# Recovering the length metric from Hausdorff measure

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 $$f_*\mathcal{H}^n_X(X)=\mathcal{H}^n_Y(Y)$$.

• Perhaps add the definition (or link) of length space. – Gerald Edgar May 1 at 14:18
• en.wikipedia.org/wiki/Intrinsic_metric – Jialong Deng May 1 at 14:40
• 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. – 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 $$\Delta_r(Y)$$.

• $f^2(\Delta_r)$ 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. – Johannes Hahn May 4 at 11:49
• @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... – Moishe Kohan May 4 at 13:08
• @ Moishe: Does it mean that a 1-Lipschitz map between Riemannian $n$-manifolds is an isometric map by the argument? – Jialong Deng May 5 at 12:22
• @JialongDeng: No, you need also the assumption that the map is measure-preserving and that it is a map from a manifold to itself. – Moishe Kohan May 5 at 13:30
• 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 – 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

• 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 ? – Jialong Deng May 5 at 11:57
• 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. – Johannes Hahn May 6 at 11:24