 7). Since we want v as the first element of Xv max, we identify it with (0, 0, 0) and then we choose the next elements as small as possible following Definition 2.5: () (0) (0, 2) (0, 2, 7) (0, 2, 8) (0, 8) (0, 8, 8) (8) (8, 8) (8, 8, 8) (0, 2, 9) (0, 8, 9) (0, 9) (0, 9, 9) (8, 8, 9) (8, 9) (8, 9, 9) (9) (9, 9) (9, 9, 9) (0, 2, 10) Therefore, under the identification discussed after Definition 2.5, we have X max v = {(0, 2, 7), (0, 2, 8), (0, 8, 8), (8, 8, 8), (0, 2, 9),...}. In general, for any k 2, X max v is constructed as follows. Definition 2.8 Let k 2 be given and suppose v = (n 1, n 2,..., n k ). First we define the E k -tree X v that will determine Xv max. X v must be a function from ω k to ω k satisfying Definition 2.5 and such that X v (0, 0,..., 0) = v. So, for m, j N +, j k, define the following auxiliary functions: f j (0) := n j and f j (m) := n k + m. Then, for t = (m 1, m 2,..., m l ) ω k set X v (t) := (f 1 (m 1 ), f 2 (m 2 ),..., f l (m l )). Finally, define Xv max to be the restriction of X v to ω k. It is routine to check that: Lemma 2.9 Let v = (n 1,..., n k ). Then X v is an E k -tree, and w = (m 1,..., m k ) ω k belongs to Xv max if and only if either m 1 > n k or else there is 1 i k such that (m 1, m 2,..., m i ) = (n 1, n 2,..., n i ), and if i < k then m i+1 > n k.

11 Banach Spaces from Barriers in High Dimensional Ellentuck Spaces 11 We use this Lemma to prove that Xv max initial value. contains all elements of AR k that have v as Proposition 2.10 Let E AR k and v = min (E). Then E X max v. Proof Let E AR k with v = min (E). Then there exists an E k -tree X such that X(0,..., 0) = (n 1,..., n k ) = v and E = r j (X) for some j. If E has more than one member, then its second element is X(0,..., 0, 1) = (n 1,..., n k 1, n k ), for some n k > n k. By Lemma 2.9, X(0,..., 0, 1) Xv max. Suppose that w = (p 1,..., p k ) is any member of E besides v. We will show that w Xv max. Let (m 1,..., m k ) ω k be the sequence such that w = X(m 1,..., m k ). Suppose first that m 1 > 0. Then (0,..., 0, 1) (m 1 ) (m 1, m 2,..., m k ). Applying X and recalling that X preserves and, we conclude that (n 1,..., n k 1, n k) (p 1 ) (p 1, p 2,..., p k ). Comparing the first two elements we see that either p 1 > n k or else p 1 = n k and p 1 > n 1. In both cases, p 1 n k > n k, which, by Lemma 2.9, gives that (p 1, p 2,..., p k ) = w X max v. Suppose now that m 1 = = m i = 0 and m i+1 > 0, where i + 1 k. Then (0,..., 0, 1) (m 1,..., m i, m i+1 ) = (0,..., 0, m i+1 ) (0,..., 0, m i+1,..., m k ). Applying X we obtain (n 1,..., n k 1, n k) (n 1,..., n i, p i+1 ) (n 1,..., n i, p i+1,..., p k ). Arguing as before, we conclude that p i+1 n k > n k, and then Lemma 2.9 yields that w X max v. Corollary 2.11 If w Xv max, then Xw max min (E) Xv max, then E Xv max. Xv max. In particular, if E AR k and Notice that the only elements in X v that are -smaller than v are the initial segments of v. Corollary 2.12 If s v and s ran( X v ) then s v.

12 12 A. Arias, N. Dobrinen, G. Girón-Garnica and J. G. Mijares 3 The Banach Spaces T k (d, θ) and T(A k d, θ) 3.1 Preliminary Definitions Set N := {0, 1,...} and N + := N\{0}. For the rest of this paper, fix d, k N +, k 2 and θ R with 0 < θ < 1. Given E, F FIN(ω k ), we write E < F (resp. E F) to denote that max (E) min (F) (resp. max (E) min (F)), and in this case we say that E and F are successive. Similarly, for v ω k, we write v < E (resp. v E) whenever {v} < E (resp. {v} E). By c 00 (ω k ) we denote the vector space of all functions x : ω k R such that the set supp (x) := {v ω k : x(v) 0} is finite. Usually we write x v instead of x(v). We can extend the orders defined above to vectors x, y c 00 (ω k ): x < y (resp. x y) iff supp (x) < supp (y) (resp. supp (x) supp (y)). From the notation in Section 2 we have that ω k = { u 1, u 2,...}. Denote by (e un ) n=1 the canonical basis of c 00 (ω k ). To simplify notation, we will usually write e n instead of e un. So, if x c 00 (ω k ), then x = n=1 x u n e un = m n=1 x u n e un for some m N +. Using the above convention, we will write x = n=1 x ne n = m n=1 x ne n. If E AR k, we put Ex := v E x ve v. 3.2 Construction of T k (d, θ) and T(A k d, θ) The Banach spaces that we introduce in this section have their roots (as all subsequent constructions [4], [25], [1], [19], [2]) in Tsirelson s fundamental discovery of a reflexive Banach space T with an unconditional basis not containing c 0 or l p with 1 p < [28]. Recall that in the construction of the Banach spaces T(S, θ) and T(A d, θ), the norm is determined using sequences of the form v 1 E 1 < v 2 E 2 < < v m E m, where m d and each set E i is some finite set of natural numbers. How one extends this method for constructing norms to high dimensional Ellentuck spaces depends on how one conceives of the sets {v 1,..., v m } and E i, 1 i m. The set {v 1,..., v m } may be thought of either as simply a set of cardinality m, or as an initial segment of a member of the barrier of rank d on the Ellentuck space. Likewise, one may think of the sets E i as simply finite sets, or as finite initial segments of members of the Ellentuck space. Thus, there are four natural ways to generalize the norm construction

