BANACH SPACES WITH THE 2-SUMMING PROPERTY. A. Arias, T. Figiel, W. B. Johnson and G. Schechtman

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1 BANACH SPACES WITH THE 2-SUMMING PROPERTY A. Arias, T. Figiel, W. B. Johnson and G. Schechtman Abstract. A Banach space X has the 2-summing property if the norm of every linear operator from X to a Hilbert space is equal to the 2-summing norm of the operator. Up to a point, the theory of spaces which have this property is independent of the scalar eld: the property is self-dual and any space with the property is a nite dimensional space of maximal distance to the Hilbert space of the same dimension. In the case of real scalars only the real line and real `2 have the 2-summing property. In the complex case there are more examples; e.g., all subspaces of complex `3 and their duals. 0. Introduction: Some important classical Banach spaces; in particular, C(K) spaces, L spaces, the disk algebra; as well as some other spaces (such as quotients of L spaces by reexive subspaces [K], [Pi]), have the property that every (bounded, linear) operator from the space into a Hilbert space is 2-summing. (Later we review equivalent formulations of the denition of 2-summing operator. Here we mention only that an operator T : X! `2 is 2-summing provided that for all operators u : `2! X the composition T u is a Hilbert-Schmidt operator; moreover, the 2-summing norm 2 (T ) of T is the supremum of the Hilbert-Schmidt norm of T u as u ranges over all norm one operators u : `2! X.) In this paper we investigate the isometric version of this property: say that a Banach space X has the 2-summing property provided that 2 (T ) = kt k for all operators T : X! `2. While the 2-summing property is a purely Banach space concept and our investigation lies purely in the realm of Banach space theory, part of the motivation for studying the 2-summing property comes from operator spaces. In [Pa], Paulsen denes for a Banach space X the parameter (X) to be the supremum of the completely bounded norm of T as T ranges over all norm one operators from X into the space B(`2) of all bounded linear operators on `2 and asks which spaces X have the property that (X) =. Paulsen's problem and study of (X) is motivated by old results of von Neumann, Sz.-Nagy, Arveson, and Parrott as well as more recent research of Misra and Sastry. The connection between Paulsen's problem and the present paper is Blecher's result [B] that (X) = implies that X has the 2-summing property. Another connection is through the property (P) introduced by 99 Mathematics Subect Classication. Primary 46B07 Secondary 47A67, 52A0, 52A5. The rst author was supported as Young Investigator, NSF DMS ; and NSF DMS ; the second and fourth authors were Workshop participants, NSF DMS ; the third author was supported by NSF DMS and ; the third and fourth authors were supported by the U.S.-Israel Binational Science Foundation. Typeset by AMS-TEX

2 2 A. ARIAS, T. FIGIEL, W. B. JOHNSON AND G. SCHECHTMAN Bagchi and Misra [BM], which Pisier noticed is equivalent to the 2-summing property. However, since we shall not investigate here (X) or property (P) directly and do not require results from operator theory, we refer the interested reader to [Pa] and [BM] for denitions, history, and references. On the other hand, since the topic we treat here is relevant for operator theorists, we repeat standard background in Banach space theory used herein for their benet. In [Pa] Paulsen asks whether (X) = only if X is a one or two dimensional C(K) or L space; in other words, ignoring the trivial one dimensional case, whether (X) = implies that X is two dimensional, and among two dimensional spaces, whether only `2 and `2 satisfy this identity. He proves that (X) = implies that dim(x) is at most 4, that (X) = (X ), and he gives another proof of Haagerup's theorem that (`2) =. Paulsen, interested in operator theory, is referring to complex Banach spaces, so `2 is not the same space as `2. From the point of view of Banach space theory, it is natural to ask which Banach spaces have the 2-summing property both in the real and the complex cases, and here we investigate both questions. Up to a point, the theory is independent of the scalar eld: In section 2 we show that the 2-summing property is self-dual, that only spaces of suciently small (nite) dimension can have the property, and that a space with the property is a maximal distance space{that is, it has maximal Banach-Mazur distance to the Hilbert space of the same dimension. The main result in section 2, Proposition 2.6, gives a useful condition for checking whether a space has the 2-summing property which takes a particularly simple form when the space is 2-dimensional (Corollary 2.7.a). The analysis in section 3 yields that the situation is very simple in the case of real scalars; namely, IR and `2 are the only spaces which have the 2-summing property. Two ingredients for proving this are Proposition 3., which says that there are many norm one operators from real `3 into `22 which have 2-summing norm larger than one, and a geometrical argument, which together with a recent lemma of Maurey implies that a maximal distance real space of dimension at least three has a two dimensional quotient whose unit ball is a regular hexagon. The complex case is a priori more complicated, since `2 and `2 both have the 2-summing property but are not isometrically isomorphic. In fact, in section 4 we show that there are many other examples of complex spaces which have the 2-summing property; in particular, `3 and all of its subspaces. The simplest way to prove that these spaces have the 2-summing property is to apply Proposition 2.6, but we also give direct proofs for `3 in section 4 and for its two dimensional subspaces in the appendix. The case of `3 itself reduces via a simple but slightly strange \abstract nonsense" argument to a calculus lemma, which, while easy, does not look familiar. (In [BM] the authors give an argument that `3 satises their property (P) which uses a variation of the calculus lemma but replaces the \abstract nonsense" with a reduction to self-adoint matrices.) We also give in Proposition 4.5 an inequality which is equivalent to the assertion that all two dimensional subspaces of complex `3 have the 2-summing property. While we do not see a simple direct proof of this inequality, we give a very simple proof of a weaker inequality which is equivalent to the assertion that every two dimensional subspace of complex `3 is of maximal distance. In section 5 we make some additional observations.

