Yet more on distortion I would like to elaborate a little bit on my previous question which can be found 
here.
Firstly, let me recall that a separable Banach space $(X, \| \cdot \|)$ is said to be
arbitrarily distortable if for every $r > 1$ there exists an equivalent norm $| \cdot |$
on $X$ such that for every infinite-dimensional subspace $Y$ of $X$ we can find a pair of
vectors $x, y$ in $Y$ such that $\|x\|=\|y\|=1$ and $|x| / |y| >r$. If $X$ has a Schauder
basis $(e_n)$, then this definition is equivalent to the following:

For every $r > 1$ there exists an equivalent norm $| \cdot |$ on $X$ such that for every
  normalized block sequence $(v_n)$ of $(e_n)$ there exist a non-empty finite subset $F$
  of $\mathbb{N}$ and a pair of vectors $x, y$ in $span\{v_n: n\in F\}$ such that $\|x\|=\|y\|=1$
  and $|x| / |y| >r$.

This equivalence gives us no hind of where the finite set $F$ is located. In other words:
if a Banach space $X$ with a Schauder basis is arbitrarily distortable, then where do
we have to search in order to find the vectors verifying that $X$ is arbitrarily distortable?
Now, there are various ways of quantifying Banach space properties and my question
is towards understanding who the "difficulty" for finding these vectors can be quantified.
The main tool will be certain families of finite subsets of $\mathbb{N}$. These families were discovered (independently) by two groups of researchers: Banach space theorists (Schreier families; see 1 below) and Ramsey theorists (uniform families; see 2 below). In particular, for the discussion below we need for every countable ordinal $\xi\geq 1$ a family $F_\xi$ of finite subsets of $\mathbb{N}$ such that:


*

*$F_\xi$ is regular (i.e. compact, hereditary and spreading; I am sorry for not giving
the precise definition of these notions but this would make the post too long; but I
will be happy to answer to any comment).

*The families are increasing (with respect to $\xi$) both in size and complexity. That is, the "order" of $F_\xi$ is at least $\xi$ and if $\zeta<\xi$ there there exists $k$ such that all subsets of $F_\zeta$ whose minimum is greater than $k$ belong to $F_\xi$.

*For $\xi=1$, let us take the Schreier family consisting of all finite subsets of
$\mathbb{N}$ whose size (or cardinality if you prefer) is less than or equal to their minimum.


There many examples of such families, all constructed using transfinite induction. 
Some of them have extra important properties. For concreteness (and to simplify things)
let us work with the Schreier families.
Now we come to the following:

Definition: Let $(X,\| \cdot \|)$ be a Banach space with a Schauder basis $(e_n)$ and $\xi$ be a countable ordinal with $\xi\geq 1$. Let us say that $X$ is $\xi$-arbitrarily
  distortable if for every $r > 1$ there exists an equivalent norm $| \cdot |$ on $X$
  such that for every normalized block sequence $(v_n)$ of $(e_n)$ there exist a non-empty
  set $F$ belonging to the family $F_\xi$ and a pair of vectors $x, y$ in $span\{v_n: n\in F\}$ such
  that $\|x\|=\|y\|=1$ and $|x| / |y| >r$.

In other words, if $X$ is $\xi$-arbitrarily distortable, then we have narrow down the
search for the critical set $F$; it has to belong to an a priori given "nice" family
of finite subsets of $\mathbb{N}$.
For every Banach space $(X,\| \cdot \|)$ with a Schauder basis $(e_n)$ define

$$ AD(X)=\min\{ \xi: X is $\xi$-arbitrarily distortable\} $$
  if $X$ is $\zeta$-arbitrarily distortable for some $1\leq \zeta< \omega_1$. Otherwise set $AD(X)=\omega_1$.

One can prove the following equivalence:
Let $X$ be a separable Banach space with a Schauder basis.
Then $X$ is arbitrarily distortable if and only if $AD(X)<\omega_1$.
This leaves open a number of interesting questions.
Question 1: Is it true that $AD(\ell_2)>1$? This is just a restatement of my 
previous question.
Question 2: Can we compute $AD(\ell_p)$ for every $1 < p < +\infty$?
Question 3: Can we find for every countable ordinal $\xi\geq 1$ an arbitrarily distortable
Banach space $X_\xi$ such that $AD(X_\xi)>\xi$. The answer is yes for $\xi=1$; any arbitrarily
distortable asymptotic $\ell_1$ space $X$ satisfies $AD(X)>1$.
Notice that an affirmative answer to Question 3 leaves no hope for a "uniform" approach
to distortion on general separable Banach spaces. 

Some references:


*

*D. Alspach and S. Argyros, Complexity of weakly null sequences, Dissertationes Math. 321 (1992), 1-44.

*P. Pudlak and V. Rodl, Partition theorems for systems of finite subsets of integers,
Discrete Math. 39 (1982), 67-73.
 A: This is not an answer to your question but it's a very similar question and an observation concerning it. 
Suppose we ask ourselves the following question: is it true that for every equivalent norm on $\ell_2$ and every ε>0 there exists a block basis $v_1&lt;v_2&lt;\dots&lt;v_n$ such that $n$ is greater than the maximum of the support of $v_1$ and $v_1,\dots,v_n$ span a subspace $(1+&epsilon;)$-isomorphic to $\ell_2^n$?
This is a bit like asking for a block basis "of size ω". Dvoretzky's theorem says we can get $n$ for arbitrarily large $n$, and the negative solution of the distortion problem says that we can't find an infinite sequence (and therefore that there must be some countable ordinal that we can't reach, in a certain obvious sense).
Now there is reason to suppose that this question could be very very hard or perhaps even undecidable (though I am less sure of the latter). The reason is that it is very similar to the Paris-Harrington theorem. The Paris-Harrington theorem asks for a set $X$ such that its cardinality exceeds its minimal element and all its subsets of size $r$ have the same colour (and it starts after $m$, say). It is known to "require" the axiom of infinity in a certain sense: one can prove it trivially if one applies the infinite Ramsey theorem, but there is no proof within PA. But with this distortion variant, we don't have the infinite Ramsey theorem to apply as it's false! So how does one go about thinking about this? I think the example of Odell and Schlumprecht shows that you can't get to $&omega;^2$.
I wonder whether your question could also be affected by concerns of this kind.
Added slightly later: I now see that this might have been more appropriate as an answer to your previous question, which is one that I have wondered about myself (as the above makes clear).
A: Nice questions. Do we know $AD(S)$ for S-Schlumprecht space?