13 Banach Spaces from Barriers in High Dimensional Ellentuck Spaces 13 to the k-dimensional Ellentuck space, E k, or for topological Ramsey spaces in general. Fixing d and letting m d, we may (a) let {v 1,..., v m } range over all m-sized subsets of ω k, or we may (b) restrict to those {v 1,..., v m } which are in AR k m. Likewise we may (i) simply let the sets E i range over all finite subsets of ω k, or (ii) we may restrict the E i to only range over members of AR k. Combination (a) and (i) yields the space T(A d, θ), so nothing new is gained by that approach. Combining (a) and (ii), we define the new space T k (d, θ). The space T(A k d, θ) is constructed using (b) and (ii). This is the most restrictive of the possible constructions. The space constructed using (b) and (i) was considered by the third author in his disseration [18]. Recalling the fact that (ω k, ) is a linear order with order type exactly that of the natural numbers, the classical notion of barrier on the natural numbers transfers to (ω k, ). We say that a subset B of FIN(ω k ) is a barrier on ω k if (a) for each infinite subset Y ω k, there is some E B which is an initial segment (in the ordering) of Y, and (b) for each pair E F in B, E F. Given a barrier B on (ω k, ), the set C 1 (B) = {F FIN(ω k ) : ( E B) F E} is a compact family on ω k. Notice that [ω k ] d is the barrier of rank d on (ω k, ), and that C 1 ([ω k ] d ) = [ω k ] d. The notion of barrier for abstract topological Ramsey spaces appears in Chapter 5 of [27]. For high dimensional Ellentuck spaces, the notion of barrier reduces to the following. Definition 3.1 A family B AR k is a barrier on E k if (a) For every Y X there exists n such that r n (Y) B; and (b) E F, for all E F in B. The barrier on E k of rank d is exactly AR k d, the set of all d-th approximations to members of E k. Barriers of infinite rank are defined recursively in a manner similarly to barriers of infinite rank on the natural numbers. However, as the precise definition takes several paragraphs and as infinite rank barriers are not used in this paper, we omit their definition here. Notation: Given a barrier B on E k, let C(B) = {F AR k : ( E B) F E}. This is the analogue for E k of a compact family determined by a barrier on the natural numbers, and we call C(B) a compact family on E k. Notice that C(AR k d ) = m d ARk m, which we denote as A k d.

14 14 A. Arias, N. Dobrinen, G. Girón-Garnica and J. G. Mijares Definition 3.2 Let F be a compact family on ω k or on E k. We say that (E i ) m AR k is F -admissible if and only if there exists {v 1, v 2,..., v m } F such that v 1 E 1 < v 2 E 2 < < v m E m. For x = n=1 x ne n c 00 (ω k ) and j N, we define a non-decreasing sequence of norms on c 00 (ω k ) as follows: x F 0 := max x n, n N + { m }} x F j+1 { x := max F j, θ max E i x F j : 1 m d, (E i ) m F-admissible. For fixed x c 00 (ω k ), the sequence ( x F j ) j N is bounded above by the l 1 (ω k )-norm of x. Therefore, we can set x F := sup x F j. j N We write Tk (d,θ) to denote the norm obtained using F = C 1([ω k ] d ), the compact family determined by the rank d barrier on (ω k, ), and write T(A k d,θ) to denote the norm obtained using F = C(AR k d ). Clearly T k (d,θ) and T(A k d,θ) are norms on c 00 (ω k ). Definition 3.3 The completion of c 00 (ω k ) with respect to the norm Tk (d,θ) is denoted by (T k (d, θ), ). Likewise, the completion of c 00 (ω k ) with respect to the norm T(A k d,θ) is denoted by (T(Ak d, θ), ). Notice that when k = 1, T 1 (d, θ) is the space Bellenot considered [4]. For v ω k and x T k (d, θ) we also write v < x whenever v < supp (x). From the preceding definition we have the following: Proposition 3.4 If F is a compact family on ω k supp (x) = n, then x F n = x F n+1 =. or on E k, x c 00 (ω k ) and Therefore, we conclude that for every x c 00 (ω k ) we have x F = max j N x F j.

15 Banach Spaces from Barriers in High Dimensional Ellentuck Spaces 15 Also, note that the formulas defining x F j do not depend on the signs of x, so the unit basis of c 00 (ω k ) is a 1-unconditional basic sequence, and by definition the unit basis generates T k (d, θ) and T(A k d, θ). Therefore we have the following: Proposition 3.5 (e n ) n=1 is a 1-unconditional basis of T k(d, θ) and of T(A k d, θ). Proposition 3.6 For x = n=1 x ne n T k (d, θ) it follows that { m }} x = max Tk(d,θ) { x, θ sup E i x : 1 m d, (E Tk(d,θ) i) m d-admissible, where x := sup n N + x n. Likewise for T(A k d, θ). 4 Subspaces of T k (d, θ) isomorphic to l N The Banach space T k (d, θ) lives in ω k, at the top of ω k. We will see that it s structure is determined by subspaces indexed by elements in the lower branches. Let s ω k. The tree generated by s and the Banach space associated to it are given by { } τ k [s] := v ω k : s v and T k [s] := span{e v : v τ k [s]}, respectively. In this section, let N N + and s 1,..., s N ω k be such that s 1 = = s N < k and s 1 s N. The following is a very useful result. Corollary 4.1 If v ω k satisfies s N v, then there is at most one i N such that Xv max τ k [s i ]. Proof Suppose that v = (n 1, n 2,..., n k ) and let w = (m 1, m 2,..., m k ) Xv max. It follows from Lemma 2.9 that either m 1 > n k or that there is 0 < i < k such that (m 1,..., m i ) = (n 1,..., n i ) and m i+1 > n k. If m 1 > n k, w does not belong to any τ k [s i ] because n k is greater than any coordinate of the s i s and as a result, none of the s i s can be an initial segment of w. Heence it follows from the other option of w that the the only way an s i is an initial segment of w is if s i is also an initial segment of v. Since all the s i s have the same length, at most one of them is an initial segment of v and the result follows.

16 16 A. Arias, N. Dobrinen, G. Girón-Garnica and J. G. Mijares It is useful to have an analogous result to the preceding corollary but related to approximations E AR k instead of special maximal elements of E k : Lemma 4.2 Suppose E AR k and set v := min (E). If s v and s v, then E τ k [s]. We will study the Banach space structure of the subspaces of T k (d, θ) of the form Z := T k [s 1 ] T k [s 2 ] T k [s N ]. Since (e un ) n=1 is 1-unconditional, we can decompose Z as F C, where (2) F = span{e v Z : v ω k, v s N } C = span{e v Z : v ω k, s N v}. and By setting k = 2, N = 4, s 1 = (4), s 2 = (6), s 3 = (8), and s 4 = (10), the following figure shows the elements of ω 2 used to generate the subspaces F (dashed outline) and C (thicker outline) in which we decompose the subspace T 2 [(4)] T 2 [(6)] T 2 [(8)] T 2 [(10)]: Figure 7: Elements of ω 2 used to generate T 2 [(4)] T 2 [(6)] T 2 [(8)] T 2 [(10)]. Applying Corollary 4.1 and Lemma 4.2 we have: Lemma 4.3 Let E AR k be such that s N min (E). If E[T k (d, θ)] := span{e w : w E}, then either E[T k (d, θ)] C =, or there is exactly one i N such that E[T k (d, θ)] C T k [s i ].