3 BANACH SPACES WITH THE 2-SUMMING PROPERTY 3. Preliminaries. Standard Banach space theory language and results can be found in [LT], [LT2], while basic results in the local theory of Banach spaces are contained in [T-J2]. However, we recall here that part of the theory and language which we think may not be well known to specialists in operator theory. Spaces are always Banach spaces, and subspaces are assumed to be closed. Operators are always bounded and linear. The [Banach-Mazur] distance between spaces X and Y is the isomorphism constant, dened as the inmum of kt k T? as T runs over all invertible operators from X onto Y. The closed unit ball of X is denoted by Ball (X). \Local theory" is loosely dened as the study of properties of innite dimensional spaces which depend only on their nite dimensional spaces, as well as the study of numerical parameters associated with nite dimensional spaces. Basic for our study and most other investigations in local theory is the fact (see [T-J2, p. 54]) that the distance from an n-dimensional space to `n2 is at most p n. One proves this by using the following consequence of F. John's theorem ([T-J2, p. 23]): If X is n-dimensional and E is the ellipsoid of minimal volume containing Ball (X), then n? 2 E Ball (X). This statement perhaps should be elaborated: Since dim(x) <, we can regard X as IR n or C n with some norm. Among all norm-increasing operators u from `n2 into X, there is by compactness one which minimizes the volume of u(ball (`n2 )); the distance assertion says that u? p n. Alternatively, if one chooses from among all norm one operators from X into `n2 one which maximizes the volume of the image of Ball (X), then the norm of the inverse of this operator is at most p n. If complex `n2 is considered as a real space, then it is isometrically isomorphic to real `2n 2. Thus the distance statement for complex spaces says that a complex space of dimension n, when considered as a real space of dimension 2n, has (real) distance to (real) `2n 2 at most p n. Actually, we need more than ust the distance consequence of John's theorem. The theorem itself [T-J2, p. 22] says that if E is the ellipsoid of minimal volume containing Ball (X), then there exist points of contact y ; : : :; y m between the unit sphere of X and the boundary of E, and there exist P positive real numbers ; : : :; m m summing to dimx so that for each x in X, x = ihx; y i iy i, where \h; i" is the scalar product which generates the ellipsoid E. The existence of many contact points between Ball (X) and E is important for the proof of Theorem 3.3. The dual concept to minimal volume ellipsoid is maximal volume ellipsoid. More precisely, an n-dimensional space can be regarded as IR n or C n under some norm kk in such a way that Ball (X) E, where E is the usual Euclidean ball and is also the ellipsoid of minimal volume containing Ball (X). Then X is naturally represented as IR n or C n under some norm, and the action of X on X is given by the usual inner product. Then E is the ellipsoid of maximal volume contained in Ball (X ). John's theorem gives many points of contact between Ball (X) and the boundary of the ellipsoid of minimal volume containing Ball (X), and many points of contact between the boundary of Ball (X) and the ellipsoid of maximal volume contained in Ball (X). It is a nuisance that these two ellipsoids are not generally homothetic (two ellipsoids are homothetic if one of them is a multiple of the other); however, the situation is better when X has the 2-summing property:

4 4 A. ARIAS, T. FIGIEL, W. B. JOHNSON AND G. SCHECHTMAN Lemma.. Assume that the real or complex n-dimensional space X has the 2- summing property and let E be the ellipsoid of minimal volume containing Ball (X). Then n? 2 E is the ellipsoid of maximal volume contained in Ball (X). Proof: Let E 2 be the ellipsoid of maximal volume contained in Ball (X) and for i = ; 2 let i be the Euclidean norm which has for its unit ball E i. Let u be the formal identity map from X to the Euclidean space (X; ), u 2 the identity map from (X; 2 ) to X, and let ; : : :; n be the s-numbers of the Hilbert space operator u u 2 (i.e., the square roots of the eigenvalues of (u u 2 ) u u 2 ). Since 2 (u ) = ku k = and ku 2 k = we have that 2 (u u 2 ). This implies that (:) n 2 : On the other hand vol(e 2 ) vol(e = p n) because n? 2 E Ball (X), so that in the case of real scalars we get, (:2:R) 2 n pn n ; and in the complex case (:2:C) n pn 2n : The only way that (.) and (.2) are true is if = 2 = = n = = p n. But this implies that E 2 = E = p n. Remark.2. B. Maurey has proved a far reaching generalization of Lemma.; namely, that if a space X does not have a unique (up to homothety) distance ellipsoid, then there is a subspace which has the same distance to a Hilbert space as the whole space and which has a unique distance ellipsoid. This implies an unpublished result due to Tomczak-Jaegermann which is stronger than Lemma.; namely, that when the distance is maximal, the minimal and maximal volume ellipsoids must be homothetic. Basic facts about 2-summing operators, and, more generally, p-summing operators, can be found in [LT] and [T-J2]. The 2-summing norm 2 (T ) of an operator Pn kt Ue ik 2 =2 from a space X to a space Y is dened to be the supremum of where the sup is over all norm one operators U from `n2, n = ; 2; : : :, into X and fe i g n is the unit vector basis for `n2. When Y is a Hilbert space, this reduces to the denition given in the rst paragraph of the introduction, and when X is also a Hilbert space, 2 (T ) is the Hilbert-Schmidt norm of T. Note that if U is an operator from `n2 to a subspace X of `, then kuk 2 = P n Ue i 2 [the absolute value is interpreted coordinatewise in `]. So if T goes from X into a space Y, 2 (T ) can be dened intrinsically by 2 (T ) 2 = supf kt x i k 2 : X n x i 2 ; xi 2 X; n = ; 2; 3; :::g;

5 BANACH SPACES WITH THE 2-SUMMING PROPERTY 5 but when Y is an N-dimensional Hilbert space, the \sup" is already achieved for n = N. (Not relevant for this paper but worth noting is that when Y is a general N-dimensional space, the \sup" is achieved for n N 2 [FLM], [T-J2, p. 4] and is estimated up to the multiplicative constant p 2 for n = N [T-J], [T-J2, p. 43].) It is easy to see that 2 is a complete norm on the space of all 2-summing operators from X to Y and that 2 has the ideal property; that is, for any dened composition T T 2 T 3 of operators, 2 (T T 2 T 3 ) kt k 2 (T 2 ) kt 3 k. The typical 2- summing operator is the formal identity mapping I ;2 from L (; ) to L 2 (; ) when is a nite measure. In this case one gets easily that 2 (I ;2 ) = () 2. That such operators are typical is a consequence of the Pietsch factorization theorem ([LT, p. 64], [T-J2, p. 47]), which says that if the space X is isometrically included in a C(K) space, and T : X! Y is 2-summing, then there is a probability measure on K and an operator S from L 2 (K; ) into Y so that T is the restriction of SI ;2 to X and ksk = 2 (T ). That is, there is a probability measure on K so that for each x in X, (:3) kt xk 2 2 (T ) 2 x (x) 2 d(x ): Of course, the converse to the Pietsch factorization theorem follows from the ideal property for 2-summing operators. The qualitative version of Dvoretzky's theorem [T-J2, p. 26] says that every innite dimensional space X contains for every n and > 0 a subspace whose distance to `n2 is less than +. In fact, for a xed n and, the same conclusion is true if dim(x) N(n; ), and the known estimates for N(n; ) are rather good. 2. General results. Here we mention some simple results about spaces which have the 2-summing property, present some motivating examples and then nd a characterization of spaces with that property. Let us say that X satises the k-dimensional 2-summing property if 2 (T ) = kt k for every operator T from X into `k2. Thus every space has the -dimensional 2-summing property, and X has the 2-summing property if X has the k-dimensional 2-summing property for every positive integer k. We introduce this denition because our techniques suggest that a space with the 2-dimensional 2-summing property has the 2-summing property, but we cannot prove this even in the case of real scalars. Throughout this section the scalars can be either IR or C unless explicitly stated otherwise. Proposition 2.. (a) If X has the 2-dimensional 2-summing property, then X is nite dimensional. (b) If X has the k-dimensional 2-summing property for some k, then so does X. (c) If X has the 2-summing property, then X is a maximal distance space. Proof: For (a), we use the fact that `m fails the 2-dimensional 2-summing property for some integer m. In fact, in the real case, m can be taken to be 3 (Example 2.3), while in the complex case, m = 4 suces (remark after example 2.3). Alternatively, one can check that a quotient mapping from ` onto `2 has 2

