I recall two definitions from Banach space theory



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> **Definition 1.** Let $E$ be a Banach space, then a basis $(e\_n)\_{n\in\mathbb{N}}$ of $E$ is called $1$-spreading if $$\left\|\sum\_{i=1}^k a\_i e\_{m\_i}\right\|\le\left\|\sum\_{i=1}^k a\_i e\_{n\_i}\right\|$$ whenever $k$ is a positive integer, $(a\_i)\_{i=0}^k$ are scalars and $n\_1<\ldots<n\_k$ and $m\_1<\dots < m\_k$.
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> **Definition 2.** Let $E$ be a Banach space and $(x\_n)\_{n\in\mathbb{N}}$ a basic sequence of $E$. A spreading model of $(x\_n)\_{n\in\mathbb{N}}$ is a normalized basic sequence $(y\_n)\_{n\in\mathbb{N}}$ in a Banach space $F$ such that for every $\epsilon > 0$ and $k\in\mathbb{N}$, there is an $N$ such that $$(1+\epsilon)^{-1}\left\|\sum\_{i=1}^k a\_i y\_i\right\|\le \left\|\sum\_{i=1}^k a\_i x\_{n\_i}\right\|\le (1+\epsilon)\left\|\sum\_{i=1}^k a\_i y\_i\right\|$$ for all $N<n\_1<\dots<n\_k$ and a sequence $(a\_i)\_{i=1}^k$ of scalars.
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Now, looking at these two definitions I get the following:


* If a basis is $1$-spreading then, fixing $n\in\mathbb{N}$, all the subspaces of dimension $n$ generated by a subset of the basis are isometrically isomorphic.
* If $(y\_n)\subset F$ is a spreading model of $(x\_n)\subset E$ then every finite dimensional subspace generated by a subset of $(y\_n)$ is $(1+\epsilon)$-isomorphic to a finite dimensional subspace of $E$.


My two related question, that I supect being almost trivial (I'm relatively new to Banach space theory), are the following:


1. If a basis is $1$-spreading then, fixing $n\in\mathbb{N}$, are all subspaces of dimension $n$ isometrically isomorphic? What about infinite dimensional subspaces? Are them all isometrically isomorphic to the space $E$ itself?
2. If $(y\_n)\subset F$ is a spreading model of $(x\_n)\subset E$, is every finite dimensional subspace of $F$ $(1+\epsilon)$-isomorphic to a finite dimensional subspace of $E$? Again what about the infinite dimensional subspaces of $F$?


Thanks!



Definition of $1$-spreading basis and spreading model
It might be more useful to read first the structure of classical $\ell\_p$ spaces before starting on spreading models. These questions are about those and has very little to do with spreading models (only exception is the pull back mentioned below).


1. The only Banach space with any one of these properties is the Hilbert space. So for instance the unit vector basis of $\ell\_p$, $p\neq 2, p<\infty$ is 1-spreading but it contains $\ell\_2^n$'s almost isometrically as well as $\ell\_p^n$'s, and many other types depending on $p$. Every non-Hilbertian Banach space contains infinite dimensional subspaces not isomorphic to the space. So the answer is always No with only one exception of the Hilbert space.
2. The first assertion is true and trivial (or routine depending on background). The finite dimensional subspace can be seen almost isomoterically as a subspace (possibly much higher dimensional) of finite dimensional spanned by basis vectors, say, $(e\_i)\_1^m$. (Take $\varepsilon$-net in the unit ball, approximate each vector by a finitely supported vector, then the union of all basis vectors used in the process gives $(e\_i)\_1^m$.) Then you pull back $(e\_i)\_1^m$into $E$.


The answer to the second assertion is negative. For instance, Tsirelson space does not contain copies of $\ell\_1$ but all of its spreading models are isomorphic to $\ell\_1$. Spreading models are usually 'nicer' than the generating space, and they are usually the starting point when looking for nice finite dimensional subspaces. But the caveat is you can't always pull back infinite dimensional information from the spreading models.


