Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $1<p<\infty$, and define an isometry of normed linear spaces to be a norm-preserving surjection. Then all isometries from $\ell^p_n(\mathbb{R})$ to itself are given by linear transformations $T$ such that Mat$(T)$ is product of a permutation matrix and a diagonal matrix with $1$'s and $-1$'s along the diagonal. However it is slightly more complicated to classify all isometries from $\ell^p_n(\mathbb{C})$ to itself since $z\mapsto \overline{z}$ is an isometry of $\mathbb{C}$.

Question: What are all isometries from $\ell^p_n(\mathbb{C})$ to itself?

share|improve this question
    
I have my doubts about your characterization, and in particular the fact that it doesn't depend on $p$. Take for example $T=\begin{bmatrix}\cos t & \sin t \\ -\sin t & \cos t\end{bmatrix}$: this is a norm-preserving surjection of $\ell^2_2(\mathbb{R})$, but I don't think it has the form you claim. And it is not an isometry for choices of $p$ other than 2. –  Martin Argerami Jan 14 '12 at 2:10

1 Answer 1

The definition of "isometry" here is inadequate. A norm-preserving surjection (which I take to mean a surjection satisfying $\|f(x)\| = \|x\|$ for all $x$) need not preserve distances, or be linear, or even be continuous. (Choose for each $r \geq 0$ an arbitrary surjection $f_r$ from $\{x: \|x\| = r\}$ to itself. Then the map on the whole space given by $x \mapsto f_{\|x\|}(x)$ is a norm-preserving surjection.) In particular, if "isometry" is interpreted in this sense, there are far more isometries of $\ell^p_n(\mathbb{R})$ than the ones listed in the question.

But one can take the list as a hint at a more appropriate definition of isometry. Here are two candidates:

  • A function on normed linear spaces $f: V \to W$ is an isometry if $f$ is linear and $\|f(x)\| = \|x\|$ for all $x \in V$.

  • A function on normed linear spaces $f: V \to W$ is an isometry if $f$ is real linear (that is, satisfies $f(x+y) = f(x) + f(y)$ and $f(tx) = t f(x)$ for all $x, y \in V$ and all $t \in \mathbb{R}$) and satisfies $\|f(x)\| = \|x\|$ for all $x \in V$.

These are the same thing when the field of scalars is taken to be $\mathbb{R}$, but the first is more restrictive than the second if the field of scalars is $\mathbb{C}$ (e.g. the conjugation $z \mapsto \overline{z}$ on $\mathbb{C}$ is not an isometry in the first sense as it is not complex linear). The main motivation for using the second definition with a complex vector space is that in strongly convex spaces at least, this condition completely characterizes the distance-preserving maps (functions $f$ satisfying $\|f(x) - f(y)\| = \|x - y\|$ for all $x,y$) that map $0$ to itself. If one wants an "isometry" to be as close to a "distance-preserving function" as possible, complex linearity is perhaps not essential. Offhand, I do not know a characterization of the isometries, in the second sense, of $\ell^p_n(\mathbb{C})$.

But if one uses the first definition and phrases things appropriately, the complex story is identical to the real story. Identify linear operators on $\ell^p_n$ (real or complex, your choice) with $n \times n$ matrices. Say a matrix is a generalized permutation matrix if it is a product of a diagonal matrix whose entries have modulus $1$ and a permutation matrix. Then when $1 < p < \infty$ and $p \neq 2$, the only isometries of $\ell^p_n$ are the generalized permutation matrices.

It is perhaps surprising that these matrices, which are rather obviously isometries of $\ell^p_n$, are the only isometries of $\ell^p_n$--- and that the set of isometries does not depend on $p$. (As Martin Argerami points out, the Hilbert space case $p=2$ is special; but the sets of isometries in this case, the so-called "orthogonal" and "unitary" matrices, are well understood.)

Assuming you know the basic duality theory of $\ell^p_n$ spaces, you can give a short and elementary proof of this fact. One approach is given (in the real case, but the complex case is almost the same) in Isometries of the $\ell^p$ norm by Chi-Kwong Li and Wasin So in the American Mathematical Monthly Vol. 101 No. 5, pp 452-53. (The authors note that their argument was previously and independently found by R. Mathias.)

This is an exceedingly special case of what is often called the Banach-Lamperti theorem, which like many results in the theory of normed linear spaces is really a family of results, varying in the generality of the hypotheses. The basic theme is that when $p \neq 2$, an isometry from one $L^p$ space to another (the measure spaces need not be finite or the same) must be the product of a suitably "nice" multiplication operator and a composition operator (ie, a map of the form $f \mapsto f \circ \sigma$, where $\sigma$ is a map on the underlying measure spaces). In general, the set of isometries does depend on $p$ and the measures used to define the $L^p$ spaces. Lamperti's original paper from 1958 is available at Project Euclid; there have been various generalizations and improvements since then, as you will find if you Google for them.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.