The metric cannot be recovered from its Hausdorff measure in general. Now, assume that $(X,d_X)$ and $(Y, d_Y)$ are connected compact length spaces and induce $n$-dimensional Hausdorff measures $\mathcal{H}^n_X$ and $\mathcal{H}^n_Y$.

Assume there exists a 1-Lipschitz map $f: (X,d_X,\mathcal{H}^n_X)\to (Y, d_Y, \mathcal{H}^n_Y)$ such that $f_*\mathcal{H}^n_X=\mathcal{H}^n_Y$, then, whether $f$ is an isometric map?

Edit: Thanks to Moishe's comment, now let $X$ and $Y$ be closed topological $n$-manifolds. In fact, I wish to show that claim is true for closed Riemannian $n$-manifolds under the condition of $f_*\mathcal{H}^n_X(X)=\mathcal{H}^n_Y(Y)$.

  • $\begingroup$ Perhaps add the definition (or link) of length space. $\endgroup$ – Gerald Edgar May 1 at 14:18
  • $\begingroup$ en.wikipedia.org/wiki/Intrinsic_metric $\endgroup$ – Jialong Deng May 1 at 14:40
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    $\begingroup$ Not in this generality. Imagine that $X$ is the unit sphere with a single hair sticking out of it and $Y$ is that sphere. Now, flatten this hair. However the claim might be true for closed topological manifolds. $\endgroup$ – Moishe Kohan May 3 at 1:35

Nan Li proved that it holds for a pair of Alexandrov spaces without boundary; in particular, it solves the problem for Riemannian manifolds. See Lipschitz-Volume rigidity in Alexandrov geometry.

It seems that his argument can be generalized quite a bit, but one cannot exect it to work for topological manifolds with intrinsic metrics. Indeed, take the standard sphere $X$. Shrink its equator by factor 2; denote the obtained space by $Y$. The induced map $X\to Y$ is measure-preserving and short.


The easiest argument I know (which works for path-metrics on topological manifolds $X$ and in even greater generality) is to consider the induced map $f^2: X^2\to Y^2$. This map also preserves the product measure. Set $$ \Delta_r(X)=\{(x,y)\in X^2: d(x,y)\le r\}. $$ By the assumption, $f^2(\Delta_r(X))\subset \Delta_r(Y)$.

If $f$ is not an isometry, there exists $r>0$ such that $f^2(\Delta_r)$ is a proper subset of $\Delta_r$, hence, by compactness, the interior of the complement $$ \Delta_r(Y) \setminus f^2(\Delta_r(X)) $$ is nonempty, hence, has positive measure. (This is the only place where I am using the manifold assumption.) A contradiction.

The right degree of generality for this proof is that we have two compact path-metric spaces $X, Y$ equipped with Borel measures, each satisfying the property that the measure of each open nonempty subset is positive. The example I gave in the comment shows that this is the right setting.

Edit. This argument works in the case of self-maps, $(X,d_X, \mu_X)=(Y,d_Y,\mu_Y)$. However, in general, it needs more work, as it is unclear why $\Delta_r(X)$ has the same mass as $\Delta_r(Y)$.

  • $\begingroup$ $f^2(\Delta_r)$ need not be a proper subset for non-isometries. Extreme counterexample: Constant maps. You need to use the volume-condition or manifold-ness somewhere to infer that. $\endgroup$ – Johannes Hahn May 4 at 11:49
  • $\begingroup$ @JohannesHahn: You are right: The argument was originally written in the case of self-maps, $X=Y$ and I did not think through the general case. I will correct... $\endgroup$ – Moishe Kohan May 4 at 13:08
  • $\begingroup$ @ Moishe: Does it mean that a 1-Lipschitz map between Riemannian $n$-manifolds is an isometric map by the argument? $\endgroup$ – Jialong Deng May 5 at 12:22
  • $\begingroup$ @JialongDeng: No, you need also the assumption that the map is measure-preserving and that it is a map from a manifold to itself. $\endgroup$ – Moishe Kohan May 5 at 13:30
  • $\begingroup$ Take the standard sphere $X$. Shrink its equator by factor 2; denote the obtained space by $Y$. The induced map $X\to Y$ is measure-preserving and short. So, something wrong with your argument. So you need to assume more about spaces. For Alexandrov spaces it was done by Nan Li arxiv.org/abs/1110.5498 $\endgroup$ – Anton Petrunin May 6 at 5:02

First, let's prove that $f$ is locally isometric, so that we can focus on small sets, say a closed ball of radius $r\ll 1$ around an arbitrary point. Because $f$ is 1-Lipschitz, the image is also contained in a ball of the same radius so that we have the situation where $f$ is a 1-Lipschitz and volume preserving map $B_r \to B_r$. Since $f(B_r)$ is compact, the complement of the image is open and because of $H_Y^n(B_r \setminus f(B_r)) = (f_\ast H_X^n)(B_r\setminus f_r(B_r) ) = H_X^n(\emptyset)$, it must be empty. Therefore $f$ is surjective.

Second, each boundary point of $B_r$ has exactly one point with distance $2r$, namely the boundary point exactly opposite to it, and all point pairs with distance $2r$ are opposite boundary points. If $y_1=f(x_1), y_2=f(x_2)$ is a pair of opposite boundary points, then $x_1$ and $x_2$ have distance at least $2r$ because $f$ is 1-Lipschitz and are therefore also boundary points. Hence $f$ maps $\partial B_r$ (and only $\partial B_r$) to $\partial B_r$, even respecting opposites, and the interior to the interior. In particular $f$ is open and isometric for points of distance $r$ from the center of the ball. Since $r$ was arbitrary, we conclude that $f$ is a local isometry.

Now let's look at the global case. We already know that $f: X\to Y$ is an open, locally isometric map. In particular it is a local homeomorphism, i.e. a covering map onto its image. By compactness, it must be a covering of finite degree $d$. But again choosing a point and a small enough ball around it, we find $d vol(B) = vol(f^{-1}(B)) = (f_\ast H_X^n)(B) = H_Y^n(B) = vol(B)$ so that $d=1$, i.e. $f$ is injective.

An injective, locally isometric map is a global isometry. QED

  • $\begingroup$ Why one have $f_*H^n_X(B_r ∖f_r(B_r))=0$, if $f$ is a covering map or $f$ is not injective at that point ? $\endgroup$ – Jialong Deng May 5 at 11:57
  • $\begingroup$ By definition $f_\ast H_X^n(A) = H_X^n(f^{-1}(A))$ and the preimage of the complement of the image is empty. $\endgroup$ – Johannes Hahn May 6 at 11:24

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