Banach Spaces from Barriers in High Dimensional Ellentuck Spaces

A new hierarchy of Banach spaces $T_k(d,\theta)$, $k$ any positive integer, is constructed using barriers in high dimensional Ellentuck spaces \cite{DobrinenJSL15} following the classical framework under which a Tsirelson type norm is defined from a barrier in the Ellentuck space \cite{Argyros/TodorcevicBK}. The following structural properties of these spaces are proved. Each of these spaces contains arbitrarily large copies of $\ell_\infty^n$, with the bound constant for all $n$. For each fixed pair $d$ and $\theta$, the spaces $T_k(d,\theta)$, $k\ge 1$, are $\ell_p$-saturated, forming natural extensions of the $\ell_p$ space, where $p$ satisfies $d\theta=d^{1/p}$. Moreover, they form a strict hierarchy over the $\ell_p$ space: For any $j<k$, the space $T_j(d,\theta)$ embeds isometrically into $T_k(d,\theta)$ as a subspace which is non-isomorphic to $T_k(d,\theta)$.


Introduction
Banach space theory is rich with applications of fronts and barriers within the framework of the Ellentuck space (see for example [8], [20], and Part B of [3]). Infinite-dimensional 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 [23], Galvin and Prikry [16], Louveau [18], and Mathias [19], culminating in Ellentuck's work in [14] introducing topological Ramsey theory on what is now called the Ellentuck space.
Building on 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 [14] (see Chapter 5 of [24]). 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 [21], [12], [13], [11], and [9]. In this context, the second author constructed a new hierarchy of topological Ramsey spaces in [9] and [10] 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 set-theoretic 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 [9].
Once constructed, it became clear that these new spaces are the natural generalizations of the Ellentuck space to higher dimensions, and hence are called highdimensional 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. What Banach spaces can be constructed by extending Tsirelson's construction method by using barriers in general topological Ramsey spaces?
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 p with 1 ≤ p < ∞ [25]. The idea of Tsirelson's construction became apparent after Figiel and Johnson [15] showed that the norm of the Tsirelson space satisfies the following equation: (1.1) n a n e n = max sup n |a n | , where the sup is taken over all sequences (E i ) m i=1 of successive finite subsets of integers with the property that m ≤ min(E 1 ) and E i ( n a n e n ) = n∈Ei 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 (1.1) replacing 1/2 by θ and using sequences (E i ) m i=1 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 ).
In this 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 1.1 (Bellenot [4]). If dθ > 1, then for every x ∈ T (A d , θ), where dθ = d 1/p and · p denotes the p -norm.
We construct new 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 p spaces. In Section 2 we review the construction of the finite-dimensional Ellentuck spaces in [9], introducing new notation and representations more suitable to the context of this paper. Our new Banach spaces T 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 p , where dθ = d 1/p , as they extends the construction of the Tsirelson type space T 1 (d, θ), which by Bellenot's Theorem 1.1 is exactly p . We prove the following structural results about the spaces T k (d, θ). The spaces T k (d, θ) contain arbitrarily large copies of n ∞ , where the bound is fixed for all n (Section 4), and that there are many natural block subspaces isomorphic to p (Section 5). Moreover, they are 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 p , where dθ = d 1/p .
As there are several different ways to naturally generalize the Tsirelson construction to high dimensional Ellentuck spaces, we consider a second, more stringent definition of norm and construct a second type of space T (A k d , θ), where the admissible sets are required to be separated by sets which are finite approximations to members of the E k (Section 9). The norms on these spaces are thus bounded by the norms on the T k (d, θ) spaces. 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.

High Dimensional Ellentuck Spaces
In [9], 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 reproduce the construction, though with slightly different notation more suited to the context of Banach spaces.
Recall that the Ellentuck space [14] 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 [9]. 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 [9], we shall use this notation. We start by defining a well-ordering 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.
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 .
Definition 2.5 (The spaces (E k , ≤, r), k ≥ 2, Dobrinen [9]). 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 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 r k n (X) 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 2.6. 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 . 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 .
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 2.7 (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: 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}.  Next we present the specifics of the structure of the space E 3 .
Example 2.8 (The space E 3 ). The well-order (ω ↓≤3 , ≺) begins as follows: 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 .

