BANACH SPACES FROM BARRIERS IN HIGH DIMENSIONAL ELLENTUCK SPACES

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1 BANACH SPACES FROM BARRIERS IN HIGH DIMENSIONAL ELLENTUCK SPACES A ARIAS, N DOBRINEN, G GIRÓN-GARNICA, AND J G MIJARES Abstract A new hierarchy of Banach spaces T k (d, θ), k any positive integer, is constructed using barriers in high dimensional Ellentuck spaces [9] 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), k 1, are l p-saturated, forming natural extensions of the l p space, where p satisfies dθ = d 1/p Moreover, they 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) 1 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 Dobrinen was partially supported by National Science Foundation Grants DMS and DMS

2 2 A ARIAS, N DOBRINEN, G GIRÓN-GARNICA, AND J G MIJARES 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 l 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: (11) { } a ne n = max sup n a n, 1 m n 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 (11) replacing 1/2 by θ and using sequences (E i ) m of finite subsets of positive integers which are F-admissible, ie, 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 Ad := {F N : F d}

3 BANACH SPACES FROM BARRIERS IN HIGH DIMENSIONAL ELLENTUCK SPACES 3 is noteworthy in the realm of Tsirelson type spaces In fact, Bellenot proved in [4] the following remarkable theorem: Theorem 11 (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 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 l 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 l 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 11 is exactly l 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 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 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 (Ak d, θ) for j < k The paper concludes with open problems for further research into the properties of these spaces 2 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

4 4 A ARIAS, N DOBRINEN, G GIRÓN-GARNICA, AND J G MIJARES 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 21 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 22 (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 23 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 22; similarly, (2) < lex (2, 2) follows by setting m = 1, as any proper initial segment of a sequence is lexicographically below that sequence Definition 24 (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 (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 Definition 25 (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 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 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

5 BANACH SPACES FROM BARRIERS IN HIGH DIMENSIONAL ELLENTUCK SPACES 5 Remark 26 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 25 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 27 (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 28 (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

6 6 A ARIAS, N DOBRINEN, G GIRÓN-GARNICA, AND J G MIJARES 21 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 : 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 25, we have that 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):

7 BANACH SPACES FROM BARRIERS IN HIGH DIMENSIONAL ELLENTUCK SPACES 7 Figure 5: r 10 (X) for a typical X E 2 22 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 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 E 2 looks like this: X max v 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 25: () (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 25, we have Xv max = {(0, 2, 7), (0, 2, 8), (0, 8, 8), (8, 8, 8), (0, 2, 9), } In general, for any k 2, Xv max is constructed as follows

8 8 A ARIAS, N DOBRINEN, G GIRÓN-GARNICA, AND J G MIJARES Definition 29 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 Xv must be a function from ω k to ω k satisfying Definition 25 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 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 210 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 We use this Lemma to prove that Xv max v as initial value contains all elements of AR k that have Proposition 211 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 210, X(0,, 0, 1) X max v 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 210, gives that (p 1, p 2,, p k ) = w Xv max 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 210 yields that w X max v Corollary 212 If w Xv max, then Xw max and min (E) Xv max, then E Xv max Xv max In particular, if E AR k Notice that the only elements in X v that are -smaller than v are the initial segments of v Corollary 213 If s v and s ran( X v ) then s v

9 BANACH SPACES FROM BARRIERS IN HIGH DIMENSIONAL ELLENTUCK SPACES 9 3 The Banach Space T k (d, θ) 31 Preliminary Definitions Set N := {0, 1, } and Z + := N \ {0} For the rest of this paper, fix d, k Z +, 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 Z + 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 32 Construction of T k (d, θ) 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 l p with 1 p < [25] Based on these constructions, we present the following definitions Definition 31 For m d, we say that (E i ) m ARk is d-admissible, or simply admissible, if and only if E 1 < E 2 < < 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 0 := max n Z + x n, { { m }} x j+1 := max x j, θ max E i x j : 1 m d, (E i ) m admissible For fixed x c 00 (ω k ), the sequence ( x j ) j N is bounded above by the l 1 (ω k )-norm of x Therefore, we can set x Tk (d,θ) Clearly, Tk (d,θ) is a norm on c 00(ω k ) := sup x j j N Definition 32 The completion of c 00 (ω k ) with respect to the norm Tk (d,θ) is denoted by (T k (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 33 If x c 00 (ω k ) and supp (x) = n, then x n = x n+1 =

