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

$(1)$ $f$ is strictly nonexpanding, i.e there exists $p,q \in M$ such that $d(f(p),f(q)) < d(p,q)$.

$(2)$ The image $f(M)$ is a submanifold of $N$. (Note I do not assume $f$ is smooth).

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

If it helps, we can assume for start $M,N$ have empty boundary.

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

$f:[0,2\pi] \to \mathbb{S}^1, f(t)=e^{it}$ is strictly nonexpanding but $\operatorname{Vol}(f([0,2\pi])=\operatorname{Vol}(\mathbb{S}^1)=\operatorname{Vol}([0,2\pi])$.

Partial result:

$(1)$ In the case where $M=N$ (as Riemannian manifolds), the answer is positive.

Assume otherwise; Then $\operatorname{Vol}(f(M))=\operatorname{Vol}(M)=\operatorname{Vol}(N)$, hence $f(M)=N$, i.e $f$ is surjective. (Otherwise $f(M)$ will be a closed subset of $N$, strictly contained in $N$, contradicting the equality of volumes).

So, $f$ is a surjective nonexpanding map from a compact metric space to itself, thus an isometry. (See Burago-Burago-Ivanov's "A course in metric geometry", theorem 1.6.15).

$(2)$ In the case the manifolds are one-dimensional, and $f$ is surjective, 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 a surjective nonexpanding map from a compact metric space to itself, hence an isometry.


Result $(1)$ suggests it might be easier to handle the case where $\operatorname{Vol}(M)=\operatorname{Vol}(N)$. The question then becomes equivalent to the following one:

Can a strictly nonexpanding map between two compact Riemannian manifolds of the same volume be surjective?

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  • $\begingroup$ For your last example, what's your distance on $\Bbb S^1$? Because $f$ is an isometry for the restriction of the euclidean distance. $\endgroup$
    – paf
    Aug 28, 2016 at 9:18
  • $\begingroup$ @paf: On $\mathbb{S}^1$ I take the distance induced by the round Riemannian metric (that is the usual arclength). Anyway, $f$ is not an isometry with respect to any distance on $\mathbb{S}^1$ since it is not injective, $f(0)=f(2\pi)$. $\endgroup$ Aug 28, 2016 at 9:30
  • $\begingroup$ the term "strictly nonexpanding" is usually called as "contractive" or "strict contraction" $\endgroup$
    – Domates
    Aug 28, 2016 at 9:37
  • $\begingroup$ @H.Ergul: I think "contraction" is generally used if $d(f(p),f(q))<d(p,q)$ for all $p,q \in M$. That is not the case here. $\endgroup$
    – shalop
    Aug 28, 2016 at 9:39
  • $\begingroup$ @Shalop: You may be right. I didn't know this difference. Thank you. $\endgroup$
    – Domates
    Aug 28, 2016 at 9:42

2 Answers 2

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Here is @AntonMalyshev counterexample in detail (for the case of manifolds with corners):

Let $M$ be the unit square $[0,1]\times [0,1]$, modulo an identification of $(0,t)\sim (1,t)$ for every $t\in [0,1/3]$. This is a manifold with corners (for example, a small enough neighborhood of the point $(0,1/3)\sim (1,1/3)$ is diffeomorphic to $[0,\infty)\times[0,\infty)$). The distance between $(0,1)$ and $(1,1)$ can easily be verified to be $1$.

Let $N$ be the unit square $[0,1]\times [0,1]$, modulo an identification of $(0,t)\sim (1,t)$ for every $t\in [0,2/3]$. Here the distance between $(0,1)$ and $(1,1)$ is at most $2/3$, as shown by the curve $$ \gamma(t) = \begin{cases} (0,1)(1-t) + (0,2/3)t & t\in[0,1] \\ (1,2/3)(2-t) + (1,1)(t-1) & t\in[1,2]. \end{cases} $$

Obviously $M$ and $N$ are diffeomorphic as manifolds with corners, and have the same volume. The trivial map $M\to N$ is surjective, volume preserving and strictly non-expending according to your definition.