2.1.
Upper Triangular Representation of ω ↓2 . 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 : 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 . Under the identification discussed after Definition 2.5, we have that (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): 2.2. 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 X max v ∈ 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.
In general, for any k ≥ 2, X max v is constructed as follows.
Definition 2.9. 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 X max v . 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 ∈ Z + , j ≤ k, define the following auxiliary functions: f j (0) := n j and f j (m) := n k +m.
It is routine to check that: We use this Lemma to prove that X max v contains all elements of AR k that have v as initial value.
Arguing as before, we conclude that p i+1 ≥ n k > n k , and then Lemma 2.10 yields Notice that the only elements in X v that are ≺-smaller than v are the initial segments of v.
, and in this case we say that E and F are successive.
From the notation in Section 2 we have that 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 un e un = m n=1 x un e un for some m ∈ Z + . Using the above convention, we will write x = ∞ n=1 x n e n = m n=1 x n e n . If E ∈ AR k , we put Ex : . The Banach spaces that we introduce in this section have their roots (as all subsequent constructions [4], [22], [1], [17], [2]) in Tsirelson's fundamental discovery of a reflexive Banach space T with an unconditional basis not containing c 0 or p with 1 ≤ p < ∞ [25]. Based on these constructions, we present the following definitions.
x n e n ∈ c 00 (ω ↓k ) and j ∈ N, we define a non-decreasing sequence of norms on c 00 (ω ↓k ) as follows: For fixed x ∈ c 00 (ω ↓k ), the sequence (|x| j ) j∈N is bounded above by the 1 (ω ↓k )-norm of x. Therefore, we can set Clearly, · T k (d,θ) is a norm on c 00 (ω ↓k ).
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: Therefore, we conclude that for every x ∈ c 00 (ω ↓k ) we have The following propositions follow by standard arguments (see Proposition 2 in [22]): 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 respectively. In this section, let N ∈ Z + 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.
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.
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 : We will study the Banach space structure of the subspaces of T k (d, θ) of the  Applying Corollary 4.1 and Lemma 4.2 we have: This lemma helps us establish the presence of arbitrarily large copies of N ∞ inside T k (d, θ): In particular, if x 1 = · · · = x N = 1, span{x 1 , . . . , x N } is isomorphic to N ∞ in a canonical way and the isomorphism constant is independent of N and of the x i 's.
Without loss of generality we assume that Repeat the argument for E 1 x. Find m ∈ {1, . . . , d} and an admissible se- . We can assume that F 1 (E 1 x) = 0, and applying Lemma 4.3 once again we conclude that for Iterating this process we conclude that 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 p (see Theorem 1.1). 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 p spaces inside T k (d, θ) for k ≥ 2: is a normalized block sequence in T k (d, θ) and that we can find a sequence (v i ) ∞ i=1 ⊂ ω ↓k such that: Notice that Corollary 2.12 implies that for every j ≥ i, v j ∈ X max vi and supp (x j ) ⊂ X max vi . Theorem 5.1 allows us to identify natural subspaces of T k (d, θ) isomorphic to p . For example, it implies that the top trees of T k (d, θ) are isomorphic to p . In section 8 we will see that the top trees are isometrically isomorphic to T 1 (d, θ). We prove Theorem 5.1 in two steps. First we prove the lower p -estimate using Bellenot's space T 1 (d, θ). Then we prove the upper 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). .
Proof. Let x = i a i t 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 j x 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 jl x 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 i x 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 j x 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 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 2.12 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

BANACH SPACES FROM BARRIERS IN HIGH DIMENSIONAL ELLENTUCK SPACES 13
Therefore, The result follows now applying Theorem 1.1.
The proof of the upper bound inequality of Theorem 5.1 is harder and we need some preliminary results.