10 10 A ARIAS, N DOBRINEN, G GIRÓN-GARNICA, AND J G MIJARES Therefore, we conclude that for every x c 00 (ω k ) we have x Tk (d,θ) = max j N x j The following propositions follow by standard arguments (see Proposition 2 in [22]): Proposition 34 (e n ) n=1 is a 1-unconditional basis of T k (d, θ) Proposition 35 For x = n=1 x ne n T k (d, θ) it follows that { { m }} x = max x, θ sup E i x : 1 m d, (E i ) m d-admissible, where x := sup n Z + x n 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 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 Corollary 41 If v ω k satisfies s N v, then there is at most one i N such that X max v 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 210 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 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 42 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 (41) F = span{e v Z : v ω k, v s N } C = span{e v Z : v ω k, s N v} and

11 BANACH SPACES FROM BARRIERS IN HIGH DIMENSIONAL ELLENTUCK SPACES 11 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 41 and Lemma 42 we have: Lemma 43 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 ] Proof Suppose that E[T k (d, θ)] C and set v := min (E) Then, by Corollary 41, 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 42 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 44 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 k [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

12 12 A ARIAS, N DOBRINEN, G GIRÓN-GARNICA, AND J G MIJARES Without loss of generality we assume that E 1 x 0, so that s N min (E 2 ) By Lemma 43, 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 43 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 5 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 11) 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 51 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: (1) v 1 x 1 < v 2 x 2 < v 3 x 3 < v 4 x 4 < (2) supp (x i ) Xv max i and v i+1 Xv max i for every i 1 Then, (x i ) is equivalent to the basis of l p Notice that Corollary 212 implies that for every j i, v j Xv max i and supp (x j ) Theorem 51 allows us to identify natural subspaces of T k (d, θ) isomorphic X max v i 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 52 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 210 that the (v i ) s and (x i ) s satisfy the hypothesis of Theorem 51 and the result follows

13 BANACH SPACES FROM BARRIERS IN HIGH DIMENSIONAL ELLENTUCK SPACES 13 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 210 to verify that the (v i ) s and (x i ) s satisfy the hypothesis of Theorem 51 and we conclude Corollary 53 D[s] = span{e vi : i 1} is isomorphic to l p, for every s ω <k We prove Theorem 51 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 54 Under the same hypotheses of Theorem 51, 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) 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 51 and by Corollary 212 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

14 14 A ARIAS, N DOBRINEN, G GIRÓN-GARNICA, AND J G MIJARES Therefore, a 1 m m it i = θ E j x i 1 θ F j z = a ix i i j=1 j=1 The result follows now applying Theorem 11 The proof of the upper bound inequality of Theorem 51 is harder and we need some preliminary results 51 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 regard, the following definition plays a key role Definition 55 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 AR k 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 Z+ }, 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 56 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 Z + 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 ) E i Then, (f i )m K n 1 with (supp (f i ))m almost admissible and f i (E ix) = f i (E i x) So, m m θ f i (E i x) = θ f i(e i x) E i x n x n ; therefore, 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, m m m f(x) = θ f i (x) θ 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 ARk such that supp (f i ) E ni, where n 1,, n m {1,, m} and n 1 < < n m So, m m θ supp (f i ) x n 1 θ E ni x n 1 x n ;

15 BANACH SPACES FROM BARRIERS IN HIGH DIMENSIONAL ELLENTUCK SPACES 15 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 57 For every x T k (d, θ), x = sup {f(x) : f K} 52 Upper Bound for Theorem 51 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 58 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) (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) 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 ) } = {f 1,, 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 (see [5]) The analogous property is true for (x j )N j=1 Proposition 59 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

16 16 A ARIAS, N DOBRINEN, G GIRÓN-GARNICA, AND J G MIJARES 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 l p -estimate that implies the upper l p -estimate of Theorem 51: Proposition 510 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: 1/p N a j x j 2 N a j p θ j=1 Proof In order to avoid confusion, we will write 1 to denote the norm on T 1 (d, θ) By Proposition 57 and Theorem 11 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 Z +, and let (K l (φ)) n l=0 be an analysis of φ (see Definition 58) 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 If f K 0 (φ), then f = e i for some i Z+ 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 57 we have e i (y) = e i (a ji x j i ) aji x j i aji 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: j=1 j=1

