# Does nonexpanding map between manifolds decrease volume?

(This question is a special case of a question I asked at SE, which got no answer there)

Let $M,N$ be diffeomorphic connected compact Riemannian manifolds, and let $f:M \to N$ be a surjective nonexpanding map (i.e Lipschitz with constant $1$).

Assume that $f$ is strictly nonexpanding, i.e there exists $p,q \in M$ such that $$d(f(p),f(q)) < d(p,q)$$

Is it true that $\operatorname{Vol}(N)<\operatorname{Vol}(M)$?

Edit:

I want $M,N$ to be smooth manifolds with or without boundary. Nik's answer gives a counter example (when $\operatorname{Vol}(N)=\operatorname{Vol}(M)$), but I am not sure it is smooth in the sense mentioned above. (The same problem comes up in this example given at S.E).

Note that if we do not assume $M,N$ are diffeomorphic then the answer is negative:

A counter-example is $f:[0,2\pi] \to \mathbb{S}^1, f(t)=e^{it}$.

Partial result:

In the case the manifolds are one-dimensional the answer is positive:

Assume otherwise. Then $\operatorname{Vol}(N)=\operatorname{Vol}(M)$. Since every two compact connected one-dimensional Riemannian manifolds with equal volumes are isometric, there exists an isometry $\phi:N \to M$. Thus, $f \circ \phi:N \to N$ is surjective and nonexpanding, hence an isometry.

• Do you continue to not assume $f$ is smooth? – Neal Sep 9 '16 at 14:56
• Yes, I do not assume the map is smooth. – Asaf Shachar Sep 9 '16 at 14:58
• @Neal: Actually, even though I am not assuming smoothness, $f$ is in fact differentiable almost everywhere (via Rademacher's theorem). It is not hard to show that at each point of differentiability $p \in M$, the singular values of the differential $df_p:T_pM \to T_{f(p)}N$ are not greater than one. – Asaf Shachar Sep 9 '16 at 15:32
• I think the answer is "yes" in the case that $f$ is actually a diffeomorphism. Compare the metric $g$ on $M$ to the pullback $h$ by $f$ of the metric on $N$. By the hypothesis that $f$ is nonexpanding, the pointwise eigenvalues of $h$ with respect to $g$ are all no greater than $1$. The hypothesis on distance implies the $h$-geodesic $\gamma$ connecting $p,q$ has shorter $h$-length than $g$-length, so by looking at the arc-length integral we should see at least one pointwise eigenvalue of $h$ wrt $g$ is strictly less than $1$. The volume inequality should follow by comparing volume forms. – Neal Sep 9 '16 at 20:55
• (This is less general than what you are asking, however.) – Neal Sep 9 '16 at 20:56

I don't think so, if "compact manifold" means "compact manifold with boundary" (as it seems to from your example) and "diffeomorphic" means "with diffeomorphic interiors". My previous example wasn't right, as Willie Wong pointed out, because after gluing the result wasn't Riemannian. But we can fix that problem by a simple modification: let $M$ be the unit disk minus the region $y > \sqrt{|x|}$, and let $f$ be the map that zips up the slit. That works, doesn't it?
• I am a bit confused. If you fold and glue the edges, don't you get an angle defect at $(1,0)$ and $(1,1)$? Are you implicitly smoothing it a bit? – Willie Wong Sep 9 '16 at 19:59
• Actually, on second thought, doesn''t your map increase distances between points that are close to the boundary of $M$? (We measure the distances intrinsically, not by the Euclidean distance function which is extrinsic) – Asaf Shachar Sep 10 '16 at 12:22
• @AsafShachar: I thought you only asked for $d(f(p),f(q)) < d(p,q)$ for some $p,q$. My map doesn't increase any distances because every geodesic in $M$ is still a geodesic in $N$. – Nik Weaver Sep 10 '16 at 14:35