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  • $\begingroup$ Thanks. I am not really sure that the resulting space is indeed a manifold with corners (I do not see why there is a neighbourhood of $[(0,\frac{1}{3})]$ which is diffeomorphic to $[0,\infty)\times[0,\infty)$). $\endgroup$ Sep 5, 2016 at 12:20
  • $\begingroup$ @AsafShachar The segment $[(0,1/3),(0,2/3))$ corresponds to $0\times[0,\infty)$ while the segment $[(1,1/3),(1,2/3))$ corresponds to $[0,\infty)\times 0$. $\endgroup$
    – C M
    Sep 5, 2016 at 13:55
  • $\begingroup$ I think that the neighbourhood you have described is not open in the quotient topology. An open neighbourhood of $(0,\frac{1}{3})$ must contain must contain $(0,\frac{1}{3}-\epsilon)$. Thus, I think the boundary of a small nehbourhood should look like $((0,1/3-\epsilon),(0,2/3))$ and $((1,1/3-\epsilon),(1,2/3))$. $\endgroup$ Sep 5, 2016 at 15:26
  • $\begingroup$ I agree, it is only homeomorphic to $[0,\infty)\times[0,\infty)$. One would like to map the segment $((0,1/3-\epsilon),(0,1/3)] \sim ((1,1/3-\epsilon),(1,1/3)]$ to (say) $\{(t,t) : t\in[0,\infty)$; the problem is that the line $((0,1/3-\epsilon),(0,2/3)$ is not smooth using this "chart". $\endgroup$
    – C M
    Sep 5, 2016 at 16:32
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If you allow boundaries and allow $\operatorname{Vol}(f(M))<\operatorname{Vol}(N),$ the answer is negative. Take $M$ to be a connected non-convex subset in $\mathbb R^2$ with smooth boundary such as an annulus, and embed it in a convex set in a scaled-up version of itself, $N.$ The embedding decreases distances between points that were not connected by straight lines in $M,$ but preserves volume.


If $M$ is complete, then the answer is positive. We don't need to assume it is diffeomorphic to $N.$

Assume $f:M\to N$ has Lipschitz constant $1,$ and $\operatorname{Vol}(f(M))=\operatorname{Vol}(M).$ The aim is to show $f$ is distance preserving. The restriction $f|_{M^o}$ to the interior of $M$ is injective and locally distance preserving by the area formula. The image is closed and open, and any path in $f(M)$ can be pulled back to $M,$ so $f$ is distance-preserving.


Now allow $M$ and $N$ to be compact manifolds with boundary, but assume they have equal volume. The answer is again positive.

As before $f|_{M^o}$ is injective and locally distance preserving, but now it also has dense image. In particular $f$ is surjective. And by flowing along inward pointing geodesics as in https://mathoverflow.net/q/253994 we find that $f$ is smooth on the whole of $M,$ preserving the metric tensor at the boundary.

We need to rule out examples like $[0,2\pi]\to S^1,$ where $[0,2\pi]$ get glued along $\{0\} \sim \{2\pi\}.$ I will argue that $f$ is a gluing map $M\to M/\pi\cong N$ where $\pi$ is a free diffeomorphic involution on a union of connected components of $\partial M.$ The map $\pi$ is defined by $\pi(x)=x'$ whenever $x,x'\in \partial M$ with $x\neq x'$ and $f(x)=f(x').$ We need to show this is well-defined and has a clopen domain.

If three distinct points $x_1,x_2,x_3$ were sent to the same point $y\in N,$ then $M^o\cap B(x_i,\epsilon)$ would be sent to, approximately, a half-sphere around $y.$ Three distinct such approximate half-spheres would have to intersect, a contradiction. So $\pi$ is well-defined.

Where two distinct points $x,x'$ are sent to the same point $y\in N,$ the point $y$ must be in the interior of $N.$ The neighborhoods of $x$ and $x'$ must end up boundary-to-boundary - the two half-spheres must completely fill the space near $y,$ because nothing else is going to. By the inverse function theorem the map $x\mapsto x'$ can locally be extended to a smooth map. And given a sequence of pairs of distinct $x_n,x'_n$ with $f(x_n)=f(x_n')$ and $x_n\to x,$ passing to a subsequence if necessary we get $x'_n\to x',$ and $x=x'$ is impossible because $f$ is locally injective (its derivative preserves the metric tensor, even at the boundary). So $\pi$ has a clopen domain.

This shows that $f$ is either injective hence an isometry, or it decreases the number of boundary components, which cannot happen if $M$ is diffeomorphic to $N.$

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