Alternative Norm.
To establish a upper 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 regard, the following definition plays a key role.
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 ∈ Z + 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).
Note that, by definition of the sets K n , each φ ∈ K has an analysis. Moreover, if f 1 ∈ K l (φ) and 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) = ∅.
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 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 (see [5]). The analogous property is true for (x j ) N j=1 . Proposition 5.9. For l ∈ {1, . . . , n} and j ∈ {1, . . . , N }, set 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 (x i ) = ∅ 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 p -estimate that implies the upper p -estimate of Theorem 5.1: 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: 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 1.1 it suffices to show that for every φ ∈ K, By unconditionality we can assume that x 1 , . . . , x N and φ are positive. Suppose φ ∈ K n \ K n−1 for some n ∈ Z + , 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 We prove the first inequality given that the other one is analogous. Let J ⊆ {1, . . . , N } and set y = j∈J a j x j .
If f ∈ K 0 (φ), then f = e * i for some i ∈ Z + . We want to prove that Suppose that e * i (y) = 0. So, there exists exactly one j i ∈ J such that e * i (x ji ) = 0. Applying Proposition 5.7 we have 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 1 + · · · + f m ), where f 1 , . . . , f m are successive elements in K l (φ) with (supp (f i )) m i=1 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: 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 However, by the induction hypothesis, 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 In this section we prove that every infinite dimensional subspace of T k (d, θ) has a subspace isomorphic to 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] =  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) We now present the main result of this section: 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 p . From Proposition 5.10 we have that n a n x n ≤ 2 θ n |a n | p 1/p .
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 p -estimate.
To this end, assume that Then we have the two cases: Case 1: ∀j ∈ Z + , Q j x n → 0.
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 ) ⊂ p1 j=1 τ k [s j ], applying Lemma 6.1 we can find q 1 > p 1 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 v2 . We now repeat the argument. Since there exists p 2 such that supp (y 2 ) ⊂ p2 j=1 τ k [s j ], applying Lemma 6.1 we can find 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 v 1 ≤ x 1 < v 2 < y 2 < v 3 < y 3 and supp (y 2 ) ⊂ X max v2 , supp (y 3 ) ⊂ X max v3 . 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 vi for i > 1.
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 se- , where for every j ∈ Z + , 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, (Q j0 x ni ) has a subsequence with a lower p -estimate, and therefore (x ni ) has a subsequence with a lower p -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 p . The result follows.

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 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 N ∞ -sequence into an 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.
We are ready to state and prove the main proposition.
If k 1 − |s| > k 2 − |t 1 |, then for every n ∈ Z + , 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 k2 [t i ] is isomorphic to p , and consequently so is On the other hand, Theorem 4.4 guarantees that T k1 [s] has arbitrarily large copies of N ∞ .
Suppose now that the result is true for m ∈ Z + 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 ∈ Z + and an isomorphism Φ : Lemma 7.2. Find N large enough and v ∈ ω ↓k1 such that s n ≺ s n+1 ≺ · · · ≺ s n+N −1 ≺ v. We will find normalized for i ≤ N and we will use Theorem 4.4 to conclude that span{x 1 , . . . , x} ≈ 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 k2 [t 1 ], and let x 1 ∈ T k1 [s n ] be such that x 1 = 1 and v < x 1 . Find a finitely supported y 1 ∈ T k2 [t 1 ] such that y 1 ≈ Φ(x 1 ). Applying Lemma 6.1 we can find Since k 1 − |s n+1 | > k 2 − |r 1 | = m we can apply the induction hypothesis. In particular, the map is not an isomorphism. As a result, there exists

