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Let $M,N$ be smooth compact oriented Riemannian manifolds with boundary. Suppose that both $M,N$ can be isometrically immersed in each other.

Must $M,N$ be isometric?

Does anything change if we also assume $\operatorname{Vol}(M)=\operatorname{Vol}(N)$?

Note: I assume $M,N$ are connected (Otherwise, as mentioned by Del, we can take $N$ to be two disjoint copies of $M$).


Of course, if both manifolds can be isometrically embedded in each other, then they are isometric.

This follows from volume considerations:

Suppose $i:M \to N,j:N \to M$ are isometric embeddings. Then, $i(M),M$ are isometric, hence $\operatorname{Vol}(M)=\operatorname{Vol}(i(M))\le \operatorname{Vol}(N)$. Similarly, $\operatorname{Vol}(N)\le \operatorname{Vol}(M)$. Thus, $\operatorname{Vol}(i(M))=\operatorname{Vol}(N)$. Since $i(M)$ is compact, it is a closed subset of $N$. Thus, if $i(M) \neq N$, then $N\setminus i(M)$ is open, and so has a positive volume, contradicting $\operatorname{Vol}(i(M))=\operatorname{Vol}(N)$. This shows $i,j$ are surjective, thus isometries.


Updades and Remarks:

$(1) \,$ If $M$, $N$ have no boundaries, the answer is positive. This follows easily from a metric argument.

Let $i:M \to N, j:N \to M$ be the given isometric immersions. Then $i(M)$ is clopen in $N$, hence $i$ is surjective. Similarly, $j$ is surjective.

A possible generalization to the case with boundaries:

Assuming that every smooth orientation preserving isometric immersion maps boundary into boundary (see this question), we know that $j \circ i(\partial M) \subseteq \partial M$, so we can imitate the above argument to this case:

First, we note $i(\partial M) \subseteq \partial N$ (since $j(N^0) \subseteq M^0$). It follows $i(M^o)$ is clopen in $N^o$, hence $i(M^o)=N^o$. Since $i(M)$ is closed in $N$, and contains the dense subset $N^o$, $i$ is surjective, and moreover $i(\partial M) = \partial N , i(M^o)= N^o$.

By symmetry, $j$ is surjective, and the same argument in the previous case imply $j \circ i:M \to M $ is a surjective nonexpanding map, hence a metric isometry. Then, the $1$-Lipschitzity of $i,j$ implies $i$ is a metric isometry. So, by the positive answer to this question $i$ is a smooth Riemannian isometry.


$(2)$ It is enough to prove that an orientation-preserving isometric immersion $M \to M$ is a Riemannian isometry. (and in particular maps $\partial M$ onto $\partial M$).

Indeed, let $i:M \to N, j:N \to M$ be the given immersions and assume the above statement holds. Then $j \circ i:M \to M$ is an isometry, and so $j \circ i(\partial M) = \partial M$. This implies that $i(\partial M) \subseteq \partial N$ (since $j(N^0) \subseteq M^0$).

Also, $j \circ i:M \to M$ is an isometry $\Rightarrow$ $i$ is injective and $j$ is surjective. By symmetry, $i,j$ are bijections.

Since we know that $i(\partial M) \subseteq \partial N , i(M^o)\subseteq N^o$, and $i$ is surjective it follows that $i(\partial M) = \partial N , i(M^o)= N^o$. Since $i$ is in particular a metric isometry, the positive answer to this question, shows $i^{-1}$ is smooth, hence $i$ is a Riemannian isometry.

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    $\begingroup$ Are they connected? Otherwise you can take one of them equal to two copies of the other $\endgroup$ – Del Nov 2 '16 at 0:45
  • $\begingroup$ It would suffice to find two hyperbolic manifolds that cover each other. I suspect this has already been done in the context of 3-manifolds before. $\endgroup$ – user98602 Nov 2 '16 at 4:06
  • $\begingroup$ @Del you are right of course, I will edit the question to address this point. $\endgroup$ – Asaf Shachar Nov 2 '16 at 7:32
  • $\begingroup$ Considering the composition of the two immersions, the question will be solved if you show the following: every isometric immersion of $M$ into itself is an isometry. Now, I don't know if this statement is true or not. However, in the case $M$ has no boundary, you have the following topological claim, which is stronger: Let $\varphi:M\to M$ be a local diffeomorphism of degree $\pm1$. Then $\varphi$ is a diffeomorphism. $\endgroup$ – Amitai Yuval Nov 5 '16 at 19:10
  • $\begingroup$ @AmitaiYuval Yes, I agree with this observation. Can you elaborate on the topological argument you are referring to? (Actually when there is no boundary, then I know $\varphi$ is surjective, since $\varphi(M)$ is clopen in $M$. Then a metric argument shows $\phi$ is injective, hence an isometry. So the only truly interesting case is when there is a non-empty boundary). Even so, I am still intereseted in your "pure" topological reasoning. $\endgroup$ – Asaf Shachar Nov 5 '16 at 19:24
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Here are some observations regarding the question at hand (I guess this would be a comment if I had enough reputation to write a comment, but one has to start somewhere...) :