6 6 A. ARIAS, T. FIGIEL, W. B. JOHNSON AND G. SCHECHTMAN 2-summing norm larger than one, which implies that `m fails the 2-dimensional 2-summing property if m is suciently large. So x a norm one operator u from `m into `22 for which 2 (u) >. By Dvoretzky's theorem, `22 is almost a quotient of every innite dimensional space, so if dimx is suciently large, then there is an operator Q from X into `22 with Ball (`22) Q[Ball (X)] but kqk < 2 (u). Pick z ; : : :; z m in Ball (X) with Qz i = ue i for i = ; 2; : : :; m and dene T from `m into X by T e i = z i, i = ; 2; : : :; m. Then u = QT and 2 (u) 2 (Q) but kqk < 2 (u). For (b), assume that X has the k-dimensional 2-summing property and let T be any norm one operator from X into `k2. It is enough to show that 2 (T u) when u is a norm one operator from `k2 into X. This brings us back to the familiar setting of Hilbert-Schmidt operators: 2 (T u) = 2 (u T ) kt k 2 (u ) = kt kku k = ; the last equality following from the hypothesis that X has the k-dimensional 2- summing property and the fact that, by (a), X is reexive. For (c), let T : X! `n2 be such that kt kkt? k = d(x; `n2); then p n = 2 (T? T ) kt? k 2 (T ) = kt? kkt k = d(x; `2 n ) p n. Therefore, d(x; `n2 p ) = n. Remark. To make the proof of (b) as simple as possible, we used (a) to reduce to the case of reexive spaces. Actually, it is well-known that if 2 (T ) C kt k for every operator from X into a Hilbert space H, then X has the same property. Indeed, it is easy to see that it is enough to consider nite rank operators from X into H and then use local reexivity (see [LT, p. 33]) and a weak approximation argument. Example 2.2. `2 has the 2-summing property. Proof: Let u : `2! `2 2, kuk =. We can assume that a u = b 0 d and that (a + b) 2 + d 2 ; or equivalently, a 2 + b 2 + d 2 + 2ab. For x = (c ; c 2 ) 2 `2 we have that (2:) kuxk 2 =ac 2 + (b 2 + d 2 )c <(ac bc 2 ) ac 2 + (b 2 + d 2 )c ab(c 2 + c 2 2 ): Set = a 2 + ab, so that? (b 2 + d 2 ) + ab. Thus (2.) gives (2:2) kuxk 2 c 2 + c 2 2 (? ): Then since (2.2) is in the form of (.3) with constant, we get that 2 (u). Remark. At least in the complex case, Proposition 2.2 follows from the fact that (`2 ) =, but we thought it desirable to give a direct proof. Another proof is given in [BM]. Proposition 2.6 provides a useful criterion for determining whether a space has the 2-summing property. All of the intuition behind Proposition 2.6 is already contained in Example 2.3:

7 BANACH SPACES WITH THE 2-SUMMING PROPERTY 7 Example 2.3. Real `3 and complex `4 do not have the 2-dimensional 2-summing property. Proof: In real `3, let x = (; 0; p 2 ); x 2 = (0; ; p 2 ) and X = span fx ; x 2 g. We denote X by X when considered as a subspace of L 3 and by X 2 when considered as a subspace of L 3 2. Also denote by I X ;2 the restriction to X of the identity I ;2 from L 3 to L 3 2 (we use the standard convention L n p = L n p() where is the probability space assigning mass n to every point). We claim that ki X k < and that ;2 2(I X ) =. ;2 For every kxk =, ki ;2 xk 2 and we have equality if and only if x is at; i.e., x is an extreme point of the unit ball of L 3. Then we verify that ki;2k X < by checking that X does not contain any one of those vectors. For the second one, dene v : `22! X by ve i = x i for i = ; 2. Then notice that kvk 2 = k x 2 + x 2 2 k =, where x 2 + x 2 2 is taken coordinatewise in L 3, and 2 (I;2) X 2 2 (I;2v) X 2 = kx k kx 2 k 2 2 =. The equality follows, since 2 (I;2) X 2 (I ;2 ) =. We have thus proved that X does not have the 2- dimensional 2-summing property. To conclude, dene u : L 3! X 2 by u = P I ;2, where P is the orthogonal proection from L 3 2 onto X 2. We claim that kuk < and that 2 (u) =. If kxk =, then ki ;2 xk 2 = i x is at, and kp xk 2 = kxk 2 i x 2 X. Since these conditions are mutually exclusive we conclude that kuk <. But = kp k 2 2 (I ;2 ) 2 2 (u) 2 2 (uv) 2 = kp x k kp x 2 k 2 2 =. The proof for complex `4 is similar: Let x = (; 0; p 2 ; p i 2 ), x 2 = (; 0; p 2 ; p 2 ) and X = span fx ; x 2 g. It is easily checked that X does not contain any at vectors and that x 2 + x 2 2 coordinatewise. Remark. We shall see in section 4 that complex `3 has the 2-summing property. Proposition 2.4. Let X be an n-dimensional subspace of C(K), K compact; u : X! `k2 a map satisfying 2 (u) = and v : `k2! X satisfying kvk = and 2 (u) = 2 (uv). Pietsch's factorization theorem gives the following diagram for some probability on K and some norm one operator : X 2! `k2: `k2 C(K) i " I;2???! L 2 (K; ) # P v?! X I;2 X???! X 2 u &. `k2 Let Y = v(`k2). Then is an isometry on Y 2. Proof: We have = 2 (uv) 2 = kx = kuve k 2 = kx = ki ;2 ve k 2 kx = ki ;2 ve k 2 kk 2 2 (I ;2 ) 2 = ; so ki ;2 ve k = ki ;2 ve k for each k. Recalling the elementary fact that if S is an operator between Hilbert spaces, then fx : ksxk = ksk kxk g is a linear subspace of the domain of S, we conclude that is an isometry on Y 2.