as in Lemma 7.2 and apply the induction hypothesis to
Now that we have a normalized x 2 ∈ T k1 [s n+1 ] that satisfies v < x 2 and P m Φ(x 2 ) ≈ 0, we find a finitely supported y 2 ∈ T k2 [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 v2 . We now repeat the argument. Use Lemma 6.1 to find v 3 ∈ τ k2 [t 1 ] such that Then we find a normalized x 3 ∈ T k1 [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 k2 [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 Since N is arbitrary, this contradicts that Φ is continuous (see equation 7.1 below) and we conclude the case M = 1.
Let M > 1 and suppose, towards a contradiction, that there exists n ∈ Z + and an isomorphism Φ : For each j ∈ Z + let Q j : i ] as in Lemma 7.2 and for each m ∈ Z + , let i ] 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 ↓k1 such that s n ≺ s n+1 ≺ · · · ≺ s n+N −1 ≺ v. Find x 1 ∈ T k1 [s n ] such that x 1 = 1 and v < x 1 and find a finitely supported be the projection onto the first blocks of each of the T k2 [t j ]'s.
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 k2 [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 ⊕T k2 [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 By Theorem 5.1, there exists C 1 > 0 independent of N such that for every j ≤ M , Using the triangle inequality for y i = M j=1 Q j (y i ), Holder's inequality p + 1 q = 1 , Theorem 5.1, and the fact that the projections Q j are contractive, we get x i stays bounded, we see that Φ cannot be bounded, contradicting our assumption.
For fixed d and θ, the spaces T k (d, θ), k ≥ 1, form a natural hierarchy in complexity over 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 [9] and explicit in the recursive construction of the finite and infinite dimensional Ellentuck spaces in [10].
The following notation will be useful. Let Φ : ω ↓k → ω ↓k+1 be defined by Φ(v) = (0) v. One can easily check that Φ preserves ≺ and <. We can naturally extend the definition of Φ to the finitely supported vectors of T k (d, θ) by Φ ( i a i e vi ) = i a i e Φ(vi) .
Proof. Let v = min ≺ (E). Proposition 2.11 says that every w ∈ E belongs to X max v .
Since Φ adds a 0 at the beginning of each sequence, the characterization of Lemma 2.10 implies that for every w ∈ E, Φ(w) belongs to X max Φ(v) . Then the initial segment } satisfies all the conditions of the Lemma.
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 Notice that this implies that |supp (E i x) | < n for every i ≤ d. By we conclude that 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) = ∅.
Proof. We will prove that if F ∈ AR k+1 and F Φ(x) = 0, then E = tr 0 (F ) ∈ AR k and Φ(Ex) = F Φ(x). The rest of the lemma follows easily from this.
Let F ∈ AR k+1 with F Φ(x) = 0. Then there exists an E k+1 -tree X such that F is an initial segment of X. Since F Φ(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 Y : ω ↓≤k → ω ↓≤k by Y (v) = tr 0 X (0) v . Since X preserves ≺ and <, it follows that Y preserves those orders as well and hence Y is an E k -tree. Since Y preserves ≺, it follows that E = tr 0 (F ) is an initial segment of Y (i.e., E ∈ AR k ).
To complete the proof of the Lemma, suppose that And from here we conclude that v ≺ w. Since the elements are arbitrary we have that E 1 · · · E m .
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. 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 Combining the previous corollaries, we obtain the following result: 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. Following the construction of Section 3, we define a sequence of norms on c 00 (ω ↓k ) by |x| 0 := max n∈Z + |x n | and and we define T (A k d , θ) to be the completion of c 00 (ω ↓k ) with norm It follows that T (A k d , θ) is 1-unconditional and where the supremum runs over all A k d -admissible families {E 1 , . . . , E m }. 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 9.2. Suppose that s 1 ≺ s 2 ≺ · · · ≺ s N belong to ω ↓<k and that In particular, if x 1 = · · · = x N = 1, span{x 1 , . . . , x N } is isomorphic to N ∞ in a canonical way and the isomorphism constant is independent of N and of the x i 's.
We now move to the results of Section 5. The main result is the same but the proof for the lower p -estimate is different.
is a normalized block sequence in T (A k d , θ) and that we can find a sequence (v i ) ∞ i=1 ⊂ ω ↓k such that: Then, (x i ) is equivalent to the basis of p .
We start with the upper p -estimate. The construction of the Alternative Norm (see Subsection 5.1) is almost identical. The main difference is the definition of almost admissible sequences.
This results in an alternative/dual description of the norm of T (A k d , θ). The sets K n are smaller than the corresponding sets for T k (d, θ), but the proofs are identical, resulting in the upper p -estimate identical to Proposition 5.10.
The lower p -estimate is harder, because we have fewer admissible sequences than in T k (d, θ). We need to have enough "room" to find A k d -admissible sequences, since we need to place the sets between an element of A k d . To do so, we will prove a lower p -estimate for the sequences (x 2n ) and (x 2n−1 ). Since the closed span of (x 2n ) and (x 2n+1 ) are complemented in the closed span of (x n ), the general result follows. We will obtain the estimate for (x 2n ). The other case is similar.
We start with the following lemma.
Lemma 9.5. Suppose that (q i ) ∞ i=1 ⊂ N is such that q 1 < q 2 < · · · . Then, there exists X = {w 1 , w 2 , . . .} ∈ E k such that for every i ∈ Z + , max(w i ) = q i and w i is a sequence all of whose terms are in the set {q 1 , q 2 , . . .}.
Proof. Following Definition 2.5 we will construct inductively an E k -tree X that determines X. Recall that for fixed k ≥ 2, the first k members of (ω ↓≤k , ≺) are s 0 = (), s 1 = (0), and in general, for 1 ≤ k ≤ k, s k is the sequence of 0's of length k . Begin by setting X( s 0 ) := (), and for each 1 ≤ k ≤ k, set X( s k ) to be the sequence of length k with all entries being q 1 . Note that u 0 is the sequence of 0's of length k, and that max( X( u 0 )) = q 0 .
Set q i := max(v 2ni−1 ). By hypothesis (1) of Theorem 9.3, it follows that q 1 < q 2 < · · · < q m . Applying Lemma 9.5, we can find {w 1 , w 2 , . . . , w m } in AR k d such that max(w i ) = q i and all the terms of w i are in {q 1 , q 2 , . . . , q m }. Consequently, By hypothesis, X max v2n i contains the support of x j for any j ≥ 2n i . Hence we define: as the initial segment of X max v2n i up to (but not including) v 2ni+1 . By construction, F i ∈ AR k , min ≺ (F i ) = v 2ni , and F i contains the supports of x 2ni , x 2(ni+1) , x 2(ni+2) , . . . , x 2(ni+1−1) .
Thus, from equation (9.1) and (9.2), we have: and we conclude that F 1 , F 2 , . . . , F m ∈ AR k is the desired A k d -admissible sequence.
With this, the proof of Proposition 5.4 applies and we obtain the lower pestimate for the normalized block sequence (x 2n ). A similar argument gives a lower p -estimate for the normalized block sequence (x 2n−1 ) and we conclude the sketch of the proof of Theorem 9.3.
Since Theorems 9.2 and 9.3 hold for T (A k d , θ), we have all the elements to show that the different T (A k d , θ)'s are not isomorphic to each other. The proof of Theorem 7.1 applies and we obtain the following: Theorem 9.7. If k 1 = k 2 , then T (A k1 d , θ) is not isomorphic to T (A k2 d , θ).