If the volumes are equal and $f:M\to N$ is surjective, then it is also injective on the interior of $M$. Indeed, by the area formula, $$ \text{Vol}M = \int_M |\det df_x|\,\text{dVol}_g(x) = \int_N |f^{-1}(y)|\,\text{dVol}_h(y) $$ where $g$ is the metric of $M$, $h$ is the metric of $N$ and $\det df$ is the intrinsic determinant (the ratio between the volume forms), which equals to $1$ for an isometric immersion. If $f$ is surjective, then the righthand side is at least $\text{Vol}N$, and therefore the volume equality shows that the set $\{ y\in N: |f^{-1}(y)|>1\}$ is of zero volume. Since $f$ is a local isometry, it follows that $f$ is injective on the interior of $M$ (otherwise if it maps two points to the same point, it would map two of their neighborhoods to the same neighborhood in contradiction). Therefore $f$ isometrically embeds $\text{int}\, M$ into $\text{int}\, N$.

This solves the problem for closed manifolds, since in this case $f(M)$ is clopen, and therefore the whole $N$, as pointed out in the comments above. For the same reason it solves the problem in case you know that $f$ maps boundary to boundary, because that $f(\text{int}\,M)$ is closed at $\text{int}\,N$.

Edit: fixed a small error (before I wrote that sujectivity implies injectivity rather than injectivity on the interior).

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Here is an alternative proof, which do not assume the volumes are equal for the case where $M$, $N$ have no boundaries.

Let $i:M \to N, j:N \to M$ be the given isometric immersions. Then $i(M)$ is clopen in $N$, hence $i$ is surjective. Similarly, $j$ is surjective.

$j \circ i$ is then a surjective nonexpanding ($1$-Lipschitz) map from $M$ to itslef. (This follows from the fact that isometric immersion cannot increase distances since it preserves lengths of paths). Thus, $j \circ i$ is a metric isometry*. Thus, $i$ is injective, hence a smooth isometric bijection. By the inverse function theorem, its inverse is also smooth, and we are done.

*Any surjective nonexpanding map from a compact metric space into itself is an isometry (This is a standard fact in "metric geometry").

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$\DeclareMathOperator{\vol}{vol}\newcommand{\Bar}[1]{\overline{#1}}$tl; dr: Yes, $M$, and $N$ are isometric, assuming only that each is connected, complete, and of finite volume.


Lemma 1: If $(M, g)$ is a complete Riemannian manifold, $(N, h)$ is a connected Riemannian manifold, and $\dim M = \dim N$, then an isometric immersion $i:(M, g) \to (N, h)$ is a surjective covering map.

Proof: The image $i(M)$ is open (because $i$ is an isometric immersion, hence a local diffeomorphism) and closed (because $(M, g)$ is complete) and non-empty, hence equal to $N$ (since $N$ is connected).

Let $(\Bar{M}, \Bar{g}) = (M, g)/i$ denote the Riemannian quotient. That is, define an equivalence relation on $M$ by $p \sim p'$ if and only if $i(p) = i(p')$. Since $i$ is an isometric immersion and $\dim M = \dim N$, the quotient acquires the structure of a smooth Riemannian manifold isometric to $(N, h)$. Let $\pi:M \to \Bar{M}$ denote the quotient map.

Let $q$ be an arbitrary point of $\Bar{M}$, and $V_{r} = V_{r}(q) \subset (\Bar{M}, \Bar{g})$ the geodesic ball of radius $r$ about $q$. Fix a point $p \in \pi^{-1}(q)$ arbitrarily, and choose $r > 0$ small enough that $U_{r}(p) \subset (M, g)$, the geodesic ball of radius $r$ about $p$, is mapped isometrically to $V_{r}$ by $\pi$.

