homogenuity of $\ell^p$ - MathOverflow

homogenuity of $\ell^p$

2011-11-24T05:41:20Z

I want to know the following:

If $x_1, x_2, \cdots, x_n, y_1,y_2, \cdots, y_n \in \ell_p$ satisfies $\|x_i-x_j\|_p=\|y_i-y_j\|_p$ for any $i,j$, then does there exist isometry $F$ of $\ell_p$ which send each $x_i$ to $y_i$ ?

Also do you know the precise description of the isometry group of $\ell_p$ ?

Answer by Alain Valette for homogenuity of $\ell^p$

2011-11-24T08:14:08Z

To answer your last question, you must combine two classical results:

1) First, the Mazur-Ulam theorem tells you that every (surjective) isometry of a real Banach space, is affine. This reduces the description of the isometry group, to the group of linear isometries.

2) By a corollary of the Banach-Lamperti theorem, every linear isometry $T$ of $\ell^p=\ell^p(\mathbb{N})$ (with $1\leq p<\infty,p\neq 2$) is of the form $T:(a_n)\mapsto (\epsilon(n)a_{\sigma(n)})$, where $\sigma$ is a permutation of $\mathbb{N}$, and $\epsilon(n)=\pm 1$ for every $n$.

Answer by Gerald Edgar for homogenuity of $\ell^p$

2011-11-24T14:15:58Z

The answer to the first question is NO. Even among norms on $\mathbb R^2$, the only ones that have this amazing property (any isometry defined on a finite set extends to an isometry defined on the whole space) are those norms that make $\mathbb R^2$ isometrically into Euclidean space.