17 Banach Spaces from Barriers in High Dimensional Ellentuck Spaces 17 Proof Suppose that E[T k (d, θ)] C and set v := min (E). Then, by Corollary 4.1, there is exactly one i {1,..., N} such that Xv max τ k [s i ] ; consequently, s i v and E τ k [s j ] = for any j {1,..., N}, j i. By hypothesis, s i s N v. Applying Lemma 4.2 we conclude that E τ k [s i ]. Hence, E[T k (d, θ)] C T k [s i ]. This lemma helps us establish the presence of arbitrarily large copies of l N inside T k (d, θ): Theorem 4.4 Suppose that s 1 s 2 s N belong to ω <k and that s 1 = = s N < k. Let v ω k with s N v and suppose that x N Tk [s i ] satifies v < x. If we decompose x as x x N with x i T k [s i ], then max x θ(d 1) i x max 1 i N 1 θ x i. 1 i N In particular, if x 1 = = x N = 1, span{x 1,..., x N } is isomorphic to l N in a canonical way and the isomorphism constant is independent of N and of the x i s. Proof Since the basis of T k (d, θ) is unconditional, max 1 i N x i x. We will check the upper bound. Let m {1,..., d} and (E i ) m ARk be an admissible sequence such that x = θ m E ix. Without loss of generality we assume that E 1 x 0, so that s N min (E 2 ). By Lemma 4.3, when j 2, we have E j x = E j x i for some i {1,..., N}. Then it follows that E j x max 1 i N x i. Consequently, x θ E 1 x + θ(d 1) max 1 i N x i. Repeat the argument for E 1 x. Find m {1,..., d} and an admissible sequence (F i ) m ARk such that E 1 x = θ m F i(e 1 x). We can assume that F 1 (E 1 x) 0, and applying Lemma 4.3 once again we conclude that for j 2, F j (E 1 x) max 1 i N x i. Then, ( ) x θ θ F 1 (E 1 x) + θ(d 1) max x i + θ(d 1) max x i. 1 i N 1 i N Iterating this process we conclude that x n=1 θ n (d 1) max x θ(d 1) i max 1 i N 1 θ x i. 1 i N

18 18 A. Arias, N. Dobrinen, G. Girón-Garnica and J. G. Mijares Example 4.5 To illustrate the previous Theorem, suppose that k = 2, that s 1 < s 2 < < s N and that s i = 1 for every i N. If we use the upper triangular representation of ω 2, the s i s represent rows. Theorem 4.4 says that if the x i s are supported in the ith row, but starting after the s N th column, then span{x 1,..., x N } is isomorphic to l N, with isomorphism constant independent of N. The image below illustrates s 1 = 0, s 1 = 2, s 2 = 3 and s 4 = 6 and the shaded horizontal lines represent the support of the x i s. Figure 8: Elements of ω 2 that generate isomorphic copies of l N. In particular, each column of the upper triangular representation of ω 2 satisfies this condition. That is, span{e (i,n) : i n} C(θ,d) l n+1, where C(d, θ) is the constant from Theorem Block Subspaces of T k (d, θ) isomorphic to l p For the rest of this paper suppose that dθ > 1 and let p (1, ) be determined by the equation dθ = d 1/p. Bellenot proved that T 1 (d, θ) is isomorphic to l p (see Theorem 1.2). The same result was then proved by Argyros and Deliyanni in [1] with different arguments which can be extended to more general cases like ours. In this section we show that we can find many copies of l p spaces inside T k (d, θ) for k 2: Theorem 5.1 Suppose that (x i ) is a normalized block sequence in T k(d, θ) and that we can find a sequence (v i ) ω k such that:

19 Banach Spaces from Barriers in High Dimensional Ellentuck Spaces 19 (1) v 1 x 1 < v 2 x 2 < v 3 x 3 < v 4 x 4 < (2) supp (x i ) X max v i and v i+1 X max v i for every i 1. Then, (x i ) is equivalent to the basis of l p. Notice that Corollary 2.11 implies that for every j i, v j X max v i and supp ( x j ) X max v i. Theorem 5.1 allows us to identify natural subspaces of T k (d, θ) isomorphic to l p. For example, it implies that the top trees of T k (d, θ) are isomorphic to l p. In section 8 we will see that the top trees are isometrically isomorphic to T 1 (d, θ). Corollary 5.2 If s ω k and s = k 1, then T k [s] is isomorphic to l p. Proof Suppose that s = (s 1,..., c k 1 ). Define v 1 = (s 1,..., c k 1, c k 1 ), v 2 = (s 1,..., c k 1, c k 1 + 1), v 3 = (s 1,..., c k 1, c k 1 + 2), and let x i = e vi. If follows from Lemma 2.9 that the (v i ) s and (x i ) s satisfy the hypothesis of Theorem 5.1 and the result follows. The diagonal subspaces of T k (d, θ) are also isomorphic to l p spaces. For s = (s 1,..., c l ) ω k with l < k, define v 1 = (s 1,..., c l, c l,..., c l ) v 2 = (s 1,..., c l, c l + 1,..., c l + 1) v 3 = (s 1,..., c l, c l + 2,..., c l + 2).. and x i = e vi for i 1. Then we use Lemma 2.9 to verify that the (v i ) s and (x i ) s satisfy the hypothesis of Theorem 5.1 and we conclude Corollary 5.3 D[s] = span{e vi : i 1} is isomorphic to l p, for every s ω <k We prove Theorem 5.1 in two steps. First we prove the lower l p -estimate using Bellenot s space T 1 (d, θ). Then we prove the upper l p -estimate in a more general case. Denote by (t i ) the canonical basis of T 1 (d, θ). In order to avoid confusion, we will write 1 to denote the norm on T 1 (d, θ). Proposition 5.4 Under the same hypotheses of Theorem 5.1, for a finitely supported z = i a ix i T k (d, θ), we have: 1 2d ( i a i p) 1/p a 1 it i a Tk ix i. i i (d,θ)

