There is a science called "Geometry of Banach spaces". I wonder if they managed to give a geometric characterization of $\ell_p$ ($p\in[1,\infty]$) up to isometric isomorphism (among all Banach spaces)?

For $p=1,\infty$ this seems to be not difficult, at least in finite dimensional case. For example, the unit ball in $\ell_\infty$ (over $\mathbb R$) of dimension $n$ has $2^n$ extreme points, while in $\ell_1$ it has $2n$ extreme points, and that is why $\ell_\infty$ and $\ell_1$ can't be isometrically isomorphic (unless $\dim\le 2$). On the other hand, for a Banach space $X$ of dimension $n$ having $2^n$ extreme points in the unit ball is not enough for being isometrically isomorphic to $\ell_\infty$ (since it is easy to construct a norm with arbitrary given (enough big, even) number of extreme points in the unit ball).

So what is the geometric explanation? I asked this in MSE without success. Is it possible that nobody considered this?

  • $\begingroup$ Of course the classification of $L_p$ spaces, both isometric and isomorphic, is very well understood. You do not say just what you want. Have you looked in any books? $\endgroup$ May 25, 2014 at 8:59
  • $\begingroup$ Bill, this classification where is it described? And what is not clear in my question? $\endgroup$ May 25, 2014 at 9:20
  • $\begingroup$ I mean, what precisely are you looking for? Is it "for what $p$ and $r$ and $n$ is $\ell_p^n$ isometrically isomorphic to $\ell_r^n$"? $\endgroup$ May 25, 2014 at 9:42
  • $\begingroup$ After all, there are books written about the isometric classification of Hilbert spaces with the category of Banach spaces. $\endgroup$ May 25, 2014 at 9:44
  • $\begingroup$ Bill, of course I mean another thing. The question is which geometric properties of (the unit ball in) a Banach space $X$ allow us to conclude that $X$ is isometrically isomorphic to $\ell_p^n$. And I don't ask about Hilbert spaces. $\endgroup$ May 25, 2014 at 9:52

1 Answer 1


Not an answer, but too long for a comment:

Look at Wells and Williams book, Embeddings and extensions in analysis, Springer Ergebnisse 84. There you find the classical description of what spaces $X$ embed isometrically in an $L_p$ space. You also need $X^*$ to embed isometrically into $L_q$, $1/p+1/q=1$. There are finite dimensional spaces other than $\ell_p$ spaces that satisfy this (e.g. Hilbert spaces), so this is not a characterization of $\ell_p$. But perhaps the only finite dimensional spaces that satisfy this are $p$ sums of Hilbert spaces. Maybe that question has been considered; I don't know.

If you are willing to take "having a $1$ symmetric basis" as a geometric condition (after all, that just says that the group of isometries is large) then the problem is much easier because for $1<p \not=2<\infty$ the space $\ell_p^n$ has a unique (up to normalizing and changing sign) $1$-symmetric basis.

  • $\begingroup$ Bill, before I find this reading, what is "1 symmetric basis"? $\endgroup$ May 25, 2014 at 11:11
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    $\begingroup$ $\|\sum \pm a_i e_{\pi(i)} \| = \|\sum a_i e_i\|$ for all permutations $\pi$ and all choices of signs. $\endgroup$ May 25, 2014 at 11:13
  • $\begingroup$ Ah, OK, thank you. Do your words mean that infinite dimensional $\ell_\infty$ were not characterised? $\endgroup$ May 25, 2014 at 11:14
  • $\begingroup$ Infinite dimensional $1$-injective spaces are just $C(K)$ with $K$ extremally disconnected, or order complete abstract $M$-spaces, or commutative von Neumann algebras, or... The $\ell_\infty$ spaces are the purely atomic lattices within the class of $1$-injective spaces. $\endgroup$ May 25, 2014 at 12:04
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    $\begingroup$ Mark was just dualizing the extreme point condition that characterizes $\ell_1^n$. I don't know an infinite dimensional version of that, although I guess you could get one by dualizing the characterization of $\ell_1$ spaces as those $L_1$ spaces whose unit ball is the closed convex hull of the extreme points. The characterizations of $\ell_\infty^n$ I mentioned that involve numerical parameters only apply to finite dimensional spaces. $\endgroup$ May 25, 2014 at 15:03

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