17 BANACH SPACES FROM BARRIERS IN HIGH DIMENSIONAL ELLENTUCK SPACES 17 I := { i m : j J with f i (x j) 0 and supp (f) supp ( x ) j supp (fi ) } and 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 59 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 + ] f(a jx j D i j), j J and consequently [ i I fi ( ) f(y) θ a j x j + f(a j x j D i j) ] j J However, by the induction hypothesis, ) f i a j x ( j D 1 j 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 1 θ a jt j j J 1 j J a jt j 1 ] 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 X max v contains arbitrary tails of the decomposition of τ k [s] Its proof follows from the

18 18 A ARIAS, N DOBRINEN, G GIRÓN-GARNICA, AND J G MIJARES definition of the E k -tree X v that determines Xv max 210) (see paragraph preceding Lemma Lemma 61 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 62 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 510 we have that a 2 ( nx n a n p) 1/p n n θ 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 T k [s j ], where for each j Z +, s s j, s j = s + 1, and s j s j+1 For each j Z + 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 Z +, Q j x n 0 Case 2: j 0 Z + such that Q j0 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 ) p 1 j=1 τ k [s j ], applying Lemma 61 we can find q 1 > p 1 τ k [s] = j=q 1 τ k [s j ] Since and v 2 τ k [s] such that v 1 x 1 < v 2 and X max v 2 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 ) p2 j=1 τ k [s j ], applying Lemma 61 we can find q 2 > p 2 and v 3 τ k [s] such that v 2 < y 2 < v 3 and Xv max 3 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 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 54, there exist C 1, C 2 R such that a ( ix ni C1 a iy ni C2 a i p) 1/p i i i

19 BANACH SPACES FROM BARRIERS IN HIGH DIMENSIONAL ELLENTUCK SPACES 19 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 T k [t j ], 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 l p -estimate, and therefore (x ni ) has a subsequence with a lower l 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 52, T k [t j ] would be isomorphic to l p The result follows 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 71 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 74 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 72 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] = T k [s i ] and for m Z +, there is a canonical projection P m : T k [s] m T k [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 73 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 210 We are ready to state and prove the main proposition Proposition 74 Let s ω k1 with s < k 1 and decompose T k1 [s] = T k1 [s i ] according to Lemma 72 Let M Z + and t 1,, t M ω k2 such that t 1 = = t M < k 2 If k 1 s > k 2 t 1, then for every n Z +, i=n T k1 [s i ] does not embed into T k2 [t 1 ] T k2 [t M ] Proof We proceed by induction For the base case we assume that k 2 t 1 = 1 < k 1 s By Corollary 52, T k2 [t i ] is isomorphic to l p, and consequently so is T k2 [t 1 ] T k2 [t M ] On the other hand, Theorem 44 guarantees that T k1 [s] has arbitrarily large copies of l N

20 20 A ARIAS, N DOBRINEN, G GIRÓN-GARNICA, AND J G MIJARES 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 : T k1 [s i ] T k2 [t 1 ] i=n Decompose T k2 [t 1 ] = j=1 T k2 [r j ] according to Lemma 72 Find N large enough and v ω k1 such that s n s n+1 s n+n 1 v We will find normalized x 1 T k1 [s n ], x 2 T k1 [s n+1 ],, x N T k1 [s n+n 1 ] such that v < x i for i N and we will use Theorem 44 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 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 61 we can find v 2 τ k2 [t 1 ] such that v 1 y 1 < v 2 and Xv max 2 τ k2 [t 1 ] = j=m τ k2 1+1 [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] : T k1 [s n+1 ] T k2 [r 1 ] T k2 [r m1 ] is not an isomorphism As a result, there exists x 2 T k1 [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 k1 [s n+1 ] = T k1 [u i ] as in Lemma 72 and apply the induction hypothesis to i=p T k1 [u i ] for p large enough 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 73 gives that supp (y 2 ) Xv max 2 We now repeat the argument Use Lemma 61 to find v 3 τ k2 [t 1 ] such that y 2 < v 3 and Xv max 3 τ k2 [t 1 ] = j=m τ k2 2+1 [r j ] 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 x i T k1 [s n+i 1 ] with v < x i, v i ω k2, and y i T k2 [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 By Theorem 51, (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 71 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 : T k1 [s i ] T k2 [t 1 ] T k2 [t 2 ] T k2 [t M ] i=n For each j Z + let Q j : M T k2 [t i ] T k2 [t j ] be the canonical projection Decompose T k2 [t j ] = T k2 [r j i ] as in Lemma 72 and for each m Z+, let Pm j : T k2 [t j ] m T k2 [r j i ] be the canonical projection onto the first m blocks