20 20 A. Arias, N. Dobrinen, G. Girón-Garnica and J. G. Mijares Proof Let x = i a it i T 1 (d, θ). Following Bellenot [4], either x 1 = max i a i, or there exist m {1,..., d} and E 1 < E 2 < < E m such that x 1 = m j=1 θ E jx 1. For each j {1,..., m}, either E j x 1 = max i { a i : i E j }, or there exist m {1,..., d} and E j1 < E j2 < < E jm subsets of E j such that E j x 1 = m l=1 θ E jlx 1. Since the sequence (a i ) has only finitely many non-zero terms, this process ends and x is normed by a tree. We will prove the result by induction on the height of the tree. If x 1 = max i a i, the result follows. Since the basis of T k (d, θ) is unconditional and the (x i ) s are a normalized block basis of T k (d, θ), we have that max i a i i a ix i. Suppose that the result is proved for elements of T 1 (d, θ) that are normed by trees of height less than or equal to h and that x is normed by a tree of height h + 1. Then, there exist m {1,..., d} and E 1 < E 2 < < E m such that x 1 = m j=1 θ E jx 1 and each E j x is normed by a tree of height less than or equal to h. We will find a corresponding admissible sequence in AR k. For each j {1,..., m}, let n j = min(e j ) and define F j = {w X max v nj : w v nj+1 } AR k. Then F 1 < F 2, < F m, and we conclude that (F j ) is d-admissible. Moreover we easily check from the hypothesis of Theorem 5.1 and by Corollary 2.11 that if i E j, then supp (x i ) F j. Hence, using the induction hypothesis and the unconditionality of the basis of T k (d, θ) we conclude that E j x 1 = a l t l l E j a l x l l E 1 j F jz. Therefore, i a it i 1 = m θ E j x 1 j=1 The result follows now applying Theorem 1.2. m θ F j z = a ix i. i The proof of the upper bound inequality of Theorem 5.1 is harder and we need some preliminary results. j=1 5.1 Alternative Norm To establish a upper l p -estimate we will adapt an alternative and useful description of the norm on T 1 (d, θ) introduced by Argyros and Deliyanni [1] to our spaces. In that

21 Banach Spaces from Barriers in High Dimensional Ellentuck Spaces 21 regard, the following definition plays a key role. Definition 5.5 Let m {1,..., d}. A sequence (F i ) m FIN(ω k ) is called almost admissible if there exists a d-admissible sequence (E n ) m n=1 in ARk such that F i E i, for i m. A standard alternative description of the norm of the space T k (d, θ), closer to the spirit of Tsirelson space, is as follows. Let K 0 := {±e i : i N + }, and for n N, K n+1 := K n {θ(f1 + + f m ) : m d, (f i ) m K n }, where (supp (f i )) m is almost admissible. Then, set K := n N K n. Now, for each n N and fixed x c 00 (ω k ), define the following non-decreasing sequence of norms: x n := max {f (x) : f K n}. Lemma 5.6 For every n N and x c 00 (ω k ) we have x n = x n. Proof Clearly, x 0 = x 0 for every x c 00(ω k ). So, let n N +. Suppose y j = y j for every j N, j < n and every y c 00 (ω k ). If x n = x n 1, then x n = x n 1 x n. Suppose x n x n 1. Let m {1,..., d} and (E i ) m ARk be an admissible sequence such that x n = θ m E ix n 1. Then, x n = θ m E ix n 1 = θ m f i(e i x) for some (f i ) m K n 1. Define, for each i {1,..., m}, a new functional f i satisfying f i (y) = f i (E i y) for every y c 00 (ω k ). This implies that supp ( f i ) ( ) = supp f i Ei. Then, (f i ) m K n 1 with (supp ( f i ) ) m almost admissible and f i (E i x) = f i (E i x). So, θ therefore, x n x n. m f i (E i x) = θ m f i (E i x) E i x n x n ; Now, let f = θ(f f m ) for some m {1,..., d} and (f i ) m K n 1 with (supp (f i )) m almost admissible. Then, f (x) = θ m f i (x) θ m m supp (f i ) x n 1 = θ supp (f i ) x n 1. Since (supp (f i )) m is almost admissible, there exists an admissible sequence (E i) d AR k such that supp (f i ) E ni, where n 1,..., n m {1,..., m} and n 1 < < n m.

22 22 A. Arias, N. Dobrinen, G. Girón-Garnica and J. G. Mijares So, m m θ supp (f i ) x n 1 θ E ni x n 1 x n ; hence, by definition of n, we conclude that x n x n. Consequently, an alternative description of the norm on T k (d, θ) is: Proposition 5.7 For every x T k (d, θ), x = sup {f (x) : f K}. 5.2 Upper Bound for Theorem 5.1 For m {1,..., d} we say that f 1,..., f m K are successive if supp (f 1 ) < supp (f 2 ) < < supp (f m ). If f K, then there exists n N such that f K n. The complexity of f increases as n increases. That is to say, for example, that the complexity of f K 1 is less than that of g K 10. This is captured in the following definition. Definition 5.8 Let n N + and φ K n \ K n 1. An analysis of φ is a sequence (K l (φ)) n l=0 of subsets of K such that: (1) K l (φ) consists of successive elements of K l and f K l (φ) supp (f ) = supp (φ). (2) If f K l+1 (φ), then either f K l (φ) or there exist m {1,..., d} and successive f 1,..., f m K l (φ) with (supp (f i )) m almost admissible and f = θ(f f m ). (3) K n (φ) = {φ}. Note that, by definition of the sets K n, each φ K has an analysis. Moreover, if f 1 K l (φ) and f 2 K l+1 (φ), then either supp (f 1 ) supp (f 2 ) or supp (f 1 ) supp (f 2 ) =. Let φ K n \ K n 1 and let (K l (φ)) n l=0 be a fixed analysis of φ. Suppose (x j) N j=1 is a finite block sequence on T k (d, θ). Following [1], for each j {1,..., N}, set l j {0,..., n 1} as the smallest integer with the property that there exists at most one g K lj +1(φ) with supp ( x j ) supp (g).

23 Banach Spaces from Barriers in High Dimensional Ellentuck Spaces 23 Then, define the initial part and final part of x j with respect to (K l (φ)) n l=0, and denote them respectively by x j and x j, as follows. Let { f Klj (φ) : supp (f ) supp ( ) } x j = {f1,..., f m }, where f 1,..., f m are successive. Set x j = (supp (f 1 ))x j and x j = ( m i=2 supp (f i ))x j. The following is a useful property of the sequence (x j )N j=1 property is true for (x j )N j=1. (see [5]). The analogous Proposition 5.9 For l {1,..., n} and j {1,..., N}, set A l (x j) := { f K l (φ) : supp (f ) supp ( x j) }. Then, there exists at most one f A l (x j ) such that supp (f ) supp ( x i) for some i j. Proof Let A l (x j ) = {f 1,..., f m }, where m 2 and f 1,..., f m are successive. Obviously, only supp (f 1 ) and supp (f m ) could intersect supp ( x i) for some i j. We will prove that it is not possible for f m. Suppose, towards a contradiction, that supp (f m ) supp ( x i) for some i > j. Given that m 2, we must have l l j. Consequently, there exists g K lj (φ) such that supp (f m ) supp (g). Since supp (g) supp ( ) x j and supp (g) supp (xi ) for some i > j, the definition of x j implies that supp (g) supp ( ) ( ) x j supp x j. Therefore, supp (f m ) supp ( x j) =, a contradiction. Following [3] and [5] we now provide an upper l p -estimate that implies the upper l p -estimate of Theorem 5.1: Proposition 5.10 Let (x j ) N j=1 be a finite normalized block basis on T k(d, θ). Denote by (t n ) n=1 the canonical basis of T 1(d, θ). Then, for any (a j ) N j=1 R, we have: N 1/p a j x j 2 N a j p θ j=1 j=1

