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 $N$?
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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 KaniaCommented May 1, 2015 at 8:23
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$\begingroup$ Aha!I am missing the Mazur-Ulam theorem.Thank you. $\endgroup$– Dongyang ChenCommented May 1, 2015 at 9:06
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$\begingroup$ Another question is : if $T$ is a Lipschitz isomorphism, is $M$ homeomorphic to $N$? $\endgroup$– Dongyang ChenCommented May 1, 2015 at 9:09
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$\begingroup$ 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 CleanerCommented Jan 24, 2016 at 19:23
1 Answer
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.
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$\begingroup$ Thank you for this update. I will have a closer look at Godard's paper. $\endgroup$ Commented Mar 9, 2016 at 9:00