To complete the proof, it suffices to show that $\pi^{-1}(V_{r})$ is a disjoint union of geodesic balls, each mapped isometrically to $V_{r}$ by $\pi$. With the notation of the preceding paragraph, $U_{r}(p) \subset \pi^{-1}(V_{r})$. Conversely, if $x$ is a point of $\pi^{-1}(V_{r})$, so that $\Bar{x} = \pi(x) \in V_{r}$, there is a minimal geodesic $\Bar{\gamma}$ joining $\Bar{x}$ to $q$. Since $\pi$ is a local isometry, the geodesic $\gamma$ that starts at $x$ and satisfies $\pi_{*}\gamma'(0) = \Bar{\gamma}'(0)$ is a lift: $\Bar{\gamma} = \pi \circ \gamma$. Consequently, $\gamma$ joins $x$ to some point $p$ in $\pi^{-1}(q)$. Since $d(x, p) = d(\Bar{x}, q) < r$, we have $x \in U_{r}(p)$.

Lemma 2: If $(M, g)$ and $(N, h)$ are Riemannian manifolds with $(M, g)$ connected, complete, and of finite volume, and if there exist isometric immersions $i:M \to N$ and $j:N \to M$, then $j \circ i:M \to M$ is an isometry

Proof: Suppose $i:M \to N$ and $j:N \to M$ are isometric immersions. (Particularly, $\dim M = \dim N$.) The composition $j \circ i:M \to M$ is an isometric immersion, hence by Lemma 1 a covering map, say with $d$ sheets, so that $\vol(M) = d\vol(M)$. Since $\vol(M)$ is finite, $d = 1$. That is, $j \circ i$ is a diffeomorphism as well as a local isometry, hence an isometry.

Corollary: If $(M, g)$ and $(N, h)$ are complete, connected, finite-volume Riemannian manifolds, and if there exist isometric immersions $i:(M, g) \to (N, h)$ and $j:(N, h) \to (M, g)$, then $i$ and $j$ are isometries.

Proof: By Lemma 2, $j \circ i$ is bijective, so $j$ is surjective and $i$ is injective. Reversing roles, $i \circ j$ is bijective, so $i$ is surjective and $j$ is injective. That is, each of $i$ and $j$ is a bijective isometric immersion, hence an isometry.

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  • $\begingroup$ A bit belated, but posting anyway in the hope the slightly weaker hypotheses are useful to posterity. $\endgroup$ – Andrew D. Hwang Nov 6 '16 at 14:23
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    $\begingroup$ 1. A geodesically-complete manifold has no boundary. (A geodesic starting at a boundary point does not generally extend to an open interval.) 2. Suppose $q$ is in the closure of $i(M)$. Fix $r > 0$ so that the geodesic ball $V_{r}(q) \subset N$ is a diffeomorphic image under $\exp$, pick a point $y$ in $V_{r}(q) \cap i(M)$ and a point $x \in i^{-1}(y)$; let $\Bar{\gamma}$ be the (minimizing) geodesic joining $y$ to $q$, and $U = U_{r}(x)$ the geodesic ball about $x$. Because $i$ is a local isometry at $x$, there is a geodesic $\gamma$ starting at $x$ with $i \circ \gamma = \Bar{\gamma}$.[...] $\endgroup$ – Andrew D. Hwang Nov 6 '16 at 19:13
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    $\begingroup$ By completeness, $\gamma$ extends; this constructs a point of $M$ that maps to $q$, showing that $q \in i(M)$. That is, the closure of $i(M)$ is contained in $i(M)$, so $i(M)$ is closed. $\endgroup$ – Andrew D. Hwang Nov 6 '16 at 19:14
  • $\begingroup$ Thanks for your great answer and patience. I have two more questions: 1) I am not sure about your claim that the smooth and Riemannian structures descend to the quotient. As far as I can see this is not a standard case of a quotient by a group action. (All the fibers have constant cardinality $k$ indeed, by degree theory, but it does not mean immediately we can construct a corresponding smooth action of $S_k$ on $M$ whose equivalence classes are the pre-images of $i$). $\endgroup$ – Asaf Shachar Nov 7 '16 at 19:42
  • $\begingroup$ I see here math.stackexchange.com/a/496628/104576 another condition which ensures the quotient is a manifold, but I am not sure it holds here. Your claim sounds plausible to me, but I would like a more rigorous justification. I guess I am supposed to use somewhere the fact $i$ is an isometry. 2) I do not see why $\pi$ is a local isometry. $\endgroup$ – Asaf Shachar Nov 7 '16 at 19:42

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