Let $(M,g)$ be a connected smooth Riemannian manifold and denote by $(M,d)$ the induced metric space following by taking topological metric to be the infimum over length of curves in the standard way. Suppose that $(M,d)$ is not complete and let $(\hat{M},d)$ denote the metric completion.

What can be said about $(\hat{M},d)$ being a smooth manifold with smooth boundary?

Edit: Take e.g. any open rectangle in Euclidean space. Then the completion will have a boundary that is not smooth (at the corners).

  • $\begingroup$ isn't $\hat{M}=M \cup \partial M$ ? $\endgroup$
    – reuns
    Feb 5, 2016 at 1:46
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    $\begingroup$ Even worse, if you delete the center point from the letter "X", (as well as the 4 end points), you have a nice smooth manifold whose metric completion is not even a manifold with boundary. If I had to, I'd guess the question is very hard to answer in general. (Also, the graph of $\sin(1/x)$ for $x > 0$ is a perfectly good Riemannian manifold whose metric completion is the graph together with $\{0\}\times [-1,1]$, so fails to be a manifold terribly.) $\endgroup$ Feb 5, 2016 at 2:24
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    $\begingroup$ @Jason DeVito, the question makes no reference to an ambient space, so I think I have an issue with your examples. For example, the graph of $\sin(1/x)$ for $x >0$ is geodesically complete (and by Hopf-Rinow, complete as a metric space), isn't it? The metric $d$ you described in your second comment is not the same metric it inherits as a subset of $\mathbb{R}^2$. $\endgroup$ Feb 5, 2016 at 4:19
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    $\begingroup$ @PhillipAndreae: You're right (on both counts)! So, I also have issues with my examples: as you say, the path metric on the graph of $\sin(1/x)$ doesn't coincide with the restriction of the standard metric on $\mathbb{R}^2$. (Likewise on $X$ with center point deleted - the path metric gives the remaining components infinite distance.) Sorry about that. Nonetheless, I still suspect that the completion of a Riemannian manifold as a metric space can fail to be (homeomorphic to) a manifold (with or without boundary). $\endgroup$ Feb 5, 2016 at 4:25
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    $\begingroup$ Why can't you take an open subset of Euclidean space whose closure is not a manifold, like the outside of the alexander Horned sphere? $\endgroup$
    – user98602
    Feb 5, 2016 at 4:51

1 Answer 1


Here is an example answering a slightly stronger question, from Jason DeVito's comment: Construct a connected Riemannian manifold whose metric completion is not homeomorphic to a topological manifold (with or without the boundary). The simplest example, I think, is:

Take 2-torus $T^2$ smoothly embedded in the unit sphere $S^3$. Now, consider the euclidean cone $C$ over $T^2$ in $R^4$. Then let $M= C- \{0\}$, equipped with the Riemannian metric $g$ induced from $R^4$. I claim that the metric completion of $(M,g)$ is homeomorphic to $C$. Indeed, if $(x_i)$ is a Cauchy sequence in $(M, d_g)$ (where $d$ is the distance function associated with $(M,g)$), then, since $d(x,y)\le |x-y|$, it follows that $(x_i)$ is also a Cauchy sequence in $R^4$. Hence, its (euclidean) limit is in the closure of $M$ in $R^4$. Thus, we obtain a map $f: \overline{(M,d)}\to C$ sending the equivalence class of $(x_i)$ to its euclidean limit. It is easy to see that $f$ is continuous. We need to check that $f$ is bijective. Surjectivity of $f$ is clear by considering a sequence $x_i= \frac{1}{i} x$, where $x\in M$ is arbitrary. Let us check injectivity. Consider two Cauchy sequences $(x_i), (y_i)$ such that $\lim x_i= \lim y_i=0\in R^4$. We need to show that $$ \lim_{i\to\infty} d(x_i, y_i)=0. $$ Let $s_i= |x_i|$, $t_i=|y_i|$. Define the point $z_i= \frac{t_i}{s_i} x_i$, hence, $|z_i|=|y_i|=t_i$. It is easy to check that $d(x_i, z_i)\to 0$. Therefore, $(z_i)$ is again a Cauchy sequence equivalent to $(x_i)$. Moreover, if $D$ is the diameter of $T^2$ with respect to its Riemannian metric, then $d(z_i, y_i)\le t_i D \to 0$. Hence, $(z_i)$ is also equivalent to $(y_i)$. Thus, $(x_i), (y_i)$ represent the same point of the metric completion of $(M,d)$. Injectivity of $f$ follows. A similar argument shows continuity of $f^{-1}$. Therefore, $f$ is a homeomorphism.

Lastly, being a cone over a torus, $C$ is not a topological manifold (since removing the origin changes its fundamental group, which cannot happen for a topological manifold).

There are other examples one can construct, for instance, removing from the euclidean plane a suitable collection of disjoint closed round disks converging to a point. But proving that the metric completion is homeomorphic to the closure in the euclidean plane (which is also not a manifold since it is not locally simply connected) is a bit more painful.

  • $\begingroup$ Could you explain (or provide a reference) for the claim that removing the origin does not change the fundamental group of a topological manifold? This seems false: Removing a point from $S^1 \subset \mathbb{R}^2$ clearly changes the fundamental group... $\endgroup$
    – AmorFati
    Aug 27, 2020 at 22:02
  • $\begingroup$ @AmorFati I was talking about manifolds of dimension 3 (or more). Of course, this fails for surfaces and curves. $\endgroup$ Aug 27, 2020 at 22:48

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