24 24 A. Arias, N. Dobrinen, G. Girón-Garnica and J. G. Mijares Proof In order to avoid confusion, we will write 1 to denote the norm on T 1 (d, θ). By Proposition 5.7 and Theorem 1.2 it suffices to show that for every φ K, N φ a j x j 2 N θ a j t j. 1 j=1 By unconditionality we can assume that x 1,..., x N and φ are positive. Suppose φ K n \ K n 1 for some n N +, and let (K l (φ)) n l=0 be an analysis of φ (see Definition 5.8). Next, split each x j into its initial and final part, x j and x j, with respect to (K l (φ)) n l=0. We will show by induction on l {0, 1,..., n} that for all J {1,..., N} and all f K l (φ) we have f a j x j j J 1 θ a j t j and j J f a j x j j J 1 θ a j t j. j J 1 1 We prove the first inequality given that the other one is analogous. Let J {1,..., N} and set y = j J a jx j. j=1 If f K 0 (φ), then f = e i for some i N +. We want to prove that e i (y) 1 θ a jt j. j J 1 Suppose that e i (y) 0. So, there exists exactly one j i J such that e i (x j i ) 0. Applying Proposition 5.7 we have e i (y) = e i (a ji x j i ) a ji x j i a ji x ji a jt j j J 1 since the basis of T k (d, θ) is unconditional, x ji = 1, and by definition max a j a jt j. j J j J 1 Now suppose that the desired inequality holds for any g K l (φ). We will prove it for K l+1 (φ). Let f K l+1 (φ) be such that f = θ(f f m ), where f 1,..., f m are successive elements in K l (φ) with (supp (f i )) m almost admissible. Then, 1 m d. Without loss of generality assume that f i (y) 0 for each i {1,..., m}. Define the following sets:

25 Banach Spaces from Barriers in High Dimensional Ellentuck Spaces 25 and I := { i m : j J with f i (x j) 0 and supp (f ) supp ( x j ) supp (fi ) } J := { j J : i {1,..., m 1} such that f i (x j) 0 and f i+1 (x j) 0 }. We claim that I + J m. Indeed, if j J, there exists i {1,..., m 1} such that f i (x j ) 0 and f i+1(x j ) 0. From the proof of Proposition 5.9 it follows that f i+1 (x h ) = 0 for every h j, which implies that i + 1 / I. Hence, we can define an injective map from J to {1,..., m} \ I and we conclude that I + J m. Finally, for each i I, set D i := { j J : supp (f ) supp ( x j) supp (fi ) }. Notice that for all i I we have D i J =. Then, [ ( ) f (y) = θ f i I i a j x j D j + ] f (a jx i j J j), and consequently [ i I fi ( ) f (y) θ a j x j D j + f (a j x i j J j) ]. However, by the induction hypothesis, ) f i a j x ( j D j 1 a j t j. i θ j D i 1 Moreover, for each j J, we have f (a j x j ) a j x j a j x j = a j = a j t j 1. Hence, [ 1 f (y) θ a j t j + θ i I 1 ] j D i 1 θ a jt j j J 1 [ 1 Di ( ) = θ θ i I a jt j + 1 ] j J 1 θ a jt j j J 1. Given that for every i I, D i J = and I + J m d, the family {D i } i I {{j}} j J is d-admissible in AR 1. So, by the definition of 1, we have [ 1 Di ( ) f (y) θ θ i I a jt j + 1 ] j J 1 θ a jt j j J 1 1 θ a jt j. j J 1

26 26 A. Arias, N. Dobrinen, G. Girón-Garnica and J. G. Mijares 6 T k (d, θ) is l p -saturated In this section we prove that every infinite dimensional subspace of T k (d, θ) has a subspace isomorphic to l p. Recall that the subspaces T k [s] for s ω k with s < k decompose naturally into countable sums. Namely, if s = (a 1, a 2,..., a l ) ω k and l < k, then τ k [s] = j=a l τ k [s j], and therefore T k [s] = j=a l T k [s j]. The next lemma tells us that we can find elements v τ k [s] such that Xv max contains arbitrary tails of the decomposition of τ k [s]. Its proof follows from the definition of the E k -tree X v that determines Xv max (see paragraph preceding Lemma 2.9). Lemma 6.1 Let s = (a 1, a 2,..., a l ) ω k with l < k. If m N with m > a l and v = s (m, m,..., m) ω k, then Xv max τ k [s] = j=m τ k [s j]. We now present the main result of this section: Theorem 6.2 Suppose that Z is an infinite dimensional subspace of T k (d, θ). Then, there exists Y Z isomorphic to l p. Proof Let Z be an infinite dimensional subspace of T k (d, θ). After a standard perturbation argument, we can assume that Z has a normalized block basic sequence (x n ). We will show that a subsequence of (x n ) is isomorphic to l p. From Proposition 5.10 we have that a 2 ( nx n n θ a n p) 1/p. n To obtain the lower bound we will find a subsequence and a projection Q onto a subspace of the form T k [s] such that ( Q ( x nj )) has a lower lp -estimate. To this end, assume that Z T k [s] for some s ω k with s < k. Decompose T k [s] = j=1 Tk [s j ], where for each j N +, s s j, s j = s + 1, and s j s j+1. For each j N + let Q j : T k [s] T k [s j ] be the projection onto T k [s j ]. Then we have the two cases: Case 1: j N +, Q j x n 0. Case 2: j 0 N + such that Q j0 x n 0.