8 8 A. ARIAS, T. FIGIEL, W. B. JOHNSON AND G. SCHECHTMAN Remark. Let X be an n-dimensional space; v : `n2! X the maximum volume ellipsoid map and set u = p n v?. It is well known that 2 (u) =, and since 2 (uv) = p n 2 (I) = we conclude that : X 2! `n2 is an isometry. Moreover, if X is of maximal distance, u is the minimal volume ellipsoid map (see Remark.2). Corollary 2.5. Let X be an n-dimensional subspace of C(K), K compact, and u : X! `n2 be an onto map satisfying 2(u) =. Suppose that for every orthogonal proection P from `n2 onto a proper subspace we have 2(P u) <. Then : X 2! `n2 is an isometry. ( is the map appearing in Proposition 2.4). Proof: Let v : `n2! X be such that kvk = and 2 (u) = 2 (uv) and let P be the orthogonal proection from `n2 onto u(v(`n2 )) = (Y 2). We clearly have that 2 (P u) 2 (P uv) = 2 (uv) =. Hence, the range of P cannot be a proper subspace of X 2 and therefore is an isometry on X 2. In the next proposition interpret 0 as 0. 0 Proposition 2.6. Let X be an n-dimensional subspace of C(K), K compact, and k n. Then 2 (u) sup kuk : u : X! `k2 = sup kp I X ;2 k : 2 P(K); P 2 = P, kp k =, rank P k, 2 (P I X ;2) = where P(K) consists of all the probability measures on K, and I ;2 is the canonical identity from C(K) to L 2 (K; ). Proof: It is clear that the left hand side dominates the right one. To prove the other inequality let u : X! `k2 be such that 2 (u) =. Then nd v : `k2! X such that kvk = and 2 (uv) =. Let Q be the orthogonal proection from `k2 onto uv(`k2 ) = (Y 2) (with the notation of Proposition 2.4), and P be the orthogonal proection from X 2 onto Y 2. Notice that Qu = P I;2: X Since is an isometry on Y 2 we have that 2 (P I;2) X = 2 (Qu) = and that kp I;2k X = kquk kuk. Therefore kp I X k 2(u) ;2 kuk : ; Proposition 2.6 has a nice form when X is 2-dimensional because then we do not need to take the orthogonal proection on X 2. Indeed, if P has rank one then it is clear that 2 (P I X ;2) = kp I X ;2k. If is a probability measure on K with support K 0, then ki X ;2k < i Ball(X) does not contain any \at" vector on K 0 ; i.e., whenever x 2 X and kxk =, then we have that x K0 6. On the other hand, 2 (I X ;2) = i there exist vectors x ; x 2 in X such that x 2 + x 2 2 on K and x 2 + x 2 2 on K 0. To see why the second statement is true, nd v : `22! X satisfying kvk = and 2 (I X ;2v) =. Then let x i = ve i for i = ; 2 and the result is easily checked for these vectors. Since every closed subset of a compact metric space is the support of some probability measure, this discussion proves:

9 BANACH SPACES WITH THE 2-SUMMING PROPERTY 9 Corollary 2.7.a. Let X be a 2-dimensional subspace of C(K), K compact metric. Then X does not have the 2-summing property if and only if there exist vectors x ; x 2 in X and a closed set K 0 K with K0 x 2 + x 2 2 such that for every x 2 X with kxk =, we have that x K0 6. If X C(K) contains a vector x, then for every probability measure on K we have ki ;2 xk = kxk; hence, ki X ;2k = and we have Corollary 2.8. Let X be a 2-dimensional subspace of C(K), K compact. If X contains a vector x, x on K then X has the 2-summing property. This applies immediately to `2 (both real and complex) and also to `2 (again real and complex) if embedded in a canonical way. It also implies that there are a continuum of pairwise nonisometric two dimensional complex spaces which have the 2-summing property. We shall see later that the real 2-summing property is quite dierent from the complex version. For the moment, take X = span f(; 0; p 2 ); (0; ; p 2 )g inside `3. We proved in Example 2.3 that real X does not have the 2-summing property. However, complex X has it. To see this, notice that (; 0; p 2 ) + i(0; ; p 2 ) is \at" and hence Corollary 2.8 implies the result. The dierence can be explained by saying that it is easier to get \at" vectors in the complex setting (see Proposition 4.4). Let us return to the discussion following Proposition 2.6. Suppose that X is an n-dimensional subspace of C(K), is a probability measure on K with support K 0 K, and P is an orthogonal proection from X 2 L 2 () onto a subspace Y 2. Notice that P 2 (P I;2) X = if and only if there exist vectors x ; x 2 ; : : :; x n in Y n with K0 x = 2 and each vector I ;2 x is in Y 2. (Keep in mind that Y 2 is relatively L 2 (){closed in X 2, hence if y 2 Y, z 2 X, and K0 y = K0 z, then also z is in Y.) Similarly, P I X ;2 = if and only if there exists a single vector x in Y with K0 x. Thus we get a version of Corollary 2.7.a for all nite dimensional spaces: Corollary 2.7.b. Let X be a nite dimensional subspace of C(K), K compact metric. Then X does not have the 2-summing property if and only P if there exist n vectors x ; x 2 ; : : :; x n in X and a closed set K 0 K with K0 x = 2 such that for every x 2 X with K0 x, we have that K0 x is not in span f K0 x ; K0 x 2 ; : : :; K0 x n g. In the complex case, the 2-summing property is not hereditary, since complex `3 has the 2-summing property but `2 is the only two dimensional subspace of it which has the 2-summing property (see Theorem 4.2 and Proposition 5.7.) Nevertheless: Proposition 2.9. Let X be a subspace of `N which has the 2-summing property. Then every subspace of X has the 2-summing property. Proof: Assume that X has a subspace which fails the 2-summing property. Write K = f; 2; : : :; Ng. Since `N = C(K), we can apply Corollary 2.7.b. There exists a P subset K 0 K for which we can nd vectors x ; x 2 ; : : :; x n in X with n K0 x = 2 such that no norm one vector in Y span fx ; : : :; x n g is unimodular on K 0. We can also assume that K 0 is maximal with respect to

