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.

I should know the answer to this (and I did some time ago, but have forgotten): If the normed linear spaces $X$ and $Y$ are isometric (there is a bijective map from $X$ to $Y$ that preserves distances), are they linearly isomorphic (there is a continuous linear bijection from $X$ to $Y$ with a continuous inverse)?

share|improve this question
    
How is Mazur-Ulam Theorem applied to the complex space? is it that the distance preserving surjective map with 0 to 0 on comlex spaces is real linear? Please show the proof. –  user55249 Jan 5 '13 at 2:16
1  
You simply forget that you can multiply vectors by complex numbers, but still remember that you can multiply by reals. Now you have a real vector space, to which the Mazur-Ulam theorem applies. And yes, the conclusion is that the map is real linear. –  user53153 Jan 5 '13 at 2:21
add comment

1 Answer 1

up vote 18 down vote accepted

This is true for real vector spaces by the Mazur-Ulam theorem which states that a surjective distance-preserving map of one real normed space onto another is an affine map. Indeed, if $f: X \to Y$ is such a map then $g(x) = f(x) - f(0)$ is a linear and onto isometry. The inverse of $g$ is of course linear and isometric, too, so, in partiular, $X$ and $Y$ are linearly isomorphic.

For a proof of this, one can't do better than refer to Väisälä's recent note which appeared in the Monthly, see here for the paywalled published version. References to the original works are given there.

Further remarks:

  1. One corollary of the Mazur-Ulam theorem is that the group of isometric and onto self-maps of a real normed space is isomorphic to $O(X) \ltimes X$, where $O(X)$ denotes the group of linear isometries and acts on $X$ in the obvious way. This generalizes the usual description of Euclidean isometry group $O(n) \ltimes \mathbb{R}^n$ nicely.

  2. If surjectivity of $f$ is dropped, the conclusion that $f$ needs to be affine is wrong. The example given by Väisälä is mentioned also this thread here: if we equip $\mathbb{R}^2$ with the max-norm and take $\phi:\mathbb{R} \to \mathbb{R}$ to be a non-linear $1$-Lipschitz map then the map $f: \mathbb{R} \to \mathbb{R}^2$ given by $f(x) = (x,\phi(x))$ is isometric but not affine.

  3. Isomorphic Banach spaces need not be isometric, as you showed yourself in this recent thread.

  4. As far as I know, the case of complex Banach spaces doesn't have such a nice formulation, since we need to ensure complex linearity in order to avoid silly counterexamples that stem from complex conjugation.

  5. Finally, the Mazur-Ulam theorem was mentioned in this MO thread, where Bill Johnson mentions a nice generalization by Figiel which shows in some sense that the counterexample given in 2. is worst possible.

  6. Tangentially relevant is also this MO thread on notions of isomorphisms of Banach spaces.

share|improve this answer
add comment

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.