27 Banach Spaces from Barriers in High Dimensional Ellentuck Spaces 27 Let us look at Case 1 first. Let v 1 be the first element of τ k [s]. Since there exists p 1 such that supp (x 1 ) p 1 j=1 τ k [s j ], applying Lemma 6.1 we can find q 1 > p 1 and v 2 τ k [s] such that v 1 x 1 < v 2 and Xv max 2 τ k [s] = j=q 1 τ k [s j ]. Since Q j x n 0 for 1 j q 1 we can find n 2 > 1 and y 2 T k [s] such that y 2 x n2 and Q j y 2 = 0 for 1 j q 1. Then we have v 1 x 1 < v 2 < y 2 and supp (y 2 ) X max v 2. We now repeat the argument. Since there exists p 2 such that supp (y 2 ) p 2 j=1 τ k [s j ], applying Lemma 6.1 we can find q 2 > p 2 and v 3 τ k [s] such that v 2 < y 2 < v 3 and τ k [s] = j=q 2 τ k [s j ]. Since Q j x n 0 for 1 j q 2, we can find n 3 > n 2 and y 3 T k [s] such that y 3 x n3 and Q j y 3 = 0 for 1 j q 2. Then we have X max v 3 v 1 x 1 < v 2 < y 2 < v 3 < y 3 and supp (y 2 ) Xv max 2, supp (y 3 ) Xv max 3. Proceeding this way we find a subsequence (x ni ) and a sequence (y i ) such that y i is close enough to x ni. Consequently, span{y i } span{x ni } and v 1 x 1 < v 2 < y 2 < v 3 < y 3 < and supp (y i ) X max v i for i > 1. By Proposition 5.4, there exist C 1, C 2 R such that i a ix ni C1 i a iy ni C2 ( i a i p) 1/p. Let us look at Case 2 now. Find a subsequence (n i ) and δ > 0 such that δ Q j0 x ni 1. Let W = span{q j0 x ni }. We now apply the argument in Case 1 to the sequence Q j0 x n1 < Q j0 x n2 < Q j0 x n3 <. That is, first decompose T k [s j0 ] = j=1 Tk [t j ], where for every j N +, s j0 t j, t j = s j0 + 1, t j t j+1. Then, look at the two cases for the sequence ( ) ( ) Q j0 x ni. If Case 1 is true, Qj0 x ni has a subsequence with a lower l p -estimate, and therefore ( ) x ni has a subsequence with a lower lp -estimate; and if Case 2 is true, we can repeat the argument for some t j that has length strictly larger than the length of s j0. If Case 1 continues to be false, after a finite number of iterations of the same argument, the length of t j will be equal to k 1, and therefore, applying Corollary 5.2, T k [t j ] would be isomorphic to l p. The result follows.

28 28 A. Arias, N. Dobrinen, G. Girón-Garnica and J. G. Mijares 7 The spaces T k (d, θ) are not isomorphic to each other In this section we prove one of the main results of the paper. Theorem 7.1 If k 1 k 2, then T k1 (d, θ) is not isomorphic to T k2 (d, θ). The proof goes by induction and it shows that when k 1 > k 2, T k1 (d, θ) does not embed in T k2 (d, θ). The idea is that if we had an isomorphic embedding, we would map an l N -sequence into an l N p -sequence for arbitrarily large N s. The induction step requires a stronger and more technical statement that appears in Proposition 7.4 below. The proof uses the notation of the trees τ k [s] and their Banach spaces T k [s] (see Section 4). We start with some lemmas. The first one is an easy consequence of the fact that the basis of T k (d, θ) is 1-unconditional. Lemma 7.2 If s ω k, and s < k, there exist s 1 s 2 s 3 such that s i = s + 1 and τ k [s] = τ k [s i ]. Consequently, we decompose T k [s] = Tk [s i ] and for m N +, there is a canonical projection P m : T k [s] m Tk [s i ]. Proof If s = (a 1,..., a l ), then s 1 = (a 1,..., a l, a l ), s 2 = (a 1,..., a l, a l + 1), s 3 = (a 1,..., a l, a l + 2),. Lemma 7.3 Let s ω k with s < k. Let v = min τ k [s]. Then τ k [s] X max v. Proof If s = (a 1,..., a l ), then we have that v = (a 1,..., a l, a l,..., a l ) and the result follows from Lemma 2.9. We are ready to state and prove the main proposition. Proposition 7.4 Let s ω k 1 with s < k 1 and decompose T k 1[s] = Tk 1[s i ] according to Lemma 7.2. Let M N + and t 1,..., t M ω k 2 such that t 1 = = t M < k 2. If k 1 s > k 2 t 1, then for every n N +, i=n Tk 1[s i ] does not embed into T k 2[t 1 ] T k 2[t M ]. Proof We proceed by induction. For the base case we assume that k 2 t 1 = 1 < k 1 s. By Corollary 5.2, T k 2[t i ] is isomorphic to l p, and consequently so is T k 2[t 1 ] T k 2[t M ]. On the other hand, Theorem 4.4 guarantees that T k 1[s] has arbitrarily large copies of l N.

29 Banach Spaces from Barriers in High Dimensional Ellentuck Spaces 29 Suppose now that the result is true for m N + and let k 2 t 1 = m + 1 < k 1 s. We will show a simpler case first, when M = 1. Suppose, towards a contradiction, that there exists n N + and an isomorphism Φ : T k 1 [s i ] T k 2 [t 1 ]. i=n Decompose T k 2[t 1 ] = j=1 Tk 2[r j ] according to Lemma 7.2. Find N large enough and v ω k 1 such that s n s n+1 s n+n 1 v. We will find normalized x 1 T k 1[s n ], x 2 T k 1[s n+1 ],..., x N T k 1[s n+n 1 ] such that v < x i for i N and we will use Theorem 4.4 to conclude that span{x 1,..., x} l N. Recall that the isomorphism constant is independent of N and of the x i s. Let v 1 be the first element of τ k 2[t 1 ], and let x 1 T k 1[s n ] be such that x 1 = 1 and v < x 1. Find a finitely supported y 1 T k 2[t 1 ] such that y 1 Φ(x 1 ). Applying Lemma 6.1 we can find v 2 τ k 2[t 1 ] such that v 1 y 1 < v 2 and Xv max 2 τ k 2[t 1 ] = j=m 1 +1 τ k 2[r j ] for some m 1 N. Since k 1 s n+1 > k 2 r 1 = m we can apply the induction hypothesis. In particular, the map P m1 Φ T k 1[sn+1 ] : Tk 1 [s n+1 ] T k 2 [r 1 ] T k 2 [r m1 ]. is not an isomorphism. As a result, there exists x 2 T k 1[s n+1 ] such that x 2 = 1 and P m1 Φ(x 2 ) 0. To add the property v < x 2, we decompose T k 1[s n+1 ] = Tk 1[u i ] as in Lemma 7.2 and apply the induction hypothesis to i=p Tk 1[u i ] for p large enough. Now that we have a normalized x 2 T k 1[s n+1 ] that satisfies v < x 2 and P m Φ(x 2 ) 0, we find a finitely supported y 2 T k 2[t 1 ] such that y 2 Φ(x 2 ) and P m1 y 2 = 0. Notice that v 1 y 1 < v 2 < y 2 and that Lemma 7.3 gives that supp (y 2 ) X max v 2. We now repeat the argument. Use Lemma 6.1 to find v 3 τ k 2[t 1 ] such that y 2 < v 3 and Xv max 3 τ k 2[t 1 ] = j=m 2 +1 τ k 2[r j ]. Then we find a normalized x 3 T k 1[s n+2 ] such that v < x 3 and P m2 Φ(x 3 ) is essentially zero. Finally, we find a finitely supported y 3 T k 2[t 1 ] such that y 3 Φ(x 3 ) and P m2 y 3 = 0. Proceeding this way, for every i N, we find normalized x i T k 1[s n+i 1 ] with v < x i, v i ω k 2, and y i T k 2[t 1 ] such that y i Φ(x i ) and v 1 y 1 < v 2 < y 2 < < v N < y N and v i+1, supp (y i ) X max v i.