10 0 A. ARIAS, T. FIGIEL, W. B. JOHNSON AND G. SCHECHTMAN this property; in particular, P n = x 2 is strictly less than one o K 0 and hence P n = K 0 x 2 <? for some > 0. On the other hand, since X has the 2-summing property, there exists a vector y 2 Y which is unimodular on K 0 (and whose restriction to K 0 agrees with the restriction to K 0 of some unit vector in X). Evidently kyk >. Thus there exists > > 0 so that z = z 2 y 2 + P n = (? ) 2 x 2 satises k K0 zk =. But then z, a square function of a system from Y, is unimodular on a set which properly contains K 0 ; this contradicts the maximality of K 0. For any real Banach space F, let F C denote the linear space F F with complex structure dened by means of the formula (a+bi)(f f 2 ) = (af?bf 2 )(af 2 +bf ). There is a natural topology on F C, namely F C is homeomorphic with the direct sum of two real Banach spaces F F. In two special cases we shall dene a norm on F C which will make it a complex Banach space. First, if E is a linear subspace of some real `k, we endow E C with the norm induced from complex `k by means of the obvious embedding. Secondly, if E = H is a Hilbert space, then H C is normed by means of the formula kh h 2 k = (kh k 2 + kh 2 k 2 ) =2. These two denitions are consistent, because H is isometric to a subspace of real `k only if dim IR H. Now, if T : F! G is a linear operator, we dene T C : F C! G C by the formula T C (f f 2 ) = (T f T f 2 ). Proposition 2.0. Let E be a subspace of real `k and S : E! `22 be a real-linear mapping from E into a 2-dimensional real Hilbert space. Then 2 (S C ) = 2 (S) = ks C k : Proof: From the discussion in Section we see that there are vectors x; y in E such that x 2 + y 2 (interpreted coordinatewise) and 2 (S) =? ksxk 2 + ksyk 2 =2 : Observe that, by our denition, kx yk EC = kx + iyk l k. It follows that (C) 2 (S C ) ks C k ks C (x y)k HC =? ksxk 2 + ksyk 2 =2 = 2 (S); hence it remains to verify that 2 (S C ) 2 (S). Take u; w in E C with u 2 + w 2 and 2 (S C ) 2 = ks C (u)k 2 + ks C (w)k 2. Interpreting real and imaginary parts of vectors in `k coordinatewise, we see that <u; =u are in E and similarly for w. Moreover, ks C (u)k 2 + ks C (w)k 2 = ks<uk 2 + ks=uk 2 + ks<wk 2 + ks=wk 2 : This last quantity is at most 2 (S) since <u 2 +=u 2 +<w 2 +=w 2 = u 2 +w 2. It is easy to determine when a complex-linear operator is the complexication of a real-linear operator:

11 BANACH SPACES WITH THE 2-SUMMING PROPERTY Proposition 2.. Let E be a real Banach space and let G be a complex Hilbert space. Let T : E C! G be a complex-linear continuous operator. The following conditions are equivalent: (i) There is a real Hilbert space H and continuous linear operators S : E! H, U : H C! G, such that T = U S C and U is an isometric embedding. (ii) For each e; e 0 2 E one has kt (e + ie 0 )k = kt (e? ie 0 )k. (iii) For each e; e 0 2 E one has =(T e; T e 0 ) = 0. (iv) There is a subset E 0 of E such that the linear span of E 0 is dense in E and for each e, e 0 in E 0, =(T e; T e 0 ) = 0. Proof: (i) implies (ii), because kt (e ie 0 )k = ks C (e ie 0 )k = (ksek 2 +kse 0 k 2 ) =2. To see that (ii) implies (i) we let H denote the closure of T (E) in G. Observe that, if x = e e 0 2 E C, then using the parallelogram identity we obtain kt e + it e 0 k 2 = 2 (kt e + it e0 k 2 + kt e? it e 0 k 2 ) = kt ek 2 + kt e 0 k 2 : Hence H ih is linearly isometric to H C and, if U denotes the natural embedding and S = T E, then we have T = U S C. It is clear that (iii) and (iv) are equivalent. On the other hand, the identity kt (e + ie 0 )k 2? kt (e? ie 0 )k 2 = (T e + it e 0 ; T e + it e 0 )? (T e? it e 0 ; T e? it e 0 ) makes it obvious that (ii) and (iii) are equivalent. = 2i((T e 0 ; T e)? (T e; T e 0 )) = 4=(T e; T e 0 ); 3. The real case. Throughout this section we deal with spaces over the reals. Example 2.3 and Proposition 2. imply that there are many norm one operators from real `3 into `22 whose 2-summing norm is larger than one. Proposition 3.. Let u be an operator from real `3 to `2 such that ue 2 ; ue 2 ; ue 3 have norm one and every two of them are linearly independent. Then 2 (u) >. Proof: Let u : `3! `22 satisfy the assumptions of Proposition 3., and set x i = ue i, i = ; 2; 3. Notice that `3 embeds isometrically via the natural evaluation mapping into C(K), where K = f(; ; ); (; ;?); (;?; ); (?; ; )g), regarded as a subset of `3 = (`3 ). Assume that 2 (u) = and consider the Pietsch factorization diagram: C(K) i " I;2 X I X ;2 u &???! L 2 (K; ) # P???! X 2. So kk = 2 (u) =. We claim that dimx 2 = 3. (This is not automatic since the support K 0 of may not be all of K.) Indeed, any two of ff ; f 2 ; f 3 g are `22

12 2 A. ARIAS, T. FIGIEL, W. B. JOHNSON AND G. SCHECHTMAN linearly independent, since this is true of ff ; f 2 ; f 3 g, so the cardinality of K 0 is at least three. But ff il g 3 is linearly independent if L is any three (or four) element subset of K. To complete the proof, ust recall that no norm one linear operator from `32 into `22 achieves its norm at three linearly independent vectors since, e. g., the set fx 2 `32 : kxk = kk kxkg is a subspace of `32. Having treated the \worst" case of `n, it is easy to formulate a version of Proposition 3. for general spaces. Corollary 3.2. Let X be a real space. (a) If T is a norm one operator from X into `22 such that for some z ; z 2 ; z 3 in Ball (X), T z ; T z 2 ; T z 3 have norm one and every two of them are linearly independent, then 2 (T ) >. (b) If there exist u : `22! X and x ; x 2 ; x 3 2 Ball (`22) such that kuk = = kux i k = for i = ; 2; 3 and every two of x ; x 2 ; x 3 are linearly independent, then X does not have the 2-dimensional 2-summing property. Proof: For (a), dene w : `3! X by we i = z i ; i 3. Then T w satises the hypothesis of Proposition 3.. Therefore 2 (T w) > ; and since kwk = we have that 2 (T ) >. For (b) it is enough to prove that X does not have the 2-dimensional 2-summing property. If x i 2 Ball (X ), i = ; 2; 3 satisfy hx i ; ux i i =, then ku x i k hu x i ; x ii = hx i ; ux ii = ; therefore, u x i = x i for i = ; 2; 3 and the previous part gives us that 2 (u ) >. We are now ready for the main result of this section: Theorem 3.3. If X is a real space of dimension at least three, then X does not have the 2-summing property. Consequently, the only real spaces which have the 2-summing property are IR and `2. Proof: The \consequently" follows from the rst statement and Proposition 2.(c) because in the real case `2 is the only 2-dimensional maximal distance space. This is an unpublished result of Davis and the second author; for a proof see Lewis' paper [L]. So assume that X is IR n, n 3, under some norm and has the 2-summing property. We can also assume that the usual Euclidean ball E is the ellipsoid of minimal volume containing Ball (X), and we use to denote the P Euclidean norm. m By John's theorem, there exist ; ; m > 0 such that i = n and y ; ; y m 2 X outside contact P points (i.e., ky i k = y i = for i = ; 2; : : :; m) such m that every x 2 X satises x = ihx; y i iy i. Recall that E= p n Ball (X), in fact, by Lemma., E= p n is the ellipsoid of maximal volume contained in Ball (X). If x is an inside contact point (i.e., kxk = and x = = p n), Milman and Wolfson [MW] proved that (3:) hx; y i i = n for every i = ; 2; : : :; m:

