7
$\begingroup$

If $T$ is a nonlinear surjective isometry from Lipschitz-free space $\mathcal{F}(M)$ to $\mathcal{F}(N)$($M,N$ are metric spaces), is $M$ homeomorphic to $N$?

$\endgroup$
  • 3
    $\begingroup$ I might be missing something but non-linearity doesn't seem to be essential here. Indeed, by the Mazur-Ulam theorem such map $T$ is affine and hence $T-T(0)I$ is a linear isometry. $\endgroup$ – Tomasz Kania May 1 '15 at 8:23
  • $\begingroup$ Aha!I am missing the Mazur-Ulam theorem.Thank you. $\endgroup$ – Dongyang Chen May 1 '15 at 9:06
  • $\begingroup$ Another question is : if $T$ is a Lipschitz isomorphism, is $M$ homeomorphic to $N$? $\endgroup$ – Dongyang Chen May 1 '15 at 9:09
  • $\begingroup$ To Dongyang Chen: It looks like Weaver has answered your question. You did not accept the answer, does this mean that you find it incomplete? $\endgroup$ – August Cleaner Jan 24 '16 at 19:23
9
$\begingroup$

No, this is false --- trivially, because if $M$ is the completion of $N$ then $M$ and $N$ have the same Arens-Eells space. But if you require $M$ and $N$ to both be complete there is a more interesting counterexample.

Let $M$ be three copies of the interval $[0,1]$, joined at the $0$'s, with path metric. (That is, a capital "Y" with path metric.) Let $N$ just be a single copy of the interval $[0,3]$. Then ${\rm Lip}_0(M)$ and ${\rm Lip}_0(N)$ are, respectively, isometrically isomorphic to $L^\infty(M)$ and $L^\infty(N)$, and are therefore isometrically isomorphic to each other, by cutting $N$ into three pieces and matching up. (To go from $L^\infty(M)$ to ${\rm Lip}_0(M)$, integrate from the origin.) The Arens-Eells spaces of $M$ and $N$ sit inside of the dual spaces ${\rm Lip}_0(M)^*$ and ${\rm Lip}_0(N)^*$ and it is straightforward to check that the dual isometric isomorphism between these spaces restricts to an isometric isomorphism of $AE(M)$ and $AE(N)$.

(If you work through this example you will see how to define the map between $AE(M)$ and $AE(N)$ directly, but I think the explanation I gave above is more illuminating as to why this isometric isomorphism exists.)

On the positive side, there are some good Banach-Stone type theorems if you impose additional conditions on $M$ and $N$; see Sections 3.7 and 3.8 of the second edition of my book on Lipschitz Algebras.


Edit: The "capital Y" example I described in this answer is not new. A. Godard (Tree metrics and their Lipschitz-free spaces, Proc. AMS 138 (2010), 4311-4320) proves that the Arens-Eells space of any separable metric tree is isometrically isomorphic to $L^1[0,1]$. Second comment: If $M$ and $N$ are compact and concave in the sense of this question and $AE(M)$ is isometrically isomorphic to $AE(N)$, then $M$ and $N$ are isometric up to a dilation. (Compactness is not needed if one assumes a uniform version of concavity.) This is Theorem 3.55 of the second edition of my book.

$\endgroup$
  • $\begingroup$ Thank you for this update. I will have a closer look at Godard's paper. $\endgroup$ – Tomasz Kania Mar 9 '16 at 9:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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