30 30 A. Arias, N. Dobrinen, G. Girón-Garnica and J. G. Mijares By Theorem 5.1, (y i ) N is isomorphic to the canonical basis of ln p. Hence, Φ maps l N isomorphically into l N p. Since N is arbitrary, this contradicts that Φ is continuous (see equation 3 below) and we conclude the case M = 1. Let M > 1 and suppose, towards a contradiction, that there exists n N + and an isomorphism Φ : T k 1 [s i ] T k 2 [t 1 ] T k 2 [t 2 ] T k 2 [t M ]. i=n For each j N + let Q j : M Tk 2[t i ] T k 2[t j ] be the canonical projection. Decompose T k 2[t j ] = Tk 2[ri] j as in Lemma 7.2 and for each m N +, let P j m : T k 2[t j ] m Tk 2[ri] j be the canonical projection onto the first m blocks. The proof is similar to the case M = 1. Find N large enough and v ω k 1 such that s n s n+1 s n+n 1 v. Find x 1 T k 1[s n ] such that x 1 = 1 and v < x 1 and find a finitely supported y 1 T k 2[t 1 ] T k 2[t M ] such that y 1 Φ(x 1 ). For each j M, let v j 1 = min (τ k 2[t j ]). Use Lemma 6.1 to find v j 2 τ k 2[t j ] such that Q j (y 1 ) < v j 2 and Xmax v 2 τ k 2[t j ] = i=m j 1 +1 τ k 2[ri] j for some m j 1 N. Let P 1 = M j=1 Pj be the projection onto the first blocks of each of the T k 2[t m j j ] s. 1 Since k 1 s n+1 > k 2 r 1 = m we can apply the induction hypothesis. In particular, the map P 1 Φ T k 1[sn+1 ] is not an isomorphism and we can find x 2 T k 2[s n+1 ] such that x 2 = 1 and P 1 Φ(x 2 ) 0. Arguing as in the case M = 1, we can also assume that v < x 2. We then find a finitely supported y 2 M j=1 Tk 2[t j ] such that y 2 Φ(x 2 ) and P 1 y 2 = 0. Proceeding this way, for every i N, we find normalized x i T k 1[s n+i 1 ] with v < x i and y i M j=1 Tk 2[t j ] such that y i Φ(x i ). Moreover, for every j M, we can find v j i ω k 2 such that v j 1 Q j(y 1 ) < v j 2 < Q j(y 2 ) < < v j N < Q j(y N ) and v j i+1, supp ( Q j (y i ) ) X max. v j i By Theorem 5.1, there exists C 1 > 0 independent of N such that for every j M, ( N ) 1 ( p 1 Q j (y i ) p N N ) 1 p Q j (y i ) C 1 Q j (y i ) p. C 1

31 Banach Spaces from Barriers in High Dimensional Ellentuck Spaces 31 Using the triangle inequality for y i = ( M j=1 Q N j(y i ), Holder s inequality a i N 1/q ( N a i p ) 1/p, 1 p + 1 q = 1 ), Theorem 5.1, and the fact that the projections Q j are contractive, we get (3) N y i N M Q j (y i ) N 1/q j=1 M j=1 ( N ) 1/p Q j (y i ) p M C 1 N 1/q N M Q j (y i ) C 1N 1/q N y i j=1 j=1 ( = C 1 N 1/q N N ) M y i C 1 N 1/q M Φ x i C 1 N 1/q N M Φ x i. Since N is arbitrary, N y i is of order N, and N x i stays bounded, we see that Φ cannot be bounded, contradicting our assumption. 8 T k (d, θ) embeds isometrically into T k+1 (d, θ) For fixed d and θ, the spaces T k (d, θ), k 1, form a natural hierarchy in complexity over l p. In this section we prove that when j < k, the Banach space T j (d, θ) embeds isomorphically into T k (d, θ). The basis for these results is the special feature that the j-dimensional Ellentuck space E j embeds into the k-dimensional Ellentuck space E k in many different ways. First, (ω j, ) embeds into (ω k, ) as a trace above any given fixed stem of length k j in ω k. Second, (ω j, ) also embeds into (ω k, ) as the projection of each member in ω k to its first j coordinates. There are many other ways to embed (ω j, ) into (ω k, ), and each of these embeddings will induce an embedding of T j (d, θ) into T k (d, θ), as these embeddings preserve both the tree structure and the order. This is implicit in the constructions of the spaces E k in [10] and explicit in the recursive construction of the finite and infinite dimensional Ellentuck spaces in [11]. The following notation will be useful. Let Φ : ω k ω k+1 be defined by Φ(v) = (0) v, where is the concatenation operation. One can easily check that Φ preserves

32 32 A. Arias, N. Dobrinen, G. Girón-Garnica and J. G. Mijares and. We can naturally extend the definition of Φ to the finitely supported vectors of T k (d, θ) by Φ ( i a ie vi ) = i a ie Φ(vi). Lemma 8.1 Let E AR k and suppose that v = min E. Then there exists F AR k+1 such that φ(e) F, min (Φ(E)) = min (F) and max (Φ(E)) = max (F). Proof Let v = min (E). Proposition 2.10 says that every w E belongs to Xv max. Since Φ adds a 0 at the beginning of each sequence, the characterization of Lemma 2.9 implies that for every w E, Φ(w) belongs to XΦ(v) max defined by F = {s Xmax X max Φ(v) Lemma.. Then the initial segment of Φ(v) : s max Φ(E)} satisfies all the conditions of the Corollary 8.2 Let x T k (d, θ) be finitely supported. Then Φ(x) Tk+1 (d,θ) x Tk (d,θ). Proof We use induction over the length of the support of x. If supp (x) = 1 the two norms are equal. Then we assume that the result is true for all vectors of T k (d, θ) that have fewer than n elements in their support and we take x T k (d, θ) with supp (x) = n. If x Tk (d,θ) = x c0, the result is obviously true. If x Tk (d,θ) > x c0, there exist E 1,..., E d AR k such that E 1 E 2 E d and x = θ d E ix Tk (d,θ). Notice that this implies that supp (E i x) < n for every i d. By Lemma 8.1, there are F 1,..., F d AR k+1 such that F 1 F 2 F d and Φ(E i ) F i for i d. Since E i x Tk (d,θ) Φ(E i x) Tk+1 (d,θ) = Φ(E i )Φ(x) Tk+1 (d,θ) F i Φ(x) Tk+1 (d,θ) we conclude that x Tk (d,θ) = θ d E i x Tk (d,θ) θ d F i Φ(x) Tk+1 (d,θ) Φ(x) Tk+1 (d,θ). To prove the reverse inequality, the following notation will be useful. For n ω, let tr n : ω k+1 ω k be defined by tr n (n 1, n 2,..., n i ) = (n 2,..., n i ) if n 1 = n and tr n (n 1, n 2,..., n i ) = if n 1 n. If E ω k+1, tr n (E) = {tr n (v) : v E, tr n (v) }. Notice that tr n (E) = if for every v E, tr n (v) =.