13 BANACH SPACES WITH THE 2-SUMMING PROPERTY 3 To see this, observe that fz 2 X : hz; xi = =ng supports E= p n at the inside contact point x, hence{draw a picture{the norm of h; xi in X is =n. Thus X m n = hx; xi = i hx; y i i 2 mx i kh; xik 2 X ky ik 2 = n : This implies that hx; y i i = n for i = ; 2; : : :; m and proves (3.). In other words, the norm one (in X ) functional h; nxi norms all of the y i 's as well as x. This implies that both convfy i : hy i ; nxi = g and convfy i : hy i ; nxi =?g are subsets of the unit sphere of X. Now we know that X has maximal distance, hence at least one inside contact point exists, whence Ball (X) has at least two \at" faces. The next step is to observe that we can nd n linearly independent outside contact points y ; ; y n and n linearly independent inside contact points x ; ; x n satisfying (3.). This will give us enough faces on Ball (X) so that a 2-dimensional section will be a hexagon and we can apply Corollary 3.2 (b). Now the John representation of the identity gives the existence of the outside contact points, and since E= p n is the ellipsoid of maximal volume contained in Ball (X), another application of John's theorem gives the inside contact points. So let y ; ; y n be linearly independent outside contact points, and x ; ; x n linearly independent inside contact points satisfying (3.), such that for the rst three of them we have, hnx ; y i = hnx ; y 2 i = hnx ; y 3 i = hnx 2 ; y i = hnx 2 ; y 2 i = hnx 2 ; y 3 i =? hnx 3 ; y i = hnx 3 ; y 2 i =? hnx 3 ; y 3 i = : (We are allowed to change signs and renumber the contact points). Let v denote the linear map P from span fy ; y 2 ; y 3 g into l 3 which takes y i to e i. Then kvyk = kyk 3 if y = iy i and either 2 0 or 3 0, hence the restriction of v to F = span f y+y2 ; y+y3 g is an isometry. But then v[ball (F )] is a regular hexagon, 2 2 which implies that the maximum volume ellipsoid for Ball (F ) touches the unit sphere of F at six points. We nish by applying Corollary 3.2 (b). Lewis [L] proved that every real maximal distance space of dimension at least two contains a subspace isometrically isomorphic to `2; that is, a subspace whose unit ball is a parallelogram. In view of Remark.2, the proof of Theorem 3.3 yields: Corollary 3.4. If X is a real maximal distance space of dimension at least three, then X has a subspace whose unit ball is a regular hexagon. 4. The complex case. As was mentioned in the introduction, the study of the 2-summing property in the complex case is a priori more complicated than in the real case even for two dimensional spaces simply because in the real case there is only one 2-dimensional maximal distance space, while in the complex space there are at least two, `2

14 4 A. ARIAS, T. FIGIEL, W. B. JOHNSON AND G. SCHECHTMAN and its dual. In fact, it is not dicult to construct other complex 2-dimensional maximal distance spaces without using Corollary 2.8. One way is to use John's representation theorem; this was done independently by Gowers and Tomczak- Jaegermann [both unpublished] a couple of years ago in order to construct real 4-dimensional maximal distance spaces whose unit ball is not strictly convex; this approach also yields maximal distance 2-dimensional complex spaces dierent from `2 and `2. However, a simpler way of seeing that there are many maximal distance 2-dimensional complex spaces is via Proposition 4.: Proposition 4.. Suppose that X is an n-dimensional complex space which, as a real space, contains a real n-dimensional subspace which has maximal distance to real `n2. Then X has maximal distance to complex `n2. Proof: Let Y be a real n-dimensional subspace of X which has maximal distance to real `n2, and suppose that T is any complex linear isomorphism from X to complex `n2. But considered as a real space, complex `n2 is ust real `2n 2, and so the restriction of T to Y is a real linear isomorphism from Y to a real n-dimensional Hilbert space, hence by the assumption on Y, so the desired conclusion follows. kt k T? TY (TY )? p n; Proposition 4. makes it easy to construct in an elementary manner 2-dimensional complex maximal distance spaces which are not isometric to either `2 or `2. For example, in complex `3, let x = (; 0; a) and y = (0; ; b) with a + b _ a? b but a + b >, and set X = span fx; yg. In Proposition 4.4 we prove that every 2-dimensional subspace of complex `3 is a maximal distance space, by proving that they all have the 2-summing property. However, in the complex setting, the 2-summing property is not restricted to two dimensional spaces. In fact, we do not know a good bound for the dimension of complex spaces which have the 2-summing property, although we suspect that dimension three is the limit. In dimension three itself, we know of only two examples, `3 and its dual. We can prove Theorem 4.2 using Proposition 2.6 and The proof of Proposition 4.4 (see Remark 4.6). The proof we give is of independent interest. Theorem 4.2. Complex `3 has the 2-summing property. Hence also `3 has the 2-summing property. Proof: The proof reduces the theorem to the following calculus lemma, which we prove after giving the reduction: Lemma 4.3. Fix arbitrary complex numbers, 2, 3 and dene a function f on the bidisk by f(z ; z 2 ) = + z + 2 z z z p? z 2p? z 2 2 : Then the maximum of f is attained at some point on the two dimensional torus; that is, when z = z 2 =. Reduction to Lemma 4.3: Let K = fgtt, where Tis the unit circle, and regard `3 as the subspace of C(K) spanned by the coordinate proections f 0; f ; f 2.