Further Directions
In this paper, we considered two different means of constructing norms on high dimensional Ellentuck spaces. One required both the admissible sets and their endpoints to be finite approximations to members of E k , and the other only required the admissible sets to be finite approximations. Theorems 7.1 and 8.5 show that this latter norm construction produces a hierarchy of Banach spaces T k (d, θ) which embed isometrically as subspaces into Banach spaces constructed from higher order Ellentuck spaces.
Question 2. For fixed d and θ and k 1 < k 2 , does T (A k1 d , θ) embed as an isometric subspace of T (A k2 d , θ)? Preliminary analysis shows that if we let d be sufficiently greater than d, we can show that the norm on T (A k1 d , θ) is bounded by the norm on the trace subspace above (0) in T (A k2 d , θ), where d is computed from d in a straightforward manner using methods from [9]. However, we have not checked whether or not this produces and isometric subspace. Question 3. For fixed d, θ, k how different are T k (d, θ) and T (A k d , θ)? We finish with some questions about the behaviors of norms over certain sequences of these spaces.
Question 4. What is the behavior of the norms on T k (B, θ), constructed using barriers B on E k of infinite rank?
Finally, we ask about Banach spaces constructed on infinite dimensional Ellentuck spaces from [10].
Question 5. What new properties of the sequence of Banach spaces emerge as we construct T α (d, θ), where α is any countable ordinal and E α is the α-dimensional Ellentuck space?
As the spaces T k (d, θ) were shown to extend the p space into a natural and very structured hierarchy, it will be interesting to see what properties emerge in these classes of new Banach spaces.