33 Banach Spaces from Barriers in High Dimensional Ellentuck Spaces 33 Lemma 8.3 Let x = N a ie vi T k+1 (d, θ). Given F 1 F m, m d an admissible sequence for T k+1 (d, θ), there is an admissible sequence E 1 E m for T k (d, θ) such that each E i = tr 0 (F i ) and Φ(E i x) = F i y. Proof We will prove that if F AR k+1 and FΦ(x) 0, then E = tr 0 (F) AR k and Φ(Ex) = FΦ(x). The rest of the lemma follows easily from this. Let F AR k+1 with FΦ(x) 0. Then there exists an E k+1 -tree X such that F is an initial segment of X. Since FΦ(x) 0, the first element of F is of the form (0, n 2,..., n k+1 ) for some n 2 n k+1. Define Ŷ : ω k ω k by Ŷ(v) = tr 0 ( X ( (0) v )). Since X preserves and, it follows that Ŷ preserves those orders as well and hence Ŷ is an E k -tree. Since Ŷ preserves, it follows that E = tr 0 (F) is an initial segment of Y (i.e., E AR k ). Since (0) v F iff v E, we have FΦ(x) = ( ) a i e Φ(vi) = Φ a i e i = Φ(Ex). (0) v i F a i e Φ(vi) = v i E To complete the proof of the Lemma, suppose that F 1 F m, m d is an admissible sequence with E i = tr 0 (F i ) for i m. Let i < j, v E i and w E j. Then (0) v F i and (0) w F j, which implies that (0) v (0) w. And from here we conclude that v w. Since the elements are arbitrary we have that E 1 E m. Corollary 8.4 Let x T k (d, θ) be finitely supported. Then Φ(x) Tk+1 (d,θ) x Tk (d,θ). v i E Proof We use induction over the length of the support of x. If supp (x) = 1 the two norms are equal. Then we assume that the result is true for all vectors of T k (d, θ) that have fewer than n elements in their support and we take x T k (d, θ) with supp (x) = n. If Φ(x) Tk+1 (d,θ) = Φ(x) c0, the result is obviously true. If Φ(x) Tk+1 (d,θ) > Φ(x) c0, there exist F 1,..., F m AR k+1, m d, such that F 1 F m and Φ(x) Tk+1 (d,θ) = θ m F iφ(x) Tk+1 (d,θ). Notice that this implies that for i m, supp (F i Φ(x)) < n. Moreover, we can assume that F i Φ(x) 0. By Lemma 8.3, there are E 1,..., E m AR k such that E 1 E m and F i Φ(E i ) = Φ(E i x). By the induction hypothesis, Φ(E i x) Tk+1 (d,θ) E i x Tk (d,θ). Then

34 34 A. Arias, N. Dobrinen, G. Girón-Garnica and J. G. Mijares m m Φ(x) Tk+1 (d,θ) = θ F i Φ(x) Tk+1 (d,θ) θ E i x Tk (d,θ) x Tk (d,θ). Combining the previous corollaries, we obtain the following result: Theorem 8.5 Φ : T k (d, θ) T k+1 (d, θ) is an isometric isomorphism. Iterating this theorem, and using the notation of the beginning of Section 4 we describe isometrically all subspaces of T k (d, θ) of the form T k [s] for s ω k and s < k. Corollary 8.6 Suppose that s ω k with s < k. Then T k [s] T k (d, θ) is isometrically isomorphic to T k s (d, θ). Remark Another way of embedding T k (d, θ) into T k+1 (d, θ) is by sending each member s = (s 1,..., s k ) ω k to Ψ(s) = (s 1,..., s k, s k ) ω k+1. One can check that Ψ maps T k (d, θ) isometrically into T k+1 (d, θ). In fact, for each k 1 < k 2, there are infinitely many different ways of embedding E k1 into E k2, and each one of these 9 The Banach space T(A k d, θ) In this section, we investigate the most complex of the natural constructions of Banach spaces using the Tsirelson methods on the high dimensional Ellentuck spaces. These spaces, T(A k d, θ), were constructed in Section 3 using admissible sets with the order of the E i s determined by elements of A k d = m d ARk m as follows: Let m {1, 2,..., d}. We say that (E i ) m ARk is A k d -admissible if and only if there exists {v 1, v 2,..., v m } A k d such that v 1 E 1 < v 2 E 2 < < v m E m. (Recall Definition 3.2.)

35 Banach Spaces from Barriers in High Dimensional Ellentuck Spaces 35 Figure 9: An A 2 7 -admissible sequence. Recall Definition 3.3 of the norm T(A k d,θ) and Propositions The Banach space T(A k d, θ) shares many properties with the Banach space T k(d, θ), their proofs being very similar. In this section we indicate what parts of the proofs from previous sections for T k (d, θ) apply also to T(A k d, θ), and we provide the proofs that are different. In particular we have the following: (1) For k 2, T(A k d, θ) has arbitrarily large copies of l n with uniform isomorphism constant. (2) T(A k d, θ) is l p -saturated. (3) If j k, then T(A j d, θ) and T(Ak d, θ) are not isomorphic. We do not know if T(A j d, θ) embeds into T(Ak d, θ) for j < k. We start with the results of Section 4. For s ω <k, define T[A k d, s] = span{e v : v τ k [s]}. Since Corollary 4.1, Lemma 4.2, and Lemma 4.3 depend only on properties of AR k, and since these are the results used in the proof of Theorem 4.4, we obtain the following result: Theorem 9.1 Suppose that s 1 s 2 s N belong to ω <k and that s 1 = = s N < k. Let v ω k with s N v and suppose that x N T[Ak d, s i] satifies v < x. If we decompose x as x x N with x i T[A k d, s i], then max x θ(d 1) i x max 1 i N 1 θ x i. 1 i N In particular, if x 1 = = x N = 1, span{x 1,..., x N } is isomorphic to l N in a canonical way and the isomorphism constant is independent of N and of the x i s.