15 BANACH SPACES WITH THE 2-SUMMING PROPERTY 5 Let u be a norm one operator from `3 into `2 (complex scalars). We want to show that 2 (u) =. That is, we want a probability on K so that for each x 2 `3 kuxk 2 x(k) 2 d(k): Notice that if we add to the uf 's mutually orthogonal vectors which are also orthogonal to the range of u, the 2 -norm of the resulting operator can only increase. Thus we assume, without loss of generality, that kuf k = for = 0; ; 2, and that uf 0 = 0 ; uf = 0 + ; uf 2 = ;, where f 0 ; ; 2 g is an orthonormal set of `2. E = span ff 0 ; f ; f 2 ; f f2 g by Dene a linear functional F on F f 0 = ; F f = F f 2 = 2 ; F (f f2 ) = : Claim: kf k = as a linear functional on (E; kk C(K) ). Assume the claim. Then by the Hahn-Banach theorem, F can be extended to a norm one linear functional, also denoted by F, on C(K). Since kf k = = F f 0, F is given by integration against a probability, say,. Now by the denition of F, the mapping v : f0; f 0 ; f ; f 2 g! `2 dened by v0 = 0, vf = uf for = 0; ; 2 is L 2 ()-to-`2 inner product preserving, hence an L 2 -isometry, whence extends to a linear isometry from (span ff 0 ; f ; f 2 g; kk L2() ) into `2. This shows that 2 (u), as desired. Proof of claim: The claim says that for all f g 3 =0 in C 4, ( ) But the left hand side is dominated by sup (z ;z 2)2TT 0 + z + 2 z z z 2 : p? 2 p? 2 2 ; so the claim follows from Lemma 4.3. Proof of Lemma 4.3: For xed z 2, you can rotate z maximum to see that at the f(z ; z 2 ) = + 2 z 2 + z + 3 z p? z 2 p? z2 2 : As z varies, (z ; p? z 2 ) varies over the unit sphere of `22, so at the maximum f(z ; z 2 ) = + 2 z 2 + p + 3 z (? z 2 2 ): In particular, z = if and only if z 2 =. The last expression can be rewritten as q + 2 z <( 3 z 2 ):

16 6 A. ARIAS, T. FIGIEL, W. B. JOHNSON AND G. SCHECHTMAN Choose so that e i 3 is purely imaginary; so <( 3 z 2 ) does not change if you add a real multiple of e i to z 2. Assume rst that 2 6= 0. Then, for any > 0, by adding either e i or?e i to z 2 you increase the rst term; this means that at the maximum z 2 =. The case where 2 = 0 can be handled in a similar way. Remarks.. A similar argument reduces the problem of whether `4 has the 2-summing property to a calculus problem; however, in the remark after Example 2.3 we give a simple direct argument that `4 fails the 2-summing property. 2. Bagchi and Misra [BM] give a dierent reduction of Theorem 4.2 to a variation of Lemma 4.3. Their argument may be more appealing to operator theorists. 3. The following Proposition is a consequence of Proposition 2.9 and Theorem 4.2. We indicate a shorter argument. Proposition 4.4. Every two dimensional subspace of `3 has the 2-summing property. Proof: Assume that X `3 is two dimensional. Applying `3 isometries we assume that X has a basis of the form f(; 0; a ); (0; ; a 2 )g where a i for i = ; 2. If a + a 2 then X is isometric to `2 and by Example 2.2 it has the 2-summing property. So assume that a +a 2 >. It is easy to nd, 0 < 2 such that a + e i a 2 =. Hence, (; 0; a ) + e i (0; ; a 2 ) is \at" and the result follows from Corollary 2.8. It is interesting to notice that Proposition 4.4 is equivalent to the following calculus formulation. Proposition 4.5. Given any complex numbers c ; c 2 ; c 3 and d ; d 2 ; d 3 with c 2 + d 2 for = ; 2; 3, suppose that ; satisfy (4:) max c + d 2 8; 2 C : =;2;3 Then To see the equivalence of Propositions 4.4 and 4.5, let X be a 2-dimensional subspace of `3 and X?!`22 u v a norm one operator. Choose `22?!X of norm one so that 2 (u) = 2 (uv). We can assume that uv is diagonal; say, uv(e ) = e and uv(e 2 ) = e 2. For = ; 2 set x = v(e ) and write x = (c ; c 2 ; c 3 ), x 2 = (d ; d 2 ; d 3 ). So 2 (u) 2 = 2 + 2, while kvk 2 = max c 2 + d 2 =. The =;2;3 implication Proposition 4.5 ) Proposition 4.4 follows by noticing that u having norm at most one is equivalent to the inequality (4.). Similar considerations yield the easier reverse implication. We do not see a really simple proof of the calculus reformulation of Proposition 4.4 without using Pietsch's factorization theorem. However, a similar reduction of the weaker statement that every 2-dimensional subspace of `3 has maximal distance to `2 2 produces a calculus statement which is very easy to prove. Indeed, given a 2-dimensional subspace X of `3, we can choose norm one operators X?!`22 u and v `22?!X so that duv = I`2 2, where d is the distance from X to `22. We can choose the orthonormal basis e ; e 2 so that = kvk = kve k and dene x ; x 2, the c i 's,and

17 BANACH SPACES WITH THE 2-SUMMING PROPERTY 7 the d i 's as in the discussion above. Since kx k =, we can assume, without loss of generality, that c =. (4.) holds with = = d, and we want to see that this implies that d p 2; i.e., that So we only need to get and of modulus one to make the right side of (4.) one. Since c =, d = 0, any such choice makes c +d =. Choose to make c 2 0; then c 2 +d 2 as long as d 2 has nonpositive real part, which happens as long as is on a certain closed semicircle. Similarly, c 3 + d 3 as long as is on another closed semicircle. Since any two closed semicircles of the unit circle intersect, the desired choice of and can be made. Remark We can use Propositions 2.6 and 4.4 to prove that `3 has the 2-summing property. Use the notation of Proposition 2.4 and let P be an orthogonal proection on L 3 2(). If the rank is one, there is nothing to prove. If the rank is three then we clearly have that ki ;2 k =, and if the rank is two, then the proof of Proposition 4.4 implies the result. 2. It is natural to ask if the only subspaces of complex L with the 2-summing property are `2 and `3. The answer is yes because a subspace of L of maximal distance is already an `k space. We prove this in the appendix, Proposition Appendix. In this section we present some related results. Proposition 5.. Every subspace of complex `3 is the complexication of a subspace of real `3 The proof of Proposition 5. follows easily from the next two lemmas. Recall that a vector in `k is said to be at if all of its coordinates are unimodular. Lemma 5.2. Suppose that the subspace X of complex `3 is not linearly isometric to l 2. Then X contains two linearly independent at vectors, say f and f 2. Moreover, each at vector in X is of the form f, where 2 f; 2g and =. Proof: Applying `3 isometries, we may assume that X is spanned by two vectors of the form x = (; 0; a) and y = (0; ; b) where a; b 2 C with a; b. Put w = x? y. For w to be at one needs that = and a? b =. Observe that, since X is not linearly isometric to l, 2 we have a + b > ;, in particular ab 6= 0. Thus 2 C should belong to the intersection of the unit circle fz : z = g and the circle fz : z? a=b = =bg, hence there are at most two solutions for. Thus it will suce to check that the two circles have a point in common and that they are not tangent at that point. Since the second circle has a bigger radius, this amounts to verifying the strict inequalities? < b a b? 0 < + These inequalities are obvious, because we have a + b >, a and b > 0. Lemma 5.3. Suppose that the 2-dimensional subspace X of complex `k is spanned by two linearly independent vectors y; z such that y = z. Then there is a linear : b