21 BANACH SPACES FROM BARRIERS IN HIGH DIMENSIONAL ELLENTUCK SPACES 21 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 y 1 T k2 [t 1 ] T k2 [t M ] such that y 1 Φ(x 1 ) For each j M, let v j 1 = min (τ k2 [t j ]) Use Lemma 61 to find v j 2 τ k2 [t j ] such that Q j (y 1 ) < v j 2 and Xmax v 2 τ k2 [t j ] = i=m j 1 +1 τ k2 [r j i ] for some mj 1 N Let P 1 = M j=1 P j be the projection onto the first blocks of each of the T k2 [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 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 x i T k1 [s n+i 1 ] with v < x i and y i M j=1 T k2 [t j ] such that y i Φ(x i ) Moreover, for every j M, we can find v j k2 i ω 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 51, there exists C 1 > 0 independent of N such that for every j M, 1 C 1 ( N Q j (y i ) p ) 1 p ( N Q j (y i ) C N 1 Q j (y i ) p ) 1 p Using the triangle inequality for y i = M j=1 Q j(y i ), Holder s inequality ( N a i N 1/q ( N a i p ) 1/p, 1 p + 1 q = 1 ), Theorem 51, and the fact that the projections Q j are contractive, we get (71) N N M y i 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)

22 22 A ARIAS, N DOBRINEN, G GIRÓN-GARNICA, AND J G MIJARES 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 ie vi ) = i a ie Φ(vi) Lemma 81 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 211 says that every w E belongs to Xv max Since Φ adds a 0 at the beginning of each sequence, the characterization of Lemma 210 implies that for every w E, Φ(w) belongs to XΦ(v) max Then the initial segment of XΦ(v) max defined by F = {s Xmax Φ(v) : s max Φ(E)} satisfies all the conditions of the Lemma Corollary 82 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 81, 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 d d x Tk (d,θ) = θ E i x Tk (d,θ) θ 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) =

23 BANACH SPACES FROM BARRIERS IN HIGH DIMENSIONAL ELLENTUCK SPACES 23 Lemma 83 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 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 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 (ie, E AR k ) Since (0) v F iff v E, we have F Φ(x) = a i e Φ(vi) = ( ) a i e Φ(vi) = Φ a i e i = Φ(Ex) (0) v i F v i E 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 84 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 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 83, 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 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 85 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

24 24 A ARIAS, N DOBRINEN, G GIRÓN-GARNICA, AND J G MIJARES Corollary 86 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 87 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) There are several natural ways of constructing Banach spaces using the Tsirelson construction and AR k on high dimensional Ellentuck spaces, depending on how one interprets what the natural generalizations of the finite sets in the constructions over the Ellentuck space should be Finite sets can be interpreted to be simply finite sets or they can be interpreted as finite approximations to members of the Ellentuck space Thus, in building Banach spaces on E k, the admissible sets can be chosen to be either finite sets or members of AR k, and likewise, the admissible sets can either be simply increasing or required to have a member of AR k separating them In this section, we produce the most complex of the natural constructions of Banach spaces using the Tsirelson methods on the high dimensional Ellentuck spaces We denote these spaces as T (A k d, θ) In T k(d, θ) we said that {E i : i m} AR k is d-admissible if m d and E 1 < E 2 < < E m The E i s were linearly ordered, and this was indicated by the index d In T (A k d, θ), the order of the E i s will be determined by elements of AR k d, which leads to the definition below 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 arbitrary large copies of ln 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 don t know if T (A j d, θ) embeds into T (Ak d, θ) for j < k Definition 91 Set A k d := d ARk i and 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

25 BANACH SPACES FROM BARRIERS IN HIGH DIMENSIONAL ELLENTUCK SPACES 25 Figure 8: An A 2 7-admissible sequence 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 { { m }} x j+1 := max x j, θ max E i x j : 1 m d, (E i ) m A k d-admissible, and we define T (A k d, θ) to be the completion of c 00(ω k ) with norm x T (A k d,θ) = sup x j j It follows that T (A k d, θ) is 1-unconditional and x T (A k d,θ) = max { x, θ sup } m E i x T (A k d,θ) 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 41, Lemma 42, and Lemma 43 depend only on properties of AR k, and since these are the results used in the proof of Theorem 44, we obtain the following result: Theorem 92 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 We now move to the results of Section 5 The main result is the same but the proof for the lower l p -estimate is different