18 8 A. ARIAS, T. FIGIEL, W. B. JOHNSON AND G. SCHECHTMAN isometry of l k such that y = z. In particular, (X) is spanned by two vectors v; w, all of whose coordinates are real and which satisfy (v 2 + w 2 ) =2 = y. Proof. Write y = (y ; y 2 ; ; y k ) and z = (z ; z 2 ; ; z k ). For = ; 2; ; k, let be a complex number with = such that y = z. Such numbers obviously exist, we may also impose the condition < y 0. Now the isometry can be dened by the formula (x ; x 2 ; ; x k ) = ( x ; 2 x 2 ; ; k x k ). Clearly, the vectors v = (y + z) and w = 2 2i (y? z) have the required property. Propositions 2.9, 2.0, and 5. suggest an alternate method for proving Proposition 4.4 since they combine to take care of the case where the operator achieves its norm at two \at" vectors: Lemma 5.4. Let X be a two-dimensional subspace of complex `3 and T a complexlinear operator from X into a Hilbert space such that kt xk = kt yk, where x; y are linearly independent vectors in X for which x = y. Then 2 (T ) = kt k. Proof: In view of Lemma 5.3 we can assume that there are vectors v, w in X all of whose coordinates are real for which v 2 + w 2 = x 2, x = v + iw, and y = v? iw. Thus if we let E be the collection of real-linear combinations of fv; wg, we can regard X as the complexication E C of E. The assumption on fx; yg means that the pair fv; wg satises condition (iv) in Proposition 2., hence condition (i) of Proposition 2. says that T is the complexication of the restriction of T to E, whence by Proposition 2.0, 2 (T ) = kt k. The next lemma takes care of the case where the operator achieves its norm at a non-at vector. Lemma 5.5. Suppose that X is a 2-dimensional subspace of complex `3 and the norm one operator X?!`2 T achieves its norm at a non-at vector x = (x 2 ; x 2 ; x 3 ) on the unit sphere of X. Then there are norm one operators X?!`2 V W and `2?!`2 2 so that T = W V. Consequently, by Example 2.2, 2 (T ) =. Proof: Since the result is trivial if T has rank one, we assume that T has rank two. This implies that two coordinates of x, say, x and x 2, are unimodular. [Indeed, suppose, for example, that x 2 and x 3 are both less than?. Take y in X with y = 0 and 0 < kyk <, so kx yk =. But since `2 is strictly convex, kt (x + y)k > kt xk for either = or =?.] We may also assume that X contains two vectors, say y; w, such that y = (; 0; y 3 ) and w = (0; ; w 3 ). (Otherwise, X is spanned by two vectors with disoint supports and the conclusion of the Lemma is obvious.) Let be the function dened for z 2 C by the formula (z) = kt (x y + zw)k 2 : Observe that (z) = kt (x y) + zt wk 2 is a quadratic function of (<z; =z), which at innity is asymptotically equal to mz 2, where m = kt wk 2 > 0. It follows that there is a number z 0 2 C such that (z) = mz? z (z 0 ) for every z 2 C. It is obvious now that either z 0 = 0, so that is constant on the unit circle, or else has a unique local maximum on the unit circle (which must also be the global maximum of on the circle). Note that at z = x 2 the function does have a

19 BANACH SPACES WITH THE 2-SUMMING PROPERTY 9 local maximum. In each case it follows that, for every z with z, we have (z) (x 2 ) =. Put V (z y + z 2 w) = (z ; z 2 ). Then V : X! l 2 and kv k. The latter property of can be restated as follows: if (z ; z 2 ) 2 C 2, z = z 2 =, and u = z =x, then kt (z y + z 2 w)k = p (z 2 =u) = maxfz ; z 2 g = kv (z y + z 2 w)k : Since W = T V? attains its norm at an extreme point, we have ust checked that kw k. Now we can give an alternate: Proof of Proposition 4.4: Suppose that the two-dimensional subspace X of complex `3 fails the 2-summing property. Let T 0 be a norm one linear mapping of X into a Hilbert space H whose 2-summing norm is maximal among all norm one linear maps of X into H. Thus 2 (T 0 ) > kt 0 k = ; in particular, T 0 is of rank >. By Lemma 5.5, T 0 does not attain its norm at any non-at vector. Hence, if f ; f 2 2 X are the two at vectors described in Lemma 5.2, then kt 0 k = maxfkt 0 f k ; kt 0 f 2 kg. Using Lemma 5.4, we rule out the possibility that kt 0 f k = kt 0 f 2 k. Assume that kt 0 k = kt 0 f k > kt 0 f 2 k. Observe that for every > 0 there is an operator T : X! H such that kt? T 0 k < ; kt f k = kt 0 f k and the inequality kt xk kt 0 xk is possible only if x = f. Since rank T is >, the latter property of T implies that 2 (T ) > 2 (T 0 ). By the maximality of 2 (T 0 ), we infer that kt k >. However, T does attain its norm somewhere and it cannot happen at any at vector, because as soon as < kt 0 f k? kt 0 f 2 k we have kt f 2 k < kt f k =. Using Lemma 5.5 again, we infer that 2 (T ) = kt k. Now, letting tend to 0, we obtain that 2 (T 0 ) = kt 0 k, which contradicts our initial assumption. To nd 3-dimensional subspaces other than `3 and `3 which have the 2-summing property, it is natural to look inside `4. However: Proposition 5.6. Let X be a three dimensional subspace of complex `4 not isometric to `3. Then X does not have the 2-dimensional 2-summing property. Proof: First notice that without loss of generality X is spanned by three vectors of the form (; 0; 0; a ); (0; ; 0; a 2 ); (0; 0; ; a 3 ) with a ; a 2 ; a 3 non-negative real numbers satisfying a ; a 2 ; a 3 and a + a 2 + a 3 >. Indeed, let (b ; b 2 ; b 3 ; b 4 ) be a non-zero vector annihilating X and assume b 4 b ; b 2 ; b 3. Applying `4 isometries we may assume that b i =b 4 are non-positive reals. Put a i =?b i =b 4. Note that, since X is not isometric to `3, a + a 2 + a 3 >. Fix ; ; ; non-negative real numbers and '; 2 C with ' = = and consider the following two vectors in X: x =(; '; ; a + 'a 2 + a 3 ) y =(; 0;